Over the weekend, I worked on a revised version of my paper on a physical interpretation of the wavefunction. However, I forgot to add the final remarks on the speed of light as an angular velocity. I know… This post is for my faithful followers only. It is dense, but let me add the missing bits here:
Tag Archives: interpretation of the wavefunction
A physical explanation for relativistic length contraction?
My last posts were all about a possible physical interpretation of the quantum-mechanical wavefunction. To be precise, we have been interpreting the wavefunction as a gravitational wave. In this interpretation, the real and imaginary component of the wavefunction get a physical dimension: force per unit mass (newton per kg). The inspiration here was the structural similarity between Coulomb’s and Newton’s force laws. They both look alike: it’s just that one gives us a force per unit charge (newton per coulomb), while the other gives us a force per unit mass.
So… Well… Many nice things came out of this – and I wrote about that at length – but last night I was thinking this interpretation may also offer an explanation of relativistic length contraction. Before we get there, let us re-visit our hypothesis.
The geometry of the wavefunction
The elementary wavefunction is written as:
ψ = a·e^{−i(E·t − p∙x)/ħ} = a·cos(p∙x/ħ – E∙t/ħ) + i·a·sin(p∙x/ħ – E∙t/ħ)
Nature should not care about our conventions for measuring the phase angle clockwise or counterclockwise and, therefore, the ψ = a·e^{i}^{[E·t − p∙x]/ħ} function may also be permitted. We know that cos(θ) = cos(–θ) and sinθ = –sin(–θ), so we can write:
ψ = a·e^{i}^{(E·t − p∙x)/ħ} = a·cos(E∙t/ħ – p∙x/ħ) + i·a·sin(E∙t/ħ – p∙x/ħ)
= a·cos(p∙x/ħ – E∙t/ħ) – i·a·sin(p∙x/ħ – E∙t/ħ)
The vectors p and x are the the momentum and position vector respectively: p = (p_{x}, p_{y}, p_{z}) and x = (x, y, z). However, if we assume there is no uncertainty about p – not about the direction nor the magnitude – then we may choose an x-axis which reflects the direction of p. As such, x = (x, y, z) reduces to (x, 0, 0), and p∙x/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis or, if p = 0, that our particle is located somewhere on the x-axis. Hence, the analysis is one-dimensional only.
The geometry of the elementary wavefunction is illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.
Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. You can check this as follows: if the origin is the (x, t) = (0, 0) point, then cos(θ) = cos(0) = 1 and sin(θ) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively. We speculated this should correspond to the two possible values for the quantum-mechanical spin of the wave: +ħ/2 or −ħ/2. The cosine and sine components for the left-handed wave are shown below. Needless to say, the cosine and sine function are the same, except for a phase difference of π/2: sin(θ) = cos(θ − π/2).
As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any wave F(kx – ωt) is given by v_{p} = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to -p and, therefore, we would get a negative phase velocity: v_{p} = ω/k = –E/p.
The de Broglie relations
E/ħ = ω gives the frequency in time (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: f = ω/2π and λ = 2π/k, which gives us the two de Broglie relations:
- E = ħ∙ω = h∙f
- p = ħ∙k = h/λ
The frequency in time is easy to interpret. The wavefunction of a particle with more energy, or more mass, will have a higher density in time than a particle with less energy.
In contrast, the second de Broglie relation is somewhat harder to interpret. According to the p = h/λ relation, the wavelength is inversely proportional to the momentum: λ = h/p. The velocity of a photon, or a (theoretical) particle with zero rest mass (m_{0} = 0), is c and, therefore, we find that p = m_{v}∙v = m_{c}∙c = m∙c (all of the energy is kinetic). Hence, we can write: p∙c = m∙c^{2} = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass, the wavelength can be written as:
λ = h/p = hc/E = h/mc
However, this is a limiting situation – applicable to photons only. Real-life matter-particles should have some mass[1] and, therefore, their velocity will never be c.[2]
Hence, if p goes to zero, then the wavelength becomes infinitely long: if p → 0 then λ → ∞. How should we interpret this inverse proportionality between λ and p? To answer this question, let us first see what this wavelength λ actually represents.
If we look at the ψ = a·cos(p∙x/ħ – E∙t/ħ) – i·a·sin(p∙x/ħ – E∙t/ħ) once more, and if we write p∙x/ħ as Δ, then we can look at p∙x/ħ as a phase factor, and so we will be interested to know for what x this phase factor Δ = p∙x/ħ will be equal to 2π. So we write:
Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ
So now we get a meaningful interpretation for that wavelength. It is the distance between the crests (or the troughs) of the wave, so to speak, as illustrated below. Of course, this two-dimensional wave has no real crests or troughs: we measure crests and troughs against the y-axis here. Hence, our definition depend on the frame of reference.
Now we know what λ actually represents for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/f = 2π·(ħ/E). Hence, we can now calculate the wave velocity:
v = λ/T = (h/p)/[2π·(ħ/E)] = E/p
Unsurprisingly, we just get the phase velocity that we had calculated already: v = v_{p} = E/p. The question remains: what if p is zero? What if we are looking at some particle at rest? It is an intriguing question: we get an infinitely long wavelength, and an infinite wave velocity.
Now, re-writing the v = E/p as v = m∙c^{2}/m∙v_{g } = c/β_{g}, in which β_{g} is the relative classical velocity[3] of our particle β_{g} = v_{g}/c) tells us that the phase velocities will effectively be superluminal (β_{g} < 1 so 1/ β_{g} > 1), but what if β_{g} approaches zero? The conclusion seems unavoidable: for a particle at rest, we only have a frequency in time, as the wavefunction reduces to:
ψ = a·e^{−i·E·t/ħ} = a·cos(E∙t/ħ) – i·a·sin(E∙t/ħ)
How should we interpret this?
A physical interpretation of relativistic length contraction?
In my previous posts, we argued that the oscillations of the wavefunction pack energy. Because the energy of our particle is finite, the wave train cannot be infinitely long. If we assume some definite number of oscillations, then the string of oscillations will be shorter as λ decreases. Hence, the physical interpretation of the wavefunction that is offered here may explain relativistic length contraction.
🙂
Yep. Think about it. 🙂
[1] Even neutrinos have some (rest) mass. This was first confirmed by the US-Japan Super-Kamiokande collaboration in 1998. Neutrinos oscillate between three so-called flavors: electron neutrinos, muon neutrinos and tau neutrinos. Recent data suggests that the sum of their masses is less than a millionth of the rest mass of an electron. Hence, they propagate at speeds that are very near to the speed of light.
[2] Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as KE = E − E_{0} = m_{v}c^{2} − m_{0}c^{2} = m_{0}γc^{2} − m_{0}c^{2} = m_{0}c^{2}(γ − 1). As v approaches c, γ approaches infinity and, therefore, the kinetic energy would become infinite as well.
[3] Because our particle will be represented by a wave packet, i.e. a superimposition of elementary waves with different E and p, the classical velocity of the particle becomes the group velocity of the wave, which is why we denote it by v_{g}.
The geometry of the wavefunction (2)
This post further builds on the rather remarkable results we got in our previous posts. Let us start with the basics once again. The elementary wavefunction is written as:
ψ = a·e^{−i[E·t − p∙x]/ħ} = a·cos(p∙x/ħ − E∙t/ħ) + i·a·sin(p∙x/ħ − E∙t/ħ)
Of course, Nature (or God, as Einstein would put it) does not care about our conventions for measuring an angle (i.e. the phase of our wavefunction) clockwise or counterclockwise and, therefore, the ψ = a·e^{i}^{[E·t − p∙x]/ħ} function is also permitted. We know that cos(θ) = cos(−θ) and sinθ = −sin(−θ), so we can write:
ψ = a·e^{i}^{[E·t − p∙x]/ħ} = a·cos(E∙t/ħ − p∙x/ħ) + i·a·sin(E∙t/ħ − p∙x/ħ)
= a·cos(p∙x/ħ − E∙t/ħ) − i·a·sin(p∙x/ħ − E∙t/ħ)
The vectors p and x are the momentum and position vector respectively: p = (p_{x}, p_{y}, p_{z}) and x = (x, y, z). However, if we assume there is no uncertainty about p – not about the direction, and not about the magnitude – then the direction of p can be our x-axis. In this reference frame, x = (x, y, z) reduces to (x, 0, 0), and p∙x/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis or, if p = 0, that our particle is located somewhere on the x-axis. So we have an analysis in one dimension only then, which facilitates our calculations. The geometry of the wavefunction is then as illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.
Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. [You can check this as follows: if the origin is the (x, t) = (0, 0) point, then cos(θ) = cos(0) = 1 and sin(θ) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively.]
Now, you will remember that we speculated the two polarizations (left- versus right-handed) should correspond to the two possible values for the quantum-mechanical spin of the wave (+ħ/2 or −ħ/2). We will come back to this at the end of this post. Just focus on the essentials first: the cosine and sine components for the left-handed wave are shown below. Look at it carefully and try to understand. Needless to say, the cosine and sine function are the same, except for a phase difference of π/2: sin(θ) = cos(θ − π/2).
As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any waveform F(kx − ωt) is given by v_{p} = ω/k. In our case, we find that v_{p} = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to −p and, therefore, we would get a negative phase velocity: v_{p} = ω/k = (E/ħ)/(−p/ħ) = −E/p.
As you know, E/ħ = ω gives the frequency in time (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). [If in doubt, check my post on essential wave math.] Now, you also know that f = ω/2π and λ = 2π/k, which gives us the two de Broglie relations:
- E = ħ∙ω = h∙f
- p = ħ∙k = h/λ
The frequency in time (oscillations or radians per second) is easy to interpret. A particle will always have some mass and, therefore, some energy, and it is easy to appreciate the fact that the wavefunction of a particle with more energy (or more mass) will have a higher density in time than a particle with less energy.
However, the second de Broglie relation is somewhat harder to interpret. Note that the wavelength is inversely proportional to the momentum: λ = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long, so we write:
If p → 0 then λ → ∞.
For the limit situation, a particle with zero rest mass (m_{0} = 0), the velocity may be c and, therefore, we find that p = m_{v}∙v = m_{c}∙c = m∙c (all of the energy is kinetic) and, therefore, p∙c = m∙c^{2} = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass (m_{0} = 0), the wavelength can be written as:
λ = h/p = hc/E = h/mc
Of course, we are talking a photon here. We get the zero rest mass for a photon. In contrast, all matter-particles should have some mass[1] and, therefore, their velocity will never equal c.[2] The question remains: how should we interpret the inverse proportionality between λ and p?
Let us first see what this wavelength λ actually represents. If we look at the ψ = a·cos(p∙x/ħ − E∙t/ħ) − i·a·sin(p∙x/ħ – E∙t/ħ) once more, and if we write p∙x/ħ as Δ, then we can look at p∙x/ħ as a phase factor, and so we will be interested to know for what x this phase factor Δ = p∙x/ħ will be equal to 2π. So we write:
Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ
So now we get a meaningful interpretation for that wavelength. It is the distance between the crests (or the troughs) of the wave, so to speak, as illustrated below. Of course, this two-dimensional wave has no real crests or troughs: they depend on your frame of reference.
So now we know what λ actually represent for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/f = 2π·(ħ/E). Hence, we can now calculate the wave velocity:
v = λ/T = (h/p)/[2π·(ħ/E)] = E/p
Unsurprisingly, we just get the phase velocity that we had calculated already: v = v_{p} = E/p. It does not answer the question: what if p is zero? What if we are looking at some particle at rest? It is an intriguing question: we get an infinitely long wavelength, and an infinite phase velocity. Now, we know phase velocities can be superluminal, but they should not be infinite. So what does the mathematical inconsistency tell us? Do these infinitely long wavelengths and infinite wave velocities tell us that our particle has to move? Do they tell us our notion of a particle at rest is mathematically inconsistent?
Maybe. But maybe not. Perhaps the inconsistency just tells us our elementary wavefunction – or the concept of a precise energy, and a precise momentum – does not make sense. This is where the Uncertainty Principle comes in: stating that p = 0, implies zero uncertainty. Hence, the σ_{p} factor in the σ_{p}∙σ_{x} ≤ ħ/2 would be zero and, therefore, σ_{p}∙σ_{x} would be zero which, according to the Uncertainty Principle, it cannot be: it can be very small, but it cannot be zero.
It is interesting to note here that σ_{p} refers to the standard deviation from the mean, as illustrated below. Of course, the distribution may be or may not be normal – we don’t know – but a normal distribution makes intuitive sense, of course. Also, if we assume the mean is zero, then the uncertainty is basically about the direction in which our particle is moving, as the momentum might then be positive or negative.
The question of natural units may pop up. The Uncertainty Principle suggests a numerical value of the natural unit for momentum and distance that is equal to the square root of ħ/2, so that’s about 0.726×10^{−17} m for the distance unit and 0.726×10^{−17} N∙s for the momentum unit, as the product of both gives us ħ/2. To make this somewhat more real, we may note that 0.726×10^{−17} m is the attometer scale (1 am = 1×10^{−18} m), so that is very small but not unreasonably small.[3]
Hence, we need to superimpose a potentially infinite number of waves with energies and momenta centered on some mean value. It is only then that we get meaningful results. For example, the idea of a group velocity – which should correspond to the classical idea of the velocity of our particle – only makes sense in the context of wave packet. Indeed, the group velocity of a wave packet (v_{g}) is calculated as follows:
v_{g} = ∂ω_{i}/∂k_{i} = ∂(E_{i}/ħ)/∂(p_{i}/ħ) = ∂(E_{i})/∂(p_{i})
This assumes the existence of a dispersion relation which gives us ω_{i} as a function of k_{i} – what amounts to the same – E_{i} as a function of p_{i}. How do we get that? Well… There are a few ways to go about it but one interesting way of doing it is to re-write Schrödinger’s equation as the following pair of equations[4]:
- Re(∂ψ/∂t) = −[ħ/(2m_{eff})]·Im(∇^{2}ψ) ⇔ ω·cos(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·cos(kx − ωt)
- Im(∂ψ/∂t) = [ħ/(2m_{eff})]·Re(∇^{2}ψ) ⇔ ω·sin(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·sin(kx − ωt)
These equations imply the following dispersion relation:
ω = ħ·k^{2}/(2m)
Of course, we need to think about the subscripts now: we have ω_{i}, k_{i}, but… What about m_{eff} or, dropping the subscript, about m? Do we write it as m_{i}? If so, what is it? Well… It is the equivalent mass of E_{i} obviously, and so we get it from the mass-energy equivalence relation: m_{i} = E_{i}/c^{2}. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too, and the two will, obviously, be related as follows: σ_{m} = σ_{E}/c^{2}. We are tempted to do a few substitutions here. Let’s first check what we get when doing the m_{i} = E_{i}/c^{2} substitution:
ω_{i} = ħ·k_{i}^{2}/(2m_{i}) = (1/2)∙ħ·k_{i}^{2}∙c^{2}/E_{i} = (1/2)∙ħ·k_{i}^{2}∙c^{2}/(ω_{i}∙ħ) = (1/2)∙ħ·k_{i}^{2}∙c^{2}/ω_{i}
⇔ ω_{i}^{2}/k_{i}^{2} = c^{2}/2 ⇔ ω_{i}/k_{i} = v_{p} = c/2 !?
We get a very interesting but nonsensical condition for the dispersion relation here. I wonder what mistake I made. 😦
Let us try another substitution. The group velocity is what it is, right? It is the velocity of the group, so we can write: k_{i} = p/ħ = m_{i} ·v_{g}. This gives us the following result:
ω_{i} = ħ·(m_{i} ·v_{g})^{2}/(2m_{i}) = ħ·m_{i}·v_{g}^{2}/2
It is yet another interesting condition for the dispersion relation. Does it make any more sense? I am not so sure. That factor 1/2 troubles us. It only makes sense when we drop it. Now you will object that Schrödinger’s equation gives us the electron orbitals – and many other correct descriptions of quantum-mechanical stuff – so, surely, Schrödinger’s equation cannot be wrong. You’re right. It’s just that… Well… When we are splitting in up in two equations, as we are doing here, then we are looking at one of the two dimensions of the oscillation only and, therefore, it’s only half of the mass that counts. Complicated explanation but… Well… It should make sense, because the results that come out make sense. Think of it. So we write this:
- Re(∂ψ/∂t) = −(ħ/m_{eff})·Im(∇^{2}ψ) ⇔ ω·cos(kx − ωt) = k^{2}·(ħ/m_{eff})·cos(kx − ωt)
- Im(∂ψ/∂t) = (ħ/m_{eff})·Re(∇^{2}ψ) ⇔ ω·sin(kx − ωt) = k^{2}·(ħ/m_{eff})·sin(kx − ωt)
We then get the dispersion relation without that 1/2 factor:
ω_{i} = ħ·k_{i}^{2}/m_{i}
The m_{i} = E_{i}/c^{2} substitution then gives us the result we sort of expected to see:
ω_{i} = ħ·k_{i}^{2}/m_{i} = ħ·k_{i}^{2}∙c^{2}/E_{i} = ħ·k_{i}^{2}∙c^{2}/(ω_{i}∙ħ) ⇔ ω_{i}/k_{i} = v_{p} = c
Likewise, the other calculation also looks more meaningful now:
ω_{i} = ħ·(m_{i} ·v_{g})^{2}/m_{i} = ħ·m_{i}·v_{g}^{2}
Sweet ! 🙂
Let us put this aside for the moment and focus on something else. If you look at the illustrations above, you see we can sort of distinguish (1) a linear velocity – the speed with which those wave crests (or troughs) move – and (2) some kind of circular or tangential velocity – the velocity along the red contour line above. We’ll need the formula for a tangential velocity: v_{t} = a∙ω.
Now, if λ is zero, then v_{t} = a∙ω = a∙E/ħ is just all there is. We may double-check this as follows: the distance traveled in one period will be equal to 2πa, and the period of the oscillation is T = 2π·(ħ/E). Therefore, v_{t} will, effectively, be equal to v_{t} = 2πa/(2πħ/E) = a∙E/ħ. However, if λ is non-zero, then the distance traveled in one period will be equal to 2πa + λ. The period remains the same: T = 2π·(ħ/E). Hence, we can write:
For an electron, we did this weird calculation. We had an angular momentum formula (for an electron) which we equated with the real-life +ħ/2 or −ħ/2 values of its spin, and we got a numerical value for a. It was the Compton radius: the scattering radius for an electron. Let us write it out:
Using the right numbers, you’ll find the numerical value for a: 3.8616×10^{−13} m. But let us just substitute the formula itself here:
This is fascinating ! And we just calculated that v_{p} is equal to c. For the elementary wavefunction, that is. Hence, we get this amazing result:
v_{t} = 2c
This tangential velocity is twice the linear velocity !
Of course, the question is: what is the physical significance of this? I need to further look at this. Wave velocities are, essentially, mathematical concepts only: the wave propagates through space, but nothing else is really moving. However, the geometric implications are obviously quite interesting and, hence, need further exploration.
One conclusion stands out: all these results reinforce our interpretation of the speed of light as a property of the vacuum – or of the fabric of spacetime itself. 🙂
[1] Even neutrinos should have some (rest) mass. In fact, the mass of the known neutrino flavors was estimated to be smaller than 0.12 eV/c^{2}. This mass combines the three known neutrino flavors.
[2] Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as KE = E − E_{0} = m_{v}c^{2} − m_{0}c^{2} = m_{0}γc^{2} − m_{0}c^{2} = m_{0}c^{2}(γ − 1). As v approaches c, γ approaches infinity and, therefore, the kinetic energy would become infinite as well.
[3] It is, of course, extremely small, but 1 am is the current sensitivity of the LIGO detector for gravitational waves. It is also thought of as the upper limit for the length of an electron, for quarks, and for fundamental strings in string theory. It is, in any case, 1,000,000,000,000,000,000 times larger than the order of magnitude of the Planck length (1.616229(38)×10^{−35} m).
[4] The m_{eff} is the effective mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write m_{eff} = m. As for the equations, they are easily derived from noting that two complex numbers a + i∙b and c + i∙d are equal if, and only if, their real and imaginary parts are the same. Now, the ∂ψ/∂t = i∙(ħ/m_{eff})∙∇^{2}ψ equation amounts to writing something like this: a + i∙b = i∙(c + i∙d). Now, remembering that i^{2} = −1, you can easily figure out that i∙(c + i∙d) = i∙c + i^{2}∙d = − d + i∙c.
The geometry of the wavefunction
My posts and article on the wavefunction as a gravitational wave are rather short on the exact geometry of the wavefunction, so let us explore that a bit here. By now, you know the formula for the elementary wavefunction by heart:
ψ = a·e^{−i[E·t − p∙x]/ħ} = a·cos(p∙x/ħ − E∙t/ħ) + i·a·sin(p∙x/ħ − E∙t/ħ)
If we assume the momentum p is all in the x-direction, then the p and x vectors will have the same direction, and p∙x/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis. The geometry of the wavefunction is illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.
Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. If the origin is the (x, t) = (0, 0) point, then cos(θ) = cos(0) = 1 and sin(θ) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively. I am convinced these correspond to the two possible values for the quantum-mechanical spin of the wave: +ħ/2 or −ħ/2. But… Well… Who am I? The cosine and sine components are shown below. Needless to say, the cosine and sine function are the same, except for a phase difference of π/2: sin(θ) = cos(θ − π/2)
Surely, Nature doesn’t care a hoot about our conventions for measuring the phase angle clockwise or counterclockwise and therefore, the ψ = a·e^{i}^{[E·t − p∙x]/ħ} function should, effectively, also be permitted. We know that cos(θ) = cos(–θ) and sinθ = –sin(–θ), so we can write:
ψ = a·e^{i}^{[E·t − p∙x]/ħ} = a·cos(E∙t/ħ − p∙x/ħ) + i·a·sin(E∙t/ħ − p∙x/ħ)
= a·cos(p∙x/ħ − E∙t/ħ) − i·a·sin(p∙x/ħ − E∙t/ħ)
E/ħ = ω gives the frequency in time (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: f = ω/2π and λ = 2π/k, which gives us the two de Broglie relations:
- E = ħ∙ω = h∙f
- p = ħ∙k = h/λ
The frequency in time is easy to interpret (a particle will always have some mass and, therefore, some energy), but the wavelength is inversely proportional to the momentum: λ = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long: if p → 0, then λ → ∞. For the limit situation, a particle with zero rest mass (m_{0} = 0), the velocity may be c and, therefore, we find that p = m_{v}∙v = m∙c and, therefore, p∙c = m∙c^{2} = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass, the wavelength can be written as:
λ = h/p = hc/E = h/mc
However, we argued that the physical dimension of the components of the wavefunction may be usefully expressed in N/kg units (force per unit mass), while the physical dimension of the electromagnetic wave are expressed in N/C (force per unit charge). This, in fact, explains the dichotomy between bosons (photons) and fermions (spin-1/2 particles). Hence, all matter-particles should have some mass.[1] But how we interpret the inverse proportionality between λ and p?
We should probably first ask ourselves what wavelength we are talking about. The wave only has a phase velocity here, which is equal to v_{p} = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, we know that, classically, the momentum will be equal to the group velocity times the mass: p = m·v_{g}. However, when p is zero, we have a division by zero once more: if p → 0, then v_{p} = E/p → ∞. Infinite wavelengths and infinite phase velocities probably tell us that our particle has to move: our notion of a particle at rest is mathematically inconsistent. If we associate this elementary wavefunction with some particle, and if we then imagine it to move, somehow, then we get an interesting relation between the group and the phase velocity:
v_{p} = ω/k = E/p = E/(m·v_{g}) = (m·c^{2})/(m·v_{g}) = c^{2}/v_{g}
We can re-write this as v_{p}·v_{g} = c^{2}, which reminds us of the relationship between the electric and magnetic constant (1/ε_{0})·(1/μ_{0}) = c^{2}. But what is the group velocity of the elementary wavefunction? Is it a meaningful concept?
The phase velocity is just the ratio of ω/k. In contrast, the group velocity is the derivative of ω with respect to k. So we need to write ω as a function of k. Can we do that even if we have only one wave? We do not have a wave packet here, right? Just some hypothetical building block of a real-life wavefunction, right? Right. So we should introduce uncertainty about E and p and build up the wave packet, right? Well… Yes. But let’s wait with that, and see how far we can go in our interpretation of this elementary wavefunction. Let’s first get that ω = ω(k) relation. You’ll remember we can write Schrödinger’s equation – the equation that describes the propagation mechanism for matter-waves – as the following pair of equations:
- Re(∂ψ/∂t) = −[ħ/(2m)]·Im(∇^{2}ψ) ⇔ ω·cos(kx − ωt) = k^{2}·[ħ/(2m)]·cos(kx − ωt)
- Im(∂ψ/∂t) = [ħ/(2m)]·Re(∇^{2}ψ) ⇔ ω·sin(kx − ωt) = k^{2}·[ħ/(2m)]·sin(kx − ωt)
This tells us that ω = ħ·k^{2}/(2m). Therefore, we can calculate ∂ω/∂k as:
∂ω/∂k = ħ·k/m = p/m = v_{g}
We learn nothing new. We are going round and round in circles here, and we always end up with a tautology: as soon as we have a non-zero momentum, we have a mathematical formula for the group velocity – but we don’t know what it represents – and a finite wavelength. In fact, using the p = ħ∙k = h/λ relation, we can write one as a function of the other:
λ = h/p = h/mv_{g} ⇔ v_{g} = h/mλ
What does this mean? It resembles the c = h/mλ relation we had for a particle with zero rest mass. Of course, it does: the λ = h/mc relation is, once again, a limit for v_{g} going to c. By the way, it is interesting to note that the v_{p}·v_{g} = c^{2} relation implies that the phase velocity is always superluminal. That’ easy to see when you re-write the equation in terms of relative velocities: (v_{p}/c)·(v_{g}/c) = β_{phase}·β_{group} = 1. Hence, if β_{group} < 1, then β_{phase} > 1.
So what is the geometry, really? Let’s look at the ψ = a·cos(p∙x/ħ – E∙t/ħ) – i·a·sin(p∙x/ħ – E∙t/ħ) formula once more. If we write p∙x/ħ as Δ, then we will be interested to know for what x this phase factor will be equal to 2π. So we write:
Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ
So now we get a meaningful interpretation for that wavelength: it’s that distance between the crests of the wave, so to speak, as illustrated below.
Can we now find a meaningful (i.e. geometric) interpretation for the group and phase velocity? If you look at the illustration above, you see we can sort of distinguish (1) a linear velocity (the speed with which those wave crests move) and (2) some kind of circular or tangential velocity (the velocity along the red contour line above). We’ll probably need the formula for the tangential velocity: v = a∙ω. If p = 0 (so we have that weird infinitesimally long wavelength), then we have two velocities:
- The tangential velocity around the a·e^{i}^{·E·t} circle, so to speak, and that will just be equal to v = a∙ω = a∙E/ħ.
- The red contour line sort of gets stretched out, like infinitely long, and the velocity becomes… What does it do? Does it go to ∞ , or to c?
Let’s think about this. For a particle at rest, we had this weird calculation. We had an angular momentum formula (for an electron) which we equated with the real-life +ħ/2 or −ħ/2 values of its spin. And so we got a numerical value for a. It was the Compton radius: the scattering radius for an electron. Let me copy it once again:
Just to bring this story a bit back to Earth, you should note the calculated value: a = 3.8616×10^{−13} m. We did then another weird calculation. We said all of the energy of the electron had to be packed in this cylinder that might of might not be there. The point was: the energy is finite, so that elementary wavefunction cannot have an infinite length in space. Indeed, assuming that the energy was distributed uniformly, we jotted down this formula, which reflects the formula for the volume of a cylinder:
E = π·a^{2}·l ⇔ l = E/(π·a^{2})
Using the value we got for the Compton scattering radius (a = 3.8616×10^{−13} m), we got an astronomical value for l. Let me write it out:
l = (8.19×10^{−14})/(π·14.9×10^{−26}) ≈ 0.175×10^{12} m
It is, literally, an astronomical value: 0.175×10^{12} m is 175 million kilometer, so that’s like the distance between the Sun and the Earth. We wrote, jokingly, that such space is too large to look for an electron and, hence, that we should really build a proper packet by making use of the Uncertainty Principle: allowing for uncertainty in the energy should, effectively, reduce the uncertainty in position.
But… Well… What if we use that value as the value for λ? We’d get that linear velocity, right? Let’s try it. The period is equal to T = T = 2π·(ħ/E) = h/E and λ = E/(π·a^{2}), so we write:We can write this as a function of m and the c and ħ constants only:
A weird formula but not necessarily nonsensical: we get a finite speed. Now, if the wavelength becomes somewhat less astronomical, we’ll get different values of course. I have a strange feeling that, with these formula, we should, somehow, be able to explain relativistic length contraction. But I will let you think about that as for now. Here I just wanted to show the geometry of the wavefunction a bit more in detail.
[1] The discussions on the mass of neutrinos are interesting in this regard. Scientists all felt the neutrino had to have some (rest) mass, so my instinct on this is theirs. In fact, only recently experimental confirmation came in, and the mass of the known neutrino flavors was estimated to be something like 0.12 eV/c^{2}. This mass combines the three known neutrino flavors. To understand this number, you should note it is the same order of magnitude of the equivalent mass of low-energy photons, like infrared or microwave radiation.
Wavefunctions as gravitational waves
This is the paper I always wanted to write. It is there now, and I think it is good – and that‘s an understatement. 🙂 It is probably best to download it as a pdf-file from the viXra.org site because this was a rather fast ‘copy and paste’ job from the Word version of the paper, so there may be issues with boldface notation (vector notation), italics and, most importantly, with formulas – which I, sadly, have to ‘snip’ into this WordPress blog, as they don’t have an easy copy function for mathematical formulas.
It’s great stuff. If you have been following my blog – and many of you have – you will want to digest this. 🙂
Abstract : This paper explores the implications of associating the components of the wavefunction with a physical dimension: force per unit mass – which is, of course, the dimension of acceleration (m/s^{2}) and gravitational fields. The classical electromagnetic field equations for energy densities, the Poynting vector and spin angular momentum are then re-derived by substituting the electromagnetic N/C unit of field strength (mass per unit charge) by the new N/kg = m/s^{2} dimension.
The results are elegant and insightful. For example, the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities, which establishes a physical normalization condition. Also, Schrödinger’s wave equation may then, effectively, be interpreted as a diffusion equation for energy, and the wavefunction itself can be interpreted as a propagating gravitational wave. Finally, as an added bonus, concepts such as the Compton scattering radius for a particle, spin angular momentum, and the boson-fermion dichotomy, can also be explained more intuitively.
While the approach offers a physical interpretation of the wavefunction, the author argues that the core of the Copenhagen interpretations revolves around the complementarity principle, which remains unchallenged because the interpretation of amplitude waves as traveling fields does not explain the particle nature of matter.
Introduction
This is not another introduction to quantum mechanics. We assume the reader is already familiar with the key principles and, importantly, with the basic math. We offer an interpretation of wave mechanics. As such, we do not challenge the complementarity principle: the physical interpretation of the wavefunction that is offered here explains the wave nature of matter only. It explains diffraction and interference of amplitudes but it does not explain why a particle will hit the detector not as a wave but as a particle. Hence, the Copenhagen interpretation of the wavefunction remains relevant: we just push its boundaries.
The basic ideas in this paper stem from a simple observation: the geometric similarity between the quantum-mechanical wavefunctions and electromagnetic waves is remarkably similar. The components of both waves are orthogonal to the direction of propagation and to each other. Only the relative phase differs : the electric and magnetic field vectors (E and B) have the same phase. In contrast, the phase of the real and imaginary part of the (elementary) wavefunction (ψ = a·e^{−i∙θ} = a∙cosθ – a∙sinθ) differ by 90 degrees (π/2).[1] Pursuing the analogy, we explore the following question: if the oscillating electric and magnetic field vectors of an electromagnetic wave carry the energy that one associates with the wave, can we analyze the real and imaginary part of the wavefunction in a similar way?
We show the answer is positive and remarkably straightforward. If the physical dimension of the electromagnetic field is expressed in newton per coulomb (force per unit charge), then the physical dimension of the components of the wavefunction may be associated with force per unit mass (newton per kg).[2] Of course, force over some distance is energy. The question then becomes: what is the energy concept here? Kinetic? Potential? Both?
The similarity between the energy of a (one-dimensional) linear oscillator (E = m·a^{2}·ω^{2}/2) and Einstein’s relativistic energy equation E = m∙c^{2} inspires us to interpret the energy as a two-dimensional oscillation of mass. To assist the reader, we construct a two-piston engine metaphor.[3] We then adapt the formula for the electromagnetic energy density to calculate the energy densities for the wave function. The results are elegant and intuitive: the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities. Schrödinger’s wave equation may then, effectively, be interpreted as a diffusion equation for energy itself.
As an added bonus, concepts such as the Compton scattering radius for a particle and spin angular, as well as the boson-fermion dichotomy can be explained in a fully intuitive way.[4]
Of course, such interpretation is also an interpretation of the wavefunction itself, and the immediate reaction of the reader is predictable: the electric and magnetic field vectors are, somehow, to be looked at as real vectors. In contrast, the real and imaginary components of the wavefunction are not. However, this objection needs to be phrased more carefully. First, it may be noted that, in a classical analysis, the magnetic force is a pseudovector itself.[5] Second, a suitable choice of coordinates may make quantum-mechanical rotation matrices irrelevant.[6]
Therefore, the author is of the opinion that this little paper may provide some fresh perspective on the question, thereby further exploring Einstein’s basic sentiment in regard to quantum mechanics, which may be summarized as follows: there must be some physical explanation for the calculated probabilities.[7]
We will, therefore, start with Einstein’s relativistic energy equation (E = mc^{2}) and wonder what it could possibly tell us.
I. Energy as a two-dimensional oscillation of mass
The structural similarity between the relativistic energy formula, the formula for the total energy of an oscillator, and the kinetic energy of a moving body, is striking:
- E = mc^{2}
- E = mω^{2}/2
- E = mv^{2}/2
In these formulas, ω, v and c all describe some velocity.[8] Of course, there is the 1/2 factor in the E = mω^{2}/2 formula[9], but that is exactly the point we are going to explore here: can we think of an oscillation in two dimensions, so it stores an amount of energy that is equal to E = 2·m·ω^{2}/2 = m·ω^{2}?
That is easy enough. Think, for example, of a V-2 engine with the pistons at a 90-degree angle, as illustrated below. The 90° angle makes it possible to perfectly balance the counterweight and the pistons, thereby ensuring smooth travel at all times. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down and provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring. Hence, we can describe it by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs.
Figure 1: Oscillations in two dimensions
If we assume there is no friction, we have a perpetuum mobile here. The compressed air and the rotating counterweight (which, combined with the crankshaft, acts as a flywheel[10]) store the potential energy. The moving masses of the pistons store the kinetic energy of the system.[11]
At this point, it is probably good to quickly review the relevant math. If the magnitude of the oscillation is equal to a, then the motion of the piston (or the mass on a spring) will be described by x = a·cos(ω·t + Δ).[12] Needless to say, Δ is just a phase factor which defines our t = 0 point, and ω is the natural angular frequency of our oscillator. Because of the 90° angle between the two cylinders, Δ would be 0 for one oscillator, and –π/2 for the other. Hence, the motion of one piston is given by x = a·cos(ω·t), while the motion of the other is given by x = a·cos(ω·t–π/2) = a·sin(ω·t).
The kinetic and potential energy of one oscillator (think of one piston or one spring only) can then be calculated as:
- K.E. = T = m·v^{2}/2 = (1/2)·m·ω^{2}·a^{2}·sin^{2}(ω·t + Δ)
- P.E. = U = k·x^{2}/2 = (1/2)·k·a^{2}·cos^{2}(ω·t + Δ)
The coefficient k in the potential energy formula characterizes the restoring force: F = −k·x. From the dynamics involved, it is obvious that k must be equal to m·ω^{2}. Hence, the total energy is equal to:
E = T + U = (1/2)· m·ω^{2}·a^{2}·[sin^{2}(ω·t + Δ) + cos^{2}(ω·t + Δ)] = m·a^{2}·ω^{2}/2
To facilitate the calculations, we will briefly assume k = m·ω^{2} and a are equal to 1. The motion of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin^{2}θ. Hence, the (instantaneous) change in kinetic energy at any point in time will be equal to:
d(sin^{2}θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ
Let us look at the second oscillator now. Just think of the second piston going up and down in the V-2 engine. Its motion is given by the sinθ function, which is equal to cos(θ−π /2). Hence, its kinetic energy is equal to sin^{2}(θ−π /2), and how it changes – as a function of θ – will be equal to:
2∙sin(θ−π /2)∙cos(θ−π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ
We have our perpetuum mobile! While transferring kinetic energy from one piston to the other, the crankshaft will rotate with a constant angular velocity: linear motion becomes circular motion, and vice versa, and the total energy that is stored in the system is T + U = ma^{2}ω^{2}.
We have a great metaphor here. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle. We know the wavefunction consist of a sine and a cosine: the cosine is the real component, and the sine is the imaginary component. Could they be equally real? Could each represent half of the total energy of our particle? Should we think of the c in our E = mc^{2} formula as an angular velocity?
These are sensible questions. Let us explore them.
II. The wavefunction as a two-dimensional oscillation
The elementary wavefunction is written as:
ψ = a·e^{−i[E·t − p∙x]/ħ} = a·e^{−i[E·t − p∙x]/ħ} = a·cos(p∙x/ħ – E∙t/ħ) + i·a·sin(p∙x/ħ – E∙t/ħ)
When considering a particle at rest (p = 0) this reduces to:
ψ = a·e^{−i∙E·t/ħ} = a·cos(–E∙t/ħ) + i·a·sin(–E∙t/ħ) = a·cos(E∙t/ħ) – i·a·sin(E∙t/ħ)
Let us remind ourselves of the geometry involved, which is illustrated below. Note that the argument of the wavefunction rotates clockwise with time, while the mathematical convention for measuring the phase angle (ϕ) is counter-clockwise.
Figure 2: Euler’s formula
If we assume the momentum p is all in the x-direction, then the p and x vectors will have the same direction, and p∙x/ħ reduces to p∙x/ħ. Most illustrations – such as the one below – will either freeze x or, else, t. Alternatively, one can google web animations varying both. The point is: we also have a two-dimensional oscillation here. These two dimensions are perpendicular to the direction of propagation of the wavefunction. For example, if the wavefunction propagates in the x-direction, then the oscillations are along the y– and z-axis, which we may refer to as the real and imaginary axis. Note how the phase difference between the cosine and the sine – the real and imaginary part of our wavefunction – appear to give some spin to the whole. I will come back to this.
Figure 3: Geometric representation of the wavefunction
Hence, if we would say these oscillations carry half of the total energy of the particle, then we may refer to the real and imaginary energy of the particle respectively, and the interplay between the real and the imaginary part of the wavefunction may then describe how energy propagates through space over time.
Let us consider, once again, a particle at rest. Hence, p = 0 and the (elementary) wavefunction reduces to ψ = a·e^{−i∙E·t/ħ}. Hence, the angular velocity of both oscillations, at some point x, is given by ω = -E/ħ. Now, the energy of our particle includes all of the energy – kinetic, potential and rest energy – and is, therefore, equal to E = mc^{2}.
Can we, somehow, relate this to the m·a^{2}·ω^{2} energy formula for our V-2 perpetuum mobile? Our wavefunction has an amplitude too. Now, if the oscillations of the real and imaginary wavefunction store the energy of our particle, then their amplitude will surely matter. In fact, the energy of an oscillation is, in general, proportional to the square of the amplitude: E µ a^{2}. We may, therefore, think that the a^{2} factor in the E = m·a^{2}·ω^{2} energy will surely be relevant as well.
However, here is a complication: an actual particle is localized in space and can, therefore, not be represented by the elementary wavefunction. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude a_{k}, and their own ω_{i} = -E_{i}/ħ. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. To calculate the contribution of each wave to the total, both a_{i} as well as E_{i} will matter.
What is E_{i}? E_{i} varies around some average E, which we can associate with some average mass m: m = E/c^{2}. The Uncertainty Principle kicks in here. The analysis becomes more complicated, but a formula such as the one below might make sense:We can re-write this as:What is the meaning of this equation? We may look at it as some sort of physical normalization condition when building up the Fourier sum. Of course, we should relate this to the mathematical normalization condition for the wavefunction. Our intuition tells us that the probabilities must be related to the energy densities, but how exactly? We will come back to this question in a moment. Let us first think some more about the enigma: what is mass?
Before we do so, let us quickly calculate the value of c^{2}ħ^{2}: it is about 1´10^{–}^{51} N^{2}∙m^{4}. Let us also do a dimensional analysis: the physical dimensions of the E = m·a^{2}·ω^{2} equation make sense if we express m in kg, a in m, and ω in rad/s. We then get: [E] = kg∙m^{2}/s^{2} = (N∙s^{2}/m)∙m^{2}/s^{2} = N∙m = J. The dimensions of the left- and right-hand side of the physical normalization condition is N^{3}∙m^{5}.
III. What is mass?
We came up, playfully, with a meaningful interpretation for energy: it is a two-dimensional oscillation of mass. But what is mass? A new aether theory is, of course, not an option, but then what is it that is oscillating? To understand the physics behind equations, it is always good to do an analysis of the physical dimensions in the equation. Let us start with Einstein’s energy equation once again. If we want to look at mass, we should re-write it as m = E/c^{2}:
[m] = [E/c^{2}] = J/(m/s)^{2} = N·m∙s^{2}/m^{2} = N·s^{2}/m = kg
This is not very helpful. It only reminds us of Newton’s definition of a mass: mass is that what gets accelerated by a force. At this point, we may want to think of the physical significance of the absolute nature of the speed of light. Einstein’s E = mc^{2} equation implies we can write the ratio between the energy and the mass of any particle is always the same, so we can write, for example:This reminds us of the ω^{2}= C^{–}^{1}/L or ω^{2} = k/m of harmonic oscillators once again.[13] The key difference is that the ω^{2}= C^{–}^{1}/L and ω^{2} = k/m formulas introduce two or more degrees of freedom.[14] In contrast, c^{2}= E/m for any particle, always. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in one physical space only: our spacetime. Hence, the speed of light c emerges here as the defining property of spacetime – the resonant frequency, so to speak. We have no further degrees of freedom here.
The Planck-Einstein relation (for photons) and the de Broglie equation (for matter-particles) have an interesting feature: both imply that the energy of the oscillation is proportional to the frequency, with Planck’s constant as the constant of proportionality. Now, for one-dimensional oscillations – think of a guitar string, for example – we know the energy will be proportional to the square of the frequency. It is a remarkable observation: the two-dimensional matter-wave, or the electromagnetic wave, gives us two waves for the price of one, so to speak, each carrying half of the total energy of the oscillation but, as a result, we get a proportionality between E and f instead of between E and f^{2}.
However, such reflections do not answer the fundamental question we started out with: what is mass? At this point, it is hard to go beyond the circular definition that is implied by Einstein’s formula: energy is a two-dimensional oscillation of mass, and mass packs energy, and c emerges us as the property of spacetime that defines how exactly.
When everything is said and done, this does not go beyond stating that mass is some scalar field. Now, a scalar field is, quite simply, some real number that we associate with a position in spacetime. The Higgs field is a scalar field but, of course, the theory behind it goes much beyond stating that we should think of mass as some scalar field. The fundamental question is: why and how does energy, or matter, condense into elementary particles? That is what the Higgs mechanism is about but, as this paper is exploratory only, we cannot even start explaining the basics of it.
What we can do, however, is look at the wave equation again (Schrödinger’s equation), as we can now analyze it as an energy diffusion equation.
IV. Schrödinger’s equation as an energy diffusion equation
The interpretation of Schrödinger’s equation as a diffusion equation is straightforward. Feynman (Lectures, III-16-1) briefly summarizes it as follows:
“We can think of Schrödinger’s equation as describing the diffusion of the probability amplitude from one point to the next. […] But the imaginary coefficient in front of the derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out along a thin tube. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”[17]
Let us review the basic math. For a particle moving in free space – with no external force fields acting on it – there is no potential (U = 0) and, therefore, the Uψ term disappears. Therefore, Schrödinger’s equation reduces to:
∂ψ(x, t)/∂t = i·(1/2)·(ħ/m_{eff})·∇^{2}ψ(x, t)
The ubiquitous diffusion equation in physics is:
∂φ(x, t)/∂t = D·∇^{2}φ(x, t)
The structural similarity is obvious. The key difference between both equations is that the wave equation gives us two equations for the price of one. Indeed, because ψ is a complex-valued function, with a real and an imaginary part, we get the following equations[18]:
- Re(∂ψ/∂t) = −(1/2)·(ħ/m_{eff})·Im(∇^{2}ψ)
- Im(∂ψ/∂t) = (1/2)·(ħ/m_{eff})·Re(∇^{2}ψ)
These equations make us think of the equations for an electromagnetic wave in free space (no stationary charges or currents):
- ∂B/∂t = –∇×E
- ∂E/∂t = c^{2}∇×B
The above equations effectively describe a propagation mechanism in spacetime, as illustrated below.
Figure 4: Propagation mechanisms
The Laplacian operator (∇^{2}), when operating on a scalar quantity, gives us a flux density, i.e. something expressed per square meter (1/m^{2}). In this case, it is operating on ψ(x, t), so what is the dimension of our wavefunction ψ(x, t)? To answer that question, we should analyze the diffusion constant in Schrödinger’s equation, i.e. the (1/2)·(ħ/m_{eff}) factor:
- As a mathematical constant of proportionality, it will quantify the relationship between both derivatives (i.e. the time derivative and the Laplacian);
- As a physical constant, it will ensure the physical dimensions on both sides of the equation are compatible.
Now, the ħ/m_{eff} factor is expressed in (N·m·s)/(N· s^{2}/m) = m^{2}/s. Hence, it does ensure the dimensions on both sides of the equation are, effectively, the same: ∂ψ/∂t is a time derivative and, therefore, its dimension is s^{–}^{1} while, as mentioned above, the dimension of ∇^{2}ψ is m^{–}^{2}. However, this does not solve our basic question: what is the dimension of the real and imaginary part of our wavefunction?
At this point, mainstream physicists will say: it does not have a physical dimension, and there is no geometric interpretation of Schrödinger’s equation. One may argue, effectively, that its argument, (p∙x – E∙t)/ħ, is just a number and, therefore, that the real and imaginary part of ψ is also just some number.
To this, we may object that ħ may be looked as a mathematical scaling constant only. If we do that, then the argument of ψ will, effectively, be expressed in action units, i.e. in N·m·s. It then does make sense to also associate a physical dimension with the real and imaginary part of ψ. What could it be?
We may have a closer look at Maxwell’s equations for inspiration here. The electric field vector is expressed in newton (the unit of force) per unit of charge (coulomb). Now, there is something interesting here. The physical dimension of the magnetic field is N/C divided by m/s.[19] We may write B as the following vector cross-product: B = (1/c)∙e_{x}×E, with e_{x} the unit vector pointing in the x-direction (i.e. the direction of propagation of the wave). Hence, we may associate the (1/c)∙e_{x}× operator, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by i also amounts to a rotation by 90° degrees. Hence, we may boldly write: B = (1/c)∙e_{x}×E = (1/c)∙i∙E. This allows us to also geometrically interpret Schrödinger’s equation in the way we interpreted it above (see Figure 3).[20]
Still, we have not answered the question as to what the physical dimension of the real and imaginary part of our wavefunction should be. At this point, we may be inspired by the structural similarity between Newton’s and Coulomb’s force laws:Hence, if the electric field vector E is expressed in force per unit charge (N/C), then we may want to think of associating the real part of our wavefunction with a force per unit mass (N/kg). We can, of course, do a substitution here, because the mass unit (1 kg) is equivalent to 1 N·s^{2}/m. Hence, our N/kg dimension becomes:
N/kg = N/(N·s^{2}/m)= m/s^{2}
What is this: m/s^{2}? Is that the dimension of the a·cosθ term in the a·e^{−iθ }= a·cosθ − i·a·sinθ wavefunction?
My answer is: why not? Think of it: m/s^{2} is the physical dimension of acceleration: the increase or decrease in velocity (m/s) per second. It ensures the wavefunction for any particle – matter-particles or particles with zero rest mass (photons) – and the associated wave equation (which has to be the same for all, as the spacetime we live in is one) are mutually consistent.
In this regard, we should think of how we would model a gravitational wave. The physical dimension would surely be the same: force per mass unit. It all makes sense: wavefunctions may, perhaps, be interpreted as traveling distortions of spacetime, i.e. as tiny gravitational waves.
V. Energy densities and flows
Pursuing the geometric equivalence between the equations for an electromagnetic wave and Schrödinger’s equation, we can now, perhaps, see if there is an equivalent for the energy density. For an electromagnetic wave, we know that the energy density is given by the following formula:E and B are the electric and magnetic field vector respectively. The Poynting vector will give us the directional energy flux, i.e. the energy flow per unit area per unit time. We write:Needless to say, the ∇∙ operator is the divergence and, therefore, gives us the magnitude of a (vector) field’s source or sink at a given point. To be precise, the divergence gives us the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. In this case, it gives us the volume density of the flux of S.
We can analyze the dimensions of the equation for the energy density as follows:
- E is measured in newton per coulomb, so [E∙E] = [E^{2}] = N^{2}/C^{2}.
- B is measured in (N/C)/(m/s), so we get [B∙B] = [B^{2}] = (N^{2}/C^{2})·(s^{2}/m^{2}). However, the dimension of our c^{2} factor is (m^{2}/s^{2}) and so we’re also left with N^{2}/C^{2}.
- The ϵ_{0} is the electric constant, aka as the vacuum permittivity. As a physical constant, it should ensure the dimensions on both sides of the equation work out, and they do: [ε_{0}] = C^{2}/(N·m^{2}) and, therefore, if we multiply that with N^{2}/C^{2}, we find that u is expressed in J/m^{3}.[21]
Replacing the newton per coulomb unit (N/C) by the newton per kg unit (N/kg) in the formulas above should give us the equivalent of the energy density for the wavefunction. We just need to substitute ϵ_{0} for an equivalent constant. We may to give it a try. If the energy densities can be calculated – which are also mass densities, obviously – then the probabilities should be proportional to them.
Let us first see what we get for a photon, assuming the electromagnetic wave represents its wavefunction. Substituting B for (1/c)∙i∙E or for −(1/c)∙i∙E gives us the following result:Zero!? An unexpected result! Or not? We have no stationary charges and no currents: only an electromagnetic wave in free space. Hence, the local energy conservation principle needs to be respected at all points in space and in time. The geometry makes sense of the result: for an electromagnetic wave, the magnitudes of E and B reach their maximum, minimum and zero point simultaneously, as shown below.[22] This is because their phase is the same.
Figure 5: Electromagnetic wave: E and B
Should we expect a similar result for the energy densities that we would associate with the real and imaginary part of the matter-wave? For the matter-wave, we have a phase difference between a·cosθ and a·sinθ, which gives a different picture of the propagation of the wave (see Figure 3).[23] In fact, the geometry of the suggestion suggests some inherent spin, which is interesting. I will come back to this. Let us first guess those densities. Making abstraction of any scaling constants, we may write:We get what we hoped to get: the absolute square of our amplitude is, effectively, an energy density !
|ψ|^{2 } = |a·e^{−i∙E·t/ħ}|^{2 }= a^{2 }= u
This is very deep. A photon has no rest mass, so it borrows and returns energy from empty space as it travels through it. In contrast, a matter-wave carries energy and, therefore, has some (rest) mass. It is therefore associated with an energy density, and this energy density gives us the probabilities. Of course, we need to fine-tune the analysis to account for the fact that we have a wave packet rather than a single wave, but that should be feasible.
As mentioned, the phase difference between the real and imaginary part of our wavefunction (a cosine and a sine function) appear to give some spin to our particle. We do not have this particularity for a photon. Of course, photons are bosons, i.e. spin-zero particles, while elementary matter-particles are fermions with spin-1/2. Hence, our geometric interpretation of the wavefunction suggests that, after all, there may be some more intuitive explanation of the fundamental dichotomy between bosons and fermions, which puzzled even Feynman:
“Why is it that particles with half-integral spin are Fermi particles, whereas particles with integral spin are Bose particles? We apologize for the fact that we cannot give you an elementary explanation. An explanation has been worked out by Pauli from complicated arguments of quantum field theory and relativity. He has shown that the two must necessarily go together, but we have not been able to find a way of reproducing his arguments on an elementary level. It appears to be one of the few places in physics where there is a rule which can be stated very simply, but for which no one has found a simple and easy explanation. The explanation is deep down in relativistic quantum mechanics. This probably means that we do not have a complete understanding of the fundamental principle involved.” (Feynman, Lectures, III-4-1)
The physical interpretation of the wavefunction, as presented here, may provide some better understanding of ‘the fundamental principle involved’: the physical dimension of the oscillation is just very different. That is all: it is force per unit charge for photons, and force per unit mass for matter-particles. We will examine the question of spin somewhat more carefully in section VII. Let us first examine the matter-wave some more.
VI. Group and phase velocity of the matter-wave
The geometric representation of the matter-wave (see Figure 3) suggests a traveling wave and, yes, of course: the matter-wave effectively travels through space and time. But what is traveling, exactly? It is the pulse – or the signal – only: the phase velocity of the wave is just a mathematical concept and, even in our physical interpretation of the wavefunction, the same is true for the group velocity of our wave packet. The oscillation is two-dimensional, but perpendicular to the direction of travel of the wave. Hence, nothing actually moves with our particle.
Here, we should also reiterate that we did not answer the question as to what is oscillating up and down and/or sideways: we only associated a physical dimension with the components of the wavefunction – newton per kg (force per unit mass), to be precise. We were inspired to do so because of the physical dimension of the electric and magnetic field vectors (newton per coulomb, i.e. force per unit charge) we associate with electromagnetic waves which, for all practical purposes, we currently treat as the wavefunction for a photon. This made it possible to calculate the associated energy densities and a Poynting vector for energy dissipation. In addition, we showed that Schrödinger’s equation itself then becomes a diffusion equation for energy. However, let us now focus some more on the asymmetry which is introduced by the phase difference between the real and the imaginary part of the wavefunction. Look at the mathematical shape of the elementary wavefunction once again:
ψ = a·e^{−i[E·t − p∙x]/ħ} = a·e^{−i[E·t − p∙x]/ħ} = a·cos(p∙x/ħ − E∙t/ħ) + i·a·sin(p∙x/ħ − E∙t/ħ)
The minus sign in the argument of our sine and cosine function defines the direction of travel: an F(x−v∙t) wavefunction will always describe some wave that is traveling in the positive x-direction (with c the wave velocity), while an F(x+v∙t) wavefunction will travel in the negative x-direction. For a geometric interpretation of the wavefunction in three dimensions, we need to agree on how to define i or, what amounts to the same, a convention on how to define clockwise and counterclockwise directions: if we look at a clock from the back, then its hand will be moving counterclockwise. So we need to establish the equivalent of the right-hand rule. However, let us not worry about that now. Let us focus on the interpretation. To ease the analysis, we’ll assume we’re looking at a particle at rest. Hence, p = 0, and the wavefunction reduces to:
ψ = a·e^{−i∙E·t/ħ} = a·cos(−E∙t/ħ) + i·a·sin(−E_{0}∙t/ħ) = a·cos(E_{0}∙t/ħ) − i·a·sin(E_{0}∙t/ħ)
E_{0} is, of course, the rest mass of our particle and, now that we are here, we should probably wonder whose time t we are talking about: is it our time, or is the proper time of our particle? Well… In this situation, we are both at rest so it does not matter: t is, effectively, the proper time so perhaps we should write it as t_{0}. It does not matter. You can see what we expect to see: E_{0}/ħ pops up as the natural frequency of our matter-particle: (E_{0}/ħ)∙t = ω∙t. Remembering the ω = 2π·f = 2π/T and T = 1/f formulas, we can associate a period and a frequency with this wave, using the ω = 2π·f = 2π/T. Noting that ħ = h/2π, we find the following:
T = 2π·(ħ/E_{0}) = h/E_{0} ⇔ f = E_{0}/h = m_{0}c^{2}/h
This is interesting, because we can look at the period as a natural unit of time for our particle. What about the wavelength? That is tricky because we need to distinguish between group and phase velocity here. The group velocity (v_{g}) should be zero here, because we assume our particle does not move. In contrast, the phase velocity is given by v_{p} = λ·f = (2π/k)·(ω/2π) = ω/k. In fact, we’ve got something funny here: the wavenumber k = p/ħ is zero, because we assume the particle is at rest, so p = 0. So we have a division by zero here, which is rather strange. What do we get assuming the particle is not at rest? We write:
v_{p} = ω/k = (E/ħ)/(p/ħ) = E/p = E/(m·v_{g}) = (m·c^{2})/(m·v_{g}) = c^{2}/v_{g}
This is interesting: it establishes a reciprocal relation between the phase and the group velocity, with c as a simple scaling constant. Indeed, the graph below shows the shape of the function does not change with the value of c, and we may also re-write the relation above as:
v_{p}/c = β_{p} = c/v_{p} = 1/β_{g} = 1/(c/v_{p})
Figure 6: Reciprocal relation between phase and group velocity
We can also write the mentioned relationship as v_{p}·v_{g} = c^{2}, which reminds us of the relationship between the electric and magnetic constant (1/ε_{0})·(1/μ_{0}) = c^{2}. This is interesting in light of the fact we can re-write this as (c·ε_{0})·(c·μ_{0}) = 1, which shows electricity and magnetism are just two sides of the same coin, so to speak.[24]
Interesting, but how do we interpret the math? What about the implications of the zero value for wavenumber k = p/ħ. We would probably like to think it implies the elementary wavefunction should always be associated with some momentum, because the concept of zero momentum clearly leads to weird math: something times zero cannot be equal to c^{2}! Such interpretation is also consistent with the Uncertainty Principle: if Δx·Δp ≥ ħ, then neither Δx nor Δp can be zero. In other words, the Uncertainty Principle tells us that the idea of a pointlike particle actually being at some specific point in time and in space does not make sense: it has to move. It tells us that our concept of dimensionless points in time and space are mathematical notions only. Actual particles – including photons – are always a bit spread out, so to speak, and – importantly – they have to move.
For a photon, this is self-evident. It has no rest mass, no rest energy, and, therefore, it is going to move at the speed of light itself. We write: p = m·c = m·c^{2}/c = E/c. Using the relationship above, we get:
v_{p} = ω/k = (E/ħ)/(p/ħ) = E/p = c ⇒ v_{g} = c^{2}/v_{p} = c^{2}/c = c
This is good: we started out with some reflections on the matter-wave, but here we get an interpretation of the electromagnetic wave as a wavefunction for the photon. But let us get back to our matter-wave. In regard to our interpretation of a particle having to move, we should remind ourselves, once again, of the fact that an actual particle is always localized in space and that it can, therefore, not be represented by the elementary wavefunction ψ = a·e^{−i[E·t − p∙x]/ħ} or, for a particle at rest, the ψ = a·e^{−i∙E·t/ħ} function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude a_{i}, and their own ω_{i} = −E_{i}/ħ. Indeed, in section II, we showed that each of these wavefunctions will contribute some energy to the total energy of the wave packet and that, to calculate the contribution of each wave to the total, both a_{i} as well as E_{i} matter. This may or may not resolve the apparent paradox. Let us look at the group velocity.
To calculate a meaningful group velocity, we must assume the v_{g} = ∂ω_{i}/∂k_{i} = ∂(E_{i}/ħ)/∂(p_{i}/ħ) = ∂(E_{i})/∂(p_{i}) exists. So we must have some dispersion relation. How do we calculate it? We need to calculate ω_{i} as a function of k_{i}_{ }here, or E_{i} as a function of p_{i}. How do we do that? Well… There are a few ways to go about it but one interesting way of doing it is to re-write Schrödinger’s equation as we did, i.e. by distinguishing the real and imaginary parts of the ∂ψ/∂t =i·[ħ/(2m)]·∇^{2}ψ wave equation and, hence, re-write it as the following pair of two equations:
- Re(∂ψ/∂t) = −[ħ/(2m_{eff})]·Im(∇^{2}ψ) ⇔ ω·cos(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·cos(kx − ωt)
- Im(∂ψ/∂t) = [ħ/(2m_{eff})]·Re(∇^{2}ψ) ⇔ ω·sin(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·sin(kx − ωt)
Both equations imply the following dispersion relation:
ω = ħ·k^{2}/(2m_{eff})
Of course, we need to think about the subscripts now: we have ω_{i}, k_{i}, but… What about m_{eff} or, dropping the subscript, m? Do we write it as m_{i}? If so, what is it? Well… It is the equivalent mass of E_{i} obviously, and so we get it from the mass-energy equivalence relation: m_{i} = E_{i}/c^{2}. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too. Here, I should refer back to Section II: E_{i} varies around some average energy E and, therefore, the Uncertainty Principle kicks in.
VII. Explaining spin
The elementary wavefunction vector – i.e. the vector sum of the real and imaginary component – rotates around the x-axis, which gives us the direction of propagation of the wave (see Figure 3). Its magnitude remains constant. In contrast, the magnitude of the electromagnetic vector – defined as the vector sum of the electric and magnetic field vectors – oscillates between zero and some maximum (see Figure 5).
We already mentioned that the rotation of the wavefunction vector appears to give some spin to the particle. Of course, a circularly polarized wave would also appear to have spin (think of the E and B vectors rotating around the direction of propagation – as opposed to oscillating up and down or sideways only). In fact, a circularly polarized light does carry angular momentum, as the equivalent mass of its energy may be thought of as rotating as well. But so here we are looking at a matter-wave.
The basic idea is the following: if we look at ψ = a·e^{−i∙E·t/ħ} as some real vector – as a two-dimensional oscillation of mass, to be precise – then we may associate its rotation around the direction of propagation with some torque. The illustration below reminds of the math here.
Figure 7: Torque and angular momentum vectors
A torque on some mass about a fixed axis gives it angular momentum, which we can write as the vector cross-product L = r×p or, perhaps easier for our purposes here as the product of an angular velocity (ω) and rotational inertia (I), aka as the moment of inertia or the angular mass. We write:
L = I·ω
Note we can write L and ω in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = I·ω (no boldface). We can now do some calculations. Let us start with the angular velocity. In our previous posts, we showed that the period of the matter-wave is equal to T = 2π·(ħ/E_{0}). Hence, the angular velocity must be equal to:
ω = 2π/[2π·(ħ/E_{0})] = E_{0}/ħ
We also know the distance r, so that is the magnitude of r in the L = r×p vector cross-product: it is just a, so that is the magnitude of ψ = a·e^{−i∙E·t/ħ}. Now, the momentum (p) is the product of a linear velocity (v) – in this case, the tangential velocity – and some mass (m): p = m·v. If we switch to scalar instead of vector quantities, then the (tangential) velocity is given by v = r·ω. So now we only need to think about what we should use for m or, if we want to work with the angular velocity (ω), the angular mass (I). Here we need to make some assumption about the mass (or energy) distribution. Now, it may or may not sense to assume the energy in the oscillation – and, therefore, the mass – is distributed uniformly. In that case, we may use the formula for the angular mass of a solid cylinder: I = m·r^{2}/2. If we keep the analysis non-relativistic, then m = m_{0}. Of course, the energy-mass equivalence tells us that m_{0} = E_{0}/c^{2}. Hence, this is what we get:
L = I·ω = (m_{0}·r^{2}/2)·(E_{0}/ħ) = (1/2)·a^{2}·(E_{0}/c^{2})·(E_{0}/ħ) = a^{2}·E_{0}^{2}/(2·ħ·c^{2})
Does it make sense? Maybe. Maybe not. Let us do a dimensional analysis: that won’t check our logic, but it makes sure we made no mistakes when mapping mathematical and physical spaces. We have m^{2}·J^{2} = m^{2}·N^{2}·m^{2} in the numerator and N·m·s·m^{2}/s^{2} in the denominator. Hence, the dimensions work out: we get N·m·s as the dimension for L, which is, effectively, the physical dimension of angular momentum. It is also the action dimension, of course, and that cannot be a coincidence. Also note that the E = mc^{2} equation allows us to re-write it as:
L = a^{2}·E_{0}^{2}/(2·ħ·c^{2})
Of course, in quantum mechanics, we associate spin with the magnetic moment of a charged particle, not with its mass as such. Is there way to link the formula above to the one we have for the quantum-mechanical angular momentum, which is also measured in N·m·s units, and which can only take on one of two possible values: J = +ħ/2 and −ħ/2? It looks like a long shot, right? How do we go from (1/2)·a^{2}·m_{0}^{2}/ħ to ± (1/2)∙ħ? Let us do a numerical example. The energy of an electron is typically 0.510 MeV » 8.1871×10^{−14} N∙m, and a… What value should we take for a?
We have an obvious trio of candidates here: the Bohr radius, the classical electron radius (aka the Thompon scattering length), and the Compton scattering radius.
Let us start with the Bohr radius, so that is about 0.×10^{−10} N∙m. We get L = a^{2}·E_{0}^{2}/(2·ħ·c^{2}) = 9.9×10^{−31} N∙m∙s. Now that is about 1.88×10^{4} times ħ/2. That is a huge factor. The Bohr radius cannot be right: we are not looking at an electron in an orbital here. To show it does not make sense, we may want to double-check the analysis by doing the calculation in another way. We said each oscillation will always pack 6.626070040(81)×10^{−34} joule in energy. So our electron should pack about 1.24×10^{−20} oscillations. The angular momentum (L) we get when using the Bohr radius for a and the value of 6.626×10^{−34} joule for E_{0} and the Bohr radius is equal to 6.49×10^{−59} N∙m∙s. So that is the angular momentum per oscillation. When we multiply this with the number of oscillations (1.24×10^{−20}), we get about 8.01×10^{−51} N∙m∙s, so that is a totally different number.
The classical electron radius is about 2.818×10^{−15} m. We get an L that is equal to about 2.81×10^{−39} N∙m∙s, so now it is a tiny fraction of ħ/2! Hence, this leads us nowhere. Let us go for our last chance to get a meaningful result! Let us use the Compton scattering length, so that is about 2.42631×10^{−12} m.
This gives us an L of 2.08×10^{−33} N∙m∙s, which is only 20 times ħ. This is not so bad, but it is good enough? Let us calculate it the other way around: what value should we take for a so as to ensure L = a^{2}·E_{0}^{2}/(2·ħ·c^{2}) = ħ/2? Let us write it out:
In fact, this is the formula for the so-called reduced Compton wavelength. This is perfect. We found what we wanted to find. Substituting this value for a (you can calculate it: it is about 3.8616×10^{−33} m), we get what we should find:
This is a rather spectacular result, and one that would – a priori – support the interpretation of the wavefunction that is being suggested in this paper.^{ }
VIII. The boson-fermion dichotomy
Let us do some more thinking on the boson-fermion dichotomy. Again, we should remind ourselves that an actual particle is localized in space and that it can, therefore, not be represented by the elementary wavefunction ψ = a·e^{−i[E·t − p∙x]/ħ} or, for a particle at rest, the ψ = a·e^{−i∙E·t/ħ} function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude a_{i}, and their own ω_{i} = −E_{i}/ħ. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. Now, we can have another wild but logical theory about this.
Think of the apparent right-handedness of the elementary wavefunction: surely, Nature can’t be bothered about our convention of measuring phase angles clockwise or counterclockwise. Also, the angular momentum can be positive or negative: J = +ħ/2 or −ħ/2. Hence, we would probably like to think that an actual particle – think of an electron, or whatever other particle you’d think of – may consist of right-handed as well as left-handed elementary waves. To be precise, we may think they either consist of (elementary) right-handed waves or, else, of (elementary) left-handed waves. An elementary right-handed wave would be written as:
ψ(θ_{i}) = a_{i}·(cosθ_{i} + i·sinθ_{i})
In contrast, an elementary left-handed wave would be written as:
ψ(θ_{i}) = a_{i}·(cosθ_{i} − i·sinθ_{i})
How does that work out with the E_{0}·t argument of our wavefunction? Position is position, and direction is direction, but time? Time has only one direction, but Nature surely does not care how we count time: counting like 1, 2, 3, etcetera or like −1, −2, −3, etcetera is just the same. If we count like 1, 2, 3, etcetera, then we write our wavefunction like:
ψ = a·cos(E_{0}∙t/ħ) − i·a·sin(E_{0}∙t/ħ)
If we count time like −1, −2, −3, etcetera then we write it as:
ψ = a·cos(−E_{0}∙t/ħ) − i·a·sin(−E_{0}∙t/ħ)= a·cos(E_{0}∙t/ħ) + i·a·sin(E_{0}∙t/ħ)
Hence, it is just like the left- or right-handed circular polarization of an electromagnetic wave: we can have both for the matter-wave too! This, then, should explain why we can have either positive or negative quantum-mechanical spin (+ħ/2 or −ħ/2). It is the usual thing: we have two mathematical possibilities here, and so we must have two physical situations that correspond to it.
It is only natural. If we have left- and right-handed photons – or, generalizing, left- and right-handed bosons – then we should also have left- and right-handed fermions (electrons, protons, etcetera). Back to the dichotomy. The textbook analysis of the dichotomy between bosons and fermions may be epitomized by Richard Feynman’s Lecture on it (Feynman, III-4), which is confusing and – I would dare to say – even inconsistent: how are photons or electrons supposed to know that they need to interfere with a positive or a negative sign? They are not supposed to know anything: knowledge is part of our interpretation of whatever it is that is going on there.
Hence, it is probably best to keep it simple, and think of the dichotomy in terms of the different physical dimensions of the oscillation: newton per kg versus newton per coulomb. And then, of course, we should also note that matter-particles have a rest mass and, therefore, actually carry charge. Photons do not. But both are two-dimensional oscillations, and the point is: the so-called vacuum – and the rest mass of our particle (which is zero for the photon and non-zero for everything else) – give us the natural frequency for both oscillations, which is beautifully summed up in that remarkable equation for the group and phase velocity of the wavefunction, which applies to photons as well as matter-particles:
(v_{phase}·c)·(v_{group}·c) = 1 ⇔ v_{p}·v_{g} = c^{2}
The final question then is: why are photons spin-zero particles? Well… We should first remind ourselves of the fact that they do have spin when circularly polarized.[25] Here we may think of the rotation of the equivalent mass of their energy. However, if they are linearly polarized, then there is no spin. Even for circularly polarized waves, the spin angular momentum of photons is a weird concept. If photons have no (rest) mass, then they cannot carry any charge. They should, therefore, not have any magnetic moment. Indeed, what I wrote above shows an explanation of quantum-mechanical spin requires both mass as well as charge.[26]
IX. Concluding remarks
There are, of course, other ways to look at the matter – literally. For example, we can imagine two-dimensional oscillations as circular rather than linear oscillations. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation – around any axis – will be some combination of a rotation around the two other axes. Hence, we may want to think of a two-dimensional oscillation as an oscillation of a polar and azimuthal angle.
Figure 8: Two-dimensional circular movement
The point of this paper is not to make any definite statements. That would be foolish. Its objective is just to challenge the simplistic mainstream viewpoint on the reality of the wavefunction. Stating that it is a mathematical construct only without physical significance amounts to saying it has no meaning at all. That is, clearly, a non-sustainable proposition.
The interpretation that is offered here looks at amplitude waves as traveling fields. Their physical dimension may be expressed in force per mass unit, as opposed to electromagnetic waves, whose amplitudes are expressed in force per (electric) charge unit. Also, the amplitudes of matter-waves incorporate a phase factor, but this may actually explain the rather enigmatic dichotomy between fermions and bosons and is, therefore, an added bonus.
The interpretation that is offered here has some advantages over other explanations, as it explains the how of diffraction and interference. However, while it offers a great explanation of the wave nature of matter, it does not explain its particle nature: while we think of the energy as being spread out, we will still observe electrons and photons as pointlike particles once they hit the detector. Why is it that a detector can sort of ‘hook’ the whole blob of energy, so to speak?
The interpretation of the wavefunction that is offered here does not explain this. Hence, the complementarity principle of the Copenhagen interpretation of the wavefunction surely remains relevant.
Appendix 1: The de Broglie relations and energy
The 1/2 factor in Schrödinger’s equation is related to the concept of the effective mass (m_{eff}). It is easy to make the wrong calculations. For example, when playing with the famous de Broglie relations – aka as the matter-wave equations – one may be tempted to derive the following energy concept:
- E = h·f and p = h/λ. Therefore, f = E/h and λ = p/h.
- v = f·λ = (E/h)∙(p/h) = E/p
- p = m·v. Therefore, E = v·p = m·v^{2}
E = m·v^{2}? This resembles the E = mc^{2} equation and, therefore, one may be enthused by the discovery, especially because the m·v^{2} also pops up when working with the Least Action Principle in classical mechanics, which states that the path that is followed by a particle will minimize the following integral:Now, we can choose any reference point for the potential energy but, to reflect the energy conservation law, we can select a reference point that ensures the sum of the kinetic and the potential energy is zero throughout the time interval. If the force field is uniform, then the integrand will, effectively, be equal to KE − PE = m·v^{2}.[27]
However, that is classical mechanics and, therefore, not so relevant in the context of the de Broglie equations, and the apparent paradox should be solved by distinguishing between the group and the phase velocity of the matter wave.
Appendix 2: The concept of the effective mass
The effective mass – as used in Schrödinger’s equation – is a rather enigmatic concept. To make sure we are making the right analysis here, I should start by noting you will usually see Schrödinger’s equation written as:This formulation includes a term with the potential energy (U). In free space (no potential), this term disappears, and the equation can be re-written as:
∂ψ(x, t)/∂t = i·(1/2)·(ħ/m_{eff})·∇^{2}ψ(x, t)
We just moved the i·ħ coefficient to the other side, noting that 1/i = –i. Now, in one-dimensional space, and assuming ψ is just the elementary wavefunction (so we substitute a·e^{−i∙[E·t − p∙x]/ħ} for ψ), this implies the following:
−a·i·(E/ħ)·e^{−}i∙^{[E·t − p∙x]/ħ} = −i·(ħ/2m_{eff})·a·(p^{2}/ħ^{2})· e^{−i∙[E·t − p∙x]/ħ }
⇔ E = p^{2}/(2m_{eff}) ⇔ m_{eff} = m∙(v/c)^{2}/2 = m∙β^{2}/2
It is an ugly formula: it resembles the kinetic energy formula (K.E. = m∙v^{2}/2) but it is, in fact, something completely different. The β^{2}/2 factor ensures the effective mass is always a fraction of the mass itself. To get rid of the ugly 1/2 factor, we may re-define m_{eff} as two times the old m_{eff} (hence, m_{eff}^{NEW} = 2∙m_{eff}^{OLD}), as a result of which the formula will look somewhat better:
m_{eff} = m∙(v/c)^{2} = m∙β^{2}
We know β varies between 0 and 1 and, therefore, m_{eff} will vary between 0 and m. Feynman drops the subscript, and just writes m_{eff} as m in his textbook (see Feynman, III-19). On the other hand, the electron mass as used is also the electron mass that is used to calculate the size of an atom (see Feynman, III-2-4). As such, the two mass concepts are, effectively, mutually compatible. It is confusing because the same mass is often defined as the mass of a stationary electron (see, for example, the article on it in the online Wikipedia encyclopedia[28]).
In the context of the derivation of the electron orbitals, we do have the potential energy term – which is the equivalent of a source term in a diffusion equation – and that may explain why the above-mentioned m_{eff} = m∙(v/c)^{2} = m∙β^{2} formula does not apply.
References
This paper discusses general principles in physics only. Hence, references can be limited to references to physics textbooks only. For ease of reading, any reference to additional material has been limited to a more popular undergrad textbook that can be consulted online: Feynman’s Lectures on Physics (http://www.feynmanlectures.caltech.edu). References are per volume, per chapter and per section. For example, Feynman III-19-3 refers to Volume III, Chapter 19, Section 3.
Notes
[1] Of course, an actual particle is localized in space and can, therefore, not be represented by the elementary wavefunction ψ = a·e^{−i∙θ} = a·e^{−i[E·t − p∙x]/ħ} = a·(cosθ – i·a·sinθ). We must build a wave packet for that: a sum of wavefunctions, each with its own amplitude a_{k} and its own argument θ_{k} = (E_{k}∙t – p_{k}∙x)/ħ. This is dealt with in this paper as part of the discussion on the mathematical and physical interpretation of the normalization condition.
[2] The N/kg dimension immediately, and naturally, reduces to the dimension of acceleration (m/s^{2}), thereby facilitating a direct interpretation in terms of Newton’s force law.
[3] In physics, a two-spring metaphor is more common. Hence, the pistons in the author’s perpetuum mobile may be replaced by springs.
[4] The author re-derives the equation for the Compton scattering radius in section VII of the paper.
[5] The magnetic force can be analyzed as a relativistic effect (see Feynman II-13-6). The dichotomy between the electric force as a polar vector and the magnetic force as an axial vector disappears in the relativistic four-vector representation of electromagnetism.
[6] For example, when using Schrödinger’s equation in a central field (think of the electron around a proton), the use of polar coordinates is recommended, as it ensures the symmetry of the Hamiltonian under all rotations (see Feynman III-19-3)
[7] This sentiment is usually summed up in the apocryphal quote: “God does not play dice.”The actual quote comes out of one of Einstein’s private letters to Cornelius Lanczos, another scientist who had also emigrated to the US. The full quote is as follows: “You are the only person I know who has the same attitude towards physics as I have: belief in the comprehension of reality through something basically simple and unified… It seems hard to sneak a look at God’s cards. But that He plays dice and uses ‘telepathic’ methods… is something that I cannot believe for a single moment.” (Helen Dukas and Banesh Hoffman, Albert Einstein, the Human Side: New Glimpses from His Archives, 1979)
[8] Of course, both are different velocities: ω is an angular velocity, while v is a linear velocity: ω is measured in radians per second, while v is measured in meter per second. However, the definition of a radian implies radians are measured in distance units. Hence, the physical dimensions are, effectively, the same. As for the formula for the total energy of an oscillator, we should actually write: E = m·a^{2}∙ω^{2}/2. The additional factor (a) is the (maximum) amplitude of the oscillator.
[9] We also have a 1/2 factor in the E = mv^{2}/2 formula. Two remarks may be made here. First, it may be noted this is a non-relativistic formula and, more importantly, incorporates kinetic energy only. Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as K.E. = E − E_{0} = m_{v}c^{2} − m_{0}c^{2} = m_{0}γc^{2} − m_{0}c^{2} = m_{0}c^{2}(γ − 1). As for the exclusion of the potential energy, we may note that we may choose our reference point for the potential energy such that the kinetic and potential energy mirror each other. The energy concept that then emerges is the one that is used in the context of the Principle of Least Action: it equals E = mv^{2}. Appendix 1 provides some notes on that.
[10] Instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft.
[11] It is interesting to note that we may look at the energy in the rotating flywheel as potential energy because it is energy that is associated with motion, albeit circular motion. In physics, one may associate a rotating object with kinetic energy using the rotational equivalent of mass and linear velocity, i.e. rotational inertia (I) and angular velocity ω. The kinetic energy of a rotating object is then given by K.E. = (1/2)·I·ω^{2}.
[12] Because of the sideways motion of the connecting rods, the sinusoidal function will describe the linear motion only approximately, but you can easily imagine the idealized limit situation.
[13] The ω^{2}= 1/LC formula gives us the natural or resonant frequency for a electric circuit consisting of a resistor (R), an inductor (L), and a capacitor (C). Writing the formula as ω^{2}= C^{–}^{1}/L introduces the concept of elastance, which is the equivalent of the mechanical stiffness (k) of a spring.
[14] The resistance in an electric circuit introduces a damping factor. When analyzing a mechanical spring, one may also want to introduce a drag coefficient. Both are usually defined as a fraction of the inertia, which is the mass for a spring and the inductance for an electric circuit. Hence, we would write the resistance for a spring as γm and as R = γL respectively.
[15] Photons are emitted by atomic oscillators: atoms going from one state (energy level) to another. Feynman (Lectures, I-33-3) shows us how to calculate the Q of these atomic oscillators: it is of the order of 10^{8}, which means the wave train will last about 10^{–8 }seconds (to be precise, that is the time it takes for the radiation to die out by a factor 1/e). For example, for sodium light, the radiation will last about 3.2×10^{–8 }seconds (this is the so-called decay time τ). Now, because the frequency of sodium light is some 500 THz (500×10^{12 }oscillations per second), this makes for some 16 million oscillations. There is an interesting paradox here: the speed of light tells us that such wave train will have a length of about 9.6 m! How is that to be reconciled with the pointlike nature of a photon? The paradox can only be explained by relativistic length contraction: in an analysis like this, one need to distinguish the reference frame of the photon – riding along the wave as it is being emitted, so to speak – and our stationary reference frame, which is that of the emitting atom.
[16] This is a general result and is reflected in the K.E. = T = (1/2)·m·ω^{2}·a^{2}·sin^{2}(ω·t + Δ) and the P.E. = U = k·x^{2}/2 = (1/2)· m·ω^{2}·a^{2}·cos^{2}(ω·t + Δ) formulas for the linear oscillator.
[17] Feynman further formalizes this in his Lecture on Superconductivity (Feynman, III-21-2), in which he refers to Schrödinger’s equation as the “equation for continuity of probabilities”. The analysis is centered on the local conservation of energy, which confirms the interpretation of Schrödinger’s equation as an energy diffusion equation.
[18] The m_{eff} is the effective mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write m_{eff} = m. Appendix 2 provides some additional notes on the concept. As for the equations, they are easily derived from noting that two complex numbers a + i∙b and c + i∙d are equal if, and only if, their real and imaginary parts are the same. Now, the ∂ψ/∂t = i∙(ħ/m_{eff})∙∇^{2}ψ equation amounts to writing something like this: a + i∙b = i∙(c + i∙d). Now, remembering that i^{2} = −1, you can easily figure out that i∙(c + i∙d) = i∙c + i^{2}∙d = − d + i∙c.
[19] The dimension of B is usually written as N/(m∙A), using the SI unit for current, i.e. the ampere (A). However, 1 C = 1 A∙s and, hence, 1 N/(m∙A) = 1 (N/C)/(m/s).
[20] Of course, multiplication with i amounts to a counterclockwise rotation. Hence, multiplication by –i also amounts to a rotation by 90 degrees, but clockwise. Now, to uniquely identify the clockwise and counterclockwise directions, we need to establish the equivalent of the right-hand rule for a proper geometric interpretation of Schrödinger’s equation in three-dimensional space: if we look at a clock from the back, then its hand will be moving counterclockwise. When writing B = (1/c)∙i∙E, we assume we are looking in the negative x-direction. If we are looking in the positive x-direction, we should write: B = -(1/c)∙i∙E. Of course, Nature does not care about our conventions. Hence, both should give the same results in calculations. We will show in a moment they do.
[21] In fact, when multiplying C^{2}/(N·m^{2}) with N^{2}/C^{2}, we get N/m^{2}, but we can multiply this with 1 = m/m to get the desired result. It is significant that an energy density (joule per unit volume) can also be measured in newton (force per unit area.
[22] The illustration shows a linearly polarized wave, but the obtained result is general.
[23] The sine and cosine are essentially the same functions, except for the difference in the phase: sinθ = cos(θ−π /2).
[24] I must thank a physics blogger for re-writing the 1/(ε_{0}·μ_{0}) = c^{2} equation like this. See: http://reciprocal.systems/phpBB3/viewtopic.php?t=236 (retrieved on 29 September 2017).
[25] A circularly polarized electromagnetic wave may be analyzed as consisting of two perpendicular electromagnetic plane waves of equal amplitude and 90° difference in phase.
[26] Of course, the reader will now wonder: what about neutrons? How to explain neutron spin? Neutrons are neutral. That is correct, but neutrons are not elementary: they consist of (charged) quarks. Hence, neutron spin can (or should) be explained by the spin of the underlying quarks.
[27] We detailed the mathematical framework and detailed calculations in the following online article: https://readingfeynman.org/2017/09/15/the-principle-of-least-action-re-visited.
[28] https://en.wikipedia.org/wiki/Electron_rest_mass (retrieved on 29 September 2017).
Playing with amplitudes
Let’s play a bit with the stuff we found in our previous post. This is going to be unconventional, or experimental, if you want. The idea is to give you… Well… Some ideas. So you can play yourself. 🙂 Let’s go.
Let’s first look at Feynman’s (simplified) formula for the amplitude of a photon to go from point a to point b. If we identify point a by the position vector r_{1} and point b by the position vector r_{2}, and using Dirac’s fancy bra-ket notation, then it’s written as:
So we have a vector dot product here: p∙r_{12} = |p|∙|r_{12}|· cosθ = p∙r_{12}·cosα. The angle here (α) is the angle between the p and r_{12} vector. All good. Well… No. We’ve got a problem. When it comes to calculating probabilities, the α angle doesn’t matter: |e^{i·θ}/r|^{2} = 1/r^{2}. Hence, for the probability, we get: P = | 〈r_{2}|r_{1}〉 |^{2} = 1/r_{12}^{2}. Always ! Now that’s strange. The θ = p∙r_{12}/ħ argument gives us a different phase depending on the angle (α) between p and r_{12}. But… Well… Think of it: cosα goes from 1 to 0 when α goes from 0 to ±90° and, of course, is negative when p and r_{12} have opposite directions but… Well… According to this formula, the probabilities do not depend on the direction of the momentum. That’s just weird, I think. Did Feynman, in his iconic Lectures, give us a meaningless formula?
Maybe. We may also note this function looks like the elementary wavefunction for any particle, which we wrote as:
ψ(x, t) = a·e^{−i}^{∙θ} = a·e^{−i}^{∙}^{(E∙t − p∙x)/ħ}= a·e^{−i}^{∙}^{(E∙t)/ħ}·e^{i}^{∙}^{(p∙x)/ħ}
The only difference is that the 〈r_{2}|r_{1}〉 sort of abstracts away from time, so… Well… Let’s get a feel for the quantities. Let’s think of a photon carrying some typical amount of energy. Hence, let’s talk visible light and, therefore, photons of a few eV only – say 5.625 eV = 5.625×1.6×10^{−19} J = 9×10^{−19} J. Hence, their momentum is equal to p = E/c = (9×10^{−19} N·m)/(3×10^{5} m/s) = 3×10^{−24} N·s. That’s tiny but that’s only because newtons and seconds are enormous units at the (sub-)atomic scale. As for the distance, we may want to use the thickness of a playing card as a starter, as that’s what Young used when establishing the experimental fact of light interfering with itself. Now, playing cards in Young’s time were obviously rougher than those today, but let’s take the smaller distance: modern cards are as thin as 0.3 mm. Still, that distance is associated with a value of θ that is equal to 13.6 million. Hence, the density of our wavefunction is enormous at this scale, and it’s a bit of a miracle that Young could see any interference at all ! As shown in the table below, we only get meaningful values (remember: θ is a phase angle) when we go down to the nanometer scale (10^{−9} m) or, even better, the angstroms scale ((10^{−9} m).
So… Well… Again: what can we do with Feynman’s formula? Perhaps he didn’t give us a propagator function but something that is more general (read: more meaningful) at our (limited) level of knowledge. As I’ve been reading Feynman for quite a while now – like three or four years 🙂 – I think… Well… Yes. That’s it. Feynman wants us to think about it. 🙂 Are you joking again, Mr. Feynman? 🙂 So let’s assume the reasonable thing: let’s assume it gives us the amplitude to go from point a to point b by the position vector along some path r. So, then, in line with what we wrote in our previous post, let’s say p·r (momentum over a distance) is the action (S) we’d associate with this particular path (r) and then see where we get. So let’s write the formula like this:
ψ = a·e^{i·θ} = (1/r)·e^{i·S/ħ} = e^{i·p∙r/ħ}/r
We’ll use an index to denote the various paths: r_{0} is the straight-line path and r_{i} is any (other) path. Now, quantum mechanics tells us we should calculate this amplitude for every possible path. The illustration below shows the straight-line path and two nearby paths. So each of these paths is associated with some amount of action, which we measure in Planck units: θ = S/ħ.
The time interval is given by t = t_{0 }= r_{0}/c, for all paths. Why is the time interval the same for all paths? Because we think of a photon going from some specific point in space and in time to some other specific point in space and in time. Indeed, when everything is said and done, we do think of light as traveling from point a to point b at the speed of light (c). In fact, all of the weird stuff here is all about trying to explain how it does that. 🙂
Now, if we would think of the photon actually traveling along this or that path, then this implies its velocity along any of the nonlinear paths will be larger than c, which is OK. That’s just the weirdness of quantum mechanics, and you should actually not think of the photon actually traveling along one of these paths anyway although we’ll often put it that way. Think of something fuzzier, whatever that may be. 🙂
So the action is energy times time, or momentum times distance. Hence, the difference in action between two paths i and j is given by:
δS = p·r_{j} − p·r_{i} = p·(r_{j} − r_{i}) = p·Δr
I’ll explain the δS < 2πħ/3 thing in a moment. Let’s first pause and think about the uncertainty and how we’re modeling it. We can effectively think of the variation in S as some uncertainty in the action: δS = ΔS = p·Δr. However, if S is also equal to energy times time (S = E·t), and we insist t is the same for all paths, then we must have some uncertainty in the energy, right? Hence, we can write δS as ΔS = ΔE·t. But, of course, E = E = m·c^{2} = p·c, so we will have an uncertainty in the momentum as well. Hence, the variation in S should be written as:
δS = ΔS = Δp·Δr
That’s just logical thinking: if we, somehow, entertain the idea of a photon going from some specific point in spacetime to some other specific point in spacetime along various paths, then the variation, or uncertainty, in the action will effectively combine some uncertainty in the momentum and the distance. We can calculate Δp as ΔE/c, so we get the following:
δS = ΔS = Δp·Δr = ΔE·Δr/c = ΔE·Δt with Δt = Δr/c
So we have the two expressions for the Uncertainty Principle here: ΔS = Δp·Δr = ΔE·Δt. Just be careful with the interpretation of Δt: it’s just the equivalent of Δr. We just express the uncertainty in distance in seconds using the (absolute) speed of light. We are not changing our spacetime interval: we’re still looking at a photon going from a to b in t seconds, exactly. Let’s now look at the δS < 2πħ/3 thing. If we’re adding two amplitudes (two arrows or vectors, so to speak) and we want the magnitude of the result to be larger than the magnitude of the two contributions, then the angle between them should be smaller than 120 degrees, so that’s 2π/3 rad. The illustration below shows how you can figure that out geometrically.Hence, if S_{0} is the action for r_{0}, then S_{1} = S_{0} + ħ and S_{2} = S_{0} + 2·ħ are still good, but S_{3} = S_{0} + 3·ħ is not good. Why? Because the difference in the phase angles is Δθ = S_{1}/ħ − S_{0}/ħ = (S_{0} + ħ)/ħ − S_{0}/ħ = 1 and Δθ = S_{2}/ħ − S_{0}/ħ = (S_{0} + 2·ħ)/ħ − S_{0}/ħ = 2 respectively, so that’s 57.3° and 114.6° respectively and that’s, effectively, less than 120°. In contrast, for the next path, we find that Δθ = S_{3}/ħ − S_{0}/ħ = (S_{0} + 3·ħ)/ħ − S_{0}/ħ = 3, so that’s 171.9°. So that amplitude gives us a negative contribution.
Let’s do some calculations using a spreadsheet. To simplify things, we will assume we measure everything (time, distance, force, mass, energy, action,…) in Planck units. Hence, we can simply write: S_{n} = S_{0} + n. Of course, n = 1, 2,… etcetera, right? Well… Maybe not. We are measuring action in units of ħ, but do we actually think action comes in units of ħ? I am not sure. It would make sense, intuitively, but… Well… There’s uncertainty on the energy (E) and the momentum (p) of our photon, right? And how accurately can we measure the distance? So there’s some randomness everywhere. 😦 So let’s leave that question open as for now.
We will also assume that the phase angle for S_{0} is equal to 0 (or some multiple of 2π, if you want). That’s just a matter of choosing the origin of time. This makes it really easy: ΔS_{n} = S_{n} − S_{0} = n, and the associated phase angle θ_{n} = Δθ_{n} is the same. In short, the amplitude for each path reduces to ψ_{n} = e^{i·n}/r_{0}. So we need to add these first and then calculate the magnitude, which we can then square to get a probability. Of course, there is also the issue of normalization (probabilities have to add up to one) but let’s tackle that later. For the calculations, we use Euler’s r·e^{i·θ} = r·(cosθ + i·sinθ) = r·cosθ + i·r·sinθ formula. Needless to say, |r·e^{i·θ}|^{2} = |r|^{2}·|e^{i·θ}|^{2} = |r|^{2}·(cos^{2}θ + sin^{2}θ) = r. Finally, when adding complex numbers, we add the real and imaginary parts respectively, and we’ll denote the ψ_{0} + ψ_{1} +ψ_{2} + … sum as Ψ.
Now, we also need to see how our ΔS = Δp·Δr works out. We may want to assume that the uncertainty in p and in r will both be proportional to the overall uncertainty in the action. For example, we could try writing the following: ΔS_{n} = Δp_{n}·Δr_{n} = n·Δp_{1}·Δr_{1}. It also makes sense that you may want Δp_{n} and Δr_{n} to be proportional to Δp_{1} and Δr_{1} respectively. Combining both, the assumption would be this:
Δp_{n} = √n·Δp_{1 }and Δr_{n} = √n·Δr_{1}
So now we just need to decide how we will distribute ΔS_{1} = ħ = 1 over Δp_{1} and Δr_{1} respectively. For example, if we’d assume Δp_{1} = 1, then Δr_{1} = ħ/Δp_{1} = 1/1 = 1. These are the calculations. I will let you analyze them. 🙂Well… We get a weird result. It reminds me of Feynman’s explanation of the partial reflection of light, shown below, but… Well… That doesn’t make much sense, does it?
Hmm… Maybe it does. 🙂 Look at the graph more carefully. The peaks sort of oscillate out so… Well… That might make sense… 🙂
Does it? Are we doing something wrong here? These amplitudes should reflect the ones that are reflected in those nice animations (like this one, for example, which is part of that’s part of the Wikipedia article on Feynman’s path integral formulation of quantum mechanics). So what’s wrong, if anything? Well… Our paths differ by some fixed amount of action, which doesn’t quite reflect the geometric approach that’s used in those animations. The graph below shows how the distance r varies as a function of n.
If we’d use a model in which the distance would increase linearly or, preferably, exponentially, then we’d get the result we want to get, right?
Well… Maybe. Let’s try it. Hmm… We need to think about the geometry here. Look at the triangle below. If b is the straight-line path (r_{0}), then ac could be one of the crooked paths (r_{n}). To simplify, we’ll assume isosceles triangles, so a equals c and, hence, r_{n} = 2·a = 2·c. We will also assume the successive paths are separated by the same vertical distance (h = h_{1}) right in the middle, so h_{b} = h_{n} = n·h_{1}. It is then easy to show the following:This gives the following graph for r_{n} = 10 and h_{1 }= 0.01.
Is this the right step increase? Not sure. We can vary the values in our spreadsheet. Let’s first build it. The photon will have to travel faster in order to cover the extra distance in the same time, so its momentum will be higher. Let’s think about the velocity. Let’s start with the first path (n = 1). In order to cover the extra distance Δr_{1}, the velocity c_{1} must be equal to (r_{0} + Δr_{1})/t = r_{0}/t + Δr_{1}/t = c + Δr_{1}/t = c_{0} + Δr_{1}/t. We can write c_{1} as c_{1} = c_{0} + Δc_{1}, so Δc_{1} = Δr_{1}/t. Now, the ratio of p_{1} and p_{0} will be equal to the ratio of c_{1} and c_{0} because p_{1}/p_{0 }= (mc_{1})/mc_{0}) = c_{1}/c_{0}. Hence, we have the following formula for p_{1}:
p_{1} = p_{0}·c_{1}/c_{0} = p_{0}·(c_{0} + Δc_{1})/c_{0} = p_{0}·[1 + Δr_{1}/(c_{0}·t) = p_{0}·(1 + Δr_{1}/r_{0})
For p_{n}, the logic is the same, so we write:
p_{n} = p_{0}·c_{n}/c_{0} = p_{0}·(c_{0} + Δc_{n})/c_{0} = p_{0}·[1 + Δr_{n}/(c_{0}·t) = p_{0}·(1 + Δr_{n}/r_{0})
Let’s do the calculations, and let’s use meaningful values, so the nanometer scale and actual values for Planck’s constant and the photon momentum. The results are shown below.
Pretty interesting. In fact, this looks really good. The probability first swings around wildly, because of these zones of constructive and destructive interference, but then stabilizes. [Of course, I would need to normalize the probabilities, but you get the idea, right?] So… Well… I think we get a very meaningful result with this model. Sweet ! 🙂 I’m lovin’ it ! 🙂 And, here you go, this is (part of) the calculation table, so you can see what I am doing. 🙂
The graphs below look even better: I just changed the h_{1}/r_{0} ratio from 1/100 to 1/10. The probability stabilizes almost immediately. 🙂 So… Well… It’s not as fancy as the referenced animation, but I think the educational value of this thing here is at least as good ! 🙂
🙂 This is good stuff… 🙂
Post scriptum (19 September 2017): There is an obvious inconsistency in the model above, and in the calculations. We assume there is a path r_{1} = , r_{2}, r_{2},etcetera, and then we calculate the action for it, and the amplitude, and then we add the amplitude to the sum. But, surely, we should count these paths twice, in two-dimensional space, that is. Think of the graph: we have positive and negative interference zones that are sort of layered around the straight-line path, as shown below.
In three-dimensional space, these lines become surfaces. Hence, rather than adding one arrow for every δ having one contribution only, we may want to add… Well… In three-dimensional space, the formula for the surface around the straight-line path would probably look like π·h_{n}·r_{1}, right? Hmm… Interesting idea. I changed my spreadsheet to incorporate that idea, and I got the graph below. It’s a nonsensical result, because the probability does swing around, but it gradually spins out of control: it never stabilizes.That’s because we increase the weight of the paths that are further removed from the center. So… Well… We shouldn’t be doing that, I guess. 🙂 I’ll you look for the right formula, OK? Let me know when you found it. 🙂
Some thoughts on the nature of reality
Some other comment on an article on my other blog, inspired me to structure some thoughts that are spread over various blog posts. What follows below, is probably the first draft of an article or a paper I plan to write. Or, who knows, I might re-write my two introductory books on quantum physics and publish a new edition soon. 🙂
Physical dimensions and Uncertainty
The physical dimension of the quantum of action (h or ħ = h/2π) is force (expressed in newton) times distance (expressed in meter) times time (expressed in seconds): N·m·s. Now, you may think this N·m·s dimension is kinda hard to imagine. We can imagine its individual components, right? Force, distance and time. We know what they are. But the product of all three? What is it, really?
It shouldn’t be all that hard to imagine what it might be, right? The N·m·s unit is also the unit in which angular momentum is expressed – and you can sort of imagine what that is, right? Think of a spinning top, or a gyroscope. We may also think of the following:
- [h] = N·m·s = (N·m)·s = [E]·[t]
- [h] = N·m·s = (N·s)·m = [p]·[x]
Hence, the physical dimension of action is that of energy (E) multiplied by time (t) or, alternatively, that of momentum (p) times distance (x). To be precise, the second dimensional equation should be written as [h] = [p]·[x], because both the momentum and the distance traveled will be associated with some direction. It’s a moot point for the discussion at the moment, though. Let’s think about the first equation first: [h] = [E]·[t]. What does it mean?
Energy… Hmm… In real life, we are usually not interested in the energy of a system as such, but by the energy it can deliver, or absorb, per second. This is referred to as the power of a system, and it’s expressed in J/s, or watt. Power is also defined as the (time) rate at which work is done. Hmm… But so here we’re multiplying energy and time. So what’s that? After Hiroshima and Nagasaki, we can sort of imagine the energy of an atomic bomb. We can also sort of imagine the power that’s being released by the Sun in light and other forms of radiation, which is about 385×10^{24} joule per second. But energy times time? What’s that?
I am not sure. If we think of the Sun as a huge reservoir of energy, then the physical dimension of action is just like having that reservoir of energy guaranteed for some time, regardless of how fast or how slow we use it. So, in short, it’s just like the Sun – or the Earth, or the Moon, or whatever object – just being there, for some definite amount of time. So, yes: some definite amount of mass or energy (E) for some definite amount of time (t).
Let’s bring the mass-energy equivalence formula in here: E = mc^{2}. Hence, the physical dimension of action can also be written as [h] = [E]·[t] = [mc]^{2}·[t] = (kg·m^{2}/s^{2})·s = kg·m^{2}/s. What does that say? Not all that much – for the time being, at least. We can get this [h] = kg·m^{2}/s through some other substitution as well. A force of one newton will give a mass of 1 kg an acceleration of 1 m/s per second. Therefore, 1 N = 1 kg·m/s^{2} and, hence, the physical dimension of h, or the unit of angular momentum, may also be written as 1 N·m·s = 1 (kg·m/s^{2})·m·s = 1 kg·m^{2}/s, i.e. the product of mass, velocity and distance.
Hmm… What can we do with that? Nothing much for the moment: our first reading of it is just that it reminds us of the definition of angular momentum – some mass with some velocity rotating around an axis. What about the distance? Oh… The distance here is just the distance from the axis, right? Right. But… Well… It’s like having some amount of linear momentum available over some distance – or in some space, right? That’s sufficiently significant as an interpretation for the moment, I’d think…
Fundamental units
This makes one think about what units would be fundamental – and what units we’d consider as being derived. Formally, the newton is a derived unit in the metric system, as opposed to the units of mass, length and time (kg, m, s). Nevertheless, I personally like to think of force as being fundamental: a force is what causes an object to deviate from its straight trajectory in spacetime. Hence, we may want to think of the quantum of action as representing three fundamental physical dimensions: (1) force, (2) time and (3) distance – or space. We may then look at energy and (linear) momentum as physical quantities combining (1) force and distance and (2) force and time respectively.
Let me write this out:
- Force times length (think of a force that is acting on some object over some distance) is energy: 1 joule (J) = 1 newton·meter (N). Hence, we may think of the concept of energy as a projection of action in space only: we make abstraction of time. The physical dimension of the quantum of action should then be written as [h] = [E]·[t]. [Note the square brackets tell us we are looking at a dimensional equation only, so [t] is just the physical dimension of the time variable. It’s a bit confusing because I also use square brackets as parentheses.]
- Conversely, the magnitude of linear momentum (p = m·v) is expressed in newton·seconds: 1 kg·m/s = 1 (kg·m/s^{2})·s = 1 N·s. Hence, we may think of (linear) momentum as a projection of action in time only: we make abstraction of its spatial dimension. Think of a force that is acting on some object during some time. The physical dimension of the quantum of action should then be written as [h] = [p]·[x]
Of course, a force that is acting on some object during some time, will usually also act on the same object over some distance but… Well… Just try, for once, to make abstraction of one of the two dimensions here: time or distance.
It is a difficult thing to do because, when everything is said and done, we don’t live in space or in time alone, but in spacetime and, hence, such abstractions are not easy. [Of course, now you’ll say that it’s easy to think of something that moves in time only: an object that is standing still does just that – but then we know movement is relative, so there is no such thing as an object that is standing still in space in an absolute sense: Hence, objects never stand still in spacetime.] In any case, we should try such abstractions, if only because of the principle of least action is so essential and deep in physics:
- In classical physics, the path of some object in a force field will minimize the total action (which is usually written as S) along that path.
- In quantum mechanics, the same action integral will give us various values S – each corresponding to a particular path – and each path (and, therefore, each value of S, really) will be associated with a probability amplitude that will be proportional to some constant times e^{−i·θ} = e^{i·(S/ħ)}. Because ħ is so tiny, even a small change in S will give a completely different phase angle θ. Therefore, most amplitudes will cancel each other out as we take the sum of the amplitudes over all possible paths: only the paths that nearly give the same phase matter. In practice, these are the paths that are associated with a variation in S of an order of magnitude that is equal to ħ.
The paragraph above summarizes, in essence, Feynman’s path integral formulation of quantum mechanics. We may, therefore, think of the quantum of action expressing itself (1) in time only, (2) in space only, or – much more likely – (3) expressing itself in both dimensions at the same time. Hence, if the quantum of action gives us the order of magnitude of the uncertainty – think of writing something like S ± ħ, we may re-write our dimensional [ħ] = [E]·[t] and [ħ] = [p]·[x] equations as the uncertainty equations:
- ΔE·Δt = ħ
- Δp·Δx = ħ
You should note here that it is best to think of the uncertainty relations as a pair of equations, if only because you should also think of the concept of energy and momentum as representing different aspects of the same reality, as evidenced by the (relativistic) energy-momentum relation (E^{2} = p^{2}c^{2} – m_{0}^{2}c^{4}). Also, as illustrated below, the actual path – or, to be more precise, what we might associate with the concept of the actual path – is likely to be some mix of Δx and Δt. If Δt is very small, then Δx will be very large. In order to move over such distance, our particle will require a larger energy, so ΔE will be large. Likewise, if Δt is very large, then Δx will be very small and, therefore, ΔE will be very small. You can also reason in terms of Δx, and talk about momentum rather than energy. You will arrive at the same conclusions: the ΔE·Δt = h and Δp·Δx = h relations represent two aspects of the same reality – or, at the very least, what we might think of as reality.
Also think of the following: if ΔE·Δt = h and Δp·Δx = h, then ΔE·Δt = Δp·Δx and, therefore, ΔE/Δp must be equal to Δx/Δt. Hence, the ratio of the uncertainty about x (the distance) and the uncertainty about t (the time) equals the ratio of the uncertainty about E (the energy) and the uncertainty about p (the momentum).
Of course, you will note that the actual uncertainty relations have a factor 1/2 in them. This may be explained by thinking of both negative as well as positive variations in space and in time.
We will obviously want to do some more thinking about those physical dimensions. The idea of a force implies the idea of some object – of some mass on which the force is acting. Hence, let’s think about the concept of mass now. But… Well… Mass and energy are supposed to be equivalent, right? So let’s look at the concept of energy too.
Action, energy and mass
What is energy, really? In real life, we are usually not interested in the energy of a system as such, but by the energy it can deliver, or absorb, per second. This is referred to as the power of a system, and it’s expressed in J/s. However, in physics, we always talk energy – not power – so… Well… What is the energy of a system?
According to the de Broglie and Einstein – and so many other eminent physicists, of course – we should not only think of the kinetic energy of its parts, but also of their potential energy, and their rest energy, and – for an atomic system – we may add some internal energy, which may be binding energy, or excitation energy (think of a hydrogen atom in an excited state, for example). A lot of stuff. 🙂 But, obviously, Einstein’s mass-equivalence formula comes to mind here, and summarizes it all:
E = m·c^{2}
The m in this formula refers to mass – not to meter, obviously. Stupid remark, of course… But… Well… What is energy, really? What is mass, really? What’s that equivalence between mass and energy, really?
I don’t have the definite answer to that question (otherwise I’d be famous), but… Well… I do think physicists and mathematicians should invest more in exploring some basic intuitions here. As I explained in several posts, it is very tempting to think of energy as some kind of two-dimensional oscillation of mass. A force over some distance will cause a mass to accelerate. This is reflected in the dimensional analysis:
[E] = [m]·[c^{2}] = 1 kg·m^{2}/s^{2} = 1 kg·m/s^{2}·m = 1 N·m
The kg and m/s^{2 }factors make this abundantly clear: m/s^{2} is the physical dimension of acceleration: (the change in) velocity per time unit.
Other formulas now come to mind, such as the Planck-Einstein relation: E = h·f = ω·ħ. We could also write: E = h/T. Needless to say, T = 1/f is the period of the oscillation. So we could say, for example, that the energy of some particle times the period of the oscillation gives us Planck’s constant again. What does that mean? Perhaps it’s easier to think of it the other way around: E/f = h = 6.626070040(81)×10^{−34} J·s. Now, f is the number of oscillations per second. Let’s write it as f = n/s, so we get:
E/f = E/(n/s) = E·s/n = 6.626070040(81)×10^{−34} J·s ⇔ E/n = 6.626070040(81)×10^{−34} J
What an amazing result! Our wavicle – be it a photon or a matter-particle – will always pack 6.626070040(81)×10^{−34} joule in one oscillation, so that’s the numerical value of Planck’s constant which, of course, depends on our fundamental units (i.e. kg, meter, second, etcetera in the SI system).
Of course, the obvious question is: what’s one oscillation? If it’s a wave packet, the oscillations may not have the same amplitude, and we may also not be able to define an exact period. In fact, we should expect the amplitude and duration of each oscillation to be slightly different, shouldn’t we? And then…
Well… What’s an oscillation? We’re used to counting them: n oscillations per second, so that’s per time unit. How many do we have in total? We wrote about that in our posts on the shape and size of a photon. We know photons are emitted by atomic oscillators – or, to put it simply, just atoms going from one energy level to another. Feynman calculated the Q of these atomic oscillators: it’s of the order of 10^{8 }(see his Lectures, I-33-3: it’s a wonderfully simple exercise, and one that really shows his greatness as a physics teacher), so… Well… This wave train will last about 10^{–8 }seconds (that’s the time it takes for the radiation to die out by a factor 1/e). To give a somewhat more precise example, for sodium light, which has a frequency of 500 THz (500×10^{12 }oscillations per second) and a wavelength of 600 nm (600×10^{–9 }meter), the radiation will lasts about 3.2×10^{–8 }seconds. [In fact, that’s the time it takes for the radiation’s energy to die out by a factor 1/e, so(i.e. the so-called decay time τ), so the wavetrain will actually last longer, but so the amplitude becomes quite small after that time.] So… Well… That’s a very short time but… Still, taking into account the rather spectacular frequency (500 THz) of sodium light, that makes for some 16 million oscillations and, taking into the account the rather spectacular speed of light (3×10^{8 }m/s), that makes for a wave train with a length of, roughly, 9.6 meter. Huh? 9.6 meter!? But a photon is supposed to be pointlike, isn’it it? It has no length, does it?
That’s where relativity helps us out: as I wrote in one of my posts, relativistic length contraction may explain the apparent paradox. Using the reference frame of the photon – so if we’d be traveling at speed c,’ riding’ with the photon, so to say, as it’s being emitted – then we’d ‘see’ the electromagnetic transient as it’s being radiated into space.
However, while we can associate some mass with the energy of the photon, none of what I wrote above explains what the (rest) mass of a matter-particle could possibly be. There is no real answer to that, I guess. You’ll think of the Higgs field now but… Then… Well. The Higgs field is a scalar field. Very simple: some number that’s associated with some position in spacetime. That doesn’t explain very much, does it? 😦 When everything is said and done, the scientists who, in 2013 only, got the Nobel Price for their theory on the Higgs mechanism, simply tell us mass is some number. That’s something we knew already, right? 🙂
The reality of the wavefunction
The wavefunction is, obviously, a mathematical construct: a description of reality using a very specific language. What language? Mathematics, of course! Math may not be universal (aliens might not be able to decipher our mathematical models) but it’s pretty good as a global tool of communication, at least.
The real question is: is the description accurate? Does it match reality and, if it does, how good is the match? For example, the wavefunction for an electron in a hydrogen atom looks as follows:
ψ(r, t) = e^{−i·(E/ħ)·t}·f(r)
As I explained in previous posts (see, for example, my recent post on reality and perception), the f(r) function basically provides some envelope for the two-dimensional e^{−i·θ} = e^{−i·(E/ħ)·t} = cosθ + i·sinθ oscillation, with r = (x, y, z), θ = (E/ħ)·t = ω·t and ω = E/ħ. So it presumes the duration of each oscillation is some constant. Why? Well… Look at the formula: this thing has a constant frequency in time. It’s only the amplitude that is varying as a function of the r = (x, y, z) coordinates. 🙂 So… Well… If each oscillation is to always pack 6.626070040(81)×10^{−34} joule, but the amplitude of the oscillation varies from point to point, then… Well… We’ve got a problem. The wavefunction above is likely to be an approximation of reality only. 🙂 The associated energy is the same, but… Well… Reality is probably not the nice geometrical shape we associate with those wavefunctions.
In addition, we should think of the Uncertainty Principle: there must be some uncertainty in the energy of the photons when our hydrogen atom makes a transition from one energy level to another. But then… Well… If our photon packs something like 16 million oscillations, and the order of magnitude of the uncertainty is only of the order of h (or ħ = h/2π) which, as mentioned above, is the (average) energy of one oscillation only, then we don’t have much of a problem here, do we? 🙂
Post scriptum: In previous posts, we offered some analogies – or metaphors – to a two-dimensional oscillation (remember the V-2 engine?). Perhaps it’s all relatively simple. If we have some tiny little ball of mass – and its center of mass has to stay where it is – then any rotation – around any axis – will be some combination of a rotation around our x- and z-axis – as shown below. Two axes only. So we may want to think of a two-dimensional oscillation as an oscillation of the polar and azimuthal angle. 🙂
Thinking again…
One of the comments on my other blog made me think I should, perhaps, write something on waves again. The animation below shows the elementary wavefunction ψ = a·e^{−iθ }= ψ = a·e^{−i·θ } = a·e^{−i}^{(}^{ω·t−k·x}^{)} = a·e^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)} .We know this elementary wavefunction cannot represent a real-life particle. Indeed, the a·e^{−i·θ }function implies the probability of finding the particle – an electron, a photon, or whatever – would be equal to P(x, t) = |ψ(x, t)|^{2} = |a·e^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)}|^{2} = |a|^{2}·|e^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)}|^{2} = |a|^{2}·1^{2}= a^{2} everywhere. Hence, the particle would be everywhere – and, therefore, nowhere really. We need to localize the wave – or build a wave packet. We can do so by introducing uncertainty: we then add a potentially infinite number of these elementary wavefunctions with slightly different values for E and p, and various amplitudes a. Each of these amplitudes will then reflect the contribution to the composite wave, which – in three-dimensional space – we can write as:
ψ(r, t) = e^{−i·(E/ħ)·t}·f(r)
As I explained in previous posts (see, for example, my recent post on reality and perception), the f(r) function basically provides some envelope for the two-dimensional e^{−i·θ} = e^{−i·(E/ħ)·t} = cosθ + i·sinθ oscillation, with r = (x, y, z), θ = (E/ħ)·t = ω·t and ω = E/ħ.
Note that it looks like the wave propagates from left to right – in the positive direction of an axis which we may refer to as the x-axis. Also note this perception results from the fact that, naturally, we’d associate time with the rotation of that arrow at the center – i.e. with the motion in the illustration, while the spatial dimensions are just what they are: linear spatial dimensions. [This point is, perhaps, somewhat less self-evident than you may think at first.]
Now, the axis which points upwards is usually referred to as the z-axis, and the third and final axis – which points towards us – would then be the y-axis, obviously. Unfortunately, this definition would violate the so-called right-hand rule for defining a proper reference frame: the figures below shows the two possibilities – a left-handed and a right-handed reference frame – and it’s the right-handed reference (i.e. the illustration on the right) which we have to use in order to correctly define all directions, including the direction of rotation of the argument of the wavefunction.Hence, if we don’t change the direction of the y– and z-axes – so we keep defining the z-axis as the axis pointing upwards, and the y-axis as the axis pointing towards us – then the positive direction of the x-axis would actually be the direction from right to left, and we should say that the elementary wavefunction in the animation above seems to propagate in the negative x-direction. [Note that this left- or right-hand rule is quite astonishing: simply swapping the direction of one axis of a left-handed frame makes it right-handed, and vice versa.]
Note my language when I talk about the direction of propagation of our wave. I wrote: it looks like, or it seems to go in this or that direction. And I mean that: there is no real traveling here. At this point, you may want to review a post I wrote for my son, which explains the basic math behind waves, and in which I also explained the animation below.
Note how the peaks and troughs of this pulse seem to move leftwards, but the wave packet (or the group or the envelope of the wave—whatever you want to call it) moves to the right. The point is: the pulse itself doesn’t travel left or right. Think of the horizontal axis in the illustration above as an oscillating guitar string: each point on the string just moves up and down. Likewise, if our repeated pulse would represent a physical wave in water, for example, then the water just stays where it is: it just moves up and down. Likewise, if we shake up some rope, the rope is not going anywhere: we just started some motion that is traveling down the rope. In other words, the phase velocity is just a mathematical concept. The peaks and troughs that seem to be traveling are just mathematical points that are ‘traveling’ left or right. That’s why there’s no limit on the phase velocity: it can – and, according to quantum mechanics, actually will – exceed the speed of light. In contrast, the group velocity – which is the actual speed of the particle that is being represented by the wavefunction – may approach – or, in the case of a massless photon, will actually equal – the speed of light, but will never exceed it, and its direction will, obviously, have a physical significance as it is, effectively, the direction of travel of our particle – be it an electron, a photon (electromagnetic radiation), or whatever.
Hence, you should not think the spin of a particle – integer or half-integer – is somehow related to the direction of rotation of the argument of the elementary wavefunction. It isn’t: Nature doesn’t give a damn about our mathematical conventions, and that’s what the direction of rotation of the argument of that wavefunction is: just some mathematical convention. That’s why we write a·e^{−i}^{(}^{ω·t−k·x}^{)} rather than a·e^{i}^{(}^{ω·t+k·x}^{)} or a·e^{i}^{(}^{ω·t−k·x}^{)}: it’s just because of the right-hand rule for coordinate frames, and also because Euler defined the counter-clockwise direction as the positive direction of an angle. There’s nothing more to it.
OK. That’s obvious. Let me now return to my interpretation of Einstein’s E = m·c^{2} formula (see my previous posts on this). I noted that, in the reference frame of the particle itself (see my basics page), the elementary wavefunction a·e^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)} reduces to a·e^{−}^{(}^{i}^{/ħ)·}^{(E’·}^{t’}^{)}: the origin of the reference frame then coincides with (the center of) our particle itself, and the wavefunction only varies with the time in the inertial reference frame (i.e. the proper time t’), with the rest energy of the object (E’) as the time scale factor. How should we interpret this?
Well… Energy is force times distance, and force is defined as that what causes some mass to accelerate. To be precise, the newton – as the unit of force – is defined as the magnitude of a force which would cause a mass of one kg to accelerate with one meter per second per second. Per second per second. This is not a typo: 1 N corresponds to 1 kg times 1 m/s per second, i.e. 1 kg·m/s^{2}. So… Because energy is force times distance, the unit of energy may be expressed in units of kg·m/s^{2}·m, or kg·m^{2}/s^{2}, i.e. the unit of mass times the unit of velocity squared. To sum it all up:
1 J = 1 N·m = 1 kg·(m/s)^{2}
This reflects the physical dimensions on both sides of the E = m·c^{2} formula again but… Well… How should we interpret this? Look at the animation below once more, and imagine the green dot is some tiny mass moving around the origin, in an equally tiny circle. We’ve got two oscillations here: each packing half of the total energy of… Well… Whatever it is that our elementary wavefunction might represent in reality – which we don’t know, of course.
Now, the blue and the red dot – i.e. the horizontal and vertical projection of the green dot – accelerate up and down. If we look carefully, we see these dots accelerate towards the zero point and, once they’ve crossed it, they decelerate, so as to allow for a reversal of direction: the blue dot goes up, and then down. Likewise, the red dot does the same. The interplay between the two oscillations, because of the 90° phase difference, is interesting: if the blue dot is at maximum speed (near or at the origin), the red dot reverses speed (its speed is, therefore, (almost) nil), and vice versa. The metaphor of our frictionless V-2 engine, our perpetuum mobile, comes to mind once more.
The question is: what’s going on, really?
My answer is: I don’t know. I do think that, somehow, energy should be thought of as some two-dimensional oscillation of something – something which we refer to as mass, but we didn’t define mass very clearly either. It also, somehow, combines linear and rotational motion. Each of the two dimensions packs half of the energy of the particle that is being represented by our wavefunction. It is, therefore, only logical that the physical unit of both is to be expressed as a force over some distance – which is, effectively, the physical dimension of energy – or the rotational equivalent of them: torque over some angle. Indeed, the analogy between linear and angular movement is obvious: the kinetic energy of a rotating object is equal to K.E. = (1/2)·I·ω^{2}. In this formula, I is the rotational inertia – i.e. the rotational equivalent of mass – and ω is the angular velocity – i.e. the rotational equivalent of linear velocity. Noting that the (average) kinetic energy in any system must be equal to the (average) potential energy in the system, we can add both, so we get a formula which is structurally similar to the E = m·c^{2} formula. But is it the same? Is the effective mass of some object the sum of an almost infinite number of quanta that incorporate some kind of rotational motion? And – if we use the right units – is the angular velocity of these infinitesimally small rotations effectively equal to the speed of light?
I am not sure. Not at all, really. But, so far, I can’t think of any explanation of the wavefunction that would make more sense than this one. I just need to keep trying to find better ways to articulate or imagine what might be going on. 🙂 In this regard, I’d like to add a point – which may or may not be relevant. When I talked about that guitar string, or the water wave, and wrote that each point on the string – or each water drop – just moves up and down, we should think of the physicality of the situation: when the string oscillates, its length increases. So it’s only because our string is flexible that it can vibrate between the fixed points at its ends. For a rope that’s not flexible, the end points would need to move in and out with the oscillation. Look at the illustration below, for example: the two kids who are holding rope must come closer to each other, so as to provide the necessary space inside of the oscillation for the other kid. 🙂The next illustration – of how water waves actually propagate – is, perhaps, more relevant. Just think of a two-dimensional equivalent – and of the two oscillations as being transverse waves, as opposed to longitudinal. See how string theory starts making sense? 🙂
The most fundamental question remains the same: what is it, exactly, that is oscillating here? What is the field? It’s always some force on some charge – but what charge, exactly? Mass? What is it? Well… I don’t have the answer to that. It’s the same as asking: what is electric charge, really? So the question is: what’s the reality of mass, of electric charge, or whatever other charge that causes a force to act on it?
If you know, please let me know. 🙂
Post scriptum: The fact that we’re talking some two-dimensional oscillation here – think of a surface now – explains the probability formula: we need to square the absolute value of the amplitude to get it. And normalize, of course. Also note that, when normalizing, we’d expect to get some factor involving π somewhere, because we’re talking some circular surface – as opposed to a rectangular one. But I’ll let you figure that out. 🙂
Re-visiting electron orbitals (III)
In my previous post, I mentioned that it was not so obvious (both from a physical as well as from a mathematical point of view) to write the wavefunction for electron orbitals – which we denoted as ψ(x, t), i.e. a function of two variables (or four: one time coordinate and three space coordinates) – as the product of two other functions in one variable only.
[…] OK. The above sentence is difficult to read. Let me write in math. 🙂 It is not so obvious to write ψ(x, t) as:
ψ(x, t) = e^{−i·(E/ħ)·t}·ψ(x)
As I mentioned before, the physicists’ use of the same symbol (ψ, psi) for both the ψ(x, t) and ψ(x) function is quite confusing – because the two functions are very different:
- ψ(x, t) is a complex-valued function of two (real) variables: x and t. Or four, I should say, because x = (x, y, z) – but it’s probably easier to think of x as one vector variable – a vector-valued argument, so to speak. And then t is, of course, just a scalar variable. So… Well… A function of two variables: the position in space (x), and time (t).
- In contrast, ψ(x) is a real-valued function of one (vector) variable only: x, so that’s the position in space only.
Now you should cry foul, of course: ψ(x) is not necessarily real-valued. It may be complex-valued. You’re right. You know the formula:Note the derivation of this formula involved a switch from Cartesian to polar coordinates here, so from x = (x, y, z) to r = (r, θ, φ), and that the function is also a function of the two quantum numbers l and m now, i.e. the orbital angular momentum (l) and its z-component (m) respectively. In my previous post(s), I gave you the formulas for Y_{l,m}(θ, φ) and F_{l,m}(r) respectively. F_{l,m}(r) was a real-valued function alright, but the Y_{l,m}(θ, φ) had that e^{i·m·φ} factor in it. So… Yes. You’re right: the Y_{l,m}(θ, φ) function is real-valued if – and only if – m = 0, in which case e^{i·m·φ} = 1. Let me copy the table from Feynman’s treatment of the topic once again:The P_{l}^{m}(cosθ) functions are the so-called (associated) Legendre polynomials, and the formula for these functions is rather horrible:Don’t worry about it too much: just note the P_{l}^{m}(cosθ) is a real-valued function. The point is the following:the ψ(x, t) is a complex-valued function because – and only because – we multiply a real-valued envelope function – which depends on position only – with e^{−i·(E/ħ)·t}·e^{i·m·φ} = e^{−i·[(E/ħ)·t − }^{m·φ]}.
[…]
Please read the above once again and – more importantly – think about it for a while. 🙂 You’ll have to agree with the following:
- As mentioned in my previous post, the e^{i·m·φ} factor just gives us phase shift: just a re-set of our zero point for measuring time, so to speak, and the whole e^{−i·[(E/ħ)·t − }^{m·φ]} factor just disappears when we’re calculating probabilities.
- The envelope function gives us the basic amplitude – in the classical sense of the word: the maximum displacement from the zero value. And so it’s that e^{−i·[(E/ħ)·t − }^{m·φ]} that ensures the whole expression somehow captures the energy of the oscillation.
Let’s first look at the envelope function again. Let me copy the illustration for n = 5 and l = 2 from a Wikimedia Commons article. Note the symmetry planes:
- Any plane containing the z-axis is a symmetry plane – like a mirror in which we can reflect one half of the shape to get the other half. [Note that I am talking the shape only here. Forget about the colors for a while – as these reflect the complex phase of the wavefunction.]
- Likewise, the plane containing both the x– and the y-axis is a symmetry plane as well.
The first symmetry plane – or symmetry line, really (i.e. the z-axis) – should not surprise us, because the azimuthal angle φ is conspicuously absent in the formula for our envelope function if, as we are doing in this article here, we merge the e^{i·m·φ} factor with the e^{−i·(E/ħ)·t}, so it’s just part and parcel of what the author of the illustrations above refers to as the ‘complex phase’ of our wavefunction. OK. Clear enough – I hope. 🙂 But why is the the xy-plane a symmetry plane too? We need to look at that monstrous formula for the P_{l}^{m}(cosθ) function here: just note the cosθ argument in it is being squared before it’s used in all of the other manipulation. Now, we know that cosθ = sin(π/2 − θ). So we can define some new angle – let’s just call it α – which is measured in the way we’re used to measuring angle, which is not from the z-axis but from the xy-plane. So we write: cosθ = sin(π/2 − θ) = sinα. The illustration below may or may not help you to see what we’re doing here.So… To make a long story short, we can substitute the cosθ argument in the P_{l}^{m}(cosθ) function for sinα = sin(π/2 − θ). Now, if the xy-plane is a symmetry plane, then we must find the same value for P_{l}^{m}(sinα) and P_{l}^{m}[sin(−α)]. Now, that’s not obvious, because sin(−α) = −sinα ≠ sinα. However, because the argument in that P_{l}^{m}(x) function is being squared before any other operation (like subtracting 1 and exponentiating the result), it is OK: [−sinα]^{2} = [sinα]^{2 }= sin^{2}α. […] OK, I am sure the geeks amongst my readers will be able to explain this more rigorously. In fact, I hope they’ll have a look at it, because there’s also that d^{l+m}/dx^{l+m} operator, and so you should check what happens with the minus sign there. 🙂
[…] Well… By now, you’re probably totally lost, but the fact of the matter is that we’ve got a beautiful result here. Let me highlight the most significant results:
- A definite energy state of a hydrogen atom (or of an electron orbiting around some nucleus, I should say) appears to us as some beautifully shaped orbital – an envelope function in three dimensions, really – which has the z-axis – i.e. the vertical axis – as a symmetry line and the xy-plane as a symmetry plane.
- The e^{−i·[(E/ħ)·t − }^{m·φ]} factor gives us the oscillation within the envelope function. As such, it’s this factor that, somehow, captures the energy of the oscillation.
It’s worth thinking about this. Look at the geometry of the situation again – as depicted below. We’re looking at the situation along the x-axis, in the direction of the origin, which is the nucleus of our atom.
The e^{i·m·φ} factor just gives us phase shift: just a re-set of our zero point for measuring time, so to speak. Interesting, weird – but probably less relevant than the e^{−i·[(E/ħ)·t} factor, which gives us the two-dimensional oscillation that captures the energy of the state.
Now, the obvious question is: the oscillation of what, exactly? I am not quite sure but – as I explained in my Deep Blue page – the real and imaginary part of our wavefunction are really like the electric and magnetic field vector of an oscillating electromagnetic field (think of electromagnetic radiation – if that makes it easier). Hence, just like the electric and magnetic field vector represent some rapidly changing force on a unit charge, the real and imaginary part of our wavefunction must also represent some rapidly changing force on… Well… I am not quite sure on what though. The unit charge is usually defined as the charge of a proton – rather than an electron – but then forces act on some mass, right? And the mass of a proton is hugely different from the mass of an electron. The same electric (or magnetic) force will, therefore, give a hugely different acceleration to both.
So… Well… My guts instinct tells me the real and imaginary part of our wavefunction just represent, somehow, a rapidly changing force on some unit of mass, but then I am not sure how to define that unit right now (it’s probably not the kilogram!).
Now, there is another thing we should note here: we’re actually sort of de-constructing a rotation (look at the illustration above once again) in two linearly oscillating vectors – one along the z-axis and the other along the y-axis. Hence, in essence, we’re actually talking about something that’s spinning. In other words, we’re actually talking some torque around the x-axis. In what direction? I think that shouldn’t matter – that we can write E or −E, in other words, but… Well… I need to explore this further – as should you! 🙂
Let me just add one more note on the e^{i·m·φ} factor. It sort of defines the geometry of the complex phase itself. Look at the illustration below. Click on it to enlarge it if necessary – or, better still, visit the magnificent Wikimedia Commons article from which I get these illustrations. These are the orbitals n = 4 and l = 3. Look at the red hues in particular – or the blue – whatever: focus on one color only, and see how how – for m = ±1, we’ve got one appearance of that color only. For m = ±1, the same color appears at two ends of the ‘tubes’ – or tori (plural of torus), I should say – just to sound more professional. 🙂 For m = ±2, the torus consists of three parts – or, in mathematical terms, we’d say the order of its rotational symmetry is equal to 3. Check that Wikimedia Commons article for higher values of n and l: the shapes become very convoluted, but the observation holds. 🙂
Have fun thinking all of this through for yourself – and please do look at those symmetries in particular. 🙂
Post scriptum: You should do some thinking on whether or not these m = ±1, ±2,…, ±l orbitals are really different. As I mentioned above, a phase difference is just what it is: a re-set of the t = 0 point. Nothing more, nothing less. So… Well… As far as I am concerned, that’s not a real difference, is it? 🙂 As with other stuff, I’ll let you think about this for yourself.
Feynman’s Lecture on Superconductivity
The ultimate challenge for students of Feynman’s iconic Lectures series is, of course, to understand his final one: A Seminar on Superconductivity. As he notes in his introduction to this formidably dense piece, the text does not present the detail of each and every step in the development and, therefore, we’re not supposed to immediately understand everything. As Feynman puts it: we should just believe (more or less) that things would come out if we would be able to go through each and every step. Well… Let’s see. It took me one long maddening day to figure out the first formula:It says that the amplitude for a particle to go from a to b in a vector potential (think of a classical magnetic field) is the amplitude for the same particle to go from a to b when there is no field (A = 0) multiplied by the exponential of the line integral of the vector potential times the electric charge divided by Planck’s constant.
Of course, after a couple of hours, I recognized the formula for the magnetic effect on an amplitude, which I described in my previous post, which tells us that a magnetic field will shift the phase of the amplitude of a particle with an amount equal to:
Hence, if we write 〈b|a〉 for A = 0 as 〈b|a〉_{A = 0} = C·e^{i}^{θ}, then 〈b|a〉 in A will, naturally, be equal to 〈b|a〉 _{in A }= C·e^{i}^{(}^{θ+φ)} = C·e^{i}^{θ}·e^{i}^{φ} = 〈b|a〉_{A = 0 }·e^{i}^{φ}, and so that explains it. 🙂 Alright… Next.
The Schrödinger equation in an electromagnetic field
Feynman then jots down Schrödinger’s equation for the same particle (with charge q) moving in an electromagnetic field that is characterized not only by a vector potential A but also by the (scalar) potential Φ:
Now where does that come from? We know the standard formula in an electric field, right? It’s the formula we used to find the energy states of electrons in a hydrogen atom:
i·ħ·∂ψ/∂t = −(1/2)·(ħ^{2}/m)∇^{2}ψ + V·ψ
Of course, it is easy to see that we replaced V by q·Φ, which makes sense: the potential of a charge in an electric field is the product of the charge (q) and the (electric) potential (Φ), because Φ is, obviously, the potential energy of the unit charge. It’s also easy to see we can re-write −ħ^{2}·∇^{2}ψ as [(ħ/i)·∇]·[(ħ/i)·∇]ψ because (1/i)·(1/i) = 1/i^{2} = 1/(−1) = −1. 🙂 Alright. So it’s just that −q·A term in the (ħ/i)∇ − q·A expression that we need to explain now.
Unfortunately, that explanation is not so easy. Feynman basically re-derives Schrödinger’s equation using his trade-mark historical argument – which did not include any magnetic field – with a vector potential. The re-derivation is rather annoying, and I didn’t have the courage to go through it myself, so you should – just like me – just believe Feynman when he says that, when there’s a vector potential – i.e. when there’s a magnetic field – then that ħ/i)·∇ operator – which is the momentum operator– ought to be replaced by a new momentum operator:
So… Well… There we are… 🙂 So far, so good.
Local conservation of probability
The title of this section in Feynman’s Lecture (yes, still the same Lecture – we’re not switching topics here) is the equation of continuity for probabilities. I find it brilliant, because it confirms my interpretation of the wave function as describing some kind of energy flow. Let me quote Feynman on his endeavor here:
“An important part of the Schrödinger equation for a single particle is the idea that the probability to find the particle at a position is given by the absolute square of the wave function. It is also characteristic of the quantum mechanics that probability is conserved in a local sense. When the probability of finding the electron somewhere decreases, while the probability of the electron being elsewhere increases (keeping the total probability unchanged), something must be going on in between. In other words, the electron has a continuity in the sense that if the probability decreases at one place and builds up at another place, there must be some kind of flow between. If you put a wall, for example, in the way, it will have an influence and the probabilities will not be the same. So the conservation of probability alone is not the complete statement of the conservation law, just as the conservation of energy alone is not as deep and important as the local conservation of energy. If energy is disappearing, there must be a flow of energy to correspond. In the same way, we would like to find a “current” of probability such that if there is any change in the probability density (the probability of being found in a unit volume), it can be considered as coming from an inflow or an outflow due to some current.”
This is it, really ! The wave function does represent some kind of energy flow – between a so-called ‘real’ and a so-called ‘imaginary’ space, which are to be defined in terms of directional versus rotational energy, as I try to point out – admittedly: more by appealing to intuition than to mathematical rigor – in that post of mine on the meaning of the wavefunction.
So what is the flow – or probability current as Feynman refers to it? Well… Here’s the formula:
Huh? Yes. Don’t worry too much about it right now. The essential point is to understand what this current – denoted by J – actually stands for:
So what’s next? Well… Nothing. I’ll actually refer you to Feynman now, because I can’t improve on how he explains how pairs of electrons start behaving when temperatures are low enough to render Boltzmann’s Law irrelevant: the kinetic energy that’s associated with temperature can no longer break up electron pairs if temperature comes close to the zero point.
Huh? What? Electron pairs? Electrons are not supposed to form pairs, are they? They carry the same charge and are, therefore, supposed to repel each other. Well… Yes and no. In my post on the electron orbitals in a hydrogen atom – which just presented Feynman’s presentation on the subject-matter in a, hopefully, somewhat more readable format – we calculated electron orbitals neglecting spin. In Feynman’s words:
“We make another approximation by forgetting that the electron has spin. […] The non-relativistic Schrödinger equation disregards magnetic effects. [However] Small magnetic effects [do] occur because, from the electron’s point-of-view, the proton is a circulating charge which produces a magnetic field. In this field the electron will have a different energy with its spin up than with it down. [Hence] The energy of the atom will be shifted a little bit from what we will calculate. We will ignore this small energy shift. Also we will imagine that the electron is just like a gyroscope moving around in space always keeping the same direction of spin. Since we will be considering a free atom in space the total angular momentum will be conserved. In our approximation we will assume that the angular momentum of the electron spin stays constant, so all the rest of the angular momentum of the atom—what is usually called “orbital” angular momentum—will also be conserved. To an excellent approximation the electron moves in the hydrogen atom like a particle without spin—the angular momentum of the motion is a constant.”
To an excellent approximation… But… Well… Electrons in a metal do form pairs, because they can give up energy in that way and, hence, they are more stable that way. Feynman does not go into the details here – I guess because that’s way beyond the undergrad level – but refers to the Bardeen-Coopers-Schrieffer (BCS) theory instead – the authors of which got a Nobel Prize in Physics in 1972 (that’s a decade or so after Feynman wrote this particular Lecture), so I must assume the theory is well accepted now. 🙂
Of course, you’ll shout now: Hey! Hydrogen is not a metal! Well… Think again: the latest breakthrough in physics is making hydrogen behave like a metal. 🙂 And I am really talking the latest breakthrough: Science just published the findings of this experiment last month! 🙂 🙂 In any case, we’re not talking hydrogen here but superconducting materials, to which – as far as we know – the BCS theory does apply.
So… Well… I am done. I just wanted to show you why it’s important to work your way through Feynman’s last Lecture because… Well… Quantum mechanics does explain everything – although the nitty-gritty of it (the Meissner effect, the London equation, flux quantization, etc.) are rather hard bullets to bite. 😦
Don’t give up ! I am struggling with the nitty-gritty too ! 🙂
An interpretation of the wavefunction
This is my umpteenth post on the same topic. 😦 It is obvious that this search for a sensible interpretation is consuming me. Why? I am not sure. Studying physics is frustrating. As a leading physicist puts it:
“The teaching of quantum mechanics these days usually follows the same dogma: firstly, the student is told about the failure of classical physics at the beginning of the last century; secondly, the heroic confusions of the founding fathers are described and the student is given to understand that no humble undergraduate student could hope to actually understand quantum mechanics for himself; thirdly, a deus ex machina arrives in the form of a set of postulates (the Schrödinger equation, the collapse of the wavefunction, etc); fourthly, a bombardment of experimental verifications is given, so that the student cannot doubt that QM is correct; fifthly, the student learns how to solve the problems that will appear on the exam paper, hopefully with as little thought as possible.”
That’s obviously not the way we want to understand quantum mechanics. [With we, I mean, me, of course, and you, if you’re reading this blog.] Of course, that doesn’t mean I don’t believe Richard Feynman, one of the greatest physicists ever, when he tells us no one, including himself, understands physics quite the way we’d like to understand it. Such statements should not prevent us from trying harder. So let’s look for better metaphors. The animation below shows the two components of the archetypal wavefunction – a simple sine and cosine. They’re the same function actually, but their phases differ by 90 degrees (π/2).
It makes me think of a V-2 engine with the pistons at a 90-degree angle. Look at the illustration below, which I took from a rather simple article on cars and engines that has nothing to do with quantum mechanics. Think of the moving pistons as harmonic oscillators, like springs.
We will also think of the center of each cylinder as the zero point: think of that point as a point where – if we’re looking at one cylinder alone – the internal and external pressure balance each other, so the piston would not move… Well… If it weren’t for the other piston, because the second piston is not at the center when the first is. In fact, it is easy to verify and compare the following positions of both pistons, as well as the associated dynamics of the situation:
Piston 1 |
Piston 2 |
Motion of Piston 1 |
Motion Piston 2 |
Top |
Center |
Compressed air will push piston down |
Piston moves down against external pressure |
Center |
Bottom |
Piston moves down against external pressure |
External air pressure will push piston up |
Bottom |
Center |
External air pressure will push piston up |
Piston moves further up and compresses the air |
Center |
Top |
Piston moves further up and compresses the air |
Compressed air will push piston down |
When the pistons move, their linear motion will be described by a sinusoidal function: a sine or a cosine. In fact, the 90-degree V-2 configuration ensures that the linear motion of the two pistons will be exactly the same, except for a phase difference of 90 degrees. [Of course, because of the sideways motion of the connecting rods, our sine and cosine function describes the linear motion only approximately, but you can easily imagine the idealized limit situation. If not, check Feynman’s description of the harmonic oscillator.]
The question is: if we’d have a set-up like this, two springs – or two harmonic oscillators – attached to a shaft through a crank, would this really work as a perpetuum mobile? We obviously talk energy being transferred back and forth between the rotating shaft and the moving pistons… So… Well… Let’s model this: the total energy, potential and kinetic, in each harmonic oscillator is constant. Hence, the piston only delivers or receives kinetic energy from the rotating mass of the shaft.
Now, in physics, that’s a bit of an oxymoron: we don’t think of negative or positive kinetic (or potential) energy in the context of oscillators. We don’t think of the direction of energy. But… Well… If we’ve got two oscillators, our picture changes, and so we may have to adjust our thinking here.
Let me start by giving you an authoritative derivation of the various formulas involved here, taking the example of the physical spring as an oscillator—but the formulas are basically the same for any harmonic oscillator.
The first formula is a general description of the motion of our oscillator. The coefficient in front of the cosine function (a) is the maximum amplitude. Of course, you will also recognize ω_{0} as the natural frequency of the oscillator, and Δ as the phase factor, which takes into account our t = 0 point. In our case, for example, we have two oscillators with a phase difference equal to π/2 and, hence, Δ would be 0 for one oscillator, and –π/2 for the other. [The formula to apply here is sinθ = cos(θ – π/2).] Also note that we can equate our θ argument to ω_{0}·t. Now, if a = 1 (which is the case here), then these formulas simplify to:
- K.E. = T = m·v^{2}/2 = m·ω_{0}^{2}·sin^{2}(θ + Δ) = m·ω_{0}^{2}·sin^{2}(ω_{0}·t + Δ)
- P.E. = U = k·x^{2}/2 = k·cos^{2}(θ + Δ)
The coefficient k in the potential energy formula characterizes the force: F = −k·x. The minus sign reminds us our oscillator wants to return to the center point, so the force pulls back. From the dynamics involved, it is obvious that k must be equal to m·ω_{0}^{2}., so that gives us the famous T + U = m·ω_{0}^{2}/2 formula or, including a once again, T + U = m·a^{2}·ω_{0}^{2}/2.
Now, if we normalize our functions by equating k to one (k = 1), then the motion of our first oscillator is given by the cosθ function, and its kinetic energy will be equal to sin^{2}θ. Hence, the (instantaneous) change in kinetic energy at any point in time will be equal to:
d(sin^{2}θ)/dθ = 2∙sinθ∙d(sinθ)/dt = 2∙sinθ∙cosθ
Let’s look at the second oscillator now. Just think of the second piston going up and down in our V-twin engine. Its motion is given by the sinθ function which, as mentioned above, is equal to cos(θ−π /2). Hence, its kinetic energy is equal to sin^{2}(θ−π /2), and how it changes – as a function of θ – will be equal to:
2∙sin(θ−π /2)∙cos(θ−π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ
We have our perpetuum mobile! While transferring kinetic energy from one piston to the other, the rotating shaft moves at constant speed. Linear motion becomes circular motion, and vice versa, in a frictionless Universe. We have the metaphor we were looking for!
Somehow, in this beautiful interplay between linear and circular motion, energy is being borrowed from one place to another, and then returned. From what place to what place? I am not sure. We may call it the real and imaginary energy space respectively, but what does that mean? One thing is for sure, however: the interplay between the real and imaginary part of the wavefunction describes how energy propagates through space!
How exactly? Again, I am not sure. Energy is, obviously, mass in motion – as evidenced by the E = m·c^{2} equation, and it may not have any direction (when everything is said and done, it’s a scalar quantity without direction), but the energy in a linear motion is surely different from that in a circular motion, and our metaphor suggests we need to think somewhat more along those lines. Perhaps we will, one day, able to square this circle. 🙂
Schrödinger’s equation
Let’s analyze the interplay between the real and imaginary part of the wavefunction through an analysis of Schrödinger’s equation, which we write as:
i·ħ∙∂ψ/∂t = –(ħ^{2}/2m)∙∇^{2}ψ + V·ψ
We can do a quick dimensional analysis of both sides:
- [i·ħ∙∂ψ/∂t] = N∙m∙s/s = N∙m
- [–(ħ^{2}/2m)∙∇^{2}ψ] = N∙m^{3}/m^{2} = N∙m
- [V·ψ] = N∙m
Note the dimension of the ‘diffusion’ constant ħ^{2}/2m: [ħ^{2}/2m] = N^{2}∙m^{2}∙s^{2}/kg = N^{2}∙m^{2}∙s^{2}/(N·s^{2}/m) = N∙m^{3}. Also note that, in order for the dimensions to come out alright, the dimension of V – the potential – must be that of energy. Hence, Feynman’s description of it as the potential energy – rather than the potential tout court – is somewhat confusing but correct: V must equal the potential energy of the electron. Hence, V is not the conventional (potential) energy of the unit charge (1 coulomb). Instead, the natural unit of charge is used here, i.e. the charge of the electron itself.
Now, Schrödinger’s equation – without the V·ψ term – can be written as the following pair of equations:
- Re(∂ψ/∂t) = −(1/2)∙(ħ/m)∙Im(∇^{2}ψ)
- Im(∂ψ/∂t) = (1/2)∙(ħ/m)∙Re(∇^{2}ψ)
This closely resembles the propagation mechanism of an electromagnetic wave as described by Maxwell’s equation for free space (i.e. a space with no charges), but E and B are vectors, not scalars. How do we get this result. Well… ψ is a complex function, which we can write as a + i∙b. Likewise, ∂ψ/∂t is a complex function, which we can write as c + i∙d, and ∇^{2}ψ can then be written as e + i∙f. If we temporarily forget about the coefficients (ħ, ħ^{2}/m and V), then Schrödinger’s equation – including V·ψ term – amounts to writing something like this:
i∙(c + i∙d) = –(e + i∙f) + (a + i∙b) ⇔ a + i∙b = i∙c − d + e+ i∙f ⇔ a = −d + e and b = c + f
Hence, we can now write:
- V∙Re(ψ) = −ħ∙Im(∂ψ/∂t) + (1/2)∙( ħ^{2}/m)∙Re(∇^{2}ψ)
- V∙Im(ψ) = ħ∙Re(∂ψ/∂t) + (1/2)∙( ħ^{2}/m)∙Im(∇^{2}ψ)
This simplifies to the two equations above for V = 0, i.e. when there is no potential (electron in free space). Now we can bring the Re and Im operators into the brackets to get:
- V∙Re(ψ) = −ħ∙∂Im (ψ)/∂t + (1/2)∙( ħ^{2}/m)∙∇^{2}Re(ψ)
- V∙Im(ψ) = ħ∙∂Re(ψ)/∂t + (1/2)∙( ħ^{2}/m)∙∇^{2}Im(ψ)
This is very interesting, because we can re-write this using the quantum-mechanical energy operator H = –(ħ^{2}/2m)∙∇^{2} + V· (note the multiplication sign after the V, which we do not have – for obvious reasons – for the –(ħ^{2}/2m)∙∇^{2} expression):
- H[Re (ψ)] = −ħ∙∂Im(ψ)/∂t
- H[Im(ψ)] = ħ∙∂Re(ψ)/∂t
A dimensional analysis shows us both sides are, once again, expressed in N∙m. It’s a beautiful expression because – if we write the real and imaginary part of ψ as r∙cosθ and r∙sinθ, we get:
- H[cosθ] = −ħ∙∂sinθ/∂t = E∙cosθ
- H[sinθ] = ħ∙∂cosθ/∂t = E∙sinθ
Indeed, θ = (E∙t − p∙x)/ħ and, hence, −ħ∙∂sinθ/∂t = ħ∙cosθ∙E/ħ = E∙cosθ and ħ∙∂cosθ/∂t = ħ∙sinθ∙E/ħ = E∙sinθ. Now we can combine the two equations in one equation again and write:
H[r∙(cosθ + i∙sinθ)] = r∙(E∙cosθ + i∙sinθ) ⇔ H[ψ] = E∙ψ
The operator H – applied to the wavefunction – gives us the (scalar) product of the energy E and the wavefunction itself. Isn’t this strange?
Hmm… I need to further verify and explain this result… I’ll probably do so in yet another post on the same topic… 🙂
Post scriptum: The symmetry of our V-2 engine – or perpetuum mobile – is interesting: its cross-section has only one axis of symmetry. Hence, we may associate some angle with it, so as to define its orientation in the two-dimensional cross-sectional plane. Of course, the cross-sectional plane itself is at right angles to the crankshaft axis, which we may also associate with some angle in three-dimensional space. Hence, its geometry defines two orthogonal directions which, in turn, define a spherical coordinate system, as shown below.
We may, therefore, say that three-dimensional space is actually being implied by the geometry of our V-2 engine. Now that is interesting, isn’t it? 🙂
All what you ever wanted to know about the photon wavefunction…
Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately read the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link.
Original post:
This post is, essentially, a continuation of my previous post, in which I juxtaposed the following images:
Both are the same, and then they’re not. The illustration on the right-hand side is a regular quantum-mechanical wavefunction, i.e. an amplitude wavefunction. You’ve seen that one before. In this case, the x-axis represents time, so we’re looking at the wavefunction at some particular point in space. ]You know we can just switch the dimensions and it would all look the same.] The illustration on the left-hand side looks similar, but it’s not an amplitude wavefunction. The animation shows how the electric field vector (E) of an electromagnetic wave travels through space. Its shape is the same. So it’s the same function. Is it also the same reality?
Yes and no. And I would say: more no than yes—in this case, at least. Note that the animation does not show the accompanying magnetic field vector (B). That vector is equally essential in the electromagnetic propagation mechanism according to Maxwell’s equations, which—let me remind you—are equal to:
- ∂B/∂t = –∇×E
- ∂E/∂t = ∇×B
In fact, I should write the second equation as ∂E/∂t = c^{2}∇×B, but then I assume we measure time and distance in equivalent units, so c = 1.
You know that E and B are two aspects of one and the same thing: if we have one, then we have the other. To be precise, B is always orthogonal to E in the direction that’s given by the right-hand rule for the following vector cross-product: B = e_{x}×E, with e_{x} the unit vector pointing in the x-direction (i.e. the direction of propagation). The reality behind is illustrated below for a linearly polarized electromagnetic wave.
The B = e_{x}×E equation is equivalent to writing B= i·E, which is equivalent to:
B = i·E = e^{i}^{(π/2)}·e^{i}^{(kx − ωt)} = cos(kx − ωt + π/2) + i·sin(kx − ωt + π/2)
= −sin((kx − ωt) + i·cos(kx − ωt)
Now, E and B have only two components: E_{y }and E_{z}, and B_{y }and B_{z}. That’s only because we’re looking at some ideal or elementary electromagnetic wave here but… Well… Let’s just go along with it. 🙂 It is then easy to prove that the equation above amounts to writing:
- B_{y }= cos(kx − ωt + π/2) = −sin(kx − ωt) = −E_{z}
- B_{z }= sin(kx − ωt + π/2) = cos(kx − ωt) = E_{y}
We should now think of E_{y} and E_{z }as the real and imaginary part of some wavefunction, which we’ll denote as ψ_{E} = e^{i}^{(kx − ωt)}. So we write:
E = (E_{y}, E_{z}) = E_{y} + i·E_{z }= cos(kx − ωt) + i∙sin(kx − ωt) = Re(ψ_{E}) + i·Im(ψ_{E}) = ψ_{E} = e^{i}^{(kx − ωt)}
What about B? We just do the same, so we write:
B = (B_{y}, B_{z}) = B_{y} + i·B_{z }= ψ_{B} = i·E = i·ψ_{E} = −sin(kx − ωt) + i∙sin(kx − ωt) = − Im(ψ_{E}) + i·Re(ψ_{E})
Now we need to prove that ψ_{E} and ψ_{B} are regular wavefunctions, which amounts to proving Schrödinger’s equation, i.e. ∂ψ/∂t = i·(ħ/m)·∇^{2}ψ, for both ψ_{E} and ψ_{B}. [Note I use the Schrödinger’s equation for a zero-mass spin-zero particle here, which uses the ħ/m factor rather than the ħ/(2m) factor.] To prove that ψ_{E} and ψ_{B} are regular wavefunctions, we should prove that:
- Re(∂ψ_{E}/∂t) = −(ħ/m)·Im(∇^{2}ψ_{E}) and Im(∂ψ_{E}/∂t) = (ħ/m)·Re(∇^{2}ψ_{E}), and
- Re(∂ψ_{B}/∂t) = −(ħ/m)·Im(∇^{2}ψ_{B}) and Im(∂ψ_{B}/∂t) = (ħ/m)·Re(∇^{2}ψ_{B}).
Let’s do the calculations for the second pair of equations. The time derivative on the left-hand side is equal to:
∂ψ_{B}/∂t = −iω·ie^{i}^{(kx − ωt) }= ω·[cos(kx − ωt) + i·sin(kx − ωt)] = ω·cos(kx − ωt) + iω·sin(kx − ωt)
The second-order derivative on the right-hand side is equal to:
∇^{2}ψ_{B }= ∂^{2}ψ_{B}/∂x^{2 }= i·k^{2}·e^{i}^{(kx − ωt) }= k^{2}·cos(kx − ωt) + i·k^{2}·sin(kx − ωt)
So the two equations for ψ_{B }are equivalent to writing:
- Re(∂ψ_{B}/∂t) = −(ħ/m)·Im(∇^{2}ψ_{B}) ⇔ ω·cos(kx − ωt) = k^{2}·(ħ/m)·cos(kx − ωt)
- Im(∂ψ_{B}/∂t) = (ħ/m)·Re(∇^{2}ψ_{B}) ⇔ ω·sin(kx − ωt) = k^{2}·(ħ/m)·sin(kx − ωt)
So we see that both conditions are fulfilled if, and only if, ω = k^{2}·(ħ/m).
Now, we also demonstrated in that post of mine that Maxwell’s equations imply the following:
- ∂B_{y}/∂t = –(∇×E)_{y} = ∂E_{z}/∂x = ∂[sin(kx − ωt)]/∂x = k·cos(kx − ωt) = k·E_{y}
- ∂B_{z}/∂t = –(∇×E)_{z} = – ∂E_{y}/∂x = – ∂[cos(kx − ωt)]/∂x = k·sin(kx − ωt) = k·E_{z}
Hence, using those B_{y }= −E_{z }and B_{z }= E_{y }equations above, we can also calculate these derivatives as:
- ∂B_{y}/∂t = −∂E_{z}/∂t = −∂sin(kx − ωt)/∂t = ω·cos(kx − ωt) = ω·E_{y}
- ∂B_{z}/∂t = ∂E_{y}/∂t = ∂cos(kx − ωt)/∂t = −ω·[−sin(kx − ωt)] = ω·E_{z}
In other words, Maxwell’s equations imply that ω = k, which is consistent with us measuring time and distance in equivalent units, so the phase velocity is c = 1 = ω/k.
So far, so good. We basically established that the propagation mechanism for an electromagnetic wave, as described by Maxwell’s equations, is fully coherent with the propagation mechanism—if we can call it like that—as described by Schrödinger’s equation. We also established the following equalities:
- ω = k
- ω = k^{2}·(ħ/m)
The second of the two de Broglie equations tells us that k = p/ħ, so we can combine these two equations and re-write these two conditions as:
ω/k = 1 = k·(ħ/m) = (p/ħ)·(ħ/m) = p/m ⇔ p = m
What does this imply? The p here is the momentum: p = m·v, so this condition implies v must be equal to 1 too, so the wave velocity is equal to the speed of light. Makes sense, because we actually are talking light here. 🙂 In addition, because it’s light, we also know E/p = c = 1, so we have – once again – the general E = p = m equation, which we’ll need!
OK. Next. Let’s write the Schrödinger wave equation for both wavefunctions:
- ∂ψ_{E}/∂t = i·(ħ/m_{E})·∇^{2}ψ_{E}, and
- ∂ψ_{B}/∂t = i·(ħ/m_{B})·∇^{2}ψ_{B}.
Huh? What’s m_{E} and m_{E}? We should only associate one mass concept with our electromagnetic wave, shouldn’t we? Perhaps. I just want to be on the safe side now. Of course, if we distinguish m_{E} and m_{B}, we should probably also distinguish p_{E} and p_{B}, and E_{E} and E_{B} as well, right? Well… Yes. If we accept this line of reasoning, then the mass factor in Schrödinger’s equations is pretty much like the 1/c^{2} = μ_{0}ε_{0} factor in Maxwell’s (1/c^{2})·∂E/∂t = ∇×B equation: the mass factor appears as a property of the medium, i.e. the vacuum here! [Just check my post on physical constants in case you wonder what I am trying to say here, in which I explain why and how c defines the (properties of the) vacuum.]
To be consistent, we should also distinguish p_{E} and p_{B}, and E_{E} and E_{B}, and so we should write ψ_{E }and ψ_{B} as:
- ψ_{E} = e^{i}^{(kEx − ωEt)}, and
- ψ_{B} = e^{i}^{(kBx − ωBt)}.
Huh? Yes. I know what you think: we’re talking one photon—or one electromagnetic wave—so there can be only one energy, one momentum and, hence, only one k, and one ω. Well… Yes and no. Of course, the following identities should hold: k_{E} = k_{B} and, likewise, ω_{E }= ω_{B}. So… Yes. They’re the same: one k and one ω. But then… Well… Conceptually, the two k’s and ω’s are different. So we write:
- p_{E} = E_{E} = m_{E}, and
- p_{B} = E_{B} = m_{B}.
The obvious question is: can we just add them up to find the total energy and momentum of our photon? The answer is obviously positive: E = E_{E} + E_{B}, p = p_{E} + p_{B} and m = m_{E} + m_{B}.
Let’s check a few things now. How does it work for the phase and group velocity of ψ_{E }and ψ_{B}? Simple:
- v_{g} = ∂ω_{E}/∂k_{E} = ∂[E_{E}/ħ]/∂[p_{E}/ħ] = ∂E_{E}/∂p_{E} = ∂p_{E}/∂p_{E} = 1
- v_{p} = ω_{E}/k_{E} = (E_{E}/ħ)/(p_{E}/ħ) = E_{E}/p_{E} = p_{E}/p_{E} = 1
So we’re fine, and you can check the result for ψ_{B }by substituting the subscript E for B. To sum it all up, what we’ve got here is the following:
- We can think of a photon having some energy that’s equal to E = p = m (assuming c = 1), but that energy would be split up in an electric and a magnetic wavefunction respectively: ψ_{E }and ψ_{B}.
- Schrödinger’s equation applies to both wavefunctions, but the E, p and m in those two wavefunctions are the same and not the same: their numerical value is the same (p_{E} =E_{E} = m_{E} = p_{B} =E_{B} = m_{B}), but they’re conceptually different. They must be: if not, we’d get a phase and group velocity for the wave that doesn’t make sense.
Of course, the phase and group velocity for the sum of the ψ_{E }and ψ_{B }waves must also be equal to c. This is obviously the case, because we’re adding waves with the same phase and group velocity c, so there’s no issue with the dispersion relation.
So let’s insert those p_{E} =E_{E} = m_{E} = p_{B} =E_{B} = m_{B} values in the two wavefunctions. For ψ_{E}, we get:
ψ_{E }= e^{i}^{[kEx − ωEt) }= e^{i}^{[(pE/ħ)·x − (EE/ħ)·t]}^{ }
You can do the calculation for ψ_{B }yourself. Let’s simplify our life a little bit and assume we’re using Planck units, so ħ = 1, and so the wavefunction simplifies to ψ_{E }= e^{i}^{·(pE·x − EE·t)}. We can now add the components of E and B using the summation formulas for sines and cosines:
1. B_{y }+ E_{y} = cos(p_{B}·x − E_{B}·t + π/2) + cos(p_{E}·x − E_{E}·t) = 2·cos[(p·x − E·t + π/2)/2]·cos(π/4) = √2·cos(p·x/2 − E·t/2 + π/4)
2. B_{z }+ E_{z} = sin(p_{B}·x − E_{B}·t+π/2) + sin(p_{E}·x − E_{E}·t) = 2·sin[(p·x − E·t + π/2)/2]·cos(π/4) = √2·sin(p·x/2 − E·t/2 + π/4)
Interesting! We find a composite wavefunction for our photon which we can write as:
E + B = ψ_{E }+ ψ_{B }= E + i·E = √2·e^{i}^{(p·x/2 − E·t/2 + π/4) }= √2·e^{i}^{(π/4)}·e^{i}^{(p·x/2 − E·t/2) }= √2·e^{i}^{(π/4)}·E
What a great result! It’s easy to double-check, because we can see the E + i·E = √2·e^{i}^{(π/4)}·E formula implies that 1 + i should equal √2·e^{i}^{(π/4)}. Now that’s easy to prove, both geometrically (just do a drawing) or formally: √2·e^{i}^{(π/4)} = √2·cos(π/4) + i·sin(π/4e^{i}^{(π/4)} = (√2/√2) + i·(√2/√2) = 1 + i. We’re bang on! 🙂
We can double-check once more, because we should get the same from adding E and B = i·E, right? Let’s try:
E + B = E + i·E = cos(p_{E}·x − E_{E}·t) + i·sin(p_{E}·x − E_{E}·t) + i·cos(p_{E}·x − E_{E}·t) − sin(p_{E}·x − E_{E}·t)
= [cos(p_{E}·x − E_{E}·t) – sin(p_{E}·x − E_{E}·t)] + i·[sin(p_{E}·x − E_{E}·t) – cos(p_{E}·x − E_{E}·t)]
Indeed, we can see we’re going to obtain the same result, because the −sinθ in the real part of our composite wavefunction is equal to cos(θ+π/2), and the −cosθ in its imaginary part is equal to sin(θ+π/2). So the sum above is the same sum of cosines and sines that we did already.
So our electromagnetic wavefunction, i.e. the wavefunction for the photon, is equal to:
ψ = ψ_{E }+ ψ_{B }= √2·e^{i}^{(p·x/2 − E·t/2 + π/4)} = √2·e^{i}^{(π/4)}·e^{i}^{(p·x/2 − E·t/2) }
What about the √2 factor in front, and the π/4 term in the argument itself? No sure. It must have something to do with the way the magnetic force works, which is not like the electric force. Indeed, remember the Lorentz formula: the force on some unit charge (q = 1) will be equal to F = E + v×B. So… Well… We’ve got another cross-product here and so the geometry of the situation is quite complicated: it’s not like adding two forces F_{1 }and F_{2 }to get some combined force F = F_{1 }and F_{2}.
In any case, we need the energy, and we know that its proportional to the square of the amplitude, so… Well… We’re spot on: the square of the √2 factor in the √2·cos product and √2·sin product is 2, so that’s twice… Well… What? Hold on a minute! We’re actually taking the absolute square of the E + B = ψ_{E }+ ψ_{B }= E + i·E = √2·e^{i}^{(p·x/2 − E·t/2 + π/4)}^{ }wavefunction here. Is that legal? I must assume it is—although… Well… Yes. You’re right. We should do some more explaining here.
We know that we usually measure the energy as some definite integral, from t = 0 to some other point in time, or over the cycle of the oscillation. So what’s the cycle here? Our combined wavefunction can be written as √2·e^{i}^{(p·x/2 − E·t/2 + π/4)}^{ }= √2·e^{i}^{(θ/2 + π/4)}, so a full cycle would correspond to θ going from 0 to 4π here, rather than from 0 to 2π. So that explains the √2 factor in front of our wave equation.
Bingo! If you were looking for an interpretation of the Planck energy and momentum, here it is. And, while everything that’s written above is not easy to understand, it’s close to the ‘intuitive’ understanding to quantum mechanics that we were looking for, isn’t it? The quantum-mechanical propagation model explains everything now. 🙂 I only need to show one more thing, and that’s the different behavior of bosons and fermions:
- The amplitudes of identitical bosonic particles interfere with a positive sign, so we have Bose-Einstein statistics here. As Feynman writes it: (amplitude direct) + (amplitude exchanged).
- The amplitudes of identical fermionic particles interfere with a negative sign, so we have Fermi-Dirac statistics here: (amplitude direct) − (amplitude exchanged).
I’ll think about it. I am sure it’s got something to do with that B= i·E formula or, to put it simply, with the fact that, when bosons are involved, we get two wavefunctions (ψ_{E }and ψ_{B}) for the price of one. The reasoning should be something like this:
I. For a massless particle (i.e. a zero-mass fermion), our wavefunction is just ψ = e^{i}^{(p·x − E·t)}. So we have no √2 or √2·e^{i}^{(π/4) }factor in front here. So we can just add any number of them – ψ_{1}_{ }+ ψ_{2}_{ }+ ψ_{3}_{ }+ … – and then take the absolute square of the amplitude to find a probability density, and we’re done.
II. For a photon (i.e. a zero-mass boson), our wavefunction is √2·e^{i}^{(π/4)}·e^{i}^{(p·x − E·t)/2}, which – let’s introduce a new symbol – we’ll denote by φ, so φ = √2·e^{i}^{(π/4)}·e^{i}^{(p·x − E·t)/2}. Now, if we add any number of these, we get a similar sum but with that √2·e^{i}^{(π/4) }factor in front, so we write: φ_{1}_{ }+ φ_{2}_{ }+ φ_{3}_{ }+ … = √2·e^{i}^{(π/4)}·(ψ_{1}_{ }+ ψ_{2}_{ }+ ψ_{3}_{ }+ …). If we take the absolute square now, we’ll see the probability density will be equal to twice the density for the ψ_{1}_{ }+ ψ_{2}_{ }+ ψ_{3}_{ }+ … sum, because
|√2·e^{i}^{(π/4)}·(ψ_{1}_{ }+ ψ_{2}_{ }+ ψ_{3}_{ }+ …)|^{2} = |√2·e^{i}^{(π/4)}|^{2}·|ψ_{1}_{ }+ ψ_{2}_{ }+ ψ_{3}_{ }+ …)|^{2} = 2·|ψ_{1}_{ }+ ψ_{2}_{ }+ ψ_{3}_{ }+ …)|^{2}
So… Well… I still need to connect this to Feynman’s (amplitude direct) ± (amplitude exchanged) formula, but I am sure it can be done.
Now, we haven’t tested the complete √2·e^{i}^{(π/4)}·e^{i}^{(p·x − E·t)/2}^{ }wavefunction. Does it respect Schrödinger’s ∂ψ/∂t = i·(1/m)·∇^{2}ψ or, including the 1/2 factor, the ∂ψ/∂t = i·[1/2m)]·∇^{2}ψ equation? [Note we assume, once again, that ħ = 1, so we use Planck units once more.] Let’s see. We can calculate the derivatives as:
- ∂ψ/∂t = −√2·e^{i}^{(π/4)}·e^{−i∙[p·x − E·t]/2}·(i·E/2)
- ∇^{2}ψ = ∂^{2}[√2·e^{i}^{(π/4)}·e^{−i∙[p·x − E·t]/2}]/∂x^{2 }= √2·e^{i}^{(π/4)}·∂[√2·e^{i}^{(π/4)}·e^{−i∙[p·x − E·t]/2}·(i·p/2)]/∂x = −√2·e^{i}^{(π/4)}·e^{−i∙[p·x − E·t]/2}·(p^{2}/4)
So Schrödinger’s equation becomes:
−i·√2·e^{i}^{(π/4)}·e^{−i∙[p·x − E·t]/2}·(i·E/2) = −i·(1/m)·√2·e^{i}^{(π/4)}·e^{−i∙[p·x − E·t]/2}·(p^{2}/4) ⇔ 1/2 = 1/4!?
That’s funny ! It doesn’t work ! The E and m and p^{2} are OK because we’ve got that E = m = p equation, but we’ve got problems with yet another factor 2. It only works when we use the 2/m coefficient in Schrödinger’s equation.
So… Well… There’s no choice. That’s what we’re going to do. The Schrödinger equation for the photon is ∂ψ/∂t = i·(2/m)·∇^{2}ψ !
It’s a very subtle point. This is all great, and very fundamental stuff! Let’s now move on to Schrödinger’s actual equation, i.e. the ∂ψ/∂t = i·(ħ/2m)·∇^{2}ψ equation.
Post scriptum on the Planck units:
If we measure time and distance in equivalent units, say seconds, we can re-write the quantum of action as:
1.0545718×10^{−34 }N·m·s = (1.21×10^{44 }N)·(1.6162×10^{−35 }m)·(5.391×10^{−44} s)
⇔ (1.0545718×10^{−34}/2.998×10^{8}) N·s^{2} = (1.21×10^{44 }N)·(1.6162×10^{−35}/2.998×10^{8} s)(5.391×10^{−44} s)
⇔ (1.21×10^{44 }N) = [(1.0545718×10^{−34}/2.998×10^{8})]/[(1.6162×10^{−35}/2.998×10^{8} s)(5.391×10^{−44} s)] N·s^{2}/s^{2}
You’ll say: what’s this? Well… Look at it. We’ve got a much easier formula for the Planck force—much easier than the standard formulas you’ll find on Wikipedia, for example. If we re-interpret the symbols ħ and c so they denote the numerical value of the quantum of action and the speed of light in standard SI units (i.e. newton, meter and second)—so ħ and c become dimensionless, or mathematical constants only, rather than physical constants—then the formula above can be written as:
F_{P} newton = (ħ/c)/[(l_{P}/c)·t_{P}] newton ⇔ F_{P} = ħ/(l_{P}·t_{P})
Just double-check it: 1.0545718×10^{−34}/(1.6162×10^{−35}·5.391×10^{−44}) = 1.21×10^{44}. Bingo!
You’ll say: what’s the point? The point is: our model is complete. We don’t need the other physical constants – i.e. the Coulomb, Boltzmann and gravitational constant – to calculate the Planck units we need, i.e. the Planck force, distance and time units. It all comes out of our elementary wavefunction! All we need to explain the Universe – or, let’s be more modest, quantum mechanics – is two numerical constants (c and ħ) and Euler’s formula (which uses π and e, of course). That’s it.
If you don’t think that’s a great result, then… Well… Then you’re not reading this. 🙂