Reality and perception

It’s quite easy to get lost in all of the math when talking quantum mechanics. In this post, I’d like to freewheel a bit. I’ll basically try to relate the wavefunction we’ve derived for the electron orbitals to the more speculative posts I wrote on how to interpret the wavefunction. So… Well… Let’s go. 🙂

If there is one thing you should remember from all of the stuff I wrote in my previous posts, then it’s that the wavefunction for an electron orbital – ψ(x, t), so that’s a complex-valued function in two variables (position and time) – can be written as the product of two functions in one variable:

ψ(x, t) = ei·(E/ħ)·t·f(x)

In fact, we wrote f(x) as ψ(x), but I told you how confusing that is: the ψ(x) and ψ(x, t) functions are, obviously, very different. To be precise, the f(x) = ψ(x) function basically provides some envelope for the two-dimensional eiθ = ei·(E/ħ)·t = cosθ + i·sinθ oscillation – as depicted below (θ = −(E/ħ)·t = ω·t with ω = −E/ħ).Circle_cos_sinWhen analyzing this animation – look at the movement of the green, red and blue dots respectively – one cannot miss the equivalence between this oscillation and the movement of a mass on a spring – as depicted below.spiral_sThe ei·(E/ħ)·t function just gives us two springs for the price of one. 🙂 Now, you may want to imagine some kind of elastic medium – Feynman’s famous drum-head, perhaps 🙂 – and you may also want to think of all of this in terms of superimposed waves but… Well… I’d need to review if that’s really relevant to what we’re discussing here, so I’d rather not make things too complicated and stick to basics.

First note that the amplitude of the two linear oscillations above is normalized: the maximum displacement of the object from equilibrium, in the positive or negative direction, which we may denote by x = ±A, is equal to one. Hence, the energy formula is just the sum of the potential and kinetic energy: T + U = (1/2)·A2·m·ω2 = (1/2)·m·ω2. But so we have two springs and, therefore, the energy in this two-dimensional oscillation is equal to E = 2·(1/2)·m·ω2 = m·ω2.

This formula is structurally similar to Einstein’s E = m·c2 formula. Hence, one may want to assume that the energy of some particle (an electron, in our case, because we’re discussing electron orbitals here) is just the two-dimensional motion of its mass. To put it differently, we might also want to think that the oscillating real and imaginary component of our wavefunction each store one half of the total energy of our particle.

However, the interpretation of this rather bold statement is not so straightforward. First, you should note that the ω in the E = m·ω2 formula is an angular velocity, as opposed to the in the E = m·c2 formula, which is a linear velocity. Angular velocities are expressed in radians per second, while linear velocities are expressed in meter per second. However, while the radian measures an angle, we know it does so by measuring a length. Hence, if our distance unit is 1 m, an angle of 2π rad will correspond to a length of 2π meter, i.e. the circumference of the unit circle. So… Well… The two velocities may not be so different after all.

There are other questions here. In fact, the other questions are probably more relevant. First, we should note that the ω in the E = m·ω2 can take on any value. For a mechanical spring, ω will be a function of (1) the stiffness of the spring (which we usually denote by k, and which is typically measured in newton (N) per meter) and (2) the mass (m) on the spring. To be precise, we write: ω2 = k/m – or, what amounts to the same, ω = √(k/m). Both k and m are variables and, therefore, ω can really be anything. In contrast, we know that c is a constant: equals 299,792,458 meter per second, to be precise. So we have this rather remarkable expression: c = √(E/m), and it is valid for any particle – our electron, or the proton at the center, or our hydrogen atom as a whole. It is also valid for more complicated atoms, of course. In fact, it is valid for any system.

Hence, we need to take another look at the energy concept that is used in our ψ(x, t) = ei·(E/ħ)·t·f(x) wavefunction. You’ll remember (if not, you should) that the E here is equal to En = −13.6 eV, −3.4 eV, −1.5 eV and so on, for = 1, 2, 3, etc. Hence, this energy concept is rather particular. As Feynman puts it: “The energies are negative because we picked our zero point as the energy of an electron located far from the proton. When it is close to the proton, its energy is less, so somewhat below zero. The energy is lowest (most negative) for n = 1, and increases toward zero with increasing n.”

Now, this is the one and only issue I have with the standard physics story. I mentioned it in one of my previous posts and, just for clarity, let me copy what I wrote at the time:

Feynman gives us a rather casual explanation [on choosing a zero point for measuring energy] in one of his very first Lectures on quantum mechanics, where he writes the following: “If we have a “condition” which is a mixture of two different states with different energies, then the amplitude for each of the two states will vary with time according to an equation like a·eiωt, with ħ·ω = E = m·c2. Hence, we can write the amplitude for the two states, for example as:

ei(E1/ħ)·t and ei(E2/ħ)·t

And if we have some combination of the two, we will have an interference. But notice that if we added a constant to both energies, it wouldn’t make any difference. If somebody else were to use a different scale of energy in which all the energies were increased (or decreased) by a constant amount—say, by the amount A—then the amplitudes in the two states would, from his point of view, be:

ei(E1+A)·t/ħ and ei(E2+A)·t/ħ

All of his amplitudes would be multiplied by the same factor ei(A/ħ)·t, and all linear combinations, or interferences, would have the same factor. When we take the absolute squares to find the probabilities, all the answers would be the same. The choice of an origin for our energy scale makes no difference; we can measure energy from any zero we want. For relativistic purposes it is nice to measure the energy so that the rest mass is included, but for many purposes that aren’t relativistic it is often nice to subtract some standard amount from all energies that appear. For instance, in the case of an atom, it is usually convenient to subtract the energy Ms·c2, where Ms is the mass of all the separate pieces—the nucleus and the electrons—which is, of course, different from the mass of the atom. For other problems, it may be useful to subtract from all energies the amount Mg·c2, where Mg is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom. So, sometimes we may shift our zero of energy by some very large constant, but it doesn’t make any difference, provided we shift all the energies in a particular calculation by the same constant.”

It’s a rather long quotation, but it’s important. The key phrase here is, obviously, the following: “For other problems, it may be useful to subtract from all energies the amount Mg·c2, where Mg is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom.” So that’s what he’s doing when solving Schrödinger’s equation. However, I should make the following point here: if we shift the origin of our energy scale, it does not make any difference in regard to the probabilities we calculate, but it obviously does make a difference in terms of our wavefunction itself. To be precise, its density in time will be very different. Hence, if we’d want to give the wavefunction some physical meaning – which is what I’ve been trying to do all along – it does make a huge difference. When we leave the rest mass of all of the pieces in our system out, we can no longer pretend we capture their energy.

So… Well… There you go. If we’d want to try to interpret our ψ(x, t) = ei·(En/ħ)·t·f(x) function as a two-dimensional oscillation of the mass of our electron, the energy concept in it – so that’s the Ein it – should include all pieces. Most notably, it should also include the electron’s rest energy, i.e. its energy when it is not in a bound state. This rest energy is equal to 0.511 MeV. […] Read this again: 0.511 mega-electronvolt (106 eV), so that’s huge as compared to the tiny energy values we mentioned so far (−13.6 eV, −3.4 eV, −1.5 eV,…).

Of course, this gives us a rather phenomenal order of magnitude for the oscillation that we’re looking at. Let’s quickly calculate it. We need to convert to SI units, of course: 0.511 MeV is about 8.2×10−14 joule (J), and so the associated frequency is equal to ν = E/h = (8.2×10−14 J)/(6.626×10−34 J·s) ≈ 1.23559×1020 cycles per second. Now, I know such number doesn’t say all that much: just note it’s the same order of magnitude as the frequency of gamma rays and… Well… No. I won’t say more. You should try to think about this for yourself. [If you do, think – for starters – about the difference between bosons and fermions: matter-particles are fermions, and photons are bosons. Their nature is very different.]

The corresponding angular frequency is just the same number but multiplied by 2π (one cycle corresponds to 2π radians and, hence, ω = 2π·ν = 7.76344×1020 rad per second. Now, if our green dot would be moving around the origin, along the circumference of our unit circle, then its horizontal and/or vertical velocity would approach the same value. Think of it. We have this eiθ = ei·(E/ħ)·t = ei·ω·t = cos(ω·t) + i·sin(ω·t) function, with ω = E/ħ. So the cos(ω·t) captures the motion along the horizontal axis, while the sin(ω·t) function captures the motion along the vertical axis. Now, the velocity along the horizontal axis as a function of time is given by the following formula:

v(t) = d[x(t)]/dt = d[cos(ω·t)]/dt = −ω·sin(ω·t)

Likewise, the velocity along the vertical axis is given by v(t) = d[sin(ω·t)]/dt = ω·cos(ω·t). These are interesting formulas: they show the velocity (v) along one of the two axes is always less than the angular velocity (ω). To be precise, the velocity approaches – or, in the limit, is equal to – the angular velocity ω when ω·t is equal to ω·= 0, π/2, π or 3π/2. So… Well… 7.76344×1020 meter per second!? That’s like 2.6 trillion times the speed of light. So that’s not possible, of course!

That’s where the amplitude of our wavefunction comes in – our envelope function f(x): the green dot does not move along the unit circle. The circle is much tinier and, hence, the oscillation should not exceed the speed of light. In fact, I should probably try to prove it oscillates at the speed of light, thereby respecting Einstein’s universal formula:

c = √(E/m)

Written like this – rather than as you know it: E = m·c2 – this formula shows the speed of light is just a property of spacetime, just like the ω = √(k/m) formula (or the ω = √(1/LC) formula for a resonant AC circuit) shows that ω, the natural frequency of our oscillator, is a characteristic of the system.

Am I absolutely certain of what I am writing here? No. My level of understanding of physics is still that of an undergrad. But… Well… It all makes a lot of sense, doesn’t it? 🙂

Now, I said there were a few obvious questions, and so far I answered only one. The other obvious question is why energy would appear to us as mass in motion in two dimensions only. Why is it an oscillation in a plane? We might imagine a third spring, so to speak, moving in and out from us, right? Also, energy densities are measured per unit volume, right?

Now that‘s a clever question, and I must admit I can’t answer it right now. However, I do suspect it’s got to do with the fact that the wavefunction depends on the orientation of our reference frame. If we rotate it, it changes. So it’s like we’ve lost one degree of freedom already, so only two are left. Or think of the third direction as the direction of propagation of the wave. 🙂 Also, we should re-read what we wrote about the Poynting vector for the matter wave, or what Feynman wrote about probability currents. Let me give you some appetite for that by noting that we can re-write joule per cubic meter (J/m3) as newton per square meter: J/m3 = N·m/m3 = N/m2. [Remember: the unit of energy is force times distance. In fact, looking at Einstein’s formula, I’d say it’s kg·m2/s2 (mass times a squared velocity), but that simplifies to the same: kg·m2/s2 = [N/(m/s2)]·m2/s2.]

I should probably also remind you that there is no three-dimensional equivalent of Euler’s formula, and the way the kinetic and potential energy of those two oscillations works together is rather unique. Remember I illustrated it with the image of a V-2 engine in previous posts. There is no such thing as a V-3 engine. [Well… There actually is – but not with the third cylinder being positioned sideways.]two-timer-576-px-photo-369911-s-original

But… Then… Well… Perhaps we should think of some weird combination of two V-2 engines. The illustration below shows the superposition of two one-dimensional waves – I think – one traveling east-west and back, and the other one traveling north-south and back. So, yes, we may to think of Feynman’s drum-head again – but combining two-dimensional waves – two waves that both have an imaginary as well as a real dimension

dippArticle-14

Hmm… Not sure. If we go down this path, we’d need to add a third dimension – so w’d have a super-weird V-6 engine! As mentioned above, the wavefunction does depend on our reference frame: we’re looking at stuff from a certain direction and, therefore, we can only see what goes up and down, and what goes left or right. We can’t see what comes near and what goes away from us. Also think of the particularities involved in measuring angular momentum – or the magnetic moment of some particle. We’re measuring that along one direction only! Hence, it’s probably no use to imagine we’re looking at three waves simultaneously!

In any case… I’ll let you think about all of this. I do feel I am on to something. I am convinced that my interpretation of the wavefunction as an energy propagation mechanism, or as energy itself – as a two-dimensional oscillation of mass – makes sense. 🙂

Of course, I haven’t answered one key question here: what is mass? What is that green dot – in reality, that is? At this point, we can only waffle – probably best to just give its standard definition: mass is a measure of inertia. A resistance to acceleration or deceleration, or to changing direction. But that doesn’t say much. I hate to say that – in many ways – all that I’ve learned so far has deepened the mystery, rather than solve it. The more we understand, the less we understand? But… Well… That’s all for today, folks ! Have fun working through it for yourself. 🙂

Post scriptum: I’ve simplified the wavefunction a bit. As I noted in my post on it, the complex exponential is actually equal to ei·[(E/ħ)·− m·φ], so we’ve got a phase shift because of m, the quantum number which denotes the z-component of the angular momentum. But that’s a minor detail that shouldn’t trouble or worry you here.

The periodic table

This post is, in essence, a continuation of my series on electron orbitals. I’ll just further tie up some loose ends and then – hopefully – have some time to show how we get the electron orbitals for other atoms than hydrogen. So we’ll sort of build up the periodic table. Sort of. 🙂

We should first review a bit. The illustration below copies the energy level diagram from Feynman’s Lecture on the hydrogen wave function. Note he uses √E for the energy scale because… Well… I’ve copied the En values for n = 1, 2, 3,… 7 next to it: the value for E(-13.6 eV) is four times the value of E(-3.4 eV).

exponential scale

How do we know those values? We discussed that before – long time back: we have the so-called gross structure of the hydrogen spectrum here. The table below gives the energy values for the first seven levels, and you can calculate an example for yourself: the difference between E2 (-3.4 eV) and E(-0.85 eV) is 2.55 eV, so that’s 4.08555×10−19 J, which corresponds to a frequency equal to = E/h = (4.08555×10−19 J)/(6.626×10−34 J·s) ≈ 0.6165872×1015 Hz. Now that frequency corresponds to a wavelength that’s equal to λ = c/= (299,792,458 m/s)/0.6165872×1015/s) ≈ 486×10−9 m. So that’s the 486 nano-meter line the so-called Balmer series, as shown in the illustration next to the table with the energy values.

So far, so good. An interesting point to note is that we only have one solution for = 1. To be precise, we have one spherical solution only: the 1s solution. Now, for n = 2, we have one 2s solution but also three 2solutions (remember the stands for principal lines). In the simplified model we’re using (we’re not discussing the fine or hyperfine structure here), these three solutions are referred to as ‘degenerate states’: they are different states with the same energy. Now, we know that any linear combination of the solutions for a differential equation must also be a solution. Therefore, any linear combination of the 2solutions will also be a stationary state of the same energy. In fact, a superposition of the 2s and one or more of the 2p states should also be a solution. There is an interesting app which visualizes how such superimposed states look like. I copy three illustrations below, but I recommend you google for stuff like this yourself: it’s really fascinating! You should, once again, pay attention to the symmetries planes and/or symmetry axes.

But we’ve written enough about the orbital of one electron now. What if there are two electrons, or three, or more. In other word, how does it work for heliumlithium, and so on? Feynman gives us a bit of an intuitive explanation here – nothing analytical, really. First, he notes Schrödinger’s equation for two electrons would look as follows:

two electronsSecond, the ψ(x) function in the ψ(x, t) = ei·(E/ħ)·t·ψ(x) function now becomes a function in six variables, which he – curiously enough – now no longer writes as ψ but as f:formulaThe rest of the text speaks for itself, although you might be disappointed by what he writes (the bold-face and/or italics are mine):

“The geometrical dependence is contained in f, which is a function of six variables—the simultaneous positions of the two electrons. No one has found an analytic solution, although solutions for the lowest energy states have been obtained by numerical methods. With 34, or 5 electrons it is hopeless to try to obtain exact solutions, and it is going too far to say that quantum mechanics has given a precise understanding of the periodic tableIt is possible, however, even with a sloppy approximation—and some fixing—to understand, at least qualitatively, many chemical properties which show up in the periodic table.

The chemical properties of atoms are determined primarily by their lowest energy states. We can use the following approximate theory to find these states and their energies. First, we neglect the electron spin, except that we adopt the exclusion principle and say that any particular electronic state can be occupied by only one electron. This means that any particular orbital configuration can have up to two electrons—one with spin up, the other with spin down.

Next we disregard the details of the interactions between the electrons in our first approximation, and say that each electron moves in a central field which is the combined field of the nucleus and all the other electrons. For neon, which has 10 electrons, we say that one electron sees an average potential due to the nucleus plus the other nine electrons. We imagine then that in the Schrödinger equation for each electron we put a V(r) which is a 1/r field modified by a spherically symmetric charge density coming from the other electrons.

In this model each electron acts like an independent particle. The angular dependence of its wave function will be just the same as the ones we had for the hydrogen atom. There will be s-states, p-states, and so on; and they will have the various possible m-values. Since V(r) no longer goes as 1/r, the radial part of the wave functions will be somewhat different, but it will be qualitatively the same, so we will have the same radial quantum numbers, n. The energies of the states will also be somewhat different.”

So that’s rather disappointing, isn’t it? We can only get some approximate – or qualitative – understanding of the periodic table from quantum mechanics – because the math is too complex: only numerical methods can give us those orbitals! Wow! Let me list some of the salient points in Feynman’s treatment of the matter:

  • For helium (He), we have two electrons in the lowest state (i.e. the 1s state): one has its spin ‘up’ and the other is ‘down’. Because the shell is filled, the ionization energy (to remove one electron) has an even larger value than the ionization energy for hydrogen: 24.6 eV! That’s why there is “practically no tendency” for the electron to be attracted by some other atom: helium is chemically inert – which explains it being part of the group of noble or inert gases.
  • For lithium (Li), two electrons will occupy the 1s orbital, and the third should go to an = 2 state. But which one? With = 0, or = 1? A 2s state or a 2p state? In hydrogen, these two = 2 states have the same energy, but in other atoms they don’t. Why not? That’s a complicated story, but the gist of the argument is as follows: a 2s state has some amplitude to be near the nucleus, while the 2p state does not. That means that a 2electron will feel some of the triple electric charge of the Li nucleus, and this extra attraction lowers the energy of the 2state relative to the 2state.

To make a long story short, the energy levels will be roughly as shown in the table below. For example, the energy that’s needed to remove the 2s electron of the lithium – i.e. the ionization energy of lithium – is only 5.4 eV because… Well… As you can see, it has a higher energy (less negative, that is) than the 1s state (−13.6 eV for hydrogen and, as mentioned above, −24.6 eV for helium). So lithium is chemically active – as opposed to helium. energy values more electrons

You should compare the table below with the table above. If you do, you’ll understand how electrons ‘fill up’ those electron shells. Note, for example, that the energy of the 4s state is slightly lower than the energy of the 3d state, so it fills up before the 3shell does. [I know the table is hard to read – just check out the original text if you want to see it better.]

periodic table

This, then, is what you learnt in high school and, of course, there are 94 naturally occurring elements – and another 24 heavier elements that have been produced in labs, so we’d need to go all the way to no. 118. Now, Feynman doesn’t do that, and so I won’t do that either. 🙂

Well… That’s it, folks. We’re done with Feynman. It’s time to move to a physics grad course now! Talk stuff like quantum field theory, for example. Or string theory. 🙂 Stay tuned!

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) = ei·(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:wavefunctionNote the derivation of this formula involved a switch from Cartesian to polar coordinates here, so from = (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 Yl,m(θ, φ) and Fl,m(r) respectively. Fl,m(r) was a real-valued function alright, but the Yl,m(θ, φ) had that ei·m·φ factor in it. So… Yes. You’re right: the Yl,m(θ, φ) function is real-valued if – and only if – m = 0, in which case ei·m·φ = 1. Let me copy the table from Feynman’s treatment of the topic once again:spherical harmonics 2The Plm(cosθ) functions are the so-called (associated) Legendre polynomials, and the formula for these functions is rather horrible:Legendre polynomialDon’t worry about it too much: just note the Plm(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 ei·(E/ħ)·t·ei·m·φ = ei·[(E/ħ)·− 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 ei·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 ei·[(E/ħ)·− 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 ei·[(E/ħ)·− 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 = 2 from 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.

n = 5

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 ei·m·φ factor with the ei·(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 Plm(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.angle 2So… To make a long story short, we can substitute the cosθ argument in the Plm(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 Plm(sinα) and Plm[sin(−α)]. Now, that’s not obvious, because sin(−α) = −sinα ≠ sinα. However, because the argument in that Plm(x) function is being squared before any other operation (like subtracting 1 and exponentiating the result), it is OK: [−sinα]2 = [sinα]sin2α. […] 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 dl+m/dxl+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:

  • 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 ei·[(E/ħ)·− 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.

spherical

The ei·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 ei·[(E/ħ)·t factor, which gives us the two-dimensional oscillation that captures the energy of the state.

Circle_cos_sin

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 ei·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 = 4 and = 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 and l: the shapes become very convoluted, but the observation holds. 🙂

l = 3

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 = ±1, ±2,…, ±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 real difference, is it? 🙂 As with other stuff, I’ll let you think about this for yourself.