**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 the more recent *exposé *on the matter that is being presented here, which you can find by clicking on the provided link. In fact, I actually made some (small) mistakes when writing the post below.

**Original post**:

I hope you find the title intriguing. A zero-mass particle? So I am talking a photon, right? Well… Yes and no. Just read this post and, more importantly, think about this story for yourself. 🙂

One of my acquaintances is a retired nuclear physicist. We mail every now and then—but he has little or no time for my questions: he usually just tells me to keep studying. I once asked him why there is never any mention of the wavefunction of a photon in physics textbooks. He bluntly told me photons don’t have a wavefunction—not in the sense I was talking at least. Photons are associated with a traveling electric and a magnetic field vector. That’s it. Full stop. Photons do *not* have a ψ or φ function. [I am using ψ and φ to refer to position or momentum wavefunction. You know both are related: if we have one, we have the other.] But then I never give up, of course. I just can’t let go out of the idea of a photon wavefunction. The *structural *similarity in the *propagation *mechanism of the electric and magnetic field vectors **E** and **B** just looks too much like the quantum-mechanical wavefunction. So I kept trying and, while I don’t think I fully solved the riddle, I feel I understand it much better now. Let me show you the why and how.

**I.** An electromagnetic wave in free space is fully described by the following two equations:

- ∂
**B**/∂t = –∇×**E** - ∂
**E**/∂t =*c*^{2}∇×**B**

We’re making abstraction here of stationary charges, and we also do *not *consider any currents here, so no *moving* charges either. So I am omitting the ∇·**E** = ρ/ε_{0} equation (i.e. the first of the set of *four *equations), and I am also omitting the **j**/ε_{0} in the second equation. So, for all practical purposes (i.e. for the purpose of this discussion), you should think of a space with no charges: ρ = 0 and **j **= **0**. It’s just a traveling electromagnetic wave. To make things even simpler, we’ll assume our time and distance units are chosen such that *c *= 1, so the equations above reduce to:

- ∂
**B**/∂t = –∇×**E** **E**/∂t = ∇×**B**

*Perfectly symmetrical! *But note the minus sign in the first equation. As for the interpretation, I should refer you to previous posts but, briefly, the ∇× operator is the *curl *operator. It’s a vector operator: it describes the (infinitesimal) rotation of a (three-dimensional) vector field. We discussed heat flow a couple of times, or the flow of a moving liquid. So… Well… If the vector field represents the flow velocity of a moving fluid, then the curl is the *circulation *density of the fluid. The direction of the curl *vector* is the axis of rotation as determined by the ubiquitous right-hand rule, and its magnitude of the curl is the magnitude of rotation. OK. Next step.

**II.** For the wavefunction, we have Schrödinger’s equation, ∂ψ/∂t = *i*·(ħ/2m)·∇^{2}ψ, which relates two *complex-valued *functions (∂ψ/∂t and ∇^{2}ψ). Complex-valued functions consist of a real and an imaginary part, and you should be able to verify this equation is equivalent to the following *set *of two equations:

*Re*(∂ψ/∂t) = −(ħ/2m)·*Im*(∇^{2}ψ)*Im*(∂ψ/∂t) = (ħ/2m)·*Re*(∇^{2}ψ)

[Two complex numbers a + *i*b and c + *i*d are equal if, and *only *if, their real and imaginary parts are the same. However, note the −*i *factor in the right-hand side of the equation, so we get: a + *i*b = −*i*·(c + *i*d) = d −*i*c.] The Schrödinger equation above also assumes free space (i.e. zero potential energy: V = 0) but, in addition – see my previous post – they also assume a zero *rest *mass of the elementary particle (E_{0} = 0). So just assume E_{0 }= V = 0 in *de Broglie’s* elementary ψ(θ) = ψ(x, t) =* e*^{−i}^{θ}* = a*·*e*^{−i[(E0 + p2/(2m) + V)·t − p∙x]/ħ} wavefunction. So, in essence, we’re looking at the wavefunction of a *massless* particle here. Sounds like nonsense, doesn’t it? But… Well… That should be the wavefunction of a photon in free space then, right? 🙂

Maybe. Maybe not. Let’s go as far as we can.

**The energy of a zero-mass particle**

What m would we use for a photon? It’s *rest *mass is zero, but it’s got energy and, hence, an equivalent *mass*. That mass is given by the m = E/*c*^{2 }mass-energy equivalence. We also know a photon has *momentum*, and it’s equal to its *energy *divided by *c*: p = m·*c* = E/*c*. [I know the notation is somewhat confusing: E is, obviously, *not *the magnitude of **E** here: it’s *energy*!] Both yield the same result. We get: m·*c* = E/*c* ⇔ m = E/*c*^{2 }⇔ E = m·*c*^{2}.

OK. Next step. Well… I’ve always been intrigued by the fact that the *kinetic *energy of a photon, using the E = m·*v*^{2}/2 = E = m·*c*^{2}/2 formula, is only half of its *total *energy E = m·*c*^{2}. Half: 1/2. That 1/2 factor is intriguing. Where’s the rest of the energy? It’s really a contradiction: our photon has no rest mass, and there’s no potential here, but its total energy is still *twice *its kinetic energy. **Quid?**

There’s only one conclusion: just because of its sheer *existence*, it must have some hidden energy, and that hidden energy is also equal to E = m·*c*^{2}/2, and so the kinetic and hidden energy add up to E = m·*c*^{2}.

** Huh? Hidden energy?** I must be joking, right?

Well… No. No joke. I am tempted to call it the *imaginary *energy, because it’s linked to the *imaginary *part of the wavefunction—but then it’s everything but imaginary: it’s as real as the imaginary part of the wavefunction. [I know that sounds a bit nonsensical, but… Well… Think about it: it *does *make sense.]

Back to that factor 1/2. You may or may not remember it popped up when we were calculating the *group *and the *phase *velocity of the wavefunction respectively, again assuming zero rest mass, and zero potential. [Note that the rest mass term is *mathematically* equivalent to the potential term in both the wavefunction as well as in Schrödinger’s equation: (E_{0}·t +V·t = (E_{0 }+ V)·t, and V·ψ + E_{0}·ψ = (V+E_{0})·ψ—obviously!]

In fact, let me quickly show you that calculation again: the *de Broglie *relations tell us that the k and the ω in the *e ^{i}*

^{(kx − ωt)}= cos(kx−ωt) +

*i*∙sin(kx−ωt) wavefunction (i.e. the spatial and temporal frequency respectively) are equal to k = p/ħ, and ω = E/ħ. If we would now use the

*kinetic*energy formula E = m·

*v*

^{2}/2 – which we can also write as E = m·

*v*·

*v*/2 = p·

*v*/2 = p·p/2m = p

^{2}/2m, with

*v*= p/m the

*classical velocity of the elementary particle*that

*Louis de Broglie*was thinking of – then we can calculate the

*group*velocity of our

*e*

^{i}^{(kx − ωt)}= cos(kx−ωt) +

*i*∙sin(kx−ωt) as:

*v*_{g} = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂[p^{2}/2m]/∂p = 2p/2m = p/m = *v*

[Don’t tell me I can’t treat m as a constant when calculating ∂ω/∂k: I can. Think about it.] Now the phase velocity. The *phase *velocity of our *e ^{i}*

^{(kx − ωt)}is only

*half*of that. Again, we get that 1/2 factor:

*v*_{p} = ω/k = (E/ħ)/(p/ħ) = E/p = (p^{2}/2m)/p = p/2m = *v*/2

Strange, isn’t it? Why would we get a *different *value for the phase velocity here? It’s not like we have *two *different frequencies here, do we? You may also note that the phase velocity turns out to be *smaller* than the group velocity, which is quite exceptional as well! So what’s the matter?

Well… The answer is: we do seem to have two frequencies here while, at the same time, it’s just one wave. There is only *one* k and ω here but, as I mentioned a couple of times already, that *e ^{i}*

^{(kx − ωt)}wavefunction seems to give you two functions for the price of one—one real and one imaginary:

*e*

^{i}^{(kx − ωt)}= cos(kx−ωt) +

*i*∙sin(kx−ωt). So

*are*we adding waves, or are we

*not*? It’s a deep question. In my previous post, I said we

*were*adding separate waves, but now I am thinking: no. We’re not. That sine and cosine are part of one and the same

*whole*. Indeed, the apparent contradiction (i.e. the different group and phase velocity) gets solved if we’d use the E = m∙

*v*

^{2}formula rather than the

*kinetic*energy E = m∙

*v*

^{2}/2. Indeed, assuming that E = m∙

*v*

^{2}formula also applies to our zero-mass particle (I mean zero

*rest*mass, of course), and measuring time and distance in

*natural*units (so

*c*= 1), we have:

E = m∙*c*^{2} = m and p = m∙*c*^{2 }= m, so we get: E = m = p

** Waw! **What a weird combination, isn’t it? But… Well… It’s OK. [

*You*tell me why it wouldn’t be OK. It’s true we’re glossing over the dimensions here, but natural units are natural units, and so

*c*=

*c*

^{2 }= 1. So… Well… No worries!] The point is: that E = m = p equality yields extremely simple but also very sensible results. For the group velocity of our

*e*

^{i}^{(kx − ωt)}wavefunction, we get:

*v*_{g} = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂p/∂p = 1

So that’s the velocity of our zero-mass particle (*c*, i.e. the speed of light) expressed in natural units once more—just like what we found before. For the phase velocity, we get:

*v*_{p} = ω/k = (E/ħ)/(p/ħ) = E/p = p/p = 1

Same result! No factor 1/2 here! ** Isn’t that great?** My ‘hidden energy theory’ makes a lot of sense. 🙂 In fact, I had mentioned a couple of times already that the E = m∙

*v*

^{2}relation comes out of the

*de Broglie*relations if we just multiply the two and use the

*v*=

*f·*λ relation:

*f*·λ = (E/h)·(h/p) = E/p*v*=*f·*λ ⇒*f*·λ =*v*= E/p ⇔ E =*v*·p =*v*·(m·*v*) ⇒ E = m·*v*^{2}

But so I had no good explanation for this. I have one now: the E = m·*v*^{2 }is the correct energy formula for our *zero-mass particle*. 🙂

**The quantization of energy and the zero-mass particle**

Let’s now think about the quantization of energy. What’s the *smallest* value for E that we could possible think of? That’s h, isn’t it? That’s the energy of *one *cycle of an oscillation according to the Planck-Einstein relation (E = h·*f*). Well… Perhaps it’s ħ? Because… Well… We saw energy levels were separated by ħ, rather than h, when studying the blackbody radiation problem. So is it ħ = h/2π? Is the natural unit a *radian *(i.e. a unit distance), rather than a *cycle*?

Neither is natural, I’d say. We also have the Uncertainty Principle, which suggests the smallest possible energy value is ħ/2, because ΔxΔp = ΔtΔE = ħ/2.

*Huh? What’s the logic here?*

Well… I am not quite sure but my intuition tells me the quantum of energy *must *be related to the quantum of time, and the quantum of distance.

*Huh? The quantum of time? The quantum of distance? What’s that? The Planck scale?*

No. Or… Well… Let me correct that: not *necessarily*. I am just thinking in terms of *logical *concepts here. *Logically*, as we think of the smallest of smallest, then our time and distance variables must become *count variables*, so they can only take on some integer value n = 0, 1, 2 etcetera. So then we’re literally *counting *in time and/or distance units. So Δx and Δt are then equal to 1. Hence, Δp and ΔE are then equal to Δp = ΔE = ħ/2. Just think of the *radian *(i.e. the unit in which we measure θ) as measuring both time as well as distance. Makes sense, no?

No? Well… Sorry. I need to move on. So the smallest possible value for m = E = p would be ħ/2. Let’s substitute that in Schrödinger’s equation, or in that *set *of equations *Re*(∂ψ/∂t) = −(ħ/2m)·*Im*(∇^{2}ψ) and *Im*(∂ψ/∂t) = (ħ/2m)·*Re*(∇^{2}ψ). We get:

*Re*(∂ψ/∂t) = −(ħ/2m)·*Im*(∇^{2}ψ) = −(2ħ/2ħ)·*Im*(∇^{2}ψ) = −*Im*(∇^{2}ψ)*Im*(∂ψ/∂t) = (ħ/2m)·*Re*(∇^{2}ψ) = (2ħ/2ħ)·*Re*(∇^{2}ψ) =*Re*(∇^{2}ψ)

** Bingo!** The

*Re*(∂ψ/∂t) = −

*Im*(∇

^{2}ψ) and

*Im*(∂ψ/∂t) =

*Re*(∇

^{2}ψ) equations were what I was looking for. Indeed, I wanted to find something that was

*structurally*similar to the ∂

**B**/∂t = –∇×

**E**and

**∂**

**E**/∂t = ∇×

**B**equations—and something that was

*exactly*similar: no coefficients in front or anything. 🙂

What about our wavefunction? Using the *de Broglie *relations once more (k = p/ħ, and ω = E/ħ), our *e ^{i}*

^{(kx − ωt)}= cos(kx−ωt) +

*i*∙sin(kx−ωt) now becomes:

*e ^{i}*

^{(kx − ωt)}=

*e*

^{i}^{(ħ·x/2 − ħ·t/2)/ħ}=

*e*

^{i}^{(x/2 − t/2) }= cos[(x−t)/2] +

*i*∙sin[(x−t)/2]

Hmm… ** Interesting!** So we’ve got that 1/2 factor now in the

*argument*of our wavefunction! I really feel I am close to squaring the circle here. 🙂 Indeed, it

*be possible to relate the ∂*

**must****B**/∂t = –∇×

**E**and ∂

**E**/∂t =

*c*

^{2}∇×

**B**to the

*Re*(∂ψ/∂t) = −

*Im*(∇

^{2}ψ) and

*Im*(∂ψ/∂t) =

*Re*(∇

^{2}ψ) equations. I am sure it’s a complicated exercise. It’s likely to involve the formula for the

*Lorentz*force, which says that the force on a

*unit*charge is equal to

**E**+

**×**

*v***B**, with

*the velocity of the charge. Why? Note the vector cross-product. Also note that ∂*

**v****B**/∂t and ∂

**E**/∂t are

*vector-valued*functions, not

*scalar-valued*functions. Hence, in that sense, ∂

**B**/∂t and ∂

**E**/∂t and

*not*like the

*Re*(∂ψ/∂t) and/or

*Im*(∂ψ/∂t) function. But… Well… For the rest, think of it:

**E**and

**B**are orthogonal vectors, and that’s how we usually interpret the real and imaginary part of a complex number as well: the real and imaginary axis are orthogonal too!

So I am almost there. Who can help me prove what I want to prove here? The two propagation mechanisms are the “same-same but different”, as they say in Asia. The difference between the two propagation mechanisms must also be related to that fundamental *dichotomy *in Nature: the distinction between bosons and fermions. Indeed, when combining two *directional *quantities (i.e. two *vectors*), we like to think there are *four *different ways of doing that, as shown below. However, when we’re only interested in the *magnitude *of the result (and *not *in *its* direction), then the first and third result below are really the same, as are the second and fourth combination. Now, we’ve got pretty much the same in quantum math: we can, in theory, combine complex-valued amplitudes in *four *different ways but, in practice, we only have *two* (rather than *four*) types of behavior only: photons versus bosons.

**Is our zero-mass particle just the electric field vector?**

Let’s analyze that *e ^{i}*

^{(x/2 − t/2) }= cos[(x−t)/2] +

*i*∙sin[(x−t)/2] wavefunction some more. It’s easy to represent it graphically. The following animation does the trick:

I am sure you’ve seen this animation before: it represents a circularly polarized electromagnetic wave… Well… Let me be precise: it presents the *electric *field vector (**E**) of such wave only. The **B** vector is *not *shown here, but you know where and what it is: orthogonal to the **E** vector, as shown below—for a *linearly *polarized wave.

Let’s think some more. What *is* that *e ^{i}*

^{(x/2 − t/2) }function? It’s subject to conceiving time and distance as

*countable*variables, right? I am tempted to say: as

*discrete*variables, but I won’t go that far—not now—because the

*countability*may be related to a particular

*interpretation*of quantum physics. So I need to think about that. In any case… The point is that x can only take on values like 0, 1, 2, etcetera. And the same goes for t. To make things easy, we’ll not consider

*negative*values for x right now (and, obviously, not for t either). So we’ve got a infinite set of

*points*like:

*e*^{i}^{(0/2 − 0/2) }= cos(0) +*i*∙sin(0)*e*^{i}^{(1/2 − 0/2) }= cos(1/2) +*i*∙sin(1/2)*e*^{i}^{(0/2 − 1/2) }= cos(−1/2) +*i*∙sin(−1/2)*e*^{i}^{(1/2 − 1/2) }= cos(0) +*i*∙sin(0)- …

Now, I quickly opened Excel and calculated those cosine and sine values for x and t going from 0 to 14 below. It’s really easy. Just five minutes of work. You should do yourself as an exercise. The result is shown below. Both graphs connect 14×14 = 196 data points, but you can see what’s going on: this does effectively, represent the elementary wavefunction of a particle traveling in spacetime. In fact, you can see its speed is equal to 1, i.e. it effectively travels at the speed of light, as it should: the wave velocity is *v* = *f*·λ = (ω/2π)·(2π/k) = ω/k = (1/2)·(1/2) = 1. The amplitude of our wave doesn’t change along the x = t diagonal. As the *Last Samurai *puts it, just before he moves to the Other World: “Perfect! They are all perfect!” 🙂

In fact, in case you wonder how the *quantum *vacuum could possibly look like, you should probably think of these discrete spacetime points, and some complex-valued wave that travels as it does in the illustration above.

Of course, that elementary wavefunction above does *not *localize our particle. For that, we’d have to add a potentially infinite number of such elementary wavefunctions, so we’d write the wavefunction as ∑ a_{j}*e*^{−iθj} functions. [I use the *j *symbol here for the subscript, rather than the more conventional *i* symbol for a subscript, so as to avoid confusion with the symbol used for the imaginary unit.] The a* _{j }*coefficients are the

*contribution*that each of these elementary wavefunctions would make to the

*composite*wave. What could they possibly be? Not sure. Let’s first look at the argument of our elementary

*component*wavefunctions. We’d inject

*uncertainty*in it. So we’d say that m = E = p is equal to

m = E = p = ħ/2 + j·ħ with j = 0, 1, 2,…

That amounts to writing: m = E = p = ħ/2, ħ, 3ħ/2, 2ħ, 5/2ħ, etcetera. * Waw! *That’s nice, isn’t it? My intuition tells me that our a

*coefficients will be smaller for higher*

_{j }*j*, so the a

*(j) function would be some*

_{j}*decreasing*function. What shape? Not sure. Let’s first sum up our thoughts so far:

- The elementary wavefunction of a zero-mass particle (again, I mean zero
*rest*mass) in free space is associated with an energy that’s equal to ħ/2. - The zero-mass particle travels at the speed of light, obviously (because it has zero rest mass), and its
*kinetic*energy is equal to E = m·*v*^{2}/2^{ }= m·*c*^{2}/2. **However,**its*total*energy is equal to E = m·*v*^{2 }= m·*c*^{2}: it has some*hidden*energy. Why? Just because it*exists*.- We may associate its kinetic energy with the
*real*part of its wavefunction, and the*hidden*energy with its imaginary part. However, you should remember that the imaginary part of the wavefunction is as essential as its real part, so the hidden energy is equally real. 🙂

So… Well… Isn’t this just ** nice**?

I think it is. Another obvious advantage of this way of looking at the elementary wavefunction is that – at first glance at least – it provides an intuitive understanding of why we need to take the (absolute) square of the wavefunction to find the probability of our particle being at some point in space and time. The *energy *of a wave is proportional to the *square *of its amplitude. Now, it is reasonable to assume the probability of finding our (point) particle would be proportional to the energy and, hence, to the *square *of the amplitude of the wavefunction, which is given by those a* _{j}*(j) coefficients.

*Huh?*

OK. You’re right. I am a bit too fast here. It’s a bit more complicated than that, of course. The argument of probability being proportional to energy being proportional to the *square *of the amplitude of the wavefunction only works for a *single *wave *a·e*^{−iθ}. The argument does not hold water for a *sum* of functions ∑ *a _{j}*

*e*

^{−iθj}. Let’s write it all out. Taking our m = E = p = ħ/2 + j·ħ = ħ/2, ħ, 3ħ/2, 2ħ, 5/2ħ,… formula into account, this sum would look like:

*a _{1}e^{i}*

^{(x − t)(1/2) }+

*a*

_{2}e^{i}^{(x − t)(2/2) }+

*a*

_{3}e^{i}^{(x − t)(3/2) }+

*a*

_{4}e^{i}^{(x − t)(4/2) }+ …

But—* Hey! *We can write this as some power series, can’t we? We just need to add

*a*

_{0}e^{i}^{(x − t)(0/2) }=

*a*, and then… Well… It’s not so easy, actually. Who can help me? I am trying to find something like this:

_{0}Or… Well… Perhaps something like this:

Whatever power series it is, we should be able to relate it to this one—I’d hope:

*Hmm…* […] It looks like I’ll need to re-visit this, but I am sure it’s going to work out. Unfortunately, I’ve got no more time today, I’ll let *you *have some fun now with all of this. 🙂 By the way, note that the result of the first power series is only valid for |x| < 1. 🙂

**Note 1:** What we should also do now is to re-insert *mass *in the equations. That should not be too difficult. It’s consistent with classical theory: the total energy of some moving mass is E = m·*c*^{2}, out of which m·*v*^{2}/2 is the classical kinetic energy. All the rest – i.e. m·*c*^{2} − m·*v*^{2}/2 – is potential energy, and so that includes the energy that’s ‘hidden’ in the imaginary part of the wavefunction. 🙂

**Note 2:** I really didn’t pay much attentions to dimensions when doing all of these manipulations above but… Well… I don’t think I did anything wrong. Just to give you some more *feel *for that wavefunction *e ^{i}*

^{(kx − ωt)}, please do a dimensional analysis of its argument. I mean, k = p/ħ, and ω = E/ħ, so check the dimensions:

- Momentum is expressed in newton·second, and we divide it by the quantum of action, which is expressed in newton·meter·second. So we get something per meter. But then we multiply it with x, so we get a
*dimensionless*number. - The same is true for the ωt term. Energy is expressed in
*joule*, i.e. newton·meter, and so we divide it by ħ once more, so we get something per second. But then we multiply it with t, so… Well… We do get a dimensionless number: a number that’s expressed in*radians*, to be precise. And so the*radian*does, indeed, integrate both the time as well as the distance dimension. 🙂

## 3 thoughts on “The wavefunction of a zero-mass particle”