# Complex wavefunctions and electromagnetic waves

1. Jul 8, 2014

### tim1608

Hi Everyone

I am wondering about something. As everyone here knows, electromagnetic waves obviously possess an electrical component and a magnetic component. Firstly, can electromagnetic waves be considered to be a sort of complex wavefunction? If yes then do the two components of electromagnetic waves correspond with the "real" and "imaginary" components of the complex wavefunction? If yes then which does the electrical component correspond with and which does the magnetic component correspond with?

Thank you very much.

Kind regards

Tim

2. Jul 8, 2014

### hiddenvariable

I think both would have to be real, since the particle has some probability, though low, of both being at X location, and from the wave frequency there is some rather high certainty as to its momentum. But I should try to reframe from answering questions, being a real beginner in QM, so my response is more to test my understanding. The real answer will follow, I'm sure :).

3. Jul 8, 2014

### bhobba

The situation is actually quite interesting, subtle, and complex (pun intended).

Maxwell's equations can be written in complex form and low and behold that is the quantum equation of a photon.

Check out:
http://oco.uoregon.edu/sites/oco.uoregon.edu/files/TTRL5_V1.pdf

There are other papers on it that Google will bring up if it interests you.

Thanks
Bill

4. Jul 9, 2014

### PhilDSP

Remember also that Maxwell often wrote those and related equations in both pre-vector form and in quaternion form. Quaternions are essentially 3 or 4 dimensional complex numbers embedded into a consistent structure with very specific rules in how each dimension relates to the others. It's a 3 dimensional expansion of a single dimension complex number.

From quaternions you directly get Pauli's matrices for quantifying the "spin" of elementary particles. Pauli's sigma basis matrices give us the mapping from vector space to quaternion space and vice versa. They are also the essential structure behind the Dirac equation.

Further still, the non-commutativity of rotations in physical space becomes plainly evident on a initial study of the application of quaternions to rotation.

Last edited: Jul 9, 2014
5. Jul 9, 2014

### kith

There is, however, an important caveat: a single photon cannot be localized in a sharply defined region, so its position is not an observable. Although the "wavefunction" ψ(x) = E(x) + iB(x) may be used to predict outcomes for valid observables, the standard interpretation of |ψ(x)|² as the probability density for the location of the particle is not possible.

6. Jul 9, 2014

### WannabeNewton

Presumably you are talking about classical electrodynamics. If so then no. We often use complex amplitudes, particularly when expressing the amplitudes in terms of a circular polarization basis, along with Fourier modes to express the electric and magnetic fields in a Fourier decomposition but in the end we always take the real part of the expressions because the electric and magnetic fields are, in classical electrodynamics, real fields that must be in and of themselves measurable or observable. The same goes for quantities computed from the electric and magnetic fields such as the time-averaged Poynting vector.

This is to be contrasted with the wave-function of QM which is not an observable in and of itself and in that light is in general a complex object once and for all. However measurable quantities computed from the wave-function, such as expectation values or fluctuations of operators on a given state, are by construction real.

7. Jul 9, 2014

### rigetFrog

I remember my professor solving gravitational ellipsoidal motion using complex notation. The real was the x and the imaginary was the y, and the transform was x+i*y = rho*exp(i*theta).

We solved both using real Cartesian and complex mapping. Standard cartesian was very tedious. The mathematics was majorly simplified by using the complex transform.

My point is, complex numbers don't necessarily mean anything weird. It's just a super nifty mathematical trick that can simplify problems.

8. Jul 9, 2014

### maline

Great question, Tim! I used to wonder about that myself.
If I understand you correctly, you're saying that because the electric & magnetic fields are perpendicular, it seems like they could correspond to the two dimensions in the wavefunction values.

First off, sorry but they don't. For example, in an ideal plane-polarized EM wave, the electric & magnetic fields have their zeros at the same set of points. The quantum wf, on the other hand, cycles about the origin in the complex plane - the modulus (distance from the origin in the complex plane) is the square root of the probability of detecting the photon there, so it doesn't depend on the phase of (either) wave. (Somebody correct me if I'm wrong on that point)

More fundamentally, the two "right angles" don't really have much in common. The directions of the field lines are actual spatial directions- the ways that a (test) positive charge or north pole would be pushed. The right angle between them is an angle in space. The complex plane, meanwhile, is purely a conceptual construct. (Don't ask me what the difference is between reality and a conceptual construct, but there is a difference!) Its directions aren't towards anywhere- they're just a way of expressing a pure quantity that is two-dimensional. The right angle expresses the fact that the dimensions are orthogonal,or independent- a real number has no imaginary component & vice-versa.

Also, in the case of quantum waves, you can't really speak about the "real and imaginary components" at all. You see, the only actual effect of wf values is to set probabilities for experimental results. These depend not on the value itself but on its modulus squared. So any two values with the same modulus, -√2 and 1+i for instance, are completely interchangeable as long as you're only dealing with one wave. When different wave elements interfere, that's when the actual complex values come in- you add them up, and they can cancel out, and then the modulus is zero, and you can't detect the photon there, and you get a black stripe on your screen. But even so, you can multiply all the wf values is your problem by any one complex number, and you haven't changed anything at all. If before two waves canceled as 1 & -1, now they'll cancel as -3+2i & 3-2i. So if we say that "ψ(x) is purely real", it's an arbitrary description, similar to choosing an "origin" for spatial coordinates.

So what's the relationship between the quantum wave of the photon and the EM wave? Well, they have the same frequency & wavelength. The direction of the fields is related to the photon's "spin", which changes only when you measure it. Beyond that I really don't know much- I haven't yet taken any actual QM courses. (maybe that's why I can still express what I do know in understandable English)

EDIT: I just read Keth's post, and it sounds like I screwed up. Keth, think you can read this post & explain my mistakes?

ψ(x)=E+iB ? Wow, really? That sounds like exactly what Tim wanted! But that must mean the value sometimes goes to zero, even within one simple wave. What causes that? Can it be observed (maybe with very-low-frequency radio waves and highly time-sensitive detectors)?

A photon can't be localized? Don't they dislodge electrons from individual atoms? And what happens if you try to trap one between mirrors and detect it later on?

Am I at least right on these two points? A: ψ(x)=E+iB must be, on one level, a convention: it could have been B-iE. B: This formula does not depend on E and B being perpendicular.

Maybe I should shut my big mouth sometimes, but posting is fun! Someone will hope fully gain from the discussion.

Last edited: Jul 9, 2014
9. Jul 9, 2014

### bhobba

Of course it can be localised - its position is an observable.

Thanks
Bill

10. Jul 10, 2014

### maline

I thought so too, but what is Kith saying in his post?

11. Jul 10, 2014

### vanhees71

A photon cannot be "localized" in a strict sense. One cannot even define a position operator in the strict sense for massless particles with spin $\geq 1$. See, Arnold Neumaier's FAQ entry on that topic:

http://www.mat.univie.ac.at/~neum/physfaq/topics/position.html

Then, the question whether you represent Schrödinger wave functions (describing one or more nonrelativistic particles with spin 0) as complex valued or as two real numbers is a question of convenience. Of course, it's way more convenient to use one complex field than two real fields, because the Schrödinger equation becomes simple in terms of the complex representation, and that's why everybody uses it. In fact, I've never seen any textbook, presenting a real formulation nor have I ever seen any necessity to do so.

It's a bit different with the electromagnetic field. The standard representation of the classical field theory (classical electrodynamics) is to use a real four-vector field of mass 0, which is necessarily a gauge field from the point of view of the representations of the Poincare group. Alternatively you can deal with it also in terms of the antisymmetric real-valued tensor field, built by the electric and magnetic field components, the Faraday tensor. Then you have a gauge-invariant formulation, but it's hard to quantize the theory in this form.

A third alternative, seen however rarely in the literature, is to use a complex-valued three-dimensional vector $\vec{F}=\frac{1}{\sqrt{2}} (\vec{E}+\mathrm{i} \vec{B})$. This representation makes sense, because the tensor-transformation properties of the Faraday tensor translate into complex rotations in the space of the complex vectors $\vec{F}$, leading to the conclusion that the proper orthochronous Lorentz group $\mathrm{SO}(1,3)^{\uparrow}$ is isomorphic to $\mathrm{SO}(3,\mathbb{C})$. For details, have a look here:

http://arxiv.org/abs/1211.1218

12. Jul 10, 2014

### bhobba

Beats me.

There may be some subtle aspects I am not aware of.

But, for example, take the double slit experiment with photons and for definiteness a photographic plate. If there is one photon at a time each photon will register a position on the plate. That seems to settle that position is an observable.

Will have a look at the links provided and see if I can nut out what they are getting at.

Thanks
Bill

13. Jul 10, 2014

### bhobba

Ahhh.

Got it now.

Check out:
'No, but you can detect the position of its interaction with another particle, the interaction destroys the photon. A photon has no rest frame wrt which you could specify its position. Theoretically it is not proven that photons move at exactly "the speed of light", so the possibility remains that a very small mass photon exists which does have a rest frame. Then we'd just need to rename the term "speed of light" to be the the speed invariant under a lorentz transformation. But in current models we talk about photon number and energy density of the em field rather than photon position.'

Yep - and it is indeed true since there is no frame where it is at rest it cant have a position.

Thanks
Bill

14. Jul 10, 2014

### tim1608

Hi Everyone

Thank you very much for all your replies. I would like to make some comments and attempt to clarify a few things.

I am not sure I entirely agree with this. As I understand it, you cannot know exactly where any particle is before the wavefunction describing its position collapses but after the wavefunction collapses upon having been "observed" (whatever that exactly means in the contexts of QM), you do know where the particle is (or was), even if it is a photon. Am I correct?

From this it looks to me as if the electrical component of electromagnetic waves is the coefficient of the "real" component of the wavefunction and the magnetic component of electromagnetic waves is the coefficient of the "imaginary" component of the wavefunction. Am I correct?
Kith, can you quote a source for this particular equation?

It looks to me as if rigetFrog has a good reply to WannabeNewton. The way I see it is that the coefficient of something imaginary does not itself have to be imaginary. The complex mathematics is just simply a sort of "skeleton" upon which completely "real-world" physics can be built. Am I correct?

Maline, you have stuck a chord with me here. I think that keeping things understandable is very important. I like your post even though I am not sure I agree with everything you said in it. I like the way you think and the way you try to find the truth.

Last edited: Jul 10, 2014
15. Jul 10, 2014

### WannabeNewton

rigetFrog's reply only compliments mine. First of all you have to specify if you are talking about classical EM or QED. There is a huge difference between the two in terms of fundamentals. As I noted above, in classical EM we use complex numbers for convenience when manipulating electric and magnetic fields but we always take the real part in the end because these are measurable quantities. The wave-function of QM is on a very different footing as I already explained.

16. Jul 10, 2014

### tim1608

I take it you were answering my question about whether or not single photons are locatable.

If a photon emitter is firing photons at a screen at a low enough regularity that you can see individual dots lighting up all over the screen then what is happening at each of those dots if it is not the arrival of an individual photon located at the dot?

17. Jul 10, 2014

### rigetFrog

Theoretically, I think one could write Schrodinger's equation without using complex notation. I.e. converting all complex numbers,x +i*y , into two element vectors, [x,y], and expressing various operators (momentum, position, energy, etc...) as matrices.

E.g. the momentum operator, i hbar d/dx, would then become a real valued skew symmetric matrix.

Perhaps this approach would shed light on the physical meaning of "imaginary numbers" in quantum mechanics.

18. Jul 10, 2014

### WannabeNewton

The statement made by Kith was not one of whether a photon is detectable. It was one of whether it makes sense to define a position operator for photons so as to ascribe to them position eigenstates of definite position eigenvalue.

The detection of the photon here is due to an interaction with particles in the screen and the photon is immediately destroyed afterwards. It is definitely not the same thing as measuring the position of the photon through localization effected by a position operator acting on the photon state.

19. Jul 10, 2014

### rigetFrog

Mathematically: the screen is a narrowly distributed potential energy in real space corresponding to a broad distribution in reciprocal space. When the photon interacts with the screen, the screen imparts a superposition of all the screens momentum (mathematically similar to the crystal momentum) to the photon. This broadens the photons wave function in reciprocal space, which corresponds to a narrowing of the wave function in real space.

20. Jul 10, 2014

### tim1608

Hi WannabeNewton

I would be very grateful if you could explain to me, in simple terms exactly what you mean by the following terms:

"position eigenstates"

"position eigenvalue"

"localization"

"position operator"

"localization effected by a position operator"

How are these things different to the normal concepts of position and location that most people understand?

Thank you very much.