# Quantum objects and Electromagnetic waves

1. Oct 24, 2014

If light is a quantum object, then what are electromagnetic waves that Maxwell predicted???

2. Oct 24, 2014

### Staff: Mentor

Classical electromagnetic waves are an approximate description of nature that "emerges" from the QED description as the number of photons becomes very large and the total energy becomes very large compared to the energy of a single photon.

3. Oct 24, 2014

### Meir Achuz

An electromagnetic wave could be considered the wave function of a photon, in the same way that the Schrödinger wave function is the wave function of an electron. In each case, the 'particle' could be considered an excitation of the quantum field.

4. Oct 24, 2014

### vanhees71

NO!!! There is no such thing as a wave function of a photon. At least it's highly misleading. As jtbell said, classical electromagnetic waves are, as seen from the point of view of quantum electrodynamics, coherent states with a large average photon number. Coherent states are superpositions of photon Fock states (states with a definite number of photons) of any number. In a coherent state the photon number is Poisson distributed, i.e., the mean photon number equals the square of its standard deviation. Thus $\sigma_N/N=1/\sqrt{N}$, i.e., the relative fluctuations in photon number (and thus of the field energy, which is $\langle E \rangle=\omega \langle{N} \rangle$) can be neglected when the photon number is large, and then the classical (Maxwell) approximation of the em. field is very good.

5. Oct 24, 2014

### f95toli

No, this is -as vanshee71 has already pointed out- flat out wrong. There are situation where one can talk about the "size" and "shape" of photons (but NOT in the usual classical sense) but in those cases these parameters have nothing to do with the wavelength of the light.

6. Oct 25, 2014

### Meir Achuz

Why?

7. Oct 25, 2014

### vanhees71

A photon does not even have a properly defined position operator. So how can it have a wave function? Also photon number is only conserved for free photons. Photons interacting with charged particles accelerate the particles and thus always more photons are produced. The only correct way to describe relativistic quantum objects is relativistic quantum field theory, and photons are as relativistic as something can get. It's a massless quantum after all, and thus has no non-relativistic limit. So the very definition and physical meaning of what you could call a "photon wave function" is very unclear.

8. Oct 25, 2014

### atyy

9. Oct 25, 2014

### Meir Achuz

"The only correct way to describe relativistic quantum objects is relativistic quantum field theory, and photons are as relativistic as something can get."
That is true of all known 'particles', including the photon.
All single particle wave functions are approximations in which particle annihilation and creation is neglected. I should have said "the Dirac wave function" to compare to the photon wave equation to keep them on an even relativistic footing. The change in particle number is true for all known particles including the photon.
Actually, a pion in Yukawa theory is closer to a photon in nature since both are Bosons.
The 'position' of any particle
is given by a wave packet .

10. Oct 25, 2014

### vanhees71

That's of course also true. The only difference between massless and massive particles is that massive particles have a non-relativistic limit and admit approximate descriptions in terms of wave functions. For photons nothing like this exists.

The classical limit of "photons" or better the quantum electromagnetic field is the classical electromagnetic field, which approximates coherent quantum states well if the mean photon number of the coherent state is large.

Concerning #8: This is a very good book about quantum optics as far as I can judge it. It's of utmost clarity, and in the very chapter 5.6, they precisely state in which sense a single-free-photon state can be described to some extent by a "wave function". It allows to calculate the detection probability of a photon in a region around $\vec{x}$ which cannot be sharper specified than a resolution around the (typical) wavelength of the photon. One should also stress that only little less than the half of the book deals with the semiclassical theory, i.e., the description of "optics" in terms of the classical electromagnetic field and quantized matter ("detectors"). This approximation leads quite far into "quantum optics". E.g., the photoelectric effect is often wrongly explained as being an experimental evidence for the existence of photons, but it can be well understood in the semiclassical theory, as you find in many textbooks like Landau-Lifshitz as an application of time-dependent perturbation theory of an electron in a harmonically time-dependent potential (dipole approximation for a classical light wave interacting with a bound electron).

11. Oct 25, 2014

### Staff: Mentor

Because position is not an observable for a photon, but why is not particularly straightforward, although it has been discussed on this forum quite a few times eg:

You may have come across papers like the following:
http://arxiv.org/ftp/quant-ph/papers/0604/0604169.pdf

'These observations imply the interpretation of the Maxwell field as akin to the Schroedinger wave function, which evolves probability amplitudes for various possible quantum events in which the electron’s position is found to be within a certain volume, rather than being a realistic description of the electron as being here or there'

The key word is akin.

Thanks
Bill

12. Oct 26, 2014

Thanks for all the replies.
But all this has gone too complicated for a first year graduate to understand. So please can you define all these in simpler (yet scientific terms) for me to understand,please.

13. Oct 26, 2014

### Staff: Mentor

But like I said this is very deep waters.

Here is the reason:
http://www.mat.univie.ac.at/~neum/physfaq/topics/position.html

Its beyond my current knowledge as well.

My answer would be QED is the theory that treats the EM field as a quantum field. The rock bottom essence of EM is gauge symmetry. To fully understand it you need to see the derivation Maxwell's equations from gauge invariance - if you can get a hold of a copy the following will explain it:
http://scitation.aip.org/content/aapt/journal/ajp/48/5/10.1119/1.12094

So apply gauge invariance to classical fields you get Maxwell's equations - to quantum fields - QED. But, just like QM and classical mechanics, if Planks constant can be neglected the quantum field becomes a classical one.

Thanks
Bill

Last edited: Oct 26, 2014
14. Oct 26, 2014

### atyy

Did you read jtbell's post #2?