# Probability Amplitude Maxwell's Equations

1. Jan 27, 2014

### Bashyboy

Hello Everyone,

I am currently reading page 20 of Townsend's Quantum Physics book. Here are a few sentences that I am unsure of:

"In general, the magnitude and phase of the probability amplitude are determined from first principles by solving Maxwell's equations. In free space, these equations can be reduced to the wave equation (1.9)."

Equation (1.9) is $\displaystyle \frac{\partial^2 E}{\partial x^2} - \frac{1}{c^2} \frac{\partial^2 E}{\partial t^2} = 0$

What are these first principles to which the author alludes; and which equations can be reduced to equation (1.9)?

2. Jan 27, 2014

### Jilang

3. Jan 27, 2014

### Bashyboy

I may be mistaken, but in the link you provided, Jilang, I don't see the probability amplitude being determined from the first principles. In fact, I searched for the word "probability" on the page, and nothing came up.

4. Jan 27, 2014

### Jilang

Ah I see what you mean. I always assumed the magnitude of E was equivalent to the amplitude of finding a photon there. Is that not right?

5. Jan 27, 2014

### Bashyboy

I am not sure. That's why I am asking the question. Even though the webpage wasn't originally what I was looking for, I thank you for the link; it looks rather interesting, and I may read it tonight.

6. Jan 27, 2014

### WannabeNewton

What's "there"? A given mode of the electromagnetic field? If you're doing QFT then $E$ and $B$ are operator fields whose coefficients are creation and annihilation operators corresponding to different modes (harmonic oscillators) and polarizations of the electromagnetic field; the creation operators generate photon states of different momenta. When such states are in eigenstates of the number operator the relative fluctuations of the electromagnetic field are quite contentious. Coherent states OTOH, wherein the relative fluctuations aren't contentious in the appropriate limit, provide a nice probability distribution for photon number in a given mode. This is all discussed very nicely in chapter 19 of Ballentine and every book on QED.

OP, could you perhaps give more context from the book? You said it's a QM book not a QFT book and if it's only 20 pages in then presumably it's talking about the classical electromagnetic field so more context would help.

7. Jan 28, 2014

Staff Emeritus
Please stay on topic, everyone. A number of off-topic posts were made and had to be removed.

8. Jan 28, 2014

### PhilDSP

9. Jan 28, 2014

### vanhees71

I hope this is not also off-topic, but it must be stressed very clearly that for photons the idea the classical electromagnetic field $(\vec{E},\vec{B})$ is interpretable as something like a probability amplitude. This is wrong from the very beginning! There is no such single-particle interpretation of "wave functions" for relativistic particles that leads to a consistent quantum theory. This fails, because in relativistic quantum theory you necessarily end up with a many-particle theory, and that's what's also well established by observations nowadays: In scattering processes at relativistic energies there's also the possibility that particles (or photons for that matter which are addtionally special because they are massless) are created and/or destroyed.

That's why the most appropriate formulation of relativistic quantum theory is in terms of a quantum field theory.

In addition to this general remarks on relativistic quantum theory for massless particles with spin $s \geq 1$ there is not even a position operator defined in the full sense as it is for massive particles of any spin. So it is not even possible to unambiguously define what you mean by "position of a photon" in the strict sense, and thus to talk about the probability to find a photon at a certain place is somewhat problematic.