# Can Maxwell's equations describe a single photon?

1. Jun 4, 2008

### Usaf Moji

One author states that "the usual Maxwell field is the quantum wave function for a single photon" - see http://arxiv.org/ftp/quant-ph/papers/0604/0604169.pdf

Is it correct that a single photon can be described using Maxwell's equations - or do the Maxwell equations only describe the behaviour of large numbers of photons?

2. Jun 4, 2008

### reilly

It is correct; basic to QED. See any QFT text; Weinberg for example.
Regards,
Reilly Atkinson

3. Jun 4, 2008

### lbrits

There is a difference between Maxwell's equations and the Maxwell field. The former are classical equations of motion, while the latter is a field that can be quantized. The quantization of this field leads to one or more photons.

There's a caveat, which applies to free fields only, that the classical equations of motion hold true as operator equations in quantum mechanics.

4. Jun 4, 2008

### pam

In this sense, classical EM is a quantum theory, with $$A^\mu$$ being the wave function of a photon.

5. Jun 4, 2008

### Fredrik

Staff Emeritus
I would interpret this to mean that the wave function of a photon is a function that assigns classical EM field components to each point in space-time. That's definitely not correct.

(Reilly must have interpreted it differently).

However, if you quantize the field and have the quantum field act on the vacuum state, the result is a one-photon state that can be called the "wave function" of a single photon. I would prefer to call it a "state vector" instead of a "wave function" though.

6. Jun 4, 2008

### jostpuur

I thought the point in QED is to calculate cross-sections without trying to answer questions like this What is the spatial probability density for the photon then? Is it possible to derive this Maxwell equation for the photon from the Schrödinger's equation of the entire system, quantized EM field?

7. Jun 4, 2008

### Fredrik

Staff Emeritus
The state of a photon is in general a superposition of states with different momenta. Let's ignore other degrees of freedom and express this as

$$\int d^3p f(\vec p)a^\dagger(\vec p)|0\rangle$$

where $a^\dagger(\vec p)$ is the creation operator that creates a one-particle state with momentum p when it acts on the vaccum. The Fourier transform of f can be interpreted as an ordinary wave function.

When the field acts on the vacuum, the Fourier transform of what corresponds to this f in that case is a delta function.

You can't derive Maxwell's equations from the Schrödinger equation. You should think of Maxwell's equations as equations satisfied by the fields and the Shrödinger equation as an equation satisfied by the time evolution operator (which can be constructed from the fields).

8. Jun 4, 2008

### jostpuur

I've been hoping that something like this could be true, but I've always got lost with the $$1/(2E_{\vec{p}})$$ factors in this business.