- #1

- 72

- 0

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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Usaf Moji
- Start date

- #1

- 72

- 0

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

reilly

Science Advisor

- 1,075

- 1

It is correct; basic to QED. See any QFT text; Weinberg for example.

Regards,

Reilly Atkinson

Regards,

Reilly Atkinson

- #3

- 410

- 0

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

- 455

- 1

- #5

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 413

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."the usual Maxwell field is the quantum wave function for a single photon"

(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

- 2,111

- 18

- #7

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 413

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

[tex]\int d^3p f(\vec p)a^\dagger(\vec p)|0\rangle[/tex]

where [itex]a^\dagger(\vec p)[/itex] 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

- 2,111

- 18

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

[tex]\int d^3p f(\vec p)a^\dagger(\vec p)|0\rangle[/tex]

where [itex]a^\dagger(\vec p)[/itex] 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.

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

Share: