# Quantization of electromagnetic field

1. Oct 8, 2011

### mritunjay

Instead of quantizing the vector potential A^μ why we do not directly quantize the B and E fields in electrodynamics.

2. Oct 8, 2011

### cmos

You can. The field Hamiltonian can be written as an integral of E2 + B2 (probably with some constants that I'm missing). Expand E and B in terms of Fourier modes and you'll end up with the Hamiltonian of the harmonic oscillator. Quantization follows.

3. Oct 8, 2011

### Bill_K

Yes, the Hamiltonian is H = ½ ∫ (E2 + B2) d3x, but this is only a shorthand for the same expression written out in terms of A:

H = ½ ∫ (A·2 + (∇ x A)2) d3x

You still have to quantize using the canonical variables A and E ≡ A·.

4. Oct 8, 2011

### tom.stoer

That's exactly the point.

Quantizing means to identify fundamental variables and to promote a classical Poisson bracket for canonical conjugate variables to a commutator. E and B are not canonical conjugate; it is not possible to express the (classical) dynamics using E and B.

5. Oct 9, 2011

### mritunjay

Thanks a lot for the answer and useful comments. But can somebody explain to me why E and B are not canonically conjugate variables?

6. Oct 9, 2011

### tom.stoer

Let's look at some action S in terms of some variables x, y, ... and their time derivatives dx/dt, dy/dt, ....

The definiton of the canonical momentum is always

$$p_x = \frac{\delta S}{\delta \dot{x}}$$

We then have the classical Poisson brackets

$$\{x,p_x\} = 1$$

In case of electrodynamics the fundamental variable x is replaced by A(x); x is something like a "continuous index". That means that the canonical momentum is defined by

$$P^i(x) = \frac{\delta S}{\delta \dot{A}^i(x)}$$

where i=1..3 is the i-th spatial direction.

Using the Lagrangian of electrodynamics one can show that P(x) corresponds to the electromagnetic field E(x). B(x) can be expressed in terms of A(x) w/o any time derivative, i.e .w/o using the canoncal conjugate momentum, so the fundamental variable is A(x) and the canonical conjugate momentum is E(x).

In addition E and B are not sufficient to define classical electrodynamics. The problem is that neither E nor B couple directly to the currents, but A does. The coupling term is

$$A_\mu j^\mu$$

which cannot be formulated using only E and B.

Note that there is one problem, namely the missing time derivative in A°(x) and therefore the missing canonical conjugate momentum which makes A°(x) a Lagrangian multiplier (instead of a dynamical field) generating the time-independent Gauss constraint. This fact is closely related to gauge symmetry.

Last edited: Oct 9, 2011