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Quantization of electromagnetic field

  1. Oct 8, 2011 #1
    Instead of quantizing the vector potential A^μ why we do not directly quantize the B and E fields in electrodynamics.
     
  2. jcsd
  3. Oct 8, 2011 #2
    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.
     
  4. Oct 8, 2011 #3

    Bill_K

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    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·.
     
  5. Oct 8, 2011 #4

    tom.stoer

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    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.
     
  6. Oct 9, 2011 #5
    Thanks a lot for the answer and useful comments. But can somebody explain to me why E and B are not canonically conjugate variables?
     
  7. Oct 9, 2011 #6

    tom.stoer

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    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

    [tex]p_x = \frac{\delta S}{\delta \dot{x}}[/tex]

    We then have the classical Poisson brackets

    [tex]\{x,p_x\} = 1[/tex]

    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

    [tex]P^i(x) = \frac{\delta S}{\delta \dot{A}^i(x)}[/tex]

    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

    [tex]A_\mu j^\mu[/tex]

    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
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