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How Fourier components of vector potential becomes operators

  1. Sep 30, 2015 #1

    goodphy

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    Gold Member

    Hello.

    I'm studying quantization of electromagnetic field (to see photon!) and on the way to reach harmonic oscillator Hamiltonian as a final stage, sudden transition that the Fourier components of vector potential A become quantum operators is observed. (See https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field)

    I guess Wikipedia doesn't describes much of the reason for this jump. It only states that the operator is for boson. I know photon is boson but how it can explain Fourier components of A should become quantum operators? And besides, I originally expected to see the prove that photon is boson in this study.

    Could anybody help me to understand this?
     
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  3. Sep 30, 2015 #2

    stevendaryl

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    It helps to see the analogy with single-particle quantum mechanics, in the case of the harmonic oscillator. You have, classically:

    [itex]E = \frac{m}{2} \dot{x}^2 + \frac{K}{2} x^2[/itex]

    Then you can turn this classical equation into a quantum equation by letting [itex]p = m \dot{x}[/itex] and rewriting it as:

    [itex]E = \frac{p^2}{2m} + \frac{K}{2} x^2[/itex]

    where [itex]p[/itex] and [itex]x[/itex] obey the commutation relation [itex][p,x] = -i \hbar[/itex]

    The next move is to rewrite it in terms of "raising and lowering" operators:

    [itex]a = \frac{m \omega}{2 \hbar} x + i \frac{1}{2m \omega \hbar}[/itex]

    [itex]a^\dagger = \frac{m \omega}{2 \hbar} x - i \frac{1}{2m \omega \hbar}[/itex]

    Now, turning to quantum fields, instead of single particles. For a massless quantum field [itex]\Phi[/itex], there is energy associated with every Fourier mode [itex]\vec{k}[/itex]. This energy is given by: (something like...)

    [itex]E(\vec{k}) = C ( |\tilde{\dot{\Phi}}|^2 + |\vec{k} \tilde{\Phi}|^2) [/itex]

    where [itex]\tilde{\Phi}[/itex] is the Fourier transform of [itex]\Phi[/itex], and [itex]\tilde{\dot{\Phi}}[/itex] is similarly the Fourier transform of [itex]\dot{\Phi}[/itex]

    To quantize this, you similarly replace [itex]\dot{\Phi}[/itex] by the canonical momentum [itex]\Pi[/itex] to get:

    [itex]E(\vec{k}) = C ( |\tilde{\Pi}|^2 + |\vec{k} \tilde{\Phi}|^2) = C ( |\tilde{\Pi}|^2 + \omega^2 |\tilde{\Phi}|^2) [/itex] (where [itex]\omega = |\vec{k}|[/itex])

    For a fixed mode [itex]\vec{k}[/itex], this looks like the hamiltonian of the harmonic oscillator. So you can similarly solve it by writing:

    [itex]a(\vec{k}) = \omega \tilde{\Phi} + i \tilde{\Pi}[/itex]
    [itex]a^\dagger(\vec{k}) = \omega \tilde{\Phi} - i \tilde{\Pi}[/itex]

    What I've described here is a scalar field. The case of the electromagnetic field is more complicated, but only because there are more components of the field.

    The point is that in QFT, you are treating [itex]\Pi[/itex] and [itex]\Phi[/itex] as operators, in the same way that you treat [itex]p[/itex] and [itex]x[/itex] as operators in the single-particle case. The point about Fourier transforms is that the hamiltonian is much simpler after the transform, and the raising and lowering operators are related to the Fourier transforms of [itex]\Phi[/itex] and [itex]\Pi[/itex]
     
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