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I Single-mode field quantization Hamiltonian

  1. May 5, 2017 #1


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    I'm having some trouble on understanding how the Hamiltonian of the e-m field in the single mode field quantization is obtained in the formalism proposed by Gerry-Knight in the book "Introductory Quantum Optics".
    (see, http://isites.harvard.edu/fs/docs/icb.topic820704.files/Lec11_Gerry_Knight.pdf )

    The electric and magnetic field are:
    [tex] E(z,t) = \hat{x} q(t) \sqrt{\frac{2\omega^2}{\epsilon_0 V}}sin(kz) [/tex]
    [tex] B(z,t) = \hat{y} \frac{1}{c^2 k} p(t) \sqrt{\frac{2\omega^2}{\epsilon_0 V}}cos(kz) [/tex]

    and the Hamiltonian corresponding to the electromagnetic energy density is:
    [tex] H=\frac{1}{2}\int_V dV [\epsilon_0 E^2 + \frac{B^2}{\mu_0}] [/tex]

    Therefore I can write:
    [tex]H=\frac{1}{2}\int_V dV [\epsilon_0 q^2(t) {\frac{2\omega^2}{\epsilon_0 V}}sin^2(kz) + \frac{\frac{1}{c^4 k^2} p(t)^2 {\frac{2\omega^2}{\epsilon_0 V}}cos^2(kz)}{\mu_0}][/tex]
    Which becomes:
    [tex]H=\omega^2 q^2 \int_v \frac{1}{V}sin^2(kz) dV + p^2 \int_v \frac{1}{V}cos^2(kz) dV[/tex]

    Now, the final result should be
    [tex]H=\frac{1}{2} (p^2+\omega^2q^2)[/tex]

    But I son't really understand how this is obtained since when I calculate the hamiltonian a get stucked in the following integral:
    [tex]\int_v\frac{1}{V}sin^2(kz) dV[/tex]

    which brings me a sin factor I'm not sure how to remove.

    Does anyone have calculated the Hamiltonian following this formalism and help me to solve my problem?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. May 5, 2017 #2

    A. Neumaier

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    Science Advisor

    Use Fourier transforms and work in frequency/momentum space then everything simplifies a lot!
  4. May 6, 2017 #3
    You might substitute kz for an angle and integrate over 0 to pi. You will need the formula for the cosine of a double angle to elimate the sin squared term. This introduces the factor of 1/2.
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