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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?

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# I Single-mode field quantization Hamiltonian

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