I Single-mode field quantization Hamiltonian

Tags:
1. May 5, 2017

ja!

Hi!
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:
$$E(z,t) = \hat{x} q(t) \sqrt{\frac{2\omega^2}{\epsilon_0 V}}sin(kz)$$
$$B(z,t) = \hat{y} \frac{1}{c^2 k} p(t) \sqrt{\frac{2\omega^2}{\epsilon_0 V}}cos(kz)$$

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

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

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

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

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. May 5, 2017

A. Neumaier

Use Fourier transforms and work in frequency/momentum space then everything simplifies a lot!

3. May 6, 2017

Jilang

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted