# Hamiltonian for single-mode field in cavity

Given that the Hamiltonian for a single-mode field is $$H = \frac{1}{2}\int dV \left[ \epsilon_0E_x^2(z,t) + \mu_0^{-1}B_y^2(z,t) \right],$$ with \begin{align} E_x(z,t) =& \sqrt{\frac{2\omega^2}{V\epsilon_0}}q(t)\sin(kz) \\ \text{and} \quad B_y(z,t) =& \left( \frac{\mu_0\epsilon_0}{k}\right) \sqrt{\frac{2\omega^2}{V\epsilon_0}}p(t)\cos(kz) \end{align} it should be possible to show that $$H=\frac{1}{2}(p^2+\omega^2 q^2)$$.

When I try, however, I can only get some of the way. More specifically $$H = \frac{1}{2} \int dV \left[ \frac{2}{V}q^2(t)\sin^2(kz) + \frac{2\omega^2}{V}p^2(t)\cos^2(kz) \right]$$ $$=\frac{1}{2} \int dV \left[\omega^2 q^2\frac{1}{V}\left(1-\cos(2kz)\right) + p^2\frac{1}{V}\left(1+\cos(2kz)\right) \right].$$

At this point I'm not sure how to move on. It looks like the integral should be solved with using $$\int \frac{1}{V} dV = \log(V)$$, but that doesn't seem to make sense. So what makes the cosine-terms disappear and how is the $$\int \frac{1}{V} dV$$ solved?

Hints:

1. The notation is confusing. In the first line $\int dV$ is an integral over the space inside a volume $V$. It would be much nicer to write $\int_V d^3r$. Or, since the example seems to be one-dimensional, $\int_V dz$.
2. The $V$ in the mode functions is is the actual volume you are integrating over. Hence it comes out of the integral. In our new notation this is easy to see, e.g. $$\int_V dz \frac{1}{V} \sin^2(kz) = \frac{1}{V} \int_V dz \sin^2(kz).$$
3. So all we have to do is compute integrals such as $\int_V dz \sin^2(kz)$ and $\int_V dz \cos^2(kz)$. This can be done if the boundary conditions of the mode are correctly specified. E.g. for a mode between 2 perfect mirrors that are separated by a distance $d$ the boundary conditions would impose $E(0)=E(d)=0$ and hence $k=\frac{n\pi}{d}$ where $n \in \mathbb{Z}$. Then, with $V=d$

$$\int_V dz \sin^2(kz) = \int_0^d dz \sin^2\left( \frac{n\pi}{d} z\right) = \frac{d}{2}.$$

Hope that helps.

I came across this in Gerry & Knight's Introductory Quantum Optics as an exercise left to the reader.

Anyways, I started with provided solutions to Maxwell's equations without a source/current $$\begin{cases}E_x(t,z) = \left(\dfrac{2\omega^2}{V\varepsilon_0} \right)^{1/2}q(t)\sin(kz) \\ B_y(t,z) = \left(\dfrac{2\omega^2}{V\varepsilon_0} \right)^{1/2}\dfrac{\mu_0\varepsilon_0}{k}\dot{q}(t)\cos(kz) \end{cases}$$

then just checked they actually satisfied the equations of motion. In doing so, I noted this requires the dynamical function (aka state of the field) $$q(t)$$ satisfy $$\partial_t \dot{q}(t) \propto q(t)$$, which makes sense here since the $$B$$-field must satisfy the wave equation $$(\partial_t^2-\partial_z^2)B=0$$ in units where the velocity $$c$$ of the thing that's waving is $$1$$. In these units, its general solution is of the form $$f(x+t)+g(x-t)$$, which is equivalent to a product of one cosine with a time argument and another with a spatial argument. If that is unclear, you can consult Feynman's Lecture Series Volume II $$\S$$ 20.1 (link) for a review.

Next up is the Hamiltonian for a single-mode field consisting of the previous $$E$$- and $$B$$-fields, $$H = \dfrac{1}{2}\int dV \left[\varepsilon_0 E^2+\dfrac{1}{\mu_0}B^2 \right],$$ where the integration is over a one-dimensional cavity along the $$z$$-axis with $$z\in[0,L]$$ and with each field vanishing at the boundaries. Choosing $$\omega = kc$$ (i.e. a single mode), recalling $$c^2 = (\varepsilon_0\mu_0)^{-1}$$, and thinking of $$\dot{q}(t)= p$$ as canonical momentum, we have $$H = \dfrac{1}{L}\left[\omega^2 q^2\int_0^L\sin^2(kz)\hspace{0.1cm}dz+p^2\int_0^L\cos^2(kz)\hspace{0.1cm} dz \right].$$ That's reduced the problem to two integrals. Following the boundary data, one can choose $$k = m\pi/L$$ with $$m\in \mathbb{N}$$ so that $$\int_0^L \sin^2\dfrac{m\pi z}{L} \hspace{0.1cm} dz = \frac{1}{2}\int_0^L\left[ 1-\cos\frac{2m\pi z}{L}\right]dz = \dfrac{L}{2}$$ as only the first term's integration survives. Similarly, $$\dfrac{L}{2}$$ is all that's left of the other integral, leaving \begin{align} H &= \frac{1}{L}\left(\omega^2 q^2 \dfrac{L}{2} + p^2 \dfrac{L}{2}\right)\\ \Rightarrow H &= \dfrac{1}{2}(p^2 + \omega^2 q^2), \end{align} showing the Hamiltonian for a single-mode field is equivalent to that of a simple harmonic oscillator with unit mass. Though this computation was entirely classical, it clearly puts one in a familiar spot to start quantizing the electromagnetic field.