How to calculate the path integral (or partition function) of a set of phonons with continuous frequencies A system of phonons with continuous frequencies has a Hamiltonian like this $H=\int_0^{k_0} h(k) a_k^\dagger a_k \mathrm{d}k$. How do I calculate the partition function of this system using path integral?
Mathematically, how do I evaluate the following expression?
$$
Z(\beta)=\int\mathcal{D}? \exp{\int_0^\beta \mathrm{d}\tau \int_0^{k_0} a_k^\dagger a_k \mathrm{d}k\,\mathrm{d}k}.
$$
If replacing $a_k^\dagger a_k$ with $\frac{p_k^2}{2m}+2m g(k)^2x^2$, it becomes
$$
Z(\beta)=\int\mathcal{D}? \exp{\int_0^\beta \mathrm{d}\tau \int_0^{k_0} \frac{p_k^2}{2m}+ \frac{1}{2}m g(k)^2x^2 \mathrm{d}k\,\mathrm{d}k}.
$$
I use question marks because I (to be honest) don't know what to be integrated.
In the discrete version of this expression, I know the ? in $\mathcal{D}?$ would be the coordinates of the oscillators.
 A: First of all your expression for the action (the argument of the exponential in the path integral) is wrong. For a general Hamiltonian, function of annihilation and creation operators $a,a^\dagger$, $
H(a,a^\dagger)$, the partition function expressed via path integral is
\begin{align}
\DeclareMathOperator{\Tr}{Tr}
Z(\beta) & =\Tr e^{-\beta H(a,a^{\dagger})}\\
 & =\int D^{2}z(\tau)e^{-S(z,z^{\ast})}\\
S(z,z^{\ast}) & =\int_{0}^{\beta}d\tau\left(z^{\ast}\partial_{\tau}z+H(z,z^{\ast})\right).
\end{align}
So the action $S$ is a functional; a function of the complex fields
$z(\tau)$, $z^{*}(\tau)$. There are also boundary boundary conditions
on such fields (which originate from the trace above). For bosons they are $z(0)=z(\beta)$ (and $z^{\ast}(0)=z^{\ast}(\beta)$).
The integration is over a not well specified space of functions $z(\tau)$
and $z^{\ast}(\tau)$. The procedure can be understood by divinding
the interval $\left[0,\beta\right]$ in $N$ parts. In this case the
integration is over $2N$complex scalars, but then one send $N\to\infty$.
For your system of non-interacting bosons with continuum label $k$ the action would be
\begin{align*}
S & =\int_{0}^{\beta}d\tau\int_{0}^{k_{0}}dk\left(z_{k}^{\ast}(\tau)\partial_{\tau}z_{k}(\tau)+z_{k}^{\ast}(\tau)h(k)z_{k}(\tau)\right)\\
 & =:z^{\ast}G^{-1}z
\end{align*}
where in the last line $z$ is a field depending on $\tau$ and
$k$: $z_{k}(\tau)$. Now this action is quadratic in the complex
fields $z,z^{*}$, and the term $e^{-S}$ is an infinite dimensional
Gaussian. The result of the (infinite dimensional) integration is
$$
Z(\beta)=\det G.
$$
This result is simply the infinite dimensional generalization of the
Gaussian integration (for complex variables).  Note that the operator
$G^{-1}$ is diagonal in the label $k$ but not in the label $\tau$
(where it acts as differentiation). Going back to the discrete fields
(with $2N$ complex scalars) one can indeed evaluate the determinant
above and one finds the familiar formula
\begin{align*}
Z(\beta) & =e^{-\beta F(\beta)}\\
F(\beta) & =\beta^{-1}\int_{0}^{k_{0}}dk\ln\left(1-e^{-\beta h(k)}\right).
\end{align*}
You can find a detailed derivation of the path integral (and also how to evaluate the determinant) in the appendices of the book by A. Auerbach, Interacting electrons and quantum magnetism.
