# 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.

• You're expression for the path integral is not correct. In particular in the action you're missing a term like $a^*\partial_\tau a$. Note also that the action is a function of fields, not operators (even though you are using the same notation).
– lcv
May 20, 2022 at 17:12
• By the way it's much simpler to evaluate this partition function in the operator formalism. Are you interested in the partition function itself or how one can evaluate it via path integral?
– lcv
May 20, 2022 at 17:14
• @lcv I am interested in how one can evaluate it via path integral. As you said, the operator formalism is more convenient for calculating the partition function. But I would like to learn how to calculate it via path integral. Sorry for the mistakes in the expression! May 20, 2022 at 18:05
• So first off you should start with the correct expression for the action. Where did you see this one (which is wrong)? Second, in general to compute the path integral you need to discretize imaginary time. But I don't think the procedure is stable as the path integral in general is not well defined. For free fermions and bosons (your example) it boils down to a certain determinant, reminiscent of the determinant that appear when you integrate a multidimensional Gaussian. This determinant must be regularized to have a proper meaning
– lcv
May 20, 2022 at 18:41
• I realized now your question is two years old. If you're still interested I can give an answer
– lcv
May 21, 2022 at 4:34

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.

• Thanks! You completed the path integral expression of the partition function by using the complex variables. If one wants to complete the integral using the coordinate $x$, do you think it is easy to do? Jun 4, 2022 at 6:29
• Yes what I presented is the coherent state path integral, where you have complex variables $z,z^*$ associated to the operators $a,a^\dagger$. This approach is very useful in many-body systems and can also be adapted to fermions. There is also the approach via real fields $x,p$ associated to the operators $\hat{x},\hat{p}$. In this case the action is the integral of the Lagrangian (as expected).
– lcv
Jun 4, 2022 at 19:06