Calculating free energy from coherent state path integral

Question

Edit: It turns out that problem encountered in this question is not limited to BdG Hamiltonians.

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I am having trouble in using the coherent state path integral approach to calculate the free energy. For example, consider a Bogoliubov-de Gennes (BdG) Hamiltonian:

$$ \mathsf{H}[b^\dagger,b] = \Delta(b^\dagger_1 b^\dagger_2 + b_2 b_1) + \lambda (b^\dagger_1 b_1 + b^\dagger_2 b_2) \quad (\lambda > \Delta > 0) $$

involving two bosons $\{b_i\}_{i=1}^2$. After diagonalizing the Hamiltonian with a Bogoliubov transformation, I can show that the free energy (up to constants independent of parameters $\Delta, \lambda$ in $\mathsf{H}$) is

$$ F = \frac{2}{\beta} \ln(1 - e^{-\beta E}) + E - \lambda, \quad E \equiv \sqrt{\lambda^2 - \Delta^2} $$

Now I describe my attempt to use the coherent state path integral. Since $\mathsf{H}$ is already normal-ordered ($b^\dagger$'s in front of $b$'s), writing down the coherent state path integral representation of the partition function is straightforward:

$$ \begin{align*} Z &= \int D[\bar{b},b] e^{-S[\bar{b},b]} \\ S[\bar{b},b] &= \int_0^\beta d\tau \, \Big[ \bar{b}_1 \partial_\tau b_1 + \bar{b}_2 \partial_\tau b_2 + \mathsf{H}[\bar{b},b] \Big] \end{align*} $$

Note that in this representation $b(\tau), \bar{b}(\tau)$ are commuting complex numbers. The integration measure $D[\bar{b},b]$ is (up to a normalization constant)

$$ D[\bar{b},b] = \prod_{0 \le \tau \le \beta} \prod_{a=1}^2 d\bar{b}_a(\tau) db_a(\tau) $$

Define the bosonic "spinor"

$$ \Phi = \begin{bmatrix} b_1 \\ \bar{b}_2 \end{bmatrix} \ \Rightarrow \ \bar{\Phi} = (\bar{b}_1, b_2) $$

I write $S[\bar{b},b]$ as

$$ \require{cancel} \begin{align*} S[\bar{b},b] &= \int_0^\beta d\tau \, \Big[ \bar{b}_1 \partial_\tau b_1 - b_2 \partial_\tau \bar{b}_2 + {\partial_\tau(\bar{b}_2 b_2)} \\ &\qquad + \Delta(\bar{b}_2 \bar{b}_1 + b_1 b_2) + \lambda (\bar{b}_1 b_1 + \bar{b}_2 b_2) \Big] \\ &= \int_0^\beta d\tau \, \bar{\Phi}(\tau) G^{-1}(\tau) \Phi(\tau) \end{align*} $$

Here I dropped total $\tau$-derivatives in $S$, and introduced the Green's function matrix

$$ G^{-1}(\tau) = \begin{bmatrix} \partial_\tau + \lambda & \Delta \\ \Delta & -\partial_\tau + \lambda \end{bmatrix} $$

The free energy is found by integrating over $b$. To deal with $\partial_\tau$, I Fourier transform to frequency space:

$$ b_a(\tau) = \frac{1}{\sqrt{\beta}} \sum_\omega b_{a\omega} e^{i\omega\tau}, \quad \bar{b}_a(\tau) = \frac{1}{\sqrt{\beta}} \sum_\omega \bar{b}_{a\omega} e^{-i\omega\tau} $$

where $\omega$ sums over boson Matsubara frequencies $\{2\pi n/\beta \ |\ n \in \mathbb{Z}\}$. This transformation is unitary, so I can change the integration measure to

$$ D[\bar{b},b] = \prod_{\omega} \prod_{a=1}^2 d\bar{b}_{a\omega} db_{a\omega} $$

The Fourier transform of $\Phi$ is

$$ \Phi = \begin{bmatrix} b_1 \\ \bar{b}_2 \end{bmatrix} = \frac{1}{\sqrt{\beta}} \sum_\omega \Phi_\omega e^{i\omega \tau}, \quad \Phi_\omega = \begin{bmatrix} b_{1\omega} \\ \bar{b}_{2,-\omega} \end{bmatrix} $$

The action becomes

$$ \begin{align*} S[\bar{b},b] &= \frac{1}{\beta} \int_0^\beta d\tau \sum_{\omega,\omega'} \bar{\Phi}_\omega e^{-i\omega\tau} G^{-1}(\tau) e^{i\omega'\tau} \Phi_{\omega'} \\ &\equiv \sum_{\omega,\omega'} \bar{\Phi}_\omega G^{-1}_{\omega \omega'} \Phi_{\omega'} \end{align*} $$

where the frequency-space Green's function is

$$ \begin{align*} G^{-1}_{\omega \omega'} &\equiv \frac{1}{\beta} \int_0^\beta d\tau \, e^{-i\omega\tau} G^{-1}(\tau) e^{i\omega'\tau} \\ &= \frac{1}{\beta} \int_0^\beta d\tau \, e^{-i\omega\tau} \begin{bmatrix} i\omega' + \lambda & \Delta \\ \Delta & -i\omega' + \lambda \end{bmatrix} e^{i\omega'\tau} \\ &= \begin{bmatrix} i\omega + \lambda & \Delta \\ \Delta & -i\omega + \lambda \end{bmatrix} \delta_{\omega \omega'} \equiv \mathcal{G}^{-1}_\omega \delta_{\omega \omega'} \end{align*} $$

Then the partition function is

$$ \begin{align*} Z &= \int D[\bar{b},b] \exp\bigg\{ - \sum_\omega \bar{\Phi}_\omega \mathcal{G}^{-1}_\omega \Phi_\omega \bigg\} \\ &= \prod_\omega \int d\bar{\Phi}_\omega \, d\Phi_\omega \exp[ - \bar{\Phi}_\omega \mathcal{G}^{-1}_\omega \Phi_\omega ] \end{align*} $$

where $d\Phi_\omega = db_{1\omega} d\bar{b}_{2,-\omega}$ and $d\bar{\Phi}_\omega = d\bar{b}_{1\omega} db_{2,-\omega}$. The integration is Gaussian, and yields

$$ Z = \text{const} \times \prod_\omega (\det \mathcal{G}^{-1}_\omega)^{-1} $$

Here and thereafter $(\text{const})$ is a constant number independent of parameters $\mu, \Delta$ in the Hamiltonian. In this formula the "const" comes from the normalization of the integration measure. Then the free energy is (define $E = \sqrt{\lambda^2 - \Delta^2}$)

$$ \begin{align*} F &= -\frac{1}{\beta} \ln Z = \frac{1}{\beta} \sum_\omega \ln \det \mathcal{G}^{-1}_\omega + \text{const} \\ &= \frac{1}{\beta} \sum_\omega \ln(\omega^2 + E^2) + \text{const} \\ &= \frac{1}{\beta} \sum_\omega \ln[\beta^2(\omega^2 + E^2)] + \text{const} \\ &= \frac{2}{\beta} \ln(1 - e^{-\beta E}) + E + \text{const} \end{align*} $$

The last equality can be found in, say, eq. (25.34) in Quantum Field Theory for the Gifted Amateurs. But the constant is independent of $\lambda$, contradicting the results from Bogoliubov transformation. This shift by $\lambda$ may look harmless, but it will cause problems if $\mathsf{H}$ is obtained by a mean field approximation, and $\Delta, \lambda$ are to be determined by minimizing the free energy. So it is important to get the $\Delta, \lambda$ dependence right in $F$.

What is the mistake I made in the derivation?

Answer 1

The extra term you are missing can be interpreted as an error in the zero point energy. This error comes from the regularization of the infinite product for the partition function.

Take for example the case of a single mode: $$ H=ha^\dagger a $$ Going to Matsubara frequencies, PI gives you formally (up to a $h$ independent factor): $$ Z\propto \prod_{n\in\mathbb Z} (i\omega_n+h)^{-1} $$ and if you reorder the product to match opposite frequencies together, you recognize the product formula of $\sinh$ to get $$ Z\propto \text{cosech}\frac{\beta h}{2} $$ which lacks the zero point energy factor $e^{-\beta h/2}$.

Generally, such formal reorganizing can only give you the result up to a non cancelling factor (typically written as an exponential like the Weierstrass factorization theorem). If you want to know what this extra factor is, you need to be careful in the regularization.

The natural regularization arises from the derivation of the PI where $[0,\beta]$ is subdivided in $N$ intervals. This gives only $N$ factors for $Z$, and $i\omega_n=\frac{2\pi n}{\beta}$ is replaced by $N\frac{\zeta^n-1}{\beta}$ with $n\in[|1,N|]$ and $\zeta=e^{i2\pi/N}$. Geometrically, the Matsubara frequencies are regularly spaced on a circle passing by the origin and center on the real line whose radius goes to infinity so that it tends to the imaginary line. This will always give the correct result, and works for the simple example and your problem, though the calculations are a bit technical.

Single mode

After discretization, the partition function becomes: $$\begin{align} Z &\propto\prod_{n=1}^N\left(\zeta^n-1+\frac{\beta h}{N}\right)^{-1} \\ &\propto \frac{1}{1-(1-\beta h/N)^N}\\ &\to \frac{1}{1-e^{-\beta h}} \end{align}$$ using the identity: $$ X^N-1=\prod_{n=1}^N(X-\zeta^n) $$

Hence the result.

BdG

After discretization, the partition function becomes:

$$\begin{align} Z^{-1}&\propto\prod_{n=1}^N\begin{vmatrix} \zeta^n-1+\frac{\beta \lambda}{N} & \frac{\beta \Delta}{N} \\ \frac{\beta \Delta}{N} & \zeta^{-n}-1+\frac{\beta \lambda}{N} \end{vmatrix} \\\\ &\propto\prod_{n=1}^N \left [\left(1-\frac{\beta \lambda}{N}\right)^2 +1-\frac{\beta \Delta}{N}^2 -2 \left(1-\frac{\beta \lambda}{N}\right)\cos\frac{2\pi n}{N}\right] \\\\ &\propto \left(1-\frac{\beta \lambda}{N}\right)^N \prod_{n=1}^N \left [\frac{1}{2}\left(1-\frac{\beta \lambda}{N}\right)+\frac{1}{2}\frac{1-\frac{\beta \Delta}{N}^2 }{1-\frac{\beta \lambda}{N}}-\cos\frac{2\pi n}{N}\right] \end{align} $$ The first factor of $Z^{-1}$ converges to $e^{-\beta\lambda}$ and gives your missing $-\lambda$ term in $F$. To finish the calculation, you can use the identity: $$ \prod_{n=1}^N\left(X-\cos\frac{2\pi n}{N}\right) =2^{-N+1}(T_N(X)-1) $$ With $T_N$ the Chebyshev polynomial of the first kind defined by: $$ T_N(\cos \theta)=\cos(N\theta) $$ The argument of the polynomial is: $$ X= \frac{1}{2}\left(1-\frac{\beta \lambda}{N}\right)+\frac{1}{2}\frac{1-\frac{\beta \Delta}{N}^2 }{1-\frac{\beta \lambda}{N}} \sim 1+\frac{1}{2}\left(\frac{\beta E}{N}\right)^2 $$ so particular: $$ \arccos X\to \pm i\frac{\beta E}{N} $$ So the second factor of $Z^{-1}$ converges to something depending only on $E$. In fact, it’s exactly what you had found: \begin{align} &\sim \cos\left(\pm Ni\frac{\beta E}{N}\right)-1 \\ &\to \cosh(\beta E)-1 \\ &\propto \left(1-e^{-\beta E}\right)^2e^{\beta E} \end{align}

Hope this helps.