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Consider two masses $m$ connected by a spring with a spring constant $k$. Each mass is also connected to the wall using the same springs. The Hamiltonian is $$ H = \frac{p_1^2 + p_2^2}{2m} + \frac{k}{2}\left[x_1^2 + x_2^2 + \left(x_1 - x_2\right)^2\right]\,. $$

The normal modes are known ($\Omega_1 = \sqrt{k/m}$ and $\Omega_2 = \sqrt{3k/m}$). This means that the vacuum energy is $\frac{\Omega_1 + \Omega_2}{2}$.

Now, let's go back to the Hamiltonian and rewrite it a bit: $$ H = \frac{p_1^2 + p_2^2}{2m} + \frac{2k}{2}\left[x_1^2 + x_2^2 - x_1 x_2\right]\,. $$

Next, we perform the usual second quantization replacement: $$ x_1 = \sqrt{\frac{1}{2m\sqrt{\frac{2k}{m}}}}\left(a^\dagger + a\right)\,, \\ p_1 = i\sqrt{\frac{m\sqrt{\frac{2k}{m}}}{2}}\left(a^\dagger - a\right)\,, $$ and the same for $x_2$ and $p_2$. This sets the Hamiltonian to $$ H = \sqrt{\frac{2k}{m}}\left(a^\dagger a + \frac{1}{2}\right) + \sqrt{\frac{2k}{m}}\left(b^\dagger b + \frac{1}{2}\right) - \frac{2k}{4m\sqrt{\frac{2k}{m}}}\left(a^\dagger + a\right)\left(b^\dagger + b\right)\,. $$

From this, the action becomes $$ S = \beta\sqrt{\frac{2k}{m}} + \sum_n \bar\phi_n\left(-i \nu_n + \sqrt{\frac{2k}{m}}\right)\phi_n + \sum_n \bar\psi_n\left(-i \nu_n + \sqrt{\frac{2k}{m}}\right)\psi_n - \sum_n\frac{1}{4} \sqrt{\frac{2k}{m}}\left(\bar\phi_n+\phi_n\right)\left(\bar\psi_n+\psi_n\right)\,. $$

Using the fact that $\phi = g + i h$ and $\psi = u + i v$, $$ S = \beta\sqrt{\frac{2k}{m}} + \sum_n \left(-i \nu_n + \sqrt{\frac{2k}{m}}\right)\left(g_n^2 + h_n^2 + u_n^2 + v_n^2\right) - \sum_n \sqrt{\frac{2k}{m}}g_n u_n\,. $$

The partition function is $e^{-S}$ integrated over the fields. This gives $$ \mathcal{Z} = e^{-\beta\sqrt{\frac{2k}{m}}} \prod_n \left[\beta\left(-i \nu_n + \sqrt{\frac{2k}{m}}\right)\right]^{-1} \prod_n \det\left[ \beta \begin{pmatrix} -i \nu_n + \sqrt{\frac{2k}{m}}&-\frac{1}{2}\sqrt{\frac{2k}{m}} \\ -\frac{1}{2}\sqrt{\frac{2k}{m}}& -i \nu_n + \sqrt{\frac{2k}{m}} \end{pmatrix} \right]^{-1/2} $$

The energy is obtained from $E = -T\ln\mathcal{Z}$: $$ E = \sqrt{\frac{2k}{m}}+ T \sum_{\nu_n}\ln\left[\beta\left(-i \nu_n + \sqrt{\frac{2k}{m}}\right)\right] + \frac{T}{2} \sum_{\nu_n}\ln\left[\beta\left(-i \nu_n +\frac{1}{2} \sqrt{\frac{2k}{m}}\right)\right] + \frac{T}{2} \sum_{\nu_n}\ln\left[\beta\left(-i \nu_n + \frac{3}{2} \sqrt{\frac{2k}{m}}\right)\right]\,. $$

The summation over the bosonic Matsubara frequencies gives $$ E = \sqrt{\frac{2k}{m}}+ T \ln\left[1 - e^{-\beta \sqrt{\frac{2k}{m}}}\right] + \frac{T}{2}\ln\left[1 - e^{-\beta \frac{1}{2}\sqrt{\frac{2k}{m}}}\right] + \frac{T}{2}\ln\left[1 - e^{-\beta \frac{3}{2}\sqrt{\frac{2k}{m}}}\right]\,. $$

As $\beta\rightarrow \infty$, the last three terms drop out, leaving only the zero-point energy, which has a different value from what's expected for this system.

The summation over the Matsubara frequencies is performed as follows.

$\ln\left[\beta\left(z + \epsilon\right)\right]$ has a branch cut on the real axis for $z < - \epsilon$. The Matsubara frequencies are the poles of of the Bose-Einstein distribution. So, we take the integral of $\ln\left[\beta\left(z + \epsilon\right)\right] n_B\left(z\right)$ over the big circle, going around the branch cut. The big circle integral vanishes and leaves us with

$$ \int_{-\infty}^\epsilon dx \ln\left[\beta\left(x +i0 + \epsilon\right)\right] n_B\left(x\right) + \int_\epsilon^{-\infty} dx \ln\left[\beta\left(x -i0 + \epsilon\right)\right] n_B\left(x\right) = \sum_{\nu_n}2\pi i \mathrm{Res}\left\{\ln\left[\beta\left(i\nu_n + \epsilon\right)\right] n_B\left(\nu_n\right)\right\} \\ = \frac{2\pi i}{\beta} \sum_{\nu_n}\ln\left[\beta\left(i\nu_n + \epsilon\right)\right] $$

Integrating by parts yields the desired expression.

Am I making some unphysical assumption somewhere?

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  • $\begingroup$ in the last formula, why is there no summation over $n$? furthermore, to obtain the ground state energy you should take the limit $\beta \rightarrow \infty (T \rightarrow 0)$. $\endgroup$ – Lorenz Mayer Mar 5 at 11:49
  • $\begingroup$ Sorry, that's a typo. There is, of course, a sum over $\nu_n$. And yes, setting $T\rightarrow 0$ will give the ground state. However, the vacuum part is still different, right? $\endgroup$ – IcyOtter Mar 5 at 11:51
  • $\begingroup$ i don't quite see how to get from the second-to-last to the last equation. i mean the sum in the former is ill-defined, so which kind of regulator are you using, or is there some subtle cancelation? also the expression is not that innocent in that it involves logarithms of complex numbers. $\endgroup$ – Lorenz Mayer Mar 5 at 12:22
  • $\begingroup$ In this part, I'm essentially following Altland and Simons for this summation. For a decoupled system, these summations give the right result: at finite T, they give the expected energies of the quanta in each oscillator. At T = 0, they give nothing as there is only zero-point energy, given by the first term. The issue that I have is that my zero-point result is wrong for a coupled system $\endgroup$ – IcyOtter Mar 5 at 12:55

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