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I am studying the notes provided by Ben Simons in this link (http://www.tcm.phy.cam.ac.uk/~bds10/tp3.html). I am currently on Lecture 16 (Applications and Connections). The corresponding textbook is Condensed Matter Field Theory (Ch 4). The first page derives the partition function for the harmonic oscillator via a parametrization $$\psi(\tau) = \sqrt{\frac{m\omega}{2 \hbar}}\left(q(\tau) + \frac{ip(\tau)}{m\omega} \right).$$ This means that $$\bar{\psi}(\tau) = \sqrt{\frac{m\omega}{2 \hbar}}\left(q(\tau) - \frac{ip(\tau)}{m\omega} \right).$$

This is confirmed in Lecture 16 (page 1) when the author shows $$\hbar \omega \bar{\psi}\psi = \frac{m\omega^2}{2}\left(q^2 + \frac{p^2}{m^2\omega^2}\right).$$

The pdf lecture claims that the partition function can be written as:

$$Z = \int D[\bar{\psi},\psi]exp\left[-\int_{0}^{\beta}(\bar{\psi}\partial_\tau\psi + \hbar \omega\bar{\psi}\psi)\right] = \int D[p,q] exp\left[-\int_{0}^\beta d\tau\left(\frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2 - \frac{ip\dot{q}}{\hbar}\right)\right].$$

My goal is to prove this equality. In order to do this, I substituted the parametrizations of $\psi$ and $\bar{\psi}$ and found that the action integral in $e^{-action}$ to be $$\int_0^\beta \frac{m\omega}{\pi \hbar} \bigg[q\dot{q} + \frac{i\dot{p}q}{m\omega} - \frac{ip\dot{q}}{m\omega} + \frac{p\dot{p}}{m^2\omega^2} + \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2\bigg]. $$

Using integration by parts on the second term I obtained:

$$\int_0^\beta \frac{m\omega}{\pi \hbar} \bigg[ \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2 - \frac{ip\dot{q}}{m\omega} - \frac{ip\dot{q}}{m\omega} + q\dot{q} + \frac{p\dot{p}}{m^2\omega^2}\bigg].$$

Some parts of this integral are equal to the expression claimed in the lecture 16 pdf. This seems to suggest that $$D[p,q] = D[\bar{\psi}, \psi]exp\left[\frac{m\omega}{\pi \hbar}\left(- \frac{ip\dot{q}}{m\omega} + q\dot{q} + \frac{p\dot{p}}{m^2\omega^2}\right)\right].$$ However I do not know how to prove this. Can anyone explain why the equality below holds true?

$$Z = \int D[\bar{\psi},\psi]exp\left[-\int_{0}^{\beta}(\bar{\psi}\partial_\tau\psi + \hbar \omega\bar{\psi}\psi)\right] = \int D[p,q] exp\left[-\int_{0}^\beta d\tau\left(\frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2 - \frac{ip\dot{q}}{\hbar}\right)\right].$$

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If I understand your problem correctly, you should notice that $q \dot q$ and $p \dot p$ are total time derivatives and as such they don't contribute to the action.

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  • $\begingroup$ Thanks; I have another question, how does the $\frac{m\omega}{\pi \hbar}$ "vanish" in the result posted in lecture 16? And also , there is an extra $\frac{\dot{p}q}{m\omega}$. $\endgroup$ – user758469 Jun 5 at 16:24
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    $\begingroup$ the overall factor is probably a matter or field redefinition, it can be absorbed rescaling them. Not sure on the extra factor 2 for $\dot p q$ but i wouldn't exclude the possibility of a typo $\endgroup$ – tbt Jun 8 at 8:25
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The lectures are dismissing boundary terms (BTs) in the action. This is strictly speaking not correct. In fact, BTs are important for the possible compatible choice of boundary conditions, cf. e.g. this related Phys.SE post.

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