# Coherent States Path Integral of Harmonic Oscillators

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?

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

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.
• 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 $\frac{\dot{p}q}{m\omega}$.
• 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