My question is related to problem 2.2.4 in Xiao-Gang Wen's book.
Problem statement: Compute the time ordered velocity correlation function for a harmonic oscillator described by $$H = \frac12 m ({\omega_0}^2 x^2 + \dot{x}^2 )$$.

This velocity correlation function is defined as:

$$ G(t) = \langle\mathcal{T} \{\dot{\hat{x}}(t), \dot{\hat{x}}(0) \} \rangle$$

I know how to compute this using Heisenberg Operators $ \dot{x}(t) = \hat{p \over m}$, but I'm at loss where to start if I want to use the Path Integral


The path integral automatically gives time-ordered operators:

$$G(t)\equiv\langle T\{\hat{\mathcal{O}}_1(t)\hat{\mathcal{O}}_2(0) \}\rangle=\int_{x(t_a)=x_a}^{x(t_b)=x_b}\mathcal{D}[x(t')] \mathcal{O}_1(t)\mathcal{O}_2(0)e^{iS[x(t')]}.$$

In order to evaluate the above, you'll need to provide the boundary value data $x(t_a)=x_a$ and $x(t_b)=x_b$ or, equivalently, the initial conditions $x(t_a)=x_a$ and $\dot{x}(t_a)=\dot{x}_a$.

Edit: the initial and final states are given by the ground state at $t=\pm\infty$. We are thus interested in computing $$ \begin{align} G(t) & \equiv\langle0;t_b=\infty|\mathcal{T}\{\hat{\mathcal{O}}_1(t)\hat{\mathcal{O}}_2(0)\}|0,t_a=-\infty\rangle \\ & = \langle0;t_b=\infty|\int_{-\infty}^\infty dx_b|x_b\rangle\langle x_b|\mathcal{T}\{\hat{\mathcal{O}}_1(t)\hat{\mathcal{O}}_2(0)\} \int_{-\infty}^\infty dx_a|x_a\rangle\langle x_a|0,t_a=-\infty\rangle\\ & = \int_{-\infty}^\infty dx_b \langle 0|x_b\rangle\int_{-\infty}^\infty dx_a\langle x_a|0\rangle\int_{x(-\infty)=x_a}^{x(\infty)=x_b}\mathcal{D}[x(t')] \mathcal{O}_1(t)\mathcal{O}_2(0)e^{iS[x(t')]}. \end{align}$$

  • $\begingroup$ Thank you. The notation I wrote $\langle \hat{O} \rangle$ technically sets the ground state as the boundary at $t = \pm \infty$, in other words: $\langle \hat{O(t)} \rangle \equiv \langle 0| U(t, \infty)\hat{O} U(-\infty, t)| 0(t = -\infty)\rangle$ $\endgroup$ – physicsdude May 31 '17 at 22:49
  • $\begingroup$ Answer edited to reflect your comment. $\endgroup$ – WAH Jun 1 '17 at 4:27
  • $\begingroup$ Sorry; dropped some terms in my original edit. I've corrected my response. $\endgroup$ – WAH Jun 1 '17 at 8:18

Here is one way to compute it without using path integral. I will try with path integral to check it:

$$\hat{\dot{x}} \equiv \hat{v} = \sqrt{\hbar \omega_0 \over 2m} (a^\dagger -a)$$

Using the heisenberg evolution $a^\dagger (t) = a^\dagger e^{i \omega_0 t}$, we get

$$ \langle 0 | \mathcal{\tau} \{ v(t) v(0) \} | 0 \rangle = {\hbar \omega_0 \over 2m} e^{-i \omega_0 |t|} $$

The problem further asks for the "time ordered current current correlation function in frequency space, namely $G_j (\omega)$"

This is the fourier transform of the velocity correlation computed above

$$ \rightarrow G_j(\omega) \equiv e^2 \int dt e^{i \omega_0 t}\langle 0 | \mathcal{\tau} \{ v(t) v(0) \} | 0 \rangle$$

$$ = {e^2 \hbar \omega_0^2 \over m (\omega^2 - \omega_0^2)}$$

Part b: Calculate the finite frequency conductance $\sigma(\omega)$, say using classical physics. You may add a small friction term

The classical physics equation of motions are:

$$ \ddot{x} + \omega_0^2 x + \gamma \dot{x} = {e \over m} E(t)$$

$$ v(\omega) = {e E(\omega) \over m} { i \omega \over \omega_0^2 - \omega^2 + i \gamma \omega}$$

$$ \rightarrow \sigma(\omega) = {e^2 \over m} {i \omega \over \omega_0^2 - \omega^2 + i \gamma \omega}$$

Part c: Show that for $\omega>0 \sigma(\omega) = C {G_j(\omega) \over \omega} $ for some C. Show that $\sigma(\omega)$ has a nonzero real part only when $\omega = \omega_0$. At this frequency, the oscillator can absorb energy by jumping to higher excited states. We see that in order to have a finite DC real conductance, we need gapless excitations

This follows trivially from the previous equations, we can see the constant C is just $\hbar \omega_0^2$.

At resonance $\omega = \omega_0$, we see that $\sigma(\omega)$ is purely real:

$\sigma(\omega_0) = {e^2 \over m \gamma}$.

Can someone maybe explain to me the cryptic comment: "We see that in order to have a real conductivity, the system must have gapless states"... That would be very helpful

Next step: Use that path integral to do the same calculation!


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