2
$\begingroup$

I am interested in how one formally takes the classical limit of:

$$ \langle A(x_0)B(x_t)\rangle = \mathrm{Tr}[e^{-\beta \hat{H}}A(\hat{x})e^{i \hat{H}t/\hbar}B(\hat{x})e^{-i \hat{H}t/\hbar}]$$ i.e. how to show that: $$\langle A(x_0)B(x_t)\rangle \to \frac{1}{2\pi\hbar} \int dx \int dp\, e^{-\beta {H}} A(x_0)B(x_t)$$

I understand that for a simple thermal expectation value:

$$\langle A(x)\rangle=\mathrm{Tr}[\exp(-\beta \hat{H})A(\hat{x})]$$

one can show simply that the classical limit arises when $\beta \to 0$ by expanding in the position basis, using the Trotter theorem and introducing the resolution of the identity in momentum to give:

$$\lim_{\beta \to 0}\mathrm{Tr}[\exp(-\beta \hat{H})A(\hat{x})]=\int dx \int dp \langle x|p\rangle\langle p | x \rangle \exp(-\beta H)A(x) \\ $$ $$ \color{white}{\lim_{\beta \to 0}\mathrm{Tr}[\exp( )A(\hat{x})]}= \frac{1}{2\pi\hbar}\int dx \int dp \exp(-\beta H)A(x)$$

I assume that, for the correlation function, one must further take the classical limit using real time path integrals by assuming that $S(x,x',t)/\hbar\to\infty$ for all $x$ and $x'$ but I cannot work out how to do this rigorously.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.