I learn ''Path integrals in quantum mechanics'' by Jean Zinn-Justin now. There is a chapter about calculating the path integral for particle on a ring (rigid rotator). So, after some calculations we can obtain something about matrix element:

$$U=\left< \theta'' \right| e^{-\beta H}\left| \theta' \right> \, .$$

Here $H$ is Hamiltonian, $\theta''$ and $\theta'$ are qoordinate of the particle in final and initial moments respectively, and $\beta$ is imaginary time, which is just $\beta = i T$, where $T$ is physical time.

We know two things:

  1. $U$ depends only on difference $\theta''-\theta':=\theta$
  2. A set of functions $e^{i \ell\theta}/\sqrt{2 \pi}$, where $\ell \in \mathbb{Z}$ forms an orthonormal basis for our Hilbert space.

Then there is said in the book, that hence we can expand $U$ to Fourier series:

$$U = \frac{1}{2 \pi}\sum\limits_{\ell = -\infty}^{+\infty}e^{i \ell \theta} e^{-\beta E_{\ell}} \, .$$

It could be fine, if instead of $e^{-\beta E_{\ell}}$ it would be something like $A_{\ell}$, but here coefficients because of some reason associated with eigenvalues of Hamiltonian $E_{\ell}$! In book author writes following:

"We have identified the coefficients of the Fourier series with the eigenvalues of $e^{- \beta H}$, because the functions $e^{i \ell\theta}/\sqrt{2 \pi}$ forms an orthonormal basis and, thus, the operator $e^{- \beta H}$ is directly diagonalized."

I completely don't understanf this move. Of course, the operator can be diagonalized, but why is it diagonal exactly in this basis? Is it correct at all?

  • $\begingroup$ Is it supposed to be a free-particle? I don't see how this claim can be true otherwise. $\endgroup$ – Dvij Mankad May 27 '19 at 17:08
  • $\begingroup$ Yeah, it is free particle, but on a ring. In coordinate representation the hamiltonian of this system is $H = -\frac{\hbar^2}{2 J} \partial_{\theta}^2$ It can be easy solved with stationary Schrodinger equation, but it is interesting to quantize this problem in terms of path integral. $\endgroup$ – Alex Goldstein May 27 '19 at 18:07

The "complete orthonormal basis" indicates $$ \langle \theta | \ell\rangle= e^{i\ell \theta}/\sqrt{2\pi} ~, \\ \int_0^{2\pi}\!\!\! d\theta ~~ \langle \ell'| \theta \rangle \langle \theta | \ell\rangle /2\pi = \delta_{\ell \ell'}~,\\ \sum_\ell \langle \theta | \ell\rangle \langle \ell| \theta '\rangle =\delta(\theta-\theta')~. $$ (Well, the last δ function is really a periodic Dirac comb function $\bbox[yellow]{\operatorname{III}_{2\pi}(\theta-\theta')}$, on account of the Poisson summation formula, but let's not fuss it here, as long as you appreciate all functions of angles are periodic.)

By the standard abuse of the Dirac notation, $$ \langle \theta | \hat L |\ell\rangle =-i\hbar \partial_\theta \langle \theta | \ell\rangle = \frac{\hbar \ell }{\sqrt{2\pi}} \langle \theta | \ell\rangle , $$ diagonal, so $$ \langle \theta | e^{-\beta \hat L ^2/2J} |\ell\rangle =e^{-\beta \hbar^2 \ell^2/2J} \langle \theta | \ell\rangle , $$ also diagonal, and $$ U=\langle \theta ''| e^{-\beta H } \sum_\ell | \ell\rangle \langle \ell| \theta'\rangle =\sum_\ell e^{-\beta \hbar^2 \ell^2/2J} e^{i\ell (\theta''-\theta')} ~/2\pi ~. $$


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.