1
$\begingroup$

I am looking for a closed form of the density operator of the quantum harmonic oscillator in thermal equilibrium, preferably in position representation. I am fairly sure it looks like a coherent state but I couldn't find it in any of my books or online sources that I skimmed. If my memory serves me well, then there exists a closed form Wigner function for it too, so I would expect that there also exists a simple pure position representation. But please inform me if I am wrong, if there is no closed form.

A short derivation is also welcome.

EDIT

I have written down what i managed to do. The density operator for a canonical ensemble with $\beta = \frac{1}{k_BT}$ should be $ \hat \rho = \exp(-\beta \hat H)$.Using the resolution of identity, we can write it as $$\begin{aligned} \hat \rho &=\exp(-\beta \hat H)\sum_{n=0}^\infty |n\rangle \langle n|\\ &= \sum_{n=0}^\infty \exp(-\beta E_n)|n\rangle \langle n|\\ &=\sum_{n=0}^\infty \exp\left(-\beta \left(\frac{1}{2}\hbar\omega(n+1) \right) \right)|n\rangle \langle n|\\ \end{aligned}$$ The eigenfunctions of the harmonic oscillator are $$ \langle x | n \rangle =\phi_n(x) = \frac{1}{\sqrt{2^nn!}} \left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}\exp(-\frac{m\omega x^2}{2\hbar})H_n\left(\frac{m\omega}{\hbar} x \right) $$

The position representation of the density operator is then $$\begin{aligned} \langle x|\hat \rho|x'\rangle &= \sum_{n=0}^\infty \exp\left(-\beta \left(\frac{1}{2}\hbar\omega(n+1) \right) \right)\langle x|n\rangle \langle n|x'\rangle\\ &= \sum_{n=0}^\infty \exp\left(-\beta \left(\frac{1}{2}\hbar\omega(n+1) \right) \right)\phi_n(x) \phi^*_n(x')\\ \end{aligned}$$ Can this sum be simplified ?

$\endgroup$
4
  • 1
    $\begingroup$ What are your attempts so far? Do you know how to write the density operator for the canonical ensemble in terms of the eigenstates of the quantum harmonic oscillator? Then you could easily write the position-representation of the density operator in terms of the position representation of the eigenstates of the quantum harmonic oscillator. It is then possible to derive a closed expression. $\endgroup$ Apr 11, 2021 at 10:30
  • $\begingroup$ @Jakob I have updated the question. $\endgroup$
    – Hans Wurst
    Apr 11, 2021 at 13:27
  • 1
    $\begingroup$ Looks correct (apart from the normalization of the density operator, i.e. the factor $1/Z$ is missing. Then you can use the equation @mike stone gave. A detailed derivation is also given here in section 5.4.3. $\endgroup$ Apr 11, 2021 at 15:06
  • $\begingroup$ That link was very helpful and contained everything that i wanted. $\endgroup$
    – Hans Wurst
    Apr 12, 2021 at 8:57

2 Answers 2

3
$\begingroup$

If you want to write $\exp\{-\beta \hat H\}$ in closed form in the position representation, you can use Mehler's formula: $$ \sum_{n=0}^\infty s^n \varphi_n(x)\varphi_n(y) =\\ \frac 1{\sqrt{\pi (1-s^2)}} \exp\left\{\frac{4xys -(x^2+y^2)(1+s^2)}{2(1-s^2)}\right\}, \quad 0\le |s|<1. $$ with $$ s= e^{-\beta(n+1/2)}. $$ Here $$ \varphi_n(x)\equiv \frac{1}{\sqrt{2^n n! \sqrt{\pi}}} H_n(x) e^{-x^2/2} $$ is the normalized harmonic oscillator wavefunction. I have set the frequency to be unity for convenience, but it is easy to generalise to arbitrary $\omega$.

$\endgroup$
1
$\begingroup$

For completeness sake and later reference i'll add the actual result here. The derivation requires besides the evaluation of the sum also a fair amount of hyperbolic trig function manipulation. Here it is,

$$\begin{aligned} \rho(x,x') &\equiv \langle x | \hat \rho |x'\rangle \\[1.0em] &= \frac{\langle x|\exp(-\beta \hat H)| x'\rangle }{Z}\\[1.0em] &=\frac{1}{\sum^\infty_{n=0}\exp(-\beta E_n)}\sum^\infty_n \phi_n(x) \phi^*_n(x') \exp(-\beta E_n)\\[1.0em] &= \sqrt{\frac{m\omega}{\hbar \pi} \tanh\left(\frac{1}{2}\beta \hbar \omega\right) } \\ &\times \exp\left( -\frac{m\omega}{4\hbar } \left( (x+x')^2\tanh\left(\frac{1}{2}\beta \hbar \omega\right) +(x-x')^2 \coth\left(\frac{1}{2}\beta \hbar \omega\right) \right) \right) \end{aligned}$$

Note that this $\hat \rho$ differs from the definition in my question and includes now the partition function which is missing in the original question.

A source with a fairly comprehensible derivation is found in

https://www.hep.phy.cam.ac.uk/theory/webber/tp2_06.pdf , Accessed 13.04.2021.

The final result is also given in an exercise in the book

Quantum Optics in Phase Space, Wolfgang P. Schleich, First Edition, page 64, Exercise 2.6.

$\endgroup$

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.