Given the differential operator: $$A=\exp(-\beta H)$$ where $$H=\frac{1}{2}\left( -\frac{d^2}{dx^2}+x^2 \right)$$ and $\beta\gt 0$ How can I get the trace of this operator? Thanks in advance.

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    $\begingroup$ In general, this Wikipedia entry may help; in the specific case, Jon's answer gives you how to compute. $\endgroup$ Commented May 24, 2013 at 15:04

1 Answer 1


The trace of this operator is easily obtained in the following way: $$ Z={\rm Tr}\exp(-\beta H). $$ that is equivalent to $$ Z=\sum_n \langle n|\exp(-\beta H)|n\rangle. $$ Assuming $H|n\rangle=E_n|n\rangle$, this is just $$ Z=\sum_n\exp(-\beta E_n). $$ Your case is the harmonic oscillator $E_n=n+\frac{1}{2}$ and the sum is just a geometric series easy to perform.

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    $\begingroup$ In general, the trace of an operator is the sum over sandwiching it between eigenstates. Thus $\text{tr} f(A)=\sum f(\lambda_n)$. $\endgroup$ Commented May 24, 2013 at 15:06
  • $\begingroup$ @Sharkos: Correct. $\endgroup$
    – Jon
    Commented May 24, 2013 at 15:08
  • $\begingroup$ Thanks, though I wasn't seeking confirmation, but merely emphasizing the underlying idea for the OP's benefit... $\endgroup$ Commented May 24, 2013 at 15:10

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