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I would like to consider the trace of the time evolution operator $e^{-\frac{i}{\hbar}\hat{H}t}$

Apparently in single-particle quantum mechanics is can be represented as

$$ tr \ e^{-\frac{i}{\hbar}\hat{H}t}= \int d^nr \left< \textbf{r}| e^{-\frac{i}{\hbar}\hat{H}t} | \textbf{r} \right>= \int d^nr K(\textbf{r},\textbf{r},t) $$

The second equality follows from the definition of a propagator but I cannot see how first equality holds.

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    $\begingroup$ That's, um, the definition of a trace. $\endgroup$
    – ACuriousMind
    Commented Mar 9, 2015 at 13:34
  • $\begingroup$ Maybe then I cannot see the reasoning behind the definition $\endgroup$
    – Trajan
    Commented Mar 9, 2015 at 13:35
  • $\begingroup$ en.wikipedia.org/wiki/Trace_(linear_algebra) $\endgroup$
    – Kyle Kanos
    Commented Mar 9, 2015 at 13:36
  • $\begingroup$ Its clear to me only in the discrete case $\endgroup$
    – Trajan
    Commented Mar 9, 2015 at 13:37

1 Answer 1

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In finite-dimensional space, given an orthonormal basis $\lvert e_i \rangle$, the trace is

$$ \mathrm{Tr}(A) := \sum_i \langle e_i \rvert A \lvert e_i \rangle $$

Informally, in infinite-dimensional spaces, the sum is replaced by an integral.

Formally, you would have to check whether the operator is trace class, and see that the integral reproduces the convergent sum over the countable basis of our separable Hilbert space, but luckily, the idea of integrating (=infinitely summing) over the uncountable position "basis" produces the correct result (if the operator is nice), so physicists seldomly bother with that.

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    $\begingroup$ It is very far from obvious. If $e^{-\tau H}$ is compact, trace class, positive (it obviously holds if $H$ is self-adjoint), and admits a $(x,y)$-continuous kernel $K(t,x,y)$ representing it, then the trace can be computed as expected $tr(e^{-\tau H})= \int K(x,x) dx$ as consequence of Mercher's theorem. Then an analytic continuation $\tau \to i t$ should extend the result the real time case... $\endgroup$ Commented Mar 9, 2015 at 13:55

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