# Why does $tr \ e^{-\frac{i}{h}\hat{H}t}= \int d^nr \left< \textbf{r}| e^{-\frac{i}{h}\hat{H}t} | \textbf{r} \right>$ hold?

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.

• That's, um, the definition of a trace. – ACuriousMind Mar 9 '15 at 13:34
• Maybe then I cannot see the reasoning behind the definition – Permian Mar 9 '15 at 13:35
• en.wikipedia.org/wiki/Trace_(linear_algebra) – Kyle Kanos Mar 9 '15 at 13:36
• Its clear to me only in the discrete case – Permian Mar 9 '15 at 13:37

In finite-dimensional space, given an orthonormal basis $\lvert e_i \rangle$, the trace is
$$\mathrm{Tr}(A) := \sum_i \langle e_i \rvert A \lvert e_i \rangle$$
• 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... – Valter Moretti Mar 9 '15 at 13:55