Firstly, we define the imaginary-time propagator $U(\tau):=e^{-H\tau/\hbar}.$ Of course, we can write the

related matrix element $\langle x^{\prime}|U(\tau)|x\rangle:=U(x^{\prime},x;\tau)=\Sigma_n\psi_n(x^{\prime})\psi^*(x)e^{-E_n\tau/\hbar},(1).$

For the simple harmonic oscillator, we can easily write down the matrix element

$U(x,x^{\prime};\tau)=\sqrt{\frac{m\omega}{2\pi\hbar sinh\omega\tau}}exp[-\frac{m\omega}{2\hbar sinh\omega\tau}((x^2+x^{\prime2})cosh\omega\tau-2xx^{\prime})],(2).$

Absolutely, U(τ) is Hermitian, and the dependence on energies in equation (1) is a decaying and

not oscillating exponential. As a result, upon acting on any initial state for a long time, U(τ) kills all

but its ground state projection: $lim_{\tau\to\infty}\bar{U(\tau)|\psi(0)\rangle}\to|\bar{0\rangle\langle0|\psi(0)\rangle}e^{-E_0\tau/\hbar}.$

This means that in order to find $|0\rangle$ ,we can take any state and hit it with U(τ→∞). In the case that

$|\psi(0)\rangle$ has no overlap with $|0\rangle,|1\rangle,\ldots,|n_0\rangle$, U will asymptotically project along $|n_0+1\rangle.$

My issue is that as $\tau \to \infty$, U( $x^{\prime }$, x; $\tau$) in equation ( 2) seems to be zero. And as I calculate this matrix

element$U(x^\prime,x;\tau)=\sum_n\psi_n(x^{\prime})\psi_n^*(x)e^{-E_n\tau/\hbar}$originally by using the fundamental QM

knowledge $\langle x|0\rangle=(\frac{m\omega}{\pi\hbar})^{1/4}exp(-\frac{m\omega x^2}{2\hbar})$ and the ground state energy $E_0=\frac12\hbar\omega$ directly, I

cannot get zero correctly. What is the problem?

Also, if I know equation (2) in advance but don't know the ground state wave function and ground

state energy of the oscillator, how can I deduce the ground state wave function and energy from

equation (2).

And as I set x= x' = o in equation (2), does it mean that the energy satisfies the relation

$e^{- E_n\tau / \hbar }= \sqrt {\frac {m\omega }{2\pi \hbar sinh\omega \tau }}$ ?