path integral about simple harmonic oscillator

Question

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

Comments

I am not sure that (2) is correct.

Anyway, many years ago we considered a similar problem, but from the perturbative approach - in order to estimate the ground state properties. I gave a brief explanation on my blog (in Russian). You may find it useful for your purposes.

https://fishers-in-the-snow.blogspot.com/2011/03/blog-post_09.html

https://www.academia.edu/6145636/Journal_of_Nuclear_Physics_1985

P.S. In ordet to write down functions, use a backslash \ in front of them: $sin(x)$ -> $\sin(x)$, $exp(x)$ -> \exp(x).

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As Feynman explains in his thesis, his method is applicable to systems whose classical analogue is described by the least action principle. I suspect that you have to use the case of a forced harmonic oscillator in order to describe it using least action.

Answer 1

Srednicki explains in his book that if you multiply H by (1 - iε) you can get a formula for the time-ordered generating functional