Firstly, we define the imaginary-time propagator U(τ):=e−Hτ/ℏ. Of course, we can write the
related matrix element ⟨x′|U(τ)|x⟩:=U(x′,x;τ)=Σnψn(x′)ψ∗(x)e−Enτ/ℏ,(1).
For the simple harmonic oscillator, we can easily write down the matrix element
U(x,x′;τ)=√mω2πℏsinhωτexp[−mω2ℏsinhωτ((x2+x′2)coshωτ−2xx′)],(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: limτ→∞¯U(τ)|ψ(0)⟩→|¯0⟩⟨0|ψ(0)⟩e−E0τ/ℏ.
This means that in order to find |0⟩ ,we can take any state and hit it with U(τ→∞). In the case that
|ψ(0)⟩ has no overlap with |0⟩,|1⟩,…,|n0⟩, U will asymptotically project along |n0+1⟩.
My issue is that as τ→∞, U( x′, x; τ) in equation ( 2) seems to be zero. And as I calculate this matrix
elementU(x′,x;τ)=∑nψn(x′)ψ∗n(x)e−Enτ/ℏoriginally by using the fundamental QM
knowledge ⟨x|0⟩=(mωπℏ)1/4exp(−mωx22ℏ) and the ground state energy E0=12ℏω 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−Enτ/ℏ=√mω2πℏsinhωτ ?