# I Energy levels from the propagator

#### PrashantGokaraju

I would like to get some information on this topic. It is not discussed in many places, so if any members here know about it, i would be interested in a brief explanation. Or any books or online documents where it is discussed.

D is the "invariant propagation function" or the "propagator". I have read somewhere that

Tr D = 1/(E - E1) + 1/(E - E2) + ...

So the energy levels are poles in the trace of the propagator.

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#### vanhees71

Gold Member
I assume the Hamiltonian is not explicitly time-dependent. Then the propgator
$$D(t-t',x,x')=\langle x|\exp[-\mathrm{i} \hat{H} (t-t')]|x' \rangle.$$
It fullfills the Schrödinger equation,
$$\mathrm{i} \partial_t D(t-t',x-x')=\hat{H} D(t-t',x-x')$$
with the initial condition
$$D(0^+,x,x')=\delta(x-x').$$
Usually you need the retarded propgator (for non-relativistic QM!), i.e., you assume $D(t-t',x,x') \propto \Theta(t-t')$.

Now insert a complete set of energy eigenstates,
$$D(t-t',x,x')=\sum_{n} \langle x|u_n \rangle \langle u_n \exp[-\mathrm{i} \hat{H} (t-t')]|x' \rangle = \sum_n u_n(x) u_n^*(x') \exp[-\mathrm{i} E_n(t-t')].$$
Now take the formal Fourier transform wrt. $t$. To make it consistent with the retardation condition you have to set $E_n=E_n-\mathrm{i}0^+$. Then you get
$$\tilde{D}(E,x,x')=\sum_n u_n(x) u_n^*(x') \frac{1}{E-E_n+\mathrm{i} 0^+}.$$
Now setting $x=x'$ and integrating over $x$, using that $\langle u_n|u_{n} \rangle=1$ to your formula (with the little addition concerning how to treat the poles!):
$$\int_{\mathbb{R}} \mathrm{d} x \tilde{D}(E,x,x)=\sum_n \frac{1}{E-E_n+\mathrm{i}0^+}.$$
Formally you can think of the integral as a trace over the retarded operator in energy representation.

"Energy levels from the propagator"

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