Given a propagator:

$$G(x_{f},x_{i};t)=\langle x_{f}|\exp\left[-i\hat Ht/\hbar\right]|x_{i}\rangle$$ Can we determine the eigenfunctions and eigenvalues?

I know that a propagator satisfies the TDSE, but how do we get the eigenfunctions from that? Since we are given the propagator already, we already know the Hamiltonian in some basis, but how do we get the eigenvalues just from $G$ ?


Inserting decomposition of unity in terms of Hamiltonian eigenstates you get $$ G(x_f,x_i;t) = \sum_{n} \langle x_f | n \rangle \langle n | x_i \rangle e^{-i E_n t}, $$ where I set $\hbar = 1$. Thus if you know $G$ you can try to represent it as an infinite sum of functions with this exponential dependence on time (think of this as a sort of Fourier transformation). If you manage to do this, eigenfunctions and energies can be read out immediately by comparing with the formula above. There are some additional tricks that are useful. The first one is considering the resolvent $\hat R(z) = \frac{i}{z- \hat H}$ ($z$ is a complex parameter, $R(z)$ is an analytic operator-valued function). If you know the resolvent, you can get eigenfunctions by looking at its poles (and branch cuts in the case of continuous spectrum). This is because we have representation $$ \hat R(z) = i \sum_n \frac{| n \rangle \langle n |}{z-E_n}. $$ Derivation is analogous to the previous one. It is useful to remember that there is a simple relation between the resolvent evaluated at $z$ with positive imaginary part and the propagator. It's given by Laplace transformation: $$ \langle x_f| \hat R( z ) | x_i \rangle = \int_0^{\infty} e^{i zt} G(x_f,x_i,t) \mathrm{d}t. $$ I encourage you to try deriving this formula yourself. Try also to explain how this formula should be changed if $ \hat R(z)$ with $\mathrm{Im} \ z <0$ is required.

There is one more trick which is especially useful for extracting the ground state and other low states (these are by far the most important eigenstates!). Namely you treat time as complex variable. It is especially useful to take $t = -i \tau$ with positive $\tau$. This is called Wick's rotation. After the rotation you have $$ G(x_f, x_i ; -i \tau) = \sum_{n} \langle x_f | n \rangle \langle n | x_i \rangle e^{- E_n \tau}. $$ States with large energy are exponentially supressed for large $\tau$. In particular if the spectrum is discrete you get $$ G(x_f, x_i ; -i \tau) \approx \sum_{n} \langle x_f | 0 \rangle \langle 0 | x_i \rangle e^{- E_0 \tau} $$ for large $\tau$. Thus if you know large $\tau$ limit of the propagator, reading out the ground state and its energy is immediate. It should be clear that some higher states could be also extracted this way.

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  • $\begingroup$ This is an excellent answer. May I ask you for some references on this subject? $\endgroup$ – QuantumBrick Jun 30 at 8:51

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