Imaginary pertubation to a Hamiltonian: how is it the same as rotation to imaginary time?

Question

I am struggling with the following affirmation found in Ryder's QFT book, page 177:

instead of rotating the time axis as we have done, the ground state contribution may be isolated by adding a small negative imaginary part to the Hamiltonian

The author refers to an effort to isolate the vacuum state in a sum over energy eigenstates: $$\langle Q|e^{-i (T-t) H}|q\rangle = \sum \phi_n(q) \phi^{*}_n(Q)\; e^{-i (T-t) E_n }$$ One option is to make time imaginary: $T \rightarrow \infty e^{- i \epsilon}$. Another, says the author, is to change the Hamiltonian by adding $-\frac{1}{2} i \epsilon q^2$. We would have then: $H^{\epsilon} = H -\frac{1}{2} i \epsilon q^2$, and:

$$\langle Q|e^{-i (T-t) H^{\epsilon}}|q\rangle = \sum \phi_n(q) \phi^{*}_n(Q)\; e^{-i (T-t) E^{\epsilon}_n }$$

I´m guessing that you could treat this as a time independent perturbation, so that the first correction to energy is (lets call the new eigenvalues $E_n^{\epsilon}$):

$$E_n^{\epsilon} = E_n -\frac{1}{2} i \epsilon \langle E_n|q^2|E_n\rangle + ...$$

That makes the new eigenvalues imaginary, but that’s not enough. What we need to have the sum dominated by the ground state is for $Im[E_n^{\epsilon}]$ to be proportional to $E_n$, so that we have a $E_n$ factor in the non-oscillatory part of the exponential.

That means $\langle E_n|q^2|E_n\rangle$ should be proportional to $E_n$. It is true for an harmonic oscilator but, can we say that in general?

Answer 1

This is not a good substitute for rotating to imaginary time. The right way to do it is to add $i\epsilon H$ to H, not $i\epsilon q^2$. To see this, normalize things so that the vacuum is not decaying:

$$ H - i(\epsilon q^2 - A)$$

Where A is the vacuum expectation value of $q^2$. Now consider a potential where some excited state has a lower value of $q^2$ than the ground state. To find such a potential, you can use the semiclassical method of computing averages described in this answer: Do stationary states with higher energy necessarily have higher position-momentum uncertainty?

The upshot of the semiclassical method is that the average of $q^2$ is the time-average of q^2 in the classical orbit to leading semiclassical order. To make a potential which has a small average q^2, you can consider a potential with deep minima at q=-A and q=A, and a shallow minimum at q=0. In the ground state, the particle is a superposition of q=A and q=-A, while in one the excited state, the shallow minimum has a big peak (since the particle is moving slowly there) and the q=-A and q=A minima have a small oscillatory wavefunction from tunneling.

With this addition to H, the excited state will not be suppressed but growing. The asymptotic value at large imaginary times will be dominated by this excited state rather than by the ground state.