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

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?

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There's nothing wrong with linking to Google Books (or any particular site) here, so don't worry about that. Also, a tip: use \langle and \rangle for matrix elements etc. (If you're doing it in real LaTeX you might want to check out the braket package, but that's not available on this site.) –  David Z Jun 21 '12 at 0:34

## 1 Answer

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.

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Yes, I noticed it would be far simpler to add $i \epsilon H$ but, when the author goes on to quantizing a scalar field (page 182) –  Forever_a_Newcomer Jun 21 '12 at 15:25
(...) he uses exactly this substitution to get $Z[J]$. Of course, the potential for a free scalar field is proportional to $q^2$, but wouldn't adding the whole $i\epsilon H$ mess up the momentum integral you need to get from the hamiltonian to the lagrangean? The bothering point here is that this $i \epsilon$ factor will be the one (on page 185) dictating the path of integration to get the propagator. –  Forever_a_Newcomer Jun 21 '12 at 15:41
@Forever_a_Newcomer: For a scalar field in perturbation theory, this is equivalent to rotating to imaginary time, because adding a little bit of imaginary mass to a Harmonic oscillator does work. He is trying to justify the $i\epsilon$ prescription, but the justification is all wrong--- the right way is to continue the particle-path path-integral, the Schwinger representation, where the mass is the action per unit proper time, and making it slightly imaginary cuts off long proper times. You should take the $i\epsilon$ prescription as the definition of which propagator you use in perturbations –  Ron Maimon Jun 21 '12 at 16:53
The full continuation of adding $i\epsilon H$ doesn't mess up the momentum integral, it makes it convergent! The most extreme case is to make time evolution be $e^{-tH}$ instead of $e^{-itH}$, and this is the full imaginary time formulation which is mathematically much more well defined because the Gaussians are decaying instead of oscillating. You should work this out in QM first, then once you understand it, the QFT will be simple. –  Ron Maimon Jun 21 '12 at 16:55