6
$\begingroup$

If a state obeys an evolution equation, we can replace $t$ by $-t$. we get another equation and it is interesting to study its solutions. If we replace $t$ by $it$ (wick rotation) we get again another equation. The Schrödinger equation becomes a thermal equation. I wonder if the rotations other than $\pi$ or $\pm \pi/2$ were studied. Do you know of any examples? I only encountered a $t + i \beta $ in the KMS condition.

$\endgroup$
0
$\begingroup$

I found that such rotations occur when we get a partial which path information. Consider the usual two holes Young experiment. Electrons are emitted by a source S at an equal distance of the holes 1 and 2. behind their screen there is (in the middle of 1 and 2) a source s of photons. when the electron passes thru the holes a photon interact with it. ccd sensors (l and r) are near each hole and register the photon. Then the electron hits the second screen at P. The electron has two choices for the holes and the photon can hit two ccds. so we have 4 possible paths each one with an associated amplitude $\phi_i(ccd) \Psi_i (P)$

To get the probability that the electron hits P we have to add things like $P_i \overline{P _j}$ where $P_i$ is a complex number associated to a given path and $\overline{P _j}$ is the conjugate associated to $P_j$ we have to add 8 terms.Due to the symmetry the sum of the first four terms reads $$(|\phi_{near}|^2 + |\phi_{far}|^2)(|\Psi_1(P)|^2 + |\Psi_2(P)|^2$$ Here $\phi_{near} = \phi_1(ccd1) = \phi_2(ccd2)$ and $\phi_{far} = \phi_1(ccd2) = \phi_2(ccd1)$

the second sum is $$(\phi_{near} \overline{ \phi_{far}} + \phi_{far} \overline{ \phi_{near}}) (\Psi_1(P) \overline{ \Psi_2(P)} + \Psi_2(P) \overline{ \Psi_1(P)})$$ we can write $\phi_{near} = r e^{i\theta}$ and $\phi_{far} = r' e^{i\theta'}$ so the second sum reads $$rr'(e^{i({\theta} - {\theta'} )} + h.c.) (\Psi_1(P) \overline{ \Psi_2(P)} + \Psi_2(P) \overline{ \Psi_1(P)})$$ This term with interferences cancels if $\theta - \theta' = \pi /2$ and if $0 \le \theta - \theta' \le \pi $ we have intermediate visibilities of the interferences.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.