1
$\begingroup$

I’ve encountered the following arguments multiple times in notes and textbooks:

$$ \lim_{t\to -\infty} || U(t)^\dagger U_0(t) |\phi_{\text{in}}(0)\rangle - |\Psi(0)\rangle || = 0 \implies |\Psi(0)\rangle = \Omega_+ |\phi_{\text{in}}(0)\rangle $$

$$ \lim_{t\to \infty} || U(t)^\dagger U_0(t) |\phi_{\text{out}}(0)\rangle - |\Psi(0)\rangle || = 0 \implies |\Psi(0)\rangle = \Omega_- |\phi_{\text{out}}(0)\rangle $$

Here:

  • $U(t)$ is the time evolution operator for the full interacting system.
  • $U_0(t)$ is the time evolution operator for the free (non-interacting) system.
  • $|\phi_{\text{in}}(0)\rangle$ is the "in" state at $t = 0$, representing the initial state of a particle in the free theory.
  • $|\Psi(0)\rangle$ is the interacting state at $t = 0$, representing the actual state of the system.
  • $\Omega_-$ is the Møller operator.

While I agree with this reasoning, there is something puzzling when computing the transition rate:

$$ \langle \phi_{\text{out}} | \Omega_-^\dagger \Omega_+ | \phi_{\text{in}} \rangle $$

From the above arguments, this expression should be equivalent to:

$$ \langle \Psi(0) | \Psi(0) \rangle $$

However, shouldn’t that always equal unity due to normalization ($\langle \Psi | \Psi \rangle = 1$)? This seems contradictory when interpreting it as a transition rate. How does one resolve this?

$\endgroup$
2
  • $\begingroup$ Please give at least one explicit reference (author, title, chapter, page and equation numbers). $\endgroup$ Commented Sep 14 at 16:20
  • $\begingroup$ First answer here for example $\endgroup$
    – MrEpsilon
    Commented Sep 14 at 17:21

1 Answer 1

0
$\begingroup$

If you want to compute the transition rate (or scattering matrix) for an arbitrary initial state, described by the in-asymptote $|\phi \rangle $ to transition into an arbitrary final state, described by the out-asymptote $|\chi \rangle$, you need to use the Møller operators individually on both states to relate them to $t=0$ and take their inner product.

The actual state at $t=0$ corresponding to the in-asymptotic state is given by
$$ |\phi +\rangle = \Omega_+ |\phi \rangle,$$ and similarly for the out-asymptotic state is $$ |\chi -\rangle = \Omega_- |\chi \rangle,$$ and so the transition probability is given by the square of the inner product $$ w(\phi \rightarrow \chi) = |\langle \chi - | \psi + \rangle|^2 = |\langle \chi | \Omega^\dagger_- \Omega_+ |\phi \rangle|^2,$$ where $S= \Omega^\dagger_-\Omega_+$ is defined as the scattering operator.

Ref: Scattering Theory: The Quantum Theory of Nonrelativistic Collisions by John R. Taylor, page 34-35.

$\endgroup$
3
  • $\begingroup$ If both $| \chi- \rangle$ and $|\phi + \rangle $ are the actual state at t=0, then why isn't their inner product is 1? Isn't the same actual normalized state ? $\endgroup$
    – MrEpsilon
    Commented Sep 14 at 17:19
  • $\begingroup$ If your in-asymptotic state $|\phi \rangle $, and out-asymptotic state $|\chi \rangle $ are such that $|\phi + \rangle = |\chi - \rangle $ (i.e. if $\Omega_{+} |\phi \rangle = \Omega_{-} |\chi \rangle $) then the transition will be one. $\endgroup$ Commented Sep 14 at 19:58
  • $\begingroup$ What you actually prepare/measure of course are the asymptotic states, where in general $|\phi + \rangle = |\chi - \rangle $ wont hold. $\endgroup$ Commented Sep 14 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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