If $H = H_0 + g H_1$ is our (free + interaction) Hamiltonian, and we assume that we have a basis of states $\{ | i \rangle \}$ under which $H_0$ is diagonal, then we may diagonalize $H$ by some unitary transformation $U$:

$$ U^{\dagger} H U = U^{\dagger} (H_0 + g H_1) U = \hat H_0$$

where $\hat H_0$ is diagonal as well. My question is, what is $U$?

I am reading a paper which claims that $U$ in this case is just $\hat U(0,\pm \infty)$ where $$ \hat U(t,t') = T\{ \exp [ -i \int_{t'}^t g \hat H_1(t) \, dt]\}$$ is the evolution operator for states in the interaction picture, $T$ is the time ordered product, and $$\hat H_1(t) = e^{iH_0 t} H_1 e^{-iH_0 t}$$ but I don't follow this claim.

  • 5
    $\begingroup$ This is the Dirac/interaction picture, which is explained in any decent QFT textbook, see e.g. Peskin & Schroeder, Section 4.2. $\endgroup$ – Qmechanic Jun 18 '18 at 8:12
  • 3
    $\begingroup$ it seems you need to pick up on your interaction picture formalism. $\endgroup$ – ZeroTheHero May 28 '20 at 2:08

The idea of the paper cited by the OP is to exploit the so-called intertwining property of the wave (or Møller) operators $\Omega^{\pm}(H,H_0)$ of scattering theory. Such operators have the property: $$H\Omega^{\pm}=\Omega^{\pm}H_0\; .$$ Therefore, given they are invertible with inverse $\Omega^{\pm}(H_0,H)$, we have that $$\Omega^{\pm}(H_0,H)H\Omega^{\pm}(H,H_0)=H_0\; .$$

However, the wave operators are not unitary in general. In addition, they do not exist if $H_0$ and $H$ have purely discrete spectrum. In fact, the wave operator $\Omega^{\pm}(A,B)$ is defined as the strong limit, when $t\to \pm \infty$, of $e^{itA}e^{-itB}$. The strong limit means that one should have $$\lim_{t\to\pm\infty}\lVert (e^{itA}e^{-itB}-\Omega^{\pm}(A,B))\psi\rVert=0$$ for any $\psi$ in the Hilbert space. However, suppose that $\psi=\psi_\lambda$ is an eigenvector of $B$ with value $\lambda$. Then the limit $t\to\pm\infty$ exists only if $\psi_\lambda$ is also an eigenvector of $A$ with the same eigenvalue. This is because if else the mutual oscillations do not cancel to give a well-defined limit. However this last condition is not realistic, since a perturbation of $H_0$ would change the spectrum.

In fact, the wave operators are usually defined with the projection on the continuous spectrum of the operator acting on the right.

Therefore, they cannot be used to define the change of basis that the OP is looking for.

  • $\begingroup$ Actually, the paper I'm talking about is this one. $\endgroup$ – Dwagg Jun 18 '18 at 12:07

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.