# What is the unitary matrix that diagonalizes the Hamiltonian?

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.

• This is the Dirac/interaction picture, which is explained in any decent QFT textbook, see e.g. Peskin & Schroeder, Section 4.2. Jun 18, 2018 at 8:12
• it seems you need to pick up on your interaction picture formalism. May 28, 2020 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.