Validity of the sudden approximation

[spaghetti3451]

Homework Statement

The Schrödinger equation is given by

$$i\hbar\ \frac{\partial}{\partial t}\ \mathcal{U}(t,t_{0})=H\ \mathcal{U}(t,t_{0}),$$

where $\mathcal{U}(t,t_{0})$ is the time evolution operator for evolution of some physical state $|\psi\rangle$ from $t_0$ to $t$.Rewriting time $t$ as $t=s\ T$, where $s$ is a dimensionless parameter and $T$ is a time scale, the Schrödinger equation becomes as

$$i\ \frac{\partial}{\partial s}\ \mathcal{U}(t,t_{0})=\frac{H}{\hbar/T}\ \mathcal{U}(t,t_{0})=\frac{H}{\hbar\ \Omega}\ \mathcal{U}(t,t_{0}),$$

where $\Omega \equiv 1/T$.

In the sudden approximation, $T \rightarrow 0$, which means that $\hbar\ \Omega \gg H$. 1. Are we allowed to redefine $H$ by adding or subtracting an arbitrary constant?
2. How does this introduce some overall phase factor in the state vectors?
3. Why does this imply that $\mathcal{U}(t,t_{0})\rightarrow 1$ as $t\rightarrow 0$?
4. How does this prove the validity of the sudden approximation?

Homework Equations

The Attempt at a Solution

1. I think that we are allowed to redfine $H$ by adding or subtracting an arbitrary constant, because $H=T-V$ and the potential $V$ can be redefined by adding or subtracting an arbitrary constant without changing the physical system.

What do you think?

 

[DrDu]

Yes, you are allowed to redefine the Hamiltonian by addition of a constant. Note however that H=T+V, unless you are using an unusual definition of V.
You also have to take in mind that $\hbar \Omega \gg H$ is meaningless as you are comparing an operator with a number. As H usually is unbound, the sudden approximation never holds for all states and convergence is non-uniform.