Validity of the sudden approximation

1. Nov 17, 2016

spaghetti3451

1. The problem statement, all variables and given/known data

The Schrodinger 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 Schrodinger 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?

2. Relevant equations

3. 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?

2. Nov 18, 2016

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.

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