# Probability Amplitude and Time Evolution Operator

1. Feb 9, 2009

### erkokite

Hello,

I am fairly new to quantum physics (I'm actually an engineer, not a physicist). I think I am getting a decent grasp on things, but I have a question.

Suppose you have two time dependent states: $$\Psi_{1}$$ and $$\Psi_{2}$$.

Also, suppose that we have a constant potential, represented in our Hamiltonian as V.

Our Hamiltonian is thus represented as: $$\hat{H}=1/(2m)\partial^2/x^2+V(x)$$

Now suppose that we want to find the probability amplitude to go from state 1 to the 2nd state in a time interval t1 to t2.

It is my understanding that the following operation is used:

$$<\Psi_{2}|exp(-i\hat{H}(t2-t1))|\Psi_{1}>$$

This is of course equal to

$$\int\Psi_{2}*exp(-i\hat{H}(t2-t1))\Psi_{1}dx$$

However, how do I evaluate the following operation?

$$exp(\hat{H})\Psi_{1}$$

Many thanks.

2. Feb 10, 2009

### nicksauce

The general idea of evaluating a function of an operator is to expand the function...

exp(H) = 1 + H + H^2/2 + H^3/6 + ...

If Psi is an eigenfunction of H with eigenvalue E, then it is not hard to show that exp(H) Psi = exp(E) Psi.