Question: If a particle is in the ground state at time t<0, use the 1st order time dependent perturbation theory to calculate the probability that the particle will still be in the ground state at time t.
Suppose we turn on the perturbation at time t=0 H(x) = ax
Suppose this particle is in an usual one dimensional infinite square well with width going from x=0 to x=a.
Then the question asks what happens to the probability of finding the particle will still be in the ground state if t tends to infinity and justify the answer? This is the bit I don't understand because somehow I have to refer this to the sum of the probabilities = 1???
The Attempt at a Solution
I get the matrix element conjugate state Fi(0) * H(x) * state Fi(0) with respect to x
and I get ax*1/2 for my matrix element.
Then I integrate this matrix element with respect to time from 0 to t. Then I take the mod square of the whole expression.
I get (ax*1/2*1/h bar * t)^2.
Is this correct or not?
And how do I answer the next part of the question?