Time-dependent Perturbation

[kakarukeys]

Given a system,
$H = H_0 + V$
V is a small perturbation that does not depend on time.

the system is in $|E_0>$ at time $t_0$
$H_0 |E_n> = E_n |E_n>$
$H_0 |E_0> = E_0 |E_0>$

Let $$|\Psi(t)>$$ be the solution of the system.
Let $$|\Phi(t)>$$ be the solution of the system without perturbation.
Let $$|u(t)> = |\Psi(t)> - |\Phi(t)>$$.

Show that $$|<E_n|u(t)>|^2 = 4 |V_{n0}|^2 [{{\sin(t - t_0)(E_n - E_0)/\hbar}\over{E_n - E_0}}]^2$$

at lowest order
No matter how many times I try, the answer I get is

$$|<E_n|u(t)>|^2 = 2 |V_{n0}|^2 [{{1 - \cos(t - t_0)(E_n - E_0)/\hbar}\over{E_n - E_0}}]^2$$

Please help!

 

[quantumworld]

I think you got the right answer, but u don't have the correct question, it should be show that ...sin(.../2), there is a divide by 2 missing in the sine function. I looked it up in my quantum book.

 

[kakarukeys]

Sorry,
I typed wrongly,

My answer was
$$|<E_n|u(t)>|^2 = 2 |V_{n0}|^2 [{{1 - \cos(t - t_0)(E_n - E_0)/\hbar}\over{(E_n - E_0)^2}}]$$

Ya, you are brilliantly right,

Since,
$$2\sin^2\theta/2 = 1 - \cos\theta$$

There should be a "divided by 2" inside the Sine

 

[koshyninan]

How can i get certain solved papers in quantum mechanics?
thanks