- #1
kakarukeys
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Given a system,
[itex]H = H_0 + V[/itex]
V is a small perturbation that does not depend on time.
the system is in [itex]|E_0>[/itex] at time [itex]t_0[/itex]
[itex]H_0 |E_n> = E_n |E_n> [/itex]
[itex]H_0 |E_0> = E_0 |E_0> [/itex]
Let [tex]|\Psi(t)>[/tex] be the solution of the system.
Let [tex]|\Phi(t)>[/tex] be the solution of the system without perturbation.
Let [tex]|u(t)> = |\Psi(t)> - |\Phi(t)>[/tex].
Show that [tex]|<E_n|u(t)>|^2 = 4 |V_{n0}|^2 [{{\sin(t - t_0)(E_n - E_0)/\hbar}\over{E_n - E_0}}]^2[/tex]
at lowest order
No matter how many times I try, the answer I get is
[tex]|<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[/tex]
Please help!
[itex]H = H_0 + V[/itex]
V is a small perturbation that does not depend on time.
the system is in [itex]|E_0>[/itex] at time [itex]t_0[/itex]
[itex]H_0 |E_n> = E_n |E_n> [/itex]
[itex]H_0 |E_0> = E_0 |E_0> [/itex]
Let [tex]|\Psi(t)>[/tex] be the solution of the system.
Let [tex]|\Phi(t)>[/tex] be the solution of the system without perturbation.
Let [tex]|u(t)> = |\Psi(t)> - |\Phi(t)>[/tex].
Show that [tex]|<E_n|u(t)>|^2 = 4 |V_{n0}|^2 [{{\sin(t - t_0)(E_n - E_0)/\hbar}\over{E_n - E_0}}]^2[/tex]
at lowest order
No matter how many times I try, the answer I get is
[tex]|<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[/tex]
Please help!