# Time dependant perturbation

Homework Helper
Gold Member
Hi,

I have a problem with the very nature of time dependant perturbation theory (TDPT). In TDPT, we consider a system of Hamiltonian H(t) = H_0 (for t<0), H(t)=H_0 + kW(t) (for t>0) [where k<<1], where H_0 is, for simplicity, discrete, non-degenerate and time-independent, and given that at t=0, the state of the system is |phi_i> (an eigenstate of H_0), we are interested in calculating P_if(t), the probability of finding the system in another eigenstate of H_0, |phi_f>, at time t.

But this does not make sense because at t, the hamiltonian is no longer H_0, so it will, in general, not have |phi_f> as an eigenstates. But we know that the result of a measurement will project the wave-function into one of the eigenfunction. So as soon as |phi_f> is not an eigenstate of H(t), the probability of finding the system in |phi_f> will be 0.

"So as soon as |phi_f> is not an eigenstate of H(t), the probability of finding the system in |phi_f> will be 0"

I'm not sure if i fully understand your question, but why does it worry you that |phi_f> will not in general be an eigenstate of H(t)? If i remember my last quarter of undergrad quantum correctly we are only worried about the probability of transition between |phi_i> and |phi_f> (both of which are eigenstates of H_0) under a time dependent pertubation that takes place for some delta t. As far as the results of measurements "forcing" the wave-function's projections onto one of the eigenstates, simply think about the TDP as having already measured the system before (hence our eigenstate |phi_i>) as well as after, (hence |phi_f>) and the TDP simply spits out the probablity of this transition within the abovementioned delta t.

Homework Helper
Gold Member
"So as soon as |phi_f> is not an eigenstate of H(t), the probability of finding the system in |phi_f> will be 0"

I'm not sure if i fully understand your question, but why does it worry you that |phi_f> will not in general be an eigenstate of H(t)? If i remember my last quarter of undergrad quantum correctly we are only worried about the probability of transition between |phi_i> and |phi_f> (both of which are eigenstates of H_0) under a time dependent pertubation that takes place for some delta t. As far as the results of measurements "forcing" the wave-function's projections onto one of the eigenstates, simply think about the TDP as having already measured the system before (hence our eigenstate |phi_i>) as well as after, (hence |phi_f>) and the TDP simply spits out the probablity of this transition within the abovementioned delta t.

Like you said, we have measured the system before the perturbation to be |phi_i>. Then the TDP is turned on and the wave function evolves according to the time dependant SE

$$i\hbar\partial_t \Psi(t)=H(t)\Psi(t)$$

with initial condition Psi(0)=phi_i. Now say we measure the energy at a time t. The wave function Psi(t) will collapse into an eigenstate of H(t). It is most likely that phi_f will not be such an eigenstate for H(t), so let's suppose for simplicity that it isn't.

Sure, mathematically, nothing stops us from expanding Psi(t) in a Fourier series in |phi_n> and we can even calculate the coefficient |<Psi(t)|phi_k>|², but it does not represent the probability of finding the system in state |phi_k>. According to the expansion postulate, this probability is 0, because phi_k is not an eigenstate of H(t).

But the books say that the probability is |<Psi(t)|phi_k>|², so this is where I'm confused, and I'm asking "where in the above am I mistaken"?

Last edited:
vanesch
Staff Emeritus