Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Setup

In the typical treatment of time-dependent perturbation theory in quantum mechanics, one arrives at the set of equations

$$ i \dot{a}^{(r + 1)}_m(t) = \sum_n \langle m |H_1(t)|n \rangle e^{i \omega_{mn} t} a^{(r)}_m(t) $$ for integers $r > 1$ and $\dot{a}^{(0)}_m(t) = 0$. Here, the Hamilton is $H(t) = H_0 + \lambda H_1(t)$, the states $\{|n\rangle\}$ satisfy $H_0 |n\rangle = \varepsilon_n |n\rangle$ and $\omega_{mn} = \varepsilon_m - \varepsilon_n$. The expansion coefficients are defined by $$ a_m(t) = \sum_{r = 0}^{\infty} \lambda^r a^{(r)}_m(t) $$ where $a_m(t)$ are the coefficients in the expansion $$ |\psi(t)\rangle = \sum_n a_n(t) e^{-i \varepsilon_n t} |n\rangle $$ where $$ i \frac{d |\psi(t)\rangle}{dt} = H(t) |\psi(t)\rangle $$ Note then that $a^{(0)}_m(t) = a^{(0)}_m(0) = a^{(0)}_m$, say.

Question

The literate then claims that if $|\psi(0)\rangle$ is known, then $a^{(0)}_m = \langle m |\psi(0)\rangle$.

Does this really follow from the above? I feel an additional assumption (perhaps a boundary condition on $t$) is required to conclude the association of the initial state with the zeroth order coefficient. (Though obviously we have $a_m(0) = \langle m |\psi(0)\rangle).$

share|improve this question

1 Answer 1

The zeroth order terms in the expansion simply correspond to the solutions for the value $\lambda=0$, the unperturbed system. That's clearly the case because all the other corrections are proportional to positive powers of $\lambda$ (or other, nonperturbative functions of it) which require $\lambda\neq 0$ for them to contribute a nonzero amount.

Now, for $\lambda=0$, you may easily solve the system, and because the basis $|n\rangle$ used in all the formulae above is obviously a basis of eigenstates of $H_0$ – as seen, for example, from the variable $\epsilon_n$ used in one of the formulae (it only makes sense to attribute a single energy to a state if it is an energy eigenstate) – it follows that their evolution in time is very simple and in the notation above, it is simply $a_n(t)=a_n(0)$.

The coefficients are constant as functions of time. The evolution by Schrödinger's equation simply adds the phase $\exp(-i\epsilon_n t)$ to each amplitude but this is already incorporated in the prescription for $|\psi(t)\rangle$ which implies that $a_n(t)$ are just constant (independent of time).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.