Feynman, Hibbs Transition Amplitudes and Energy

  • #1
I’m currently self-studying from Feynman & Hibbs Quantum Mechanics and Path Integrals, but having trouble with a statement in the chapter on time-dependent perturbations.

Background: They define
$$V_{mn}(t_c) = \int_{-\infty}^\infty \phi_m^*(x_c)V(x_c,t_c)\phi_n(x_c)\,dx_c,$$
where [itex]V(x,t)[/itex] is the perturbation potential, and [itex]\phi_{m/n}(x_c)[/itex] are unperturbed eigenfunctions of energy [itex]E_{m/n}[/itex]. By looking at the first-order expansion for [itex]V[/itex], we see that [itex]-(i/\hbar)V_{mn}(t)\,dt[/itex] corresponds to the amplitude for state [itex]n[/itex] to transition to state [itex]m[/itex] during time interval [itex]t\rightarrow t+dt[/itex], giving (eventually)
$$P(n\rightarrow m)=\frac{|V_{mn}|^2}{(E_m-E_n)^2}\left[4\sin^2\frac{(E_m-E_n)T}{2\hbar}\right],$$
as the probability that state [itex]n[/itex] transitions to state [itex]m[/itex] during time interval [itex]T[/itex] due to a single scattering event. This is all fine.

Problem: Where I’m struggling is when they go on to talk about situations where the energy spectrum is continuous, defining [itex]\rho(E)\,dE[/itex] as the distribution of states in the range of energy [itex]E[/itex] to [itex]E+dE[/itex], and looking at the total probability of transition
$$P=\int P(n\rightarrow m)\rho(E_m)\,dE_m.$$
Assuming [itex]E_n[/itex] and [itex]E_m[/itex] to be similar, they simplify this integral by treating [itex]|V_{mn}|[/itex] and [itex]\rho(E_m)[/itex] as effectively constant, leading eventually to the claim (p151) “[...] we obtain the result that the probability for a transition to some state in the continuum is
$$P(n\rightarrow m) = \frac{2\pi}{\hbar}|V_{mn}|^2\rho(E_n)T$$
and that the energy in the final state is the same as the energy in the initial state.” (emphasis mine).

Again, the result of the integration is fine, but the statement about the energy in the two states seems to have come out of nowhere, and doesn’t seem (to me) to chime at all with the idea of the continuum of energy states. They go on to give the probability for a transition per unit time with a [itex]\delta(E_m-E_n)[/itex] term, reinforcing the above statement, but again, I can’t see where it has come from. I’ve considered the possibility that the statement is subject to approximations, but the text leading up to this point has been quite deliberate about indicating where approximations have been made.

Can anyone shine some light?
 

Answers and Replies

  • #3
Interesting. I can see in your derivation of the photoelectric effect that the delta function arises from [itex]\sin^{(2)}(Tx)/x^{(2)}[/itex] as [itex]T\rightarrow\infty[/itex]. In the text, the quoted result comes from an integral like
$$\int \frac{\sin^2[(E_m-E_n)T/2\hbar]}{(E_m-E_n)^2}\,dE_m,$$
and noting that [itex]\int_{-\infty}^\infty[(\sin^2x)/x^2]\,dx=\pi[/itex] the result follows. So I can see how this follows the form of your piece, and it suggests that the authors may have intended to take [itex]T\rightarrow\infty[/itex] as you did. It seems odd that they would have elided this, as, like I said, they have been quite careful about that sort of thing in the text leading up to this point, but there we go.
 
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