Feynman, Hibbs Transition Amplitudes and Energy

In summary, the conversation is discussing an equation in Feynman & Hibbs Quantum Mechanics and Path Integrals that defines the perturbation potential V(x,t) and the unperturbed eigenfunctions \phi_{m/n}(x_c). The authors then go on to talk about continuous energy spectra and the total probability of transition between states. They simplify the integral and make a statement about the energy in the two states being the same, which seems to come out of nowhere. The conversation also references an Insights article on the photoelectric effect which sheds some light on the derivation of the formula.
  • #1
crossword.bob
11
4
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?
 
Physics news on Phys.org
  • #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.
 
  • Like
Likes vanhees71

1. What is the significance of Feynman-Hibbs transition amplitudes?

Feynman-Hibbs transition amplitudes are used to calculate the probability of a physical system transitioning from one state to another. This is important in understanding the behavior of quantum systems and predicting their future states.

2. How are Feynman-Hibbs transition amplitudes calculated?

Feynman-Hibbs transition amplitudes are calculated using the path integral formulation of quantum mechanics. This involves summing over all possible paths that a system can take in order to transition from one state to another.

3. What is the relationship between Feynman-Hibbs transition amplitudes and energy?

Feynman-Hibbs transition amplitudes are directly related to the energy of a system. In fact, they can be used to calculate the energy spectrum of a quantum system and determine the allowed energy levels.

4. Can Feynman-Hibbs transition amplitudes be applied to all physical systems?

Yes, Feynman-Hibbs transition amplitudes can be applied to all physical systems, as long as they are described by quantum mechanics. This includes atoms, molecules, and even larger systems such as solids and liquids.

5. How have Feynman-Hibbs transition amplitudes impacted the field of quantum mechanics?

Feynman-Hibbs transition amplitudes have had a significant impact on the field of quantum mechanics. They have allowed for more accurate calculations of physical quantities and have provided a deeper understanding of the behavior of quantum systems. They have also been instrumental in the development of quantum computing and other advanced technologies.

Similar threads

Replies
5
Views
1K
Replies
7
Views
955
Replies
1
Views
463
  • Quantum Physics
Replies
6
Views
1K
  • Quantum Physics
Replies
4
Views
1K
  • Quantum Physics
Replies
15
Views
2K
Replies
6
Views
822
Replies
16
Views
1K
Replies
1
Views
1K
Replies
10
Views
2K
Back
Top