Fermi's Golden Rule: Derivation with Discrete & Continuous Spectrum

[giova7_89]

When I read about the derivation of this formula from Cohen Tannoudji or similar books, I couldn't understand one thing. He starts off assuming that the unperturbed Hamiltonian has a non-degenerate discrete spectrum, writes down the formulas for the evolution of an initial state using the resolution of the identity corresponding to this spectrum, and then assumes that the initial state is an eigenvector of the unperturbed Hamiltonian. After all of this he calculates the probability amplitude (at first order) that the evolved state at time t is found in another eigenstate of the unperturbed Hamiltonian (different from the original one he started from). So far everything's ok. But the he says that he wants to calculate the probability amplitude for the initial state to be found in a continuum (I hope my expression is correct) of eigenstates of the original hamiltonian and uses exactly the same formulas for the evolution of the initial state! That is, the resolution of the identity he uses is the same as before! Then my question is: if the unperturbed Hamiltonian had a continuous and discrete spectrum, shouldn't both of them enter in the resolution fo the identity? The formulas would be different this way when I calculate the evolution of an arbitrary initial state.

Long story short, why doesn't one use Dyson series and then (as it is done in presence of a simple non-degenerate discrete spectrum) insert in this series of operators a lot of resolutions of the identity that account for the case in which there's a continuous spectrum too? That would be the most general formula, IMHO...

 

[dextercioby]

When I read about the derivation of this formula from Cohen Tannoudji or similar books, I couldn't understand one thing. He starts off assuming that the unperturbed Hamiltonian has a non-degenerate discrete spectrum, writes down the formulas for the evolution of an initial state using the resolution of the identity corresponding to this spectrum, and then assumes that the initial state is an eigenvector of the unperturbed Hamiltonian. After all of this he calculates the probability amplitude (at first order) that the evolved state at time t is found in another eigenstate of the unperturbed Hamiltonian (different from the original one he started from). So far everything's ok. But the he says that he wants to calculate the probability amplitude for the initial state to be found in a continuum (I hope my expression is correct) of eigenstates of the original hamiltonian and uses exactly the same formulas for the evolution of the initial state! That is, the resolution of the identity he uses is the same as before! Then my question is: if the unperturbed Hamiltonian had a continuous and discrete spectrum, shouldn't both of them enter in the resolution fo the identity? The formulas would be different this way when I calculate the evolution of an arbitrary initial state.

Yes, but the initial state is always assumed to be an eigenstate. That's a simplifying assumption, of course, but it makes his heuristic discussion in section C, subsection 3 of chapter 13 valid.

I haven't seen a discussion of standard perturbative treatments in terms of general eigenstates for operators with mixed spectrum.

 

[giova7_89]

Ok... I tried to write down the Dyson series for an unperturbed Hamiltonian with mixed spectrum: the formulas are absurdly long but not difficult to understand. I made an initial eigenstate (belonging to the discrete spectrum) evolve, assuming that the "scalar product" between "discrete eigenstates" and "continuous eigenstates" is 0, and then I calculated the scalar product of this evolved state with an eigenstate of the continuous spectrum.

If I understood quantum mechanics (...) I evaluate the modulus squared of this complex number and get the probability density for the initial state to be in an interval of continuous eigenstates of the unperturbed Hamiltonian at time t. The problem is that (again: the crucial point seems that I assumed that the "scalar product" between "discrete eigenstates" and "continuous eigenstates" is 0) this probability density comes off as 0. So I can't find a way to obtain Fermi's golden rule with what I thought to be a more rigorous approach...

My morale is so close to the ground right now...

 

[Bill_K]

It depends on how much mathematical rigor you want to use in treating the continuum states, and people's opinion on this vary. Cohen-Tannoudji tends to use more rigor than most. In my opinion, rigor can get in the way, obscuring ideas that are otherwise simple. Most textbooks unblushingly confine their wavefunctions to "a box of side L", derive the result they want, and in the end let L → ∞. This may seem crude but is after all much closer to physical reality. In these terms the transition is from one discrete state to a group of states, which are discrete but closely spaced.

In terms of the continuum, we're not asking this time for the scalar product of a discrete one with a continuous one. We're asking for a matrix element of the perturbation between discrete and continuous states, which is nonzero, like δ(E + hω - E₀).