Time-dependant perturbation theory & transitions

[elemental09]

Time-dependant perturbation theory & "transitions"

I'm studying approximation methods, and something is really bothering me about the standard treatment of time-dependant perturbation theory.

In lecture, the prof introduced time-dependant perturbation theory with the following motivation: suppose we have a system with an initially time-independent Hamiltonian, initially in some eigenstate |i> of said Hamiltonian. At some time time t, a perturbation is turned on, which may be static or time-dependant itself, e.g. some fixed jump in the potential at time t, or a classical EM field applied at time t0. The question we ask is: what is the probability of finding the system at time t in another eigenstate |f> of the initial, unperturbed time-independant Hamiltonian? In other words, what is the probability of the system undergoing a transition from i to f?
He then goes on the develop the standard iterative method of determining said probability, involving the mag-squared inner product of |f> with the new, time-dependent state psi(t):

My problem is with the very question being asked. It seems to me ill-posed. Once the Hamiltonian is changed by the perturbation, unless the perturbation is again zero after some time, then "the probability of finding the system in state |f>", where |f> is an eigenstate of the original Hamiltonian, is (in general) zero.
My reasoning is: once the Hamiltonian is perturbed, its eigenstates change with time, and the state |f> is generally no longer an eigenstate. So a measurement of energy at time t would not collapse the system into state |f>, but rather some eigenstate of the new Hamiltonian at time t. Further, even if by chance |f> is an eigenstate at some time t, the measured energy eigenvalue could be different, if for example the perturbation at that time t was a nonzero constant.

I have consulted three standard texts: Shankar, Sakurai and Desai, but each of them states the question in essentially the same way. Finally, after turning to a great (but for some reason non-standard) book by Ballentine, I find some discussion of this:

When problems of this sort are discussed formally, it is common to speak
of the perturbation as causing transitions between the eigenstates H0. If
this means only that the system has absorbed from the perturbing field (or
emitted to it) the energy difference
ωfi = εf − εi, and so has changed
its energy, there is no harm in such language. But if the statement is
interpreted to mean that the state has changed from its initial value of
|Ψ(0) = |i to a final value of |Ψ(T ) = |f, then it is incorrect. The perturbation leads to a final state |Ψ(t), for t ≥ T , that is of the form
(12.46) with an(t) replaced by an(T ). It is not a stationary state, but
rather it is a coherent superposition of eigenstates of H0. The interference
between the terms in (12.46) is detectable, though of course it has no effect
on the probability |af (T )|2 for the final energy to be E = εf. The spinflip
neutron interference experiments of Badurek et al . (1983), which were
discussed in Sec. 12.4, provide a very clear demonstration that the effect of
a time-dependent perturbation is to produce a nonstationary state, rather
than to cause a jump from one stationary state to another. The ambiguity
of the informal language lies in its confusion between the two statements,
“the energy is εf ” and “the state is |f”. If the state vector |Ψ is of the
form (12.46) it is correct to say that the probability of the energy being
εf is |af |2. In the formal notation this becomes Prob(E = εf |Ψ) = |af |2,
which is a correct formula of quantum theory. But it is nonsense to speak
of the probability of the state being |f when in fact the state is |Ψ.

(Leslie E. Ballentine, Quantum Mechanics: A Modern Development (1998), pp. 351-352

This basically confirms my suspicions. It's neat to find out that the distinction has been made experimentally. However, I still have a problem: Ballentine addresses the case where the perturbation is zero outside a closed interval in time. In other texts, and in my course, we are also looking at the case where the perturbation remains in place for all time after t0. Certainly the system may then evolve to have a nonzero component along the |f> subspace, but the fact remains that a measurement of the energy will generally collapse the system to |f>, nor will it return an energy eigenvalue the same as |f>'s, in the context of H0.

Sorry if that was long or hard to follow. Let me know if I need to clarify.

Any thoughts?

 

[SpectraCat]

Well, I think the issue is with the approximations that are being used not being stated clearly (or perhaps with you reading over them without understanding their significance). Somewhere in there it probably says that you are assuming the perturbation is small with respect to the energy separation between the eigenstates of the zero-order Hamiltonian. In such cases, the eigenfunctions of the zero-order Hamiltonian remain good approximations to the true eigenfunctions of the system.

So, your point is absolutely correct, and is well-taken ... the wave-function at time t *is* a superposition of the original eigenstates. However, for small perturbations, only one of the coefficients (the one for the "final" state in the usual expression) in the expansion will be non-zero (above some appropriately low threshold value).

 

[elemental09]

Thanks for the reply.

Indeed, the perturbation is implied to be small, so I guess it's reasonable to assume the
perturbed, time-dependant eigenstates won't stray too far from their original form. Your next point is less clear to me. Why would only only one coefficient be appreciably nonzero? I suppose I haven't studied this well enough, but it seems possible that the state vector could evolve at some time to, say c1|i> + c2|f> where c1 and c2 are both not close to zero.

I wonder, does this effect of the eigenstates and eigenenergies of the Hamiltonian changing with an applied time-dependant perturbation manifest itself as some sort of broadening effect in the context of spectroscopy? So, for example, would the spectral lines coming from a stimulated process (maybe lasing) be broader than those corresponding to spontaneous emission, holding all other broadening factors constant?

 

[SpectraCat]

Thanks for the reply.

Indeed, the perturbation is implied to be small, so I guess it's reasonable to assume the
perturbed, time-dependant eigenstates won't stray too far from their original form. Your next point is less clear to me. Why would only only one coefficient be appreciably nonzero?

Because that condition is mathematically identical to saying that the eigenstates of the zero-order Hamiltonian remain good approximations to the true eigenstates in the presence of the perturbation.

I suppose I haven't studied this well enough, but it seems possible that the state vector could evolve at some time to, say c1|i> + c2|f> where c1 and c2 are both not close to zero.

I wonder, does this effect of the eigenstates and eigenenergies of the Hamiltonian changing with an applied time-dependant perturbation manifest itself as some sort of broadening effect in the context of spectroscopy? So, for example, would the spectral lines coming from a stimulated process (maybe lasing) be broader than those corresponding to spontaneous emission, holding all other broadening factors constant?

Yes, these factors can contribute to the *homogeneous* broadening of spectral lines. Spontaneous and stimulated emission have different rates, implying different lifetimes of the excited states for the two processes. It is the lifetime of the excited state that determines the homogeneous linewidth of a transition. Of course, there may be other factors contributing to the linewidth ... for example the lines could be heterogeneously broadened, or there could be another factor (such as collisions) which shorten the excited state lifetime relative to the expected value for stimulated or spontaneous emission.

 

[DrDu]

Because that condition is mathematically identical to saying that the eigenstates of the zero-order Hamiltonian remain good approximations to the true eigenstates in the presence of the perturbation.

That's indeed a very delicate point. Usually, the perturbation converts discrete eigenvalues into a continuous spectrum. There exist some mathematical tricks (e.g. complex scaling, absorbing complex potentials) to render the problem at least mathematically reasonable. However, personally, I think it makes much more sense to view time dependent perturbation theory as a problem of scattering theory, i.e. the determination of resonance widths.

 

[A. Neumaier]

That's indeed a very delicate point. Usually, the perturbation converts discrete eigenvalues into a continuous spectrum. There exist some mathematical tricks (e.g. complex scaling, absorbing complex potentials) to render the problem at least mathematically reasonable..

It depends on the situation. Discrete eigenstates will remain discrete under small perturbations except when they are embedded into the continuum, in which case they will turn into a resonance.

 

[elemental09]

Because that condition is mathematically identical to saying that the eigenstates of the zero-order Hamiltonian remain good approximations to the true eigenstates in the presence of the perturbation.

Sorry it's taken me so long to reply to this.

I still fail to understand why this is the case. I understand that time-dependent perturbation theory normally fails unless the transition probability being calculated is << 1, but I don't see why only one coefficient (I presume you mean only one coefficient, excluding that of the initial state) on the excited states can be appreciable. What's more, I don't understand the mathematical equivalence to the statement of the eigenstates of H_0 remaining good approximations.

If it is something obvious I'm missing, do please explain. If a demonstration of the equivalence is more involved, I don't wish to take up your time with this. It's likely things will become clearer as I continue to study. The next two weeks are dedicated to studying for my quantum theory final, so that's the only thing on my mind right now :)

 

[DrDu]

I read your initial question again. I think the argumentation is the following: In deed, in principle you could consider the transition probabilities to the adiabatic eigenstates $|\tilde{f}(t)\rangle$ of H(t).
However, in zeroth order in the perturbation, these coincide with the unperturbed eigenstates $|f \rangle$ of $H_0$. As the perturbation has been taken into account already in the formulation of the perturbed eigenstate $\psi(t)$, this difference between the final states does not appear in lowest order.

 

[DrDu]

You are completely right with your assumption that this procedure is not always correct!
In the case of the sudden potential jump, the initial state does not change with time, i.e.
$|\psi(t)\rangle =|i>=\mathrm{const.}$. Hence you must use the new adiabatic final states $|\tilde{f}\rangle$ instead of the old ones $|f\rangle$ to describe the transition.

The procedure you were criticising is only correct if the perturbation is switched on infinitely slowly and you are considering transitions to a (quasi-) continuum of final states. You then arrive at Fermi's golden rule which can also be arrived at using different formalisms.