Time-dependent perturbation theory

1. Nov 20, 2009

Niles

Hi all

Please look at this link (Search for the phrase "The quantum state at each instant can be expressed as a linear combination of the eigenbasis"): http://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)

If we write the wavefunction for the perturbed system as a (time-dependent) linear combination of the solutions to the unperturbed system, then do we assume that the perturbation is so weak that it does not change the energy-levels? If not, then I do not understand why we can write the perturbed wavefunction as a linear combination of the unperturbed wavefunctions. Because the probability of finding the particle in some state A is the square of the amplitude in front of the wavefunctions for the state A, but all the states in the linear combination are the "old" unperturbed wavefunctions.

Last edited: Nov 20, 2009
2. Nov 20, 2009

xepma

Yes, the new wavefunction is a linear combination of the "old" (unperturbed) energy eigenfunctions. But the energy levels always change if you have a time-dependent Hamiltonian. So although we have an explicit linear combination for the wavefunction, it is actually not expanded in terms of eigenfunctions of the (time-dependent) Hamiltonian. You can still interpret the coefficients in front as "the amplitude that you find the particle in that state". But the state is simply not an energy eigenstate (which is ok).

However, you can imagine that the perturbation is turned on and off again (slowly). In that case you do end up with an expansion in terms of eigenfunctions of the (new) Hamiltonian.

3. Nov 20, 2009

Bob_for_short

One-time measurement does not say anything about the quantum state - you have to perform multiple measurements. Then you will obtain an average energy which does not coincide with any "old" eigenvalue. Even with constant coefficients such a superposition is a state without certain energy. An example is a coherent state of photons.

4. Nov 20, 2009

Niles

But what is the idea behind this? So we find out that the particle is in some state A, which is only an eigenstate of the unperturbed Hamiltonian. Thus we don't know anything?

5. Nov 20, 2009

xepma

Yes, you do actually. For instance, you can calculate transition rates from one state to another state. Take for example electron orbitals of an atom in the presence of an oscillating electromagnetic field (a light pulse or a laser beam). We can assume we have one electron which starts in the lowest state. At one point the light pulse is turned on (weakly). The orbitals are no longer the energy eigenfunctions. Using this technique we can calculate what the overlap of the wavefunction with the other orbitals is as a function of time. You will notice that the electron starts to oscillate between the ground state and the first excited states (depending on the energy beam though -- other orbitals can be involved as well)

You can go as far as to determine how long the laser beam has to be turned on in order for the electron to fully transfer from the ground state to the first excited state. Sounds pretty practical to me!

A more general result which follows from perturbation theory is known as Fermi's Golden rule. It tells you the transition rate from one energy eigenstate to other eigenstates -- very useful for optical purposes, but far more general than that.

6. Nov 20, 2009

Niles

Ok, so the above oscillations mean that the true (perturbed) eigenfunctions are a superposition of the old (unperturbed) eigenfunctions? And also that these new, perturbed eigenfunctions change over time?

And thank you!

7. Nov 20, 2009

xepma

Yes, that statement is actually true in general when you add a peturbation: the eigenfunctions are a complete basis of the Hilbert space. You can view the new eigenfunctions as a change of basis, so naturally this basis is expandable in terms of the old one (the Hilbert space stays the same).

Indeed they do. The point is that you do not know what the exact expression is of these time-dependent eigenfunctions. We simply can't solve the corresponding equations of motion. This is where perturbation theory comes into play.

The number of Hamiltonians for which you can determine the exact form of the eigenfunctions is remarkably small. The whole point of perturbation theory is that we can still make useful statements about the system using these solved Hamiltonians.

8. Nov 20, 2009

Niles

Have a nice weekend.

9. Nov 20, 2009

Bob_for_short

The perturbation theory gives approximate time-dependent coefficients, the exact time-dependent wave function contains exact time-dependent coefficients at the non-perturbed basis wave-functions.

Last edited: Nov 20, 2009
10. Nov 20, 2009

dextercioby

I have a question: is the perturbed (time-dependent) hamiltonian still a self-adjoint operator ?

11. Nov 20, 2009

Niles

By the way, I thought of a question: Let us say that at some time t0 we have the perturbed wavefunction given by the following linear combination of unperturbed eigenstates

$$\Psi(x,t_0) = 0.4\psi_1+0.6\psi_2$$.

In this case, what do the amplitudes 0.4² and 0.6² represent? Nothing, right? It is only in the case when the coefficients are 1 and 0 respectively that it is interesting, because then we can say with 100% which state the particle is in, correct?

12. Nov 20, 2009

Bob_for_short

No, you are wrong. The ratio (0.4/0.6)2 is the relative population of the corresponding states (your wave function is not normalized). Yes, the wave function can be a superposition of different eigenstates. The energy is not certain is such a state: there is an average energy and its dispersion. For example, after scattering atomic wave function becomes a superposition of different excited states.

13. Nov 20, 2009

Niles

Regarding normalization: Yes, my fault. Sorry.

I don't follow this argument. If e.g. we have $\Psi(x,t_0) = \sqrt{0.05}\psi_1+\sqrt{0.95}\psi_2$, and we look at one particle, then what can I say about this particle? That upon measurement, then there is 95% chance of the wavefunction collapsing to the perturbed state described by unperturbed state $\psi_2$?

14. Nov 20, 2009

Bob_for_short

No. In QM one looks at a quantum state, not at a particle.

QM does not describe "a single particle" but a "single" quantum state Ψ. Ψ can be represented as a superposition of some basis wave-functions. Different measuring devices give different measured basis states. Think of an interference pattern. It consists of many points. The screen "measures" the particle coordinates. They may be different since the wave function is not an eigenstate of coordinate operator. The whole pattern describes an electron/photon quantum state behind the double-slitted diaphragm. There is no interest in one point on the screen. It is the pattern which is interesting to predict. The wave
function does it.

One point on a screen is like one letter in a text - it does not carry much information. We are interested in the whole text which is an ensemble of letters. This gives the description of a quantum state.

15. Nov 20, 2009

Niles

I see your point. But I must admit, I do not have any idea of what those probability amplitudes mean, since they "belong" to the unperturbed states. That was why I wanted to look at the single quantum state: To make things more "down to earth".

16. Nov 20, 2009

Bob_for_short

OK, it's easy: take an atom in its ground state Ψin (an eigenstate of non-perturbed Hamiltonian). When I say: "Take", it means your preparatory device supplies (or contains as a target) the atoms only in this state. You make sure by multiple measurements.

Then you scatter fast electrons off such atoms. After scattering the target atom can transfer into another state Ψout which is a superposition of the ground and all possible excited states allowed with the energy conservation law: Ψout = ∑ Cnψn. The excited atoms radiate. Observing the spectral lines you note that they belong to different final states ψn. So Ψout is not a state with certain atomic energy. It is so because the projectile energy loss is distributed between the kinetic energy of the final atom and its "internal" energetic state ψn.

Similarly, if you act with a variable electric filed on atom, its state cease to be stationary but becomes time dependent. Again it is a superposition of the basis wave functions but with time-dependent coefficients. It is natural: in a variable external field the energy is not conserved. So the energy measurements (spectral lines) give spread values (populations of certain excited eigenstates change with time). A resonant external filed may populate (pump) certain eigenstates and deplete the initial state.

Last edited: Nov 20, 2009
17. Nov 20, 2009

xepma

You're approaching it too practical. Try to look at it from a more formal point of view: we would like to able to keep track of the wavefunction: given a the state at some time, how does it evolve? Sure, we have the Schrodinger equation which dictates how states evolve, but that doesn't mean we're able to solve this equation.

The method described here shows us how we can use a different basis for our Hilbert space and describe how the system evolves with respect to that state -- even though it's not the true eigenbasis of the Hamiltonian. This is why "solved" Hamiltonians are so useful, and are not just simple pedagogical examples. Try to appreciate that statement, cause it's quite a big one ;)

Having said that: recall that in time-independent perturbation theory (present on the same wiki page you linked to) there is a treatment to solve the new eigenfunctions in terms of the old ones. So we can actually switch to the true eigenbasis at any timeslice (I'm not saying it's easy or the right approach, but it can be done, at least approximately). This only holds at that particular timeslice though.

18. Nov 20, 2009

Niles

The reason why I am looking at it this way is because there is a plot in my book showing the square modulus of the two coefficients (they oscillate). And if I choose some time t0 where P2=0.8 and P1=1-P2=0.2, then I really can't see what the physical interpretation of these probabilities is.

It might be that I am missing some point in both your replies, but I still cannot interpret the probabilities.

19. Nov 20, 2009

Bob_for_short

Look at an interference pattern. Its picture is still, the probabilities do not depend on time. Then look at the interference picture while one varies the slit widths and the distance between the slits. The picture will be varying like in a movie. The probabilities depend on time now. Any time slice gives different picture.

20. Nov 21, 2009

Niles

I must admit, the analogy with the interference pattern does not help me much. I want to keep this practical, so I will use $\Psi(x,t_0) = \sqrt{0.05}\psi_1+\sqrt{0.95}\psi_2$ again. Does this wave equation mean that 95% of the quantum states are in Ψ2, or that 95% are in the perturbed eigenstates, which can be written as Ψ2?

21. Nov 21, 2009

Bob_for_short

Yes, but the second state is not perturbed but excited. It is an eigenstate of the non-perturbed Hamiltonian. Initially its coefficient is zero. Then an external force supply some energy to the system and excited states become possible.

In case of variable external force the coefficients may depend on time.

22. Nov 21, 2009

Niles

I do not wish to be rude, but I do not think you answered my question. Ok, look at this wavefunction: $\Psi(x,t_0) = \sqrt{0.05}\psi_1+\sqrt{0.95}\psi_2$. This wavefunction is the wavefunction at a specific time t0, when the external perturbation is turned on.

Now my question is: How should I interpret the modulus squared of the probability amplitudes at the time t0? If I perform a measurement (and please neglect any errors in my terminilogy - I am only interested in (for now) understanding the interpretation of the probability amplitudes), then there is 95% chance of finding a quantum state in $\psi_2$. Now, this $\psi_2$ is the eigenstate of the unperturbed Hamiltonian, but we are now dealing with the perturbed Hamiltonian, so $\psi_2$ is not an eigenstate anymore.

So are 95% of the quantum states in the unperturbed state $\psi_2$ or if not, which state are the 95% quantum states in?

23. Nov 21, 2009

Bob_for_short

You interpretation is right. It is the standard QM interpretation. You have 95% of chances to find the system in the state $\psi_2$ and 5% in the state $\psi_1$.

24. Nov 21, 2009

Niles

But $\psi_2$ is not an eigenstate anymore, so which energy is associated with it?

25. Nov 21, 2009

Bob_for_short

Yes, it is still an eigenstate of the old Hamiltonian. It has the energy E2. If it is an atom, you see it as a spectral line with hf = E1-E2 (if the state 2 emits a photon later on). It is the total wave function which is not an eigenstate of the old Hamiltonian. It becomes a superposition of those and measurements give different energies. It is an average E which replaces the initial E1 now.