Time-dependent perturbation theory

[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.

 

[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.

 

[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.

 

[Niles]

But the state is simply not an energy eigenstate (which is ok).

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?

 

[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.

 

[Niles]

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!

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!

 

[xepma]

Ok, so the above oscillations mean that the true (perturbed) eigenfunctions are a superposition of the old (unperturbed) eigenfunctions?

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).

And also that these new, perturbed eigenfunctions change over time?

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.

 

[Niles]

Thank you very much. Your answers cleared things up for me.

Have a nice weekend.

 

[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.

 

[dextercioby]

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

 

[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

[tex]\Psi(x,t_0) = 0.4\psi_1+0.6\psi_2[/tex].

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?

 

[Bob_for_short]

Let us say that at some time t0 we have the perturbed wavefunction given by the following linear combination of unperturbed eigenstates

[tex]\Psi(x,t_0) = 0.4\psi_1+0.6\psi_2[/tex].

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?

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.

 

[Niles]

Regarding normalization: Yes, my fault. Sorry.

The ratio (0.4/0.6)² is the relative population of the corresponding states (your wave function is not normalized).

I don't follow this argument. If e.g. we have [itex]\Psi(x,t_0) = \sqrt{0.05}\psi_1+\sqrt{0.95}\psi_2[/itex], 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 [itex]\psi_2[/itex]?

 

[Bob_for_short]

I don't follow this argument. If e.g. we have [itex]\Psi(x,t_0) = \sqrt{0.05}\psi_1+\sqrt{0.95}\psi_2[/itex], 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 [itex]\psi_2[/itex]?

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.

 

[Niles]

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.

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".

 

[Bob_for_short]

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".

OK, it's easy: take an atom in its ground state Ψ_in (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 Ψ_out which is a superposition of the ground and all possible excited states allowed with the energy conservation law: Ψ_out = ∑ C_nψ_n. The excited atoms radiate. Observing the spectral lines you note that they belong to different final states ψ_n. So Ψ_out 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 ψ_n.

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.

 

[xepma]

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

[tex]\Psi(x,t_0) = 0.4\psi_1+0.6\psi_2[/tex].

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?

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.

 

[Niles]

You're approaching it too practical.

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.

 

[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.

 

[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 [itex]
\Psi(x,t_0) = \sqrt{0.05}\psi_1+\sqrt{0.95}\psi_2
[/itex] 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?

 

[Bob_for_short]

I must admit, the analogy with the interference pattern does not help me much. I want to keep this practical, so I will use [itex]
\Psi(x,t_0) = \sqrt{0.05}\psi_1+\sqrt{0.95}\psi_2
[/itex] 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?

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.

 

[Niles]

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.

I do not wish to be rude, but I do not think you answered my question. Ok, look at this wavefunction: [itex]

\Psi(x,t_0) = \sqrt{0.05}\psi_1+\sqrt{0.95}\psi_2

[/itex]. 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 [itex]\psi_2[/itex]. Now, this [itex]\psi_2[/itex] is the eigenstate of the unperturbed Hamiltonian, but we are now dealing with the perturbed Hamiltonian, so [itex]\psi_2[/itex] is not an eigenstate anymore.

So are 95% of the quantum states in the unperturbed state [itex]\psi_2[/itex] or if not, which state are the 95% quantum states in?

 

[Bob_for_short]

I do not wish to be rude, but I do not think you answered my question. Ok, look at this wavefunction: [itex]

\Psi(x,t_0) = \sqrt{0.05}\psi_1+\sqrt{0.95}\psi_2

[/itex]. 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 [itex]\psi_2[/itex]. Now, this [itex]\psi_2[/itex] is the eigenstate of the unperturbed Hamiltonian, but we are now dealing with the perturbed Hamiltonian, so [itex]\psi_2[/itex] is not an eigenstate anymore.

So are 95% of the quantum states in the unperturbed state [itex]\psi_2[/itex] or if not, which state are the 95% quantum states in?

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

 

[Niles]

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

But [itex]\psi_2[/itex] is not an eigenstate anymore, so which energy is associated with it?

 

[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.

 

[Niles]

Yes, it is still an eigenstate of the old Hamiltonian.

But it is not an eigenstate of the new Hamiltonian, and that is the one we are interested in?

It is an average E which replaces the initial E₁ now.

So this means that the unperturbed state [itex]\psi_2[/itex] is a superposition of the perturbed states, and that the unperturbed energy E₂ is some average of the perturbed energies?

 

[Bob_for_short]

But it is not an eigenstate of the new Hamiltonian, and that is the one we are interested in?
No, it is not. Generally a new Hamiltonian may have its own eigenstates or may not if time dependent. It's OK since after switching off the external filed we return to the original Hamiltonian. The final state will be a superposition of its eigenstates, not a single eigenstate.
The average energy is calculated as sum of all measured energies divided by the number of measurements. It results into E_av. = ∑ |C_n|²E_n where C_n are the superposition coefficients.

Look at this in the following way (it is a strict way). The old Hamiltonian has its eigenstates that can serve as a basis in a space of functions. Initially you prepare your physical system in a single state Ψ(t=0) = ψ₁. In other words, the initial state Ψ is along one axis in this basis. Then the external force takes the system state and starts to turn in the space of functions. The vector Ψ(t) now is not along the initial basis vector but "looks" at somewhere else (it has another direction). Of course it has now non-zero projections on the other basis vectors. The projections are called amplitudes and their squares are probabilities to find such a state while measuring. (Your measuring device is capable of measuring the old basis vectors.) You use the old, know basis to learn the state Ψ evolution.

 

[Niles]

The projections are called amplitudes and their squares are probabilities to find such a state while measuring. (Your measuring device is capable of measuring the old basis vectors.) You use the old, know basis to learn the state Ψ evolution.

Ok, this is my last question (I think I almost understand it now). If our measuring device measures that the quantum system at some time t₀ is in the (unperturbed) state Ψ₁ with energy E₁, then the next instant later t₂ the quantum system will not be in (the unperturbed state) Ψ₁, since it is not a stationary state?

 

[Bob_for_short]

It depends on the external force. The system may stay in the same state if there is no external force (perturbation). In case of some perturbation the probability to find the system intact decreases. You still may find it in the same state but with less chance. In particular if during t₂ - t₁ the perturbation persists, the total wave function evolves - it is not stationary.

 

[Niles]

It depends on the external force. The system may stay in the same state if there is no external force (perturbation). In case of some perturbation the probability to find the system intact decreases. You still may find it in the same state but with less chance. In particular if during t₂ - t₁ the perturbation persists, the total wave function evolves - it is not stationary.

Thanks. This is definitely my last question (you guys have been patient): Why is it always that is says a small time-dependent perturbation?

 

[Bob_for_short]

...Why is it always that is says a small time-dependent perturbation?

Actually the external force or field can be strong, not small. The time-dependent solution exists and can still be represented as a superposition of old basis vectors with time-dependent coefficients.

In case of small perturbation these exact coefficients are well approximated with the perturbation series (a la Taylor series). The perturbation theory technique is developing these coefficients in Taylor series in powers of perturbation. Thus smallness of V(t) guarantees good accuracy of a truncated series (practical convergence).