# Expanding a ket in the basis of different Hamiltonians

1. Aug 8, 2011

### JK423

I have the following very basic question, and i'd really like your help!

If we have a system, that is described by a Hamiltonian H, then we can expand the state of the system to the basis of H. And we say that, if we measure the observable H the state will collapse to one of H's eigenstates.
Is it correct to expand the state ket to the basis of a different, random Hamiltonian H'?
Remember that the system is being described by H and not H'!
If this is possible, then when measuring the energy of the system, what are we going to measure? We are going to measure the eigenvalues of H, or the eigenvalues of H'?

2. Aug 8, 2011

### dextercioby

I don't understand, if there's one system, there's one Hamiltonian. H' is who then ?

3. Aug 8, 2011

### JK423

It's a random Hamiltonian, it doesnt matter which.
Mathematically i think it's consistent to do so, if the basis of this random Hamiltonian covers up all Hillbert space. But physically, it seems weird.

For the origin of my question, check the thread 'Time-dependent perturbation theory'. It's one of the new threads.

4. Aug 8, 2011

### mikeph

I think so, as long as H' is chosen wisely...

Your (mathematically inconvenient) choice of basis will not affect the physically measurable energy eigenvalues, however. Perhaps a simple example would help make the question/answer clearer.

5. Aug 8, 2011

### JK423

Well.. the observable of the system called energy is uniquely defined. You cannot measure 2 different energies of a system (like H and H').
If we expand the state to the H' eigenstates, then a measurement would collapse the state to one of H' eigenstates?
If the answer is yes, then we are going to measure different energies! Because the eigenvalues of H and H' are different. (Remember that H is that hamiltonian that describes the system's dynamics).

Do you have such a simple example in mind?

6. Aug 8, 2011

### genneth

Your mistake is in thinking that H' has anything to do with energy. In general, for a "good" measurement operator, its states will span the physical Hilbert space and you can express anything in that basis. It's just a basis. There is only one preferred Hamiltonian operator however, and that's the one which generates time translation.

7. Aug 8, 2011

### JK423

If this is the case then please explain to me the following.
Suppose your system is described initially by the hamiltonian H0, and then you add an external time-dependent potential V(t), with the new Hamiltonian of the system being
H(t)=H0+V(t).
What textbooks do to solve this problem (usually in the context of time-dependent perturbation theory) is to expand the time-dependent ket of the system in the basis of H0`s eigenstates. And then they try to find the time-dependent coefficients (probability amplitudes), either exact solutions (its possible for 2-state systems) or perturbative.
The point is that the Hamiltonian that describes the system is H(t) and not H0.

So, what's the point of doing that if H0 has nothing to do with energy?

8. Aug 9, 2011

### nnnm4

The idea is that basically that the new eigenstates of the perturbed hamiltonian can be described by a series of eigenkets of the original unperturbed hamiltonian. However, you can't in general get the exact eigenstates of the new hamiltonian just approximations (the series). The old eigenstates do have nothing to do with the new energies (well if the perturbation is small then the old eigenenergies may be close to the new ones), but they are a CONVENIENT basis to express the new eigenstates.

9. Aug 9, 2011

### JK423

First of all, the state of the system at time t is not an eigenstate of the perturbed Hamiltonian. The time evolution of the initial unperturbed via the evolution operator doesnt give you an H's eigenstate.
So, you say that we expand on the unperturbed kets just for convenience and that the unperturbed energies have no physical significance. It seems a little weird because i'm studying Sakurai at the moment, and he gives great attention in analyzing the physical meaning of the probability amplitudes of the unperturbed kets (e.g Nuclear magnetic resonance where we have spin-flips, absorption and emission of radiaton of a 2-level system etc.). If the unperturbed kets had no physical significance, i wouldnt care extracting the physical meaning that they.. dont have!

What comes to my mind is that maybe we expand on the unperturbed eigenkets, because are going to measure the system while it's not interacting (V(t)=0 at the time of the measurement), so what we're going to measure is Ho eigenenergies/states. For example, in the case of absorption and emission of radiation of an atom: The atom interacts with the E/M field but discontinuously. The time of measurement of the atom, it's free so we are measuring the Ho observable.
You think this line of thinking is correct?

Last edited: Aug 9, 2011
10. Aug 9, 2011

### dextercioby

Yes, we can't <probe> the system while it's interacting, that's why we can compute only transition probabilities among eigenstates of the unperturbed Hamiltonian. We introduce the time-dependent interactions adiabatically, so they are 0 at t=-infinity and return to 0 at t=+infinity. From an experimental perspective, we have the measurements taken at t=+infinity. The theory gives us, to certain degrees of accuracy, the impact of the perturbation onto the initial system, and the initial system is always supposed to have a time-independent Hamiltonian in the Schrödinger picture and whose spectral problem is always assumed as being completely solved.

11. Aug 9, 2011

### mikeph

Why do you think H' are eigenstates when H' is not the Hamiltonian?

12. Aug 9, 2011

### JK423

I like this explanation, and it also seems logical. I hope it's correct as well. If anyone has any objections on this, please say.
I don't understand the question. Every hermitian operator has eigenstates and eigenvalues.

13. Aug 10, 2011

### mikeph

But why would you think these eigenstates (of H') have anything to do with the system, when H' was chosen arbitrarily by yourself?

14. Aug 10, 2011

### JK423

That was my question actually. As i said before, i didnt understand why -in the case of a time dependent perturbation- we expand the ket in the basis of the unperturbed Hamiltonian since it has nothinf to do with the system! Why bother talking about transitions between unperturbed eigenkets when they have nothing to do with the energy of the system!
The answer to this, seems to be that the measurement takes place when the system doesnt interact, and thats why we care about the unperturbed eigenkets!
I think that textbooks should elaborate a little more on this, because its not at all obvious..

15. Aug 10, 2011

### mikeph

It has everything to do with the system. For a time-dependant perturbation H = H0 + V(t), H0 is still important because it's the static part of the Hamiltonian, so if V(t) -> 0 the system will return to one of the eigenstates of H0. When you increase V(t) you're adding/removing energy, and causing the system to move between higher and lower eigenstates of H0. They are still useful. V(t) is a boost of energy which will induce transitions between eigenstates of H0.

Your initial question in this thread was different- you suggested that H' had nothing to do with H, where in fact H0 is still fundamentally linked to H.

I'm not sure that solution is correct, since in a sense, the perturbation itself IS a measurement.

16. Aug 10, 2011

### JK423

I agree that for V(t) -> 0 the system's Hamiltonian will be H0. But we're not talking about this limit. In H = H0 + V(t), V(t) can be large, not just a perturbation. In this sense, H is the Hamiltonian, and H0 is just ANOTHER hamiltonian.
As for an example, you can take the hamiltonian of a harmonic oscillator, and consider H0 to be the p^2/2m free particle hamiltonian, and V(x) to be the interaction potential. In this example, H0 has nothing to do with our system because it's just a different system, even thought for V(x)->0 we get H->H0!

I disagree. If you consider H(t) to be the hamiltonian of the system, then H0 has nothing to do with the energy of the system! The system is an open quantum system, which means that in order to know its energy you have to know what the 'enviroment' is doing as well. H0 is defined only by the system's variables, so some potential energy isnt considered. As an example you can think of two harmonic oscillators interacting with a potential energy V~x*y, where x and y are the position variables of each oscillator. The energy of each oscillator seperately cannot be defined without specifying where the other oscillator is.
In this case, H0=p2/2m+kx^2/2 is not the energy of an oscillator.

So, when the oscillator (or a general open quantum system) is interacting (V other than zero), H0 eigenvalues are not the system's energy since energy cannot be defined only by H0 without considering the interaction potential (that is the FULL hamiltonian).
That's what i mean when i say that H0 has nothing to do with the system when the perturbation is on. It's just NOT the system's energy! So there is not meaning in talking about transitions to different energy levels of H0!

Yes, maybe i didnt highlighted that connection. But for following the arguements above, as H0 is not the system's energy (that is, its not the systems Hamiltonian), H' is not the system's energy either. This'symmetry' between the two that i wanted to hightlight, none of the two represent the system's energy.

Im not sure either, but if V(t) is acting on the system, then it's energy cannot be defined..

17. Aug 10, 2011

### mikeph

Post #14 you said "in the case of a time-dependant perturbation". That is V(t), not V(x).

Until that post I've seen no discussion at all about space dependant (and, note that now in your example, it is time-independent!) perturbations. Obviously, if you completely alter the system your old Hamiltonian is now worthless. But that is not what you asked.

One final point that frustrates me, I never said H0 was the system's energy.

18. Aug 10, 2011

### f95toli

No, V(t) can't be arbitrarily large. Perturbation theory (obviously) only works for perturbations. If V(t) is large and you drive the system "hard" most analytical methods fail miserably and you start getting effects that can not be predicted by perturbation theory. There are other approaches such as Floquet theory etc, but generally speaking you need to use numerical methods. Both in theory and experimentally (using for example Rydberg atoms) you will start seeing things like multi-photons transitions via ladder states and so on. It gets messy very quickly, even if you are only using the hamiltonian of for example a weakly anharmonic oscillator.

19. Aug 10, 2011

### SpectraCat

So what? You can always expand any state in any arbitrary basis. In some cases, when the time-dependent perturbation is small, then the unperturbed eigenstates of H0 will be VERY good approximations to the actual eigenstates of the time-dependent Hamiltonian ... there will be small corrections to the energies, but the states themselves will be largely unchanged. Remember this is perturbation theory, which is an approximate mathematical treatment of the system. In cases where the perturbation is large, then you need to do more work, and the answer will be expressed as a superposition of the H0 states. However, you can still calculate the energy, or the transition probability, or any other observable using such expansions of the wavefunction. The benefit to using the basis of H0 states is that you know what they are and can work with them.

This case is no different, although it is not a time-dependent example, and you have described the situation in non-QM terms ... remember that you cannot really know the positions of the oscillators in the classical sense that you suggest. Anyway, if the perturbation is small with respect to the harmonic frequencies of the unperturbed oscillators, then the unperturbed eigenstates will remain good approximations for the states. However, if the perturbation becomes large with respect to the level spacing then the states become strongly mixed, and you will have to either solve the full Hamiltonian explicitly (if possible), or work with expansions in the basis sets of the unperturbed oscillators.

I am not sure why you are so hung up on this point ... all that is being done is a simple basis expansion in the unperturbed states. It is not correct to say that H0 has "nothing to do with the system when the perturbation is on" ... assuming that you have made an intelligent choice of H0, it should represent the unperturbed system, the eigenstates of which are likely to a (very) good choice for a basis to expand in, because they will be close to the right answers as long as the perturbation does not become very large. Furthermore, the transition probabilities between the unperturbed states ARE relevant to the perturbed system, since you can express the states of the full Hamiltonian as superpositions of unperturbed states, and calculate transition probabilities between them using the summed transition probabilities for the unperturbed kets.

Once again, remember that perturbation theory is an APPROXIMATE treatment that is designed to work best when the perturbations are small with respect to the natural energy scale of the problem (i.e. the state spacing of the unperturbed Hamiltonian). As the perturbations become larger, this approximation becomes less and less useful, although the mathematics remains valid and you can still use it if you have no other option .. this has happened to me occasionally with spectroscopic problems.

I am not sure what you are trying to say here ....

Last edited: Aug 10, 2011
20. Aug 10, 2011

### JK423

What example are you talking about? The two oscillators? If you consider both the oscillators in the Hamiltonian, you will get a time independent interaction potential. But if you 'forget' one of them, then the hamiltonian of the other will get time-dependence.

In particular you said:

If H0 is not the energy of the system. what is the point of saying that V(t) will induce transitions between eigenstates of H0 when the system's energy isn't one of the H0 eigenstates? I really dont understand this. That's the point of my original question, *why shouldnt i expand the ket to a basis of a random hamiltonian*, since that random hamiltonian as well will not have anything to do with the energy of the system, like H0 doesnt have to do wiith the energy of the system!

Please elaborate a little because im confused