Time-dependent perturbation theory

[JK423]

I'm studying Sakurai at the moment, time-dependent perturbation theory (TDPT). I'm having a problem in understanding a basic concept here.
According to Sakurai we have the following problem:
Let a system be described initially by a known hamiltonian H₀, being in one of its eigenstates |i>. Then, a time-dependent perturbation (V) is added to the system, with the total hamiltonian now being H=H₀+V. Now Sakurai asks, what is the probability of finding the system, at time t, in the energy eigenstate |n> of H₀.

Here is where my problem is.. We have a system being described by a hamiltonian H (the total hamiltonian), also being in an eigenstate of H at time t, and we want to know in which eigenstate of another hamiltonian H₀ the state of the system will collapse if we measure it! Can we do that? For example, if i have the electron of H₁ at the ground state, am i able to expand this eigenstate to the basis of another hamiltonian -like the one of a harmonic oscillator- and then say that I am going to measure in which state (and in which energy) of the harmonic oscillator the electron of the hydrogen atom is going to be??

I must have been missing something very crucial here...

 

[Dickfore]

We are not considering the system to be in an 'instantaneous eigenstate' of the full time-dependent Hamiltonian [itex]H(t)[/itex]. That would correspond to the adiabatic approximation which is valid only for very 'slowly' varying perturabations (slowly meaning that if the characteristic time is over which the perturbation changes appriciably is [itex]\tau[/itex], or for periodic perturbations, if the characteristic frequency is [itex]\tau^{-1}[/itex] and the nearest spacing of energy levels for the unperturbed hamiltonian is [itex]\Delta E[/itex], then [itex]\tau \gg \hbar/\Delta E[/itex]).

 

[JK423]

And what are we considering then? Its not in an Ho eigenstate either! What i know is that my system is described by H, whichever it's state is. Why do i expand in the basis of Ho? In a real situation, if i'd measure the energy of the system i would get a specific number. Why would that number be an eigenvalue of Ho?

 

[Dickfore]

We consider the time evolution of a general state ket [itex]|\Psi(t)\rangle[/itex] under the full hamiltonian, but we expand the general ket in the complete orthonormal basis of eigenkets of the unperturbed hamiltonian [itex]H_{0}[/itex]. Furthemore, to compensate for the time dependence [itex]e^{-i E_{n} t/\hbar}[/itex] that would be there even if there is no perturbation, we use the ansatz:

[tex]
|\Psi(t)\rangle = \sum_{n}{c_{n}(t) \, e^{-\frac{i E_{n} t}{\hbar}} \, |n^{(0)}\rangle}
[/tex]

This is called the interaction picture. The equations of motion of the coefficients [itex]c_{n}(t)[/itex] is what is derived from the wave equation.

 

[JK423]

Yes, i understand what we do. I do not understand why do what we do! Why do we expand in the basis of Ho since it doesn't describe our system (H describes it). In the same sense, i would be able to expand the state in the basis of a random Hamiltonian that i i'd like. Do we have such a freedom? When i'll measure the energy of the system, why would i find an eigenvalue of a Hamiltonian that doesn't describe the system??

 

[Dickfore]

The point is, the perturbation [itex]V(t)[/itex] is considered ''small'', hence we use perturbation theory.

 

[JK423]

The fact that V(t) is small has nothing to do wth it. This just determines a way to calculate the coefficients Cn in the state
|Ψ(t)⟩=∑nCn(t)e−iEnt|n⟩
that you wrote, and if V is small the way is via perturbation theory.
So, the fact that we are able to expand in the {|n>} basis of Ho has nothing to do with the fact that V is small. It must be something else

 

[Dickfore]

The fact that V(t) is small has nothing to do wth it. This just determines a way to calculate the coefficients Cn in the state
|Ψ(t)⟩=∑nCn(t)e−iEnt|n⟩
that you wrote, and if V is small the way is via perturbation theory.
So, the fact that we are able to expand in the {|n>} basis of Ho has nothing to do with the fact that V is small. It must be something else

Oh, but it does. The equations for the coefficients are unsolvable exactly, just like the problem with the total Hamiltonian is unsolvable exactly. Therefore, we solve these equations iteratively, in powers of the perturbation. We can only stop to the first one or two iterations just because the perturbation is small.

 

[kof9595995]

The fact that V(t) is small has nothing to do wth it. This just determines a way to calculate the coefficients Cn in the state
|Ψ(t)⟩=∑nCn(t)e−iEnt|n⟩
that you wrote, and if V is small the way is via perturbation theory.
So, the fact that we are able to expand in the {|n>} basis of Ho has nothing to do with the fact that V is small. It must be something else

A. Neumaier told me the precise condition for the expansion is "The right assumption is that H_0 and V are defined on the same dense domain of the Hilbert space, and that V is relatively compact with respect to H_0. "
see it here:https://www.physicsforums.com/showthread.php?t=462519

 

[JK423]

The fact that WE can't solve the equations exactly is not a matter of principle. For example, in 2-state problems which can be solved exactly, V(t) can also be large with perturbation theory not being applicable. But there is a solution, an exact one, and we can find it. In these problems as well, where V(t) is large, we also expand the state in the {|n>} basis of Ho, so what you suggest is not correct.
See p.320 in Sakurai for the 2-state problems.

We come back to the original question, which is:
Why do we expand the state in the basis of Ho, since Ho is NOT the hamiltonian that describes the system?

 

[Dickfore]

That is only to explain what the meaning of the "eigenkets of the unperturbed Hamiltonian being a complete basis" actually means. It gives you the ability to expand your state ket in this basis. However, it does not justify the applicability of the perturbative approximate solution.

It seems to me that you have the misunderstanding that perturbation theory gives you an exact solution? Am I right in assuming this?

 

[Dickfore]

We expand it in the basis of the unperturbed hamiltonian, because, if you rewrite the full hamiltonian as:

[tex]
H_{\lambda}(t) = H_{0} + \lambda \, V(t)
[/tex]

then these eigenkets are the correct eigenkets in the limit [itex]\lambda \rightarrow 0[/itex] and the coefficients [itex]c_{n}(\lambda, t) \rightarrow c_{n}[/itex] become time-independent.

 

[Dickfore]

in 2-state problems which can be solved exactly, V(t) can also be large with perturbation theory not being applicable. But there is a solution, an exact one, and we can find it.

Please solve this time-dependent problem exactly and post your solution here.

 

[JK423]

A. Neumaier told me the precise condition for the expansion is "The right assumption is that H_0 and V are defined on the same dense domain of the Hilbert space, and that V is relatively compact with respect to H_0. "
see it here:https://www.physicsforums.com/showthread.php?t=462519

As Dickfore said, this only justifies the fact that you can expand in that basis. My question was that, we have many available basis to expand, which one is the correct one..

We expand it in the basis of the unperturbed hamiltonian, because, if you rewrite the full hamiltonian as:

[tex]
H_{\lambda}(t) = H_{0} + \lambda \, V(t)
[/tex]

then these eigenkets are the correct eigenkets in the limit [itex]\lambda \rightarrow 0[/itex] and the coefficients [itex]c_{n}(\lambda, t) \rightarrow c_{n}[/itex] become time-independent.

Hmm, this answer seems more reasonable and logical.. I'll think about it a little more. Thanks!

Please solve this time-dependent problem exactly and post your solution here.

I answered in the previous post, check p. 320 of Sakurai for the exact solution of 2-state problems.

 

[Dickfore]

I answered in the previous post, check p. 320 of Sakurai for the exact solution of 2-state problems.

In the version I have, p.320 discusses the Zeeman effect. Could you tell us what section you are referring to.

 

[JK423]

Oh sorry, its:
Chapter 5 - Approximation methods
Section 5.6 - Time dependent perturbation theory.
He deals with 2-state problems somewhere in that section.

 

[Dickfore]

Oh sorry, its:
Chapter 5 - Approximation methods
Section 5.6 - Time dependent perturbation theory.
He deals with 2-state problems somewhere in that section.

I don't see a discussion on 2-level problems anywhere in that section. What equation are you referring to as an exact solution of the time-dependent 2-level problem?

 

[JK423]

I'm referring to "Modern Quantum Mechanics, Revised Edition".

[tex]
|\Psi(t)\rangle = \sum_{n}{c_{n}(t) \, e^{-\frac{i E_{n} t}{\hbar}} \, |n^{(0)}\rangle}
[/tex]

He calculates the coefficients Cn(t) of the above expansion to the {|n>} basis exactly, for time dependent potentials that are not necesserily weak, in contrast to what you suggest.

 

[Dickfore]

I'm referring to "Modern Quantum Mechanics, Revised Edition".

[tex]
|\Psi(t)\rangle = \sum_{n}{c_{n}(t) \, e^{-\frac{i E_{n} t}{\hbar}} \, |n^{(0)}\rangle}
[/tex]

He calculates the coefficients Cn(t) of the above expansion to the {|n>} basis exactly, for time dependent potentials that are not necesserily weak as you suggest.

But, how is this a 2-level problem? Sure, he finds equations for the coefficients, but does he solve them? Could you state the exact equation where he gives an exact explicit solution for [itex]c_{1}(t)[/itex] and [itex]c_{2}(t)[/itex]?

 

[JK423]

The exact problem is:
Ho=E1 |1><1|+ E2 |2><2|
V(t)=γexp(iωt) |1><2| + γexp(-iωt) |2><1|.

Now take equations of motion for the coefficients Cn (n=1,2) -your version of Sakurai includes them for sure- and see that you can very easily get an exact solution of the equations without assuming that your potential is weak (without using perturbation theory).

 

[JK423]

I made a mistake on the section i was reffering to, the correct one is
Chapter 5 - Approximation methods
Section 5.5 - Time-dependent potentials: The interaction picture.

 

[Dickfore]

The exact problem is:
Ho=E1 |1><1|+ E2 |2><2|
V(t)=γexp(iωt) |1><2| + γexp(-iωt) |2><1|.

Now take equations of motion for the coefficients Cn (n=1,2) -your version of Sakurai includes them for sure- and see that you can very easily get an exact solution of the equations without assuming that your potential is weak (without using perturbation theory).

But, this is a very particular type of perturbation. You said you can do it for an arbitrary perturbation.

 

[JK423]

The fact that V(t) is small has nothing to do wth it. This just determines a way to calculate the coefficients Cn in the state
|Ψ(t)⟩=∑nCn(t)e−iEnt|n⟩
that you wrote, and if V is small the way is via perturbation theory.
So, the fact that we are able to expand in the {|n>} basis of Ho has nothing to do with the fact that V is small. It must be something else

Oh, but it does. The equations for the coefficients are unsolvable exactly, just like the problem with the total Hamiltonian is unsolvable exactly. Therefore, we solve these equations iteratively, in powers of the perturbation. We can only stop to the first one or two iterations just because the perturbation is small.

With the 2-state problems, i prove to you that the fact that we are able to expand to the basis of Ho has nothing to do with the strength of the potential V as you suggest in your post that i quoted above.
I think you have the misunderstanding that we expand to the zeroth-order eigenkets of the unperturbed Hamiltonian? And that is just a first order approximation? No, that is not correct.

 

[Dickfore]

But, this is a very particular type of perturbation. You said you can do it for an arbitrary perturbation.

Please show us the solution for a general perturbation and not a simple harmonic one.

 

[JK423]

When did i suggest that i can find the solution for a general perturbation? Quote my post please?
You disagree with what i told you above?

 

[Dickfore]

Here:

The fact that WE can't solve the equations exactly is not a matter of principle. For example, in 2-state problems which can be solved exactly, V(t) can also be large with perturbation theory not being applicable. But there is a solution, an exact one, and we can find it. In these problems as well, where V(t) is large, we also expand the state in the {|n>} basis of Ho, so what you suggest is not correct.
See p.320 in Sakurai for the 2-state problems.

We come back to the original question, which is:
Why do we expand the state in the basis of Ho, since Ho is NOT the hamiltonian that describes the system?

 

[JK423]

I still don't see where in that post i said that i can find a solution for a general perturbation? Can you bold it please?
You suggested that the expansion on the Ho basis is correct only if V(t) is small, as i quote in post #23. And in the post that you quoted above, i prove to you that this is not correct. In 2-state problems where exact solutions can be found WE STILL expand in the basis of Ho. That expansion HAS NOTHING to do with the strength of V(t).
What do you not understand of what I'm saying?

 

[Dickfore]

We expand in the eigenkets of the unpertrubed hamiltonian because those are the solutions in the zeroth order approximation.

We want the perturbation to be 'small', so that the perturbation procedure (which does not mean expanding the state ket in any basis, but it means expanding it in powers of the parameter [itex]\lambda[/itex] I introduced in an earlier post) is mathematically justified and higher order corrections give negligible contributions to the lower order contributions.

 

[JK423]

We expand in the eigenkets of the unpertrubed hamiltonian because those are the solutions in the zeroth order approximation.

We want the perturbation to be 'small', so that the perturbation procedure (which does not mean expanding the state ket in any basis, but it means expanding it in powers of the parameter [itex]\lambda[/itex] I introduced in an earlier post) is mathematically justified and higher order corrections give negligible contributions to the lower order contributions.

No, that is not correct. The fact that we expand in the Ho basis is not a part of a perturbation procedure!
The perturbation procedure comes into play only in the calculation of the coefficients Cn(t)!
The iteration method that you reffered to, involves the evolution operator (that means the coefficients Cn(t)) and not the basis {|n>} that we have expanded. The {|n>} are not the zeroth order kets of an 'exact ket'!
Thats why i wonder why we expand to the {|n>} basis. If it just was a perturbation procedure i would have understanded it..

 

[Dickfore]

yes, they are.

 

[JK423]

Anyway, i will not insist more. Study Sakurai carefully and you'll see that what you say is wrong.

 

[Dickfore]

Maybe you are referring to this:

EDIT:

Let us give the perturbative expansion of the wave equation:

[tex]
i \, \hbar \, \frac{\partial |\Psi_{\lambda}(t)\rangle}{\partial t} = \left(H_{0} + \lambda \, V(t)\right) \, |\Psi_{\lambda}(t)\rangle
[/tex]

in a basis-free form. Namely, we expand:

[tex]
|\Psi_{\lambda}(t) \rangle = \sum_{n = 0}^{\infty}{|\Psi^{(n)}(t)\rangle \, \lambda^{n}}
[/tex]

[tex]
n = 0 \ i \, \hbar \, \frac{\partial |\Psi^{(0)}(t)}{\partial t} = H_{0} \, |\Psi^{(0)}(t)\rangle
[/tex]

[tex]
n > 0 \ i \, \hbar \, \frac{\partial |\Psi^{(n)}(t)}{\partial t} = H_{0} \, |\Psi^{(n)}(t)\rangle + V(t) \, |\Psi^{(n - 1)}(t)\rangle
[/tex]

As you can see the l.h.s. of the zeroth order equation does not contain the time dependent perturbation. That is why we can use separation of variables:
[tex]
|\Psi^{(0)}(t)\rangle = T(t) \, |\psi^{(0)}\rangle
[/tex]

Substituting this gives:
[tex]
H_{0} \, |\psi^{(0)}\rangle = \frac{i \, \hbar \, T'(t)}{T(t)} \, |\psi^{(0)}\rangle = E \, |\psi^{(0)}\rangle, T(t) = c \, \exp{\left(-\frac{i \, E \, t}{\hbar}\right)}
[/tex]

As you can see, the separation constant E is an eigenvalue of the unperturbed Hamiltonian. Therefore, according to the principle of superposition, the general unperturbed ket is:
[tex]
|\Psi^{(0)}(t)\rangle = \sum_{\alpha}{c_{\alpha} \, \exp{\left(-\frac{i \, E_{\alpha} \, t}{\hbar}\right)} \, |E_{\alpha}\rangle}
[/tex]

Notice that the coefficients [itex]c_{\alpha}[/itex] are time-independent and are determined by the initial state ket of the problem. This, in general, is not an eigenket of the unperturbed Hamiltonian. Nevertheless, every term of the sum is. Because of the linearity of the wave equation, it is sufficient to consider the perturbative expansion of each summation component separately (then the coefficients become functions of time; this corresponds to the method of variation of constants). Then, the true ket is given by the same sum.

 

[JK423]

Really? Can you justify what you say? In Sakurai,and anywhere else, there is no mention of 'zeroth order expansion'. The expansion is an EXACT equation, not a perturbative one.

I challenge you to prove your arguments.

 

[Dickfore]

Really? Can you justify what you say? In Sakurai,and anywhere else, there is no mention of 'zeroth order expansion'. The expansion is an EXACT equation, not a perturbative one.

I challenge you to prove your arguments.

I edited my last post. Please go through it.

 

[JK423]

Maybe you are referring to this:

EDIT:

Let us give the perturbative expansion of the wave equation:

[tex]
i \, \hbar \, \frac{\partial |\Psi_{\lambda}(t)\rangle}{\partial t} = \left(H_{0} + \lambda \, V(t)\right) \, |\Psi_{\lambda}(t)\rangle
[/tex]

in a basis-free form. Namely, we expand:

[tex]
|\Psi_{\lambda}(t) \rangle = \sum_{n = 0}^{\infty}{|\Psi^{(n)}(t)\rangle \, \lambda^{n}}
[/tex]

[tex]
n = 0 \ i \, \hbar \, \frac{\partial |\Psi^{(0)}(t)}{\partial t} = H_{0} \, |\Psi^{(0)}(t)\rangle
[/tex]

[tex]
n > 0 \ i \, \hbar \, \frac{\partial |\Psi^{(n)}(t)}{\partial t} = H_{0} \, |\Psi^{(n)}(t)\rangle + V(t) \, |\Psi^{(n - 1)}(t)\rangle
[/tex]

As you can see the l.h.s. of the zeroth order equation does not contain the time dependent perturbation. That is why we can use separation of variables:
[tex]
|\Psi^{(0)}(t)\rangle = T(t) \, |\psi^{(0)}\rangle
[/tex]

Substituting this gives:
[tex]
H_{0} \, |\psi^{(0)}\rangle = \frac{i \, \hbar \, T'(t)}{T(t)} \, |\psi^{(0)}\rangle = E \, |\psi^{(0)}\rangle, T(t) = c \, \exp{\left(-\frac{i \, E \, t}{\hbar}\right)}
[/tex]

As you can see, the separation constant E is an eigenvalue of the unperturbed Hamiltonian. Therefore, according to the principle of superposition, the general unperturbed ket is:
[tex]
|\Psi^{(0)}(t)\rangle = \sum_{\alpha}{c_{\alpha} \, \exp{\left(-\frac{i \, E_{\alpha} \, t}{\hbar}\right)} \, |E_{\alpha}\rangle}
[/tex]

Notice that the coefficients [itex]c_{\alpha}[/itex] are time-independent and are determined by the initial state ket of the problem. This, in general, is not an eigenket of the unperturbed Hamiltonian. Nevertheless, every term of the sum is. Because of the linearity of the wave equation, it is sufficient to consider the perturbative expansion of each summation component separately (then the coefficients become functions of time; this corresponds to the method of variation of constants). Then, the true ket is given by the same sum.

In the above you follow a different perturbation approach than Sakurai's, and is similar to what we do in the time-independent perturbation formalism.

First, let's make sure that we agree on the following. Sakurai does the following. The exact ket at time t is:
|Ψ(t)> = U |i>,
where U the evolution operator of the total Hamiltonian and |i> the initial eigenket of the unperturbed Hamiltonian Ho.
Since: 1=Σ_n |n><n| , |Ψ(t)> becomes
|Ψ(t)> = Σ_n <n|U |i> |n> (1)
This Ket is the exact ket of the system at time t, its not perturbative since in what i have done i haven't used the fact that V(t) is small or large. What i did (i follow Sakurai) is GENERAL.
So we have to agree that ket (1) is exact!

Do you argue that (1) is just an approximation? Elaborate on this please