Time-dependent perturbation theory

[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:

<br /> i \, \hbar \, \frac{\partial |\Psi_{\lambda}(t)\rangle}{\partial t} = \left(H_{0} + \lambda \, V(t)\right) \, |\Psi_{\lambda}(t)\rangle<br />

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

<br /> |\Psi_{\lambda}(t) \rangle = \sum_{n = 0}^{\infty}{|\Psi^{(n)}(t)\rangle \, \lambda^{n}}<br />

<br /> n = 0 \ i \, \hbar \, \frac{\partial |\Psi^{(0)}(t)}{\partial t} = H_{0} \, |\Psi^{(0)}(t)\rangle<br />

<br /> n &gt; 0 \ i \, \hbar \, \frac{\partial |\Psi^{(n)}(t)}{\partial t} = H_{0} \, |\Psi^{(n)}(t)\rangle + V(t) \, |\Psi^{(n - 1)}(t)\rangle<br />

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:
<br /> |\Psi^{(0)}(t)\rangle = T(t) \, |\psi^{(0)}\rangle<br />

Substituting this gives:
<br /> H_{0} \, |\psi^{(0)}\rangle = \frac{i \, \hbar \, T&#039;(t)}{T(t)} \, |\psi^{(0)}\rangle = E \, |\psi^{(0)}\rangle, T(t) = c \, \exp{\left(-\frac{i \, E \, t}{\hbar}\right)}<br />

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:
<br /> |\Psi^{(0)}(t)\rangle = \sum_{\alpha}{c_{\alpha} \, \exp{\left(-\frac{i \, E_{\alpha} \, t}{\hbar}\right)} \, |E_{\alpha}\rangle}<br />

Notice that the coefficients c_{\alpha} 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:

<br /> i \, \hbar \, \frac{\partial |\Psi_{\lambda}(t)\rangle}{\partial t} = \left(H_{0} + \lambda \, V(t)\right) \, |\Psi_{\lambda}(t)\rangle<br />

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

<br /> |\Psi_{\lambda}(t) \rangle = \sum_{n = 0}^{\infty}{|\Psi^{(n)}(t)\rangle \, \lambda^{n}}<br />

<br /> n = 0 \ i \, \hbar \, \frac{\partial |\Psi^{(0)}(t)}{\partial t} = H_{0} \, |\Psi^{(0)}(t)\rangle<br />

<br /> n &gt; 0 \ i \, \hbar \, \frac{\partial |\Psi^{(n)}(t)}{\partial t} = H_{0} \, |\Psi^{(n)}(t)\rangle + V(t) \, |\Psi^{(n - 1)}(t)\rangle<br />

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:
<br /> |\Psi^{(0)}(t)\rangle = T(t) \, |\psi^{(0)}\rangle<br />

Substituting this gives:
<br /> H_{0} \, |\psi^{(0)}\rangle = \frac{i \, \hbar \, T&#039;(t)}{T(t)} \, |\psi^{(0)}\rangle = E \, |\psi^{(0)}\rangle, T(t) = c \, \exp{\left(-\frac{i \, E \, t}{\hbar}\right)}<br />

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:
<br /> |\Psi^{(0)}(t)\rangle = \sum_{\alpha}{c_{\alpha} \, \exp{\left(-\frac{i \, E_{\alpha} \, t}{\hbar}\right)} \, |E_{\alpha}\rangle}<br />

Notice that the coefficients c_{\alpha} 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

 

[Dickfore]

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

Of course not.

 

[JK423]

Since you agree that ket (1) is exact, why do you say that its expansion on the {|n>} basis of Ho depends on the potential V(t) being a perturbation? This is the expansion:
|Ψ(t)> = Σ_n <n|U |i> |n> ,
and you agree that no approximation has been made! We expanded |Ψ(t)> on {|n>} without specifying what V(t) is!

 

[Dickfore]

Since you agree that ket (1) is exact, why do you say that its expansion on the {|n>} basis of Ho depends on the potential V(t) being a perturbation? This is the expansion:
|Ψ(t)> = Σ_n <n|U |i> |n> ,
and you agree that no approximation has been made! We expanded |Ψ(t)> on {|n>} without specifying what V(t) is!

Ok, now let me start from the beginning:

What do you want to know about time-dependent pertubation theory? The sentences you had written in the OP don't make any sense.

 

[JK423]

Ok, i'll express it differently.
Lets go back to |Ψ(t)> = U |i>.
Sakurai uses the fact that: 1=Σ_n |n><n|, where {|n>} are the eigenkets of H₀.
And in this way we get,
|Ψ(t)> = Σ_n <n|U |i> |n> . (1)
Let me do something different. I use: 1=Σ_n |n>><<n|
where {|n>>} are the eigenkets of a random Hamiltonian H'. And i get:
|Ψ(t)> = Σ_n <<n|U |i> |n>>. (2)

In (1), (2) we have expanded |Ψ(t)> in two different basis of different Hamiltonians.

Which is the correct one, and why?

 

[Dickfore]

Both, you are not doing pertubation theory at this step. The point is you cannot calculate \langle n | U(t) | i \rangle exactly for a time dependent Hamiltonian.

 

[JK423]

You're right that the title of the OP is misleading because this question is not about perturbation theory, but i naively named it that way because i faced this problem while studying perturbation theory.. Stupid i know.

I do not care about calculating that matrix element of U.
You're saying that both (1) and (2) are correct. However, when i'll measure the energy of the system I am going to find a specific energy eigenvalue. What is this eigenvalue going to be? Its going to be an eigenvalue of Ho or an eigenvalue of the random(!) Hamiltonian that i have used??

 

[Dickfore]

You're right that the title of the OP is misleading because this question is not about perturbation theory, but i naively named it that way because i faced this problem while studying perturbation theory.. Stupid i know.

I do not care about calculating that matrix element of U.
You're saying that both (1) and (2) are correct. However, when i'll measure the energy of the system I am going to find a specific energy eigenvalue. What is this eigenvalue going to be? Its going to be an eigenvalue of Ho or an eigenvalue of the random(!) Hamiltonian that i have used??

Ok, first of all, there is no quantum system with a time dependent Hamiltonian. The perturbation is due to an external force that changes with time in a controlled manner. So, a system with a time dependent Hamiltonian is an open system whose energy is not conserved. So, I don't know what you mean by 'measure the energy' of the system. The very process of measurement in QM means you interact with the system through an extra interaction Hamiltonian, so you change its state.

 

[genneth]

This seems to be a weird argument. I think the question the OP has is "why choose the basis of H_0 rather than some other basis?", to which the answer is "why not?". Usually, we assume that we can solve H_0 exactly, so that gives us a convenient basis in which to expand things (i.e. anything). As Dickfore then tried to say, this is especially convenient if V is small, so the states of H_0 are "good" approximations, which is to say that one does not need many states out of H_0's eigenstates to express the actual state. But the OP is correct that the exact convergence questions are subtle and do not depend on (neither necessary nor sufficient) that V is "small" in some sense.

Another way to put it: suppose you tried to expand in eigenstates of H instead. But you don't know what these are! The point is that to actually do calculations one needs to pick a basis, and H_0 gives you a convenient basis.

 

[Dickfore]

But, this argument holds for time-independent perturbaiton theory. Solving:
<br /> H(t) \, |\varphi(t)\rangle = E(t) \, |\varphi(t)\rangle<br />
in which time is treated as a parameter, does not give anything! Neither is E(t) the energy of the system, nor is |\varphi(t)\rangle the time dependent state ket of the system.

 

[JK423]

Ok, first of all, there is no quantum system with a time dependent Hamiltonian. The perturbation is due to an external force that changes with time in a controlled manner. So, a system with a time dependent Hamiltonian is an open system whose energy is not conserved. So, I don't know what you mean by 'measure the energy' of the system. The very process of measurement in QM means you interact with the system through an extra interaction Hamiltonian, so you change its state.

I don't think its proper to discuss the "how the measurement is really accomplished in a microscopic level" issue. Basic postulate of QM is the superposition principle, and that in principle you can measure an observable with the ket collapsing on an eigenstate of the observable in mind.
Here we have an open system as you say, whose energy is not conserved. I agree, but if i measure it's energy i will find a specific number (eg 5 joules). It doesn't matter if the system is open or closed!
So, if i expand the ket in two different basis of two different Hamiltonians, in an eigenstate of which Hamiltonian will the state of the system collapse if i measure its energy?

 

[Dickfore]

The point I am trying to make is that an open system is not characterized with an observable called 'energy'.

 

[JK423]

It doesn't have a specific energy, i agree, it's in a superposition of energy eigenstates. But ofcourse if you measure the system's energy you will measure SOMETHING! The question is, what?

Also if what you say was true, then Sakurai wouldn't bother expanding the ket in the Ho basis and finding the coefficients Cn(t) that give the probability to find the system in the |n> eigenstate of Ho..

 

[Dickfore]

Ok, so now we arrive at the crucial step. A quantum system is described by the pure Hamiltonian H_{0}. The eigenvalues of this (hermitian) operator give the possible energy levels of the system E_{\alpha}.

When this system is subject to an external force, modeled by the interaction Hamiltonian V(t), then the energy eigenstates are no longer stationary states of the system, i.e. the probability of being in any given energy eigenstate is no longer a constant in time. This is the effect of the external time-dependent potential. It makes possible for the quantum system to undergo transitions from one energy level into another. This would not be possible for an isolated system, where the expansion coefficients c_{\alpha} that I had defined earlier are independent of time.

For a time-dependent potential, the expansion coefficients become functions of time and generally obey the equation (exact as you also pointed out):
<br /> \dot{c}_{\alpha} = -\frac{i}{\hbar} \, \sum_{\beta}{e^{i \, \omega_{\alpha \beta} \, t} \, V_{\alpha \beta}(t) \, c_{\beta}(t)}, \ \omega_{\alpha \beta} = -\omega_{\beta \alpha} = \frac{E_{\alpha} - E_{\beta}}{\hbar}<br />
These equations describe the transition from one energy level to another.

 

[JK423]

I think we are going in circles...
Yes, i agree with you that we will have transitions from one energy level to another. But only if we expand the ket in the Ho basis! If we expand it in the basis of another (random) Hamiltonian H', then we will have transitions from one energy level to another of H'!

The basic question is:
Can we expand a ket simultaneously in different basis of different Hamiltonians?

This questions has, actually, little to do with the time-depedence of the Hamiltonian . It's something general. From example, if a system is described by a time-inderpendent Hamiltonian then can we expand it's state ket to the basis of another (random) Hamiltonian?

If can anyone else help, please do so.

 

[Dickfore]

Well, I suggest you go through Sakurai sections 1.3 and 1.5. I think you had not read what I wrote about the significance of the eigenvalues of H_{0}.

 

[JK423]

I read those sections and i didnt find anything usefull. Please if you know the answer to the question just tell me..
I started another thread that is more relevant with my question.