Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Time-dependent perturbation theory

  1. Aug 7, 2011 #1

    JK423

    User Avatar
    Gold Member

    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 im 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...
     
  2. jcsd
  3. Aug 7, 2011 #2
    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]).
     
  4. Aug 7, 2011 #3

    JK423

    User Avatar
    Gold Member

    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?
     
  5. Aug 7, 2011 #4
    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.
     
    Last edited: Aug 7, 2011
  6. Aug 7, 2011 #5

    JK423

    User Avatar
    Gold Member

    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 doesnt 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 doesnt describe the system??
     
  7. Aug 7, 2011 #6
    The point is, the perturbation [itex]V(t)[/itex] is considered ''small'', hence we use perturbation theory.
     
  8. Aug 7, 2011 #7

    JK423

    User Avatar
    Gold Member

    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
     
  9. Aug 7, 2011 #8
    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.
     
  10. Aug 7, 2011 #9
    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
     
  11. Aug 7, 2011 #10

    JK423

    User Avatar
    Gold Member

    The fact that WE cant 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?
     
  12. Aug 7, 2011 #11
    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?
     
  13. Aug 7, 2011 #12
    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.
     
  14. Aug 7, 2011 #13
    Please solve this time-dependent problem exactly and post your solution here.
     
  15. Aug 7, 2011 #14

    JK423

    User Avatar
    Gold Member

    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..

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

    I answered in the previous post, check p. 320 of Sakurai for the exact solution of 2-state problems.
     
  16. Aug 7, 2011 #15
    In the version I have, p.320 discusses the Zeeman effect. Could you tell us what section you are referring to.
     
  17. Aug 7, 2011 #16

    JK423

    User Avatar
    Gold Member

    Oh sorry, its:
    Chapter 5 - Approximation methods
    Section 5.6 - Time dependent perturbation theory.
    He deals with 2-state problems somewhere in that section.
     
  18. Aug 7, 2011 #17
    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?
     
  19. Aug 7, 2011 #18

    JK423

    User Avatar
    Gold Member

    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.
     
  20. Aug 7, 2011 #19
    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]?
     
  21. Aug 7, 2011 #20

    JK423

    User Avatar
    Gold Member

    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook