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Perturbation Theory

  1. Jan 26, 2008 #1


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    I'm looking at the beginning of of Chapter 6 of the 2nd edition of Griffiths Introduction to Quantum Mechanics.

    He starts out by writing the hamiltonian for a system we'd like to solve as the sum of a hamiltonian with a known solution and a small perturbation:

    [tex] H^0 + \lambda H^\prime [/tex]

    He says that lambda is meant to be a small number for now, but later it will be set to 1, and when it is, that will be the true Hamiltonian. My first question is huh? Why? Then later he says that the eigenfunctions and eigenvalues can be written as a power series in lambda. E.g.:

    [tex] \psi_n = \psi_n^0 + \lambda \psi_n^1 + \lambda^2 \psi_n^2 + \cdots [/tex]

    Again, huh? How can can just write this down? How does he know that [itex] \psi_n [/itex] can be written in this form? Why does the lambda from the perturbation appear in the corrections to the wavefunction and allowed energies in this way? It just seems to come from nowhere. Am I missing something?

    Also, what is meant by "first order correction" and "second order correction?" First order in what? The above equation would seem to indicate that the answer is "in lambda." But later on, he claims that lambda is just a "device for keeping track of the orders" and sets it to 1, effectively eliminating it from the equation. So what is it intrinsically about [itex] \psi_n^1 [/itex] that makes it "first order" for example?

    I'm sorry for being such a dunce, but I am having a hard time understanding where he's coming from.
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  3. Jan 26, 2008 #2


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    What he means by "lambda is meant to be a small number for now, but later it will be set to 0", is that he has not said anything about the H' in comparison with [tex]H^0[/tex]. (if I remember the text of griffiths) In order to get perturbation to "work" it must be weak perturbation, so lamda is a small number. Later when we do examples, [tex]H'[/tex] is given (often) and one sees that it is weak in comparison with [tex]H^0[/tex]. When lamda is = 1, then the perturbation is "on" 100%.

    So when does derivations, one often keeps the lamda to track the order of expansion, how many times the perturbation enters.

    I dont remember, but I dont think it is a derviation or proof that you can write the solution to the perturbed hamiltoian as [tex] \psi_n = \psi_n^0 + \lambda \psi_n^1 + \lambda^2 \psi_n^2 + \cdots [/tex]
    And for an introductory course, you dont have to know "why". You can look up more detalied derivations in more advanced textbooks if you like (like Sakurai: modern quantum mechanics).

    The "order of correction" is just how careful one have to be, compare with taylor series. [tex] \psi_n^0 [/tex] is the wavefunction satisfying the eigenvalue equation for [tex]H^0[/tex].

    So if you like, get Sakurai and look in chapter 5. I first started with Gritthis to learn how to use this machinery. Then when taking advanced courses in QM, using sakurai, I learned WHY.
    Last edited: Jan 26, 2008
  4. Jan 26, 2008 #3


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    But i can at least write this down for you:

    The solution to the full hamiltonian [tex] H = H' + H^0 [/tex] is [tex] |n >[/tex], the solution to [tex]H^0[/tex] is [tex] |n^0> [/tex]

    [tex] |n > = |n^0> + \dfrac{\phi _n}{E^0_n-H^0} (\lambda H' - \Delta _n ) |n> [/tex] (takes a couple of pages to get this eq.)

    Where [tex] \Delta _n = E_n - E_^0 [/tex]
    and [tex] \phi _n = 1 - |n^0 ><n^0 | [/tex]

    (btw, I hope you know bra-ket notation: <x|n> = psi_n(x))

    So you see that the solution is both on the LHS and RHS. So what you do it to stick the LHS in to the |n> on the RHS etc. etc. etc. So now you see why the lambda was quite important.

    Then you might argue: "But then the soulution is not known exactly" You are right, thats why this only works for weak perturbations, so you can truncate the series somewhere (usally around the 1st or 2nd order)
    Last edited: Jan 26, 2008
  5. Jan 26, 2008 #4


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    firstly, I never studied Griffiths in details but learnt my time-independent perturbation theory via Merzbacher. But still, Griffith's book are usually great

    why? isn't 1 a small number? setting it to one simply gives you the H you want (which is supposedly some perturbed version of H^0.) actually it doesn't matter.. we can assume it to be anything from 0 to 1 (0 being no perturbation and 1 being full on) if you want you choose use any real number, it doesn't matter. H' is the perturbation not \lambda

    the ability to expand your energy level and function psi as such is a fundamental assumption of perturbation theory. Note: the expansion (into powers of \lambda) of psi is not unique. But all physically observable quantities like energy levels calculated from your favorite expansion of psi would have unique expansions in \lambda. Remember you always have to go back to the eigenvalue equation
    [tex]H^0\Psi_n^{(0)} = E^{(0)}_n \Psi^{(0)}_n[/tex]
    when working these things out

    I think malawi_glenn may have explained this, but let me say something anyway

    Ok, firstly, I think we agree that \lambda = 1 is when you recover the actual H you want (ie. the one that is close to H^0 but not quite). The key here is that in perturbation theory we assume that if E = E0 + E' then E' should be small when we turn on H'. Turning on H' fully means setting \lambda to 1. Now it is convenient to write \lambda rather than 1 because 1^2=1 etc. and makes it hard to keep track of things.

    so what is intrinsic about [tex]\psi_n^1[/tex] that makes it "first order" is that we have defined it so. Through the assumption of perturbation theory we can say we can approximate our function with
    f = f0 + f1 + f2 + ....

    it is a bit like say y = x + k a
    [tex]y^2 = x^2 + 2xa\; k + O(k^2)[/tex]

    when the perturbation a is small, we can approximate the function y^2 by just
    x^2 (unperturbed)
    x^2 + 2xa (first order, 2xa is your [tex]\psi_n^1[/tex] )

    if you set k=1 at the start it is harder to keep track of which is which later on
    or course this is an oversimplified example, but I am sure you 've got it.

  6. Jan 26, 2008 #5
    Unfortunately I have not had a chance to read David Griffith's book, nor have I really studied QM yet, but I think I've encountered this type of perturbation before. I agree that it's a weird way of presenting things, and my professor did not give many hints. I apologize if any of what I say is obvious or tangential.

    When the system is perturbed, the value of [tex]\psi_n[/tex] also changes. The perturbation equation

    [tex] \psi_n = \psi_n^0 + \lambda \psi_n^1 + \lambda^2 \psi_n^2 + \cdots [/tex]

    is simply a way of representing [tex] \psi_n [/tex] in terms of the degree to which the system has been perturbed, [tex] \lambda [/tex]. The quantity [tex]\psi_n^0[/tex] represents the unperturbed value of [tex]\psi_n[/tex], i.e. the old value. Thus [tex]\lambda \psi_n^1[/tex] is the "first order correction" to the unperturbed [tex]\psi_n^0[/tex] value. The higher-order corrections give smaller values, as [tex]\lambda < 1[/tex]. When the system is fully perturbed, [tex]\lambda = 1[/tex], so [tex] \psi_n = \psi_n^0 + \psi_n^1 + \psi_n^2 + \cdots [/tex].

    I'm not sure what you want to solve for in the end, but if you wanted to solve for the values of [tex]\psi_n^1, \psi_n^2,[/tex] etc., it would be difficult to do so from this equation alone. But setting [tex]\lambda < 1[/tex] allows you to "keep track of the orders," in the sense that by taking derivatives of the perturbation equation with respect to [tex]\lambda[/tex] should allow you to solve for the correction factors. The variable [tex]\lambda[/tex] thus gives you something to work with.

    If you would like, I can ask David next week for an explanation. His answer would probably be much nicer than mine.
  7. Jan 26, 2008 #6


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    But the problem the OP also have is WHY the solution to the full hamiltonian looks as it does, as a power series expansion, you must study perturbation theory on a higher level than griffiths. In post #3 I wrote down how the solution looks, and that sheds some light on "why" you expand the solution as you does. But as I wrote, how you get to that solution requires some work, and is often written in more advanced books in QM.
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