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

Perturbation theory (the math)

  1. Mar 11, 2014 #1
    My study of Quantum Mechanics have brought me to perturbation theory. I'm here talking about the non-degenerate type.

    My questions relate to the math behind it, and the power series expansion that we do.

    [itex]H = H^0 + \lambda H'[/itex] (Eq. 1)

    Question 1:
    So in equation 1 I think I understand that, in order for us to write this expression, we assume that the hamiltonian would to a "good" approximation be able to be written as a known hamiltonian, plus a small perturbation.

    Or is it in fact, from a mathematical perspective, such that it is actually possible to precisely write any hamiltonian in the form of equation 1. Meaning that it is not an approximation, but it is a true equal sign?

    Question 2:
    Moving on from the previous, when they in the books (Griffiths in my case), write the following power series in lambda:

    [itex] E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \cdots[/itex] (Eq. 2)
    [itex] \psi_n = \psi_n^{(0)} + \lambda \psi_n^{(1)} + \lambda^2 \psi_n^{(2)} + \cdots[/itex] (Eq. 3)

    This question is actually quite the same as the previous. Is this an exact representation of the energy E_n and wavefunction Psi_n? Or can we only approximate these in this sense? Mathematically speaking, as I see it, it should be perfectly possible to precisely write ANY energy (which is just a number) in the form of one known number, plus a second number?

    The same goes with the functions psi. Is it a precise representation or an approximation?

    Note: I see the relationship with the taylor expansion, and from wikipedia I think I saw that perhaps the validity of equation 2 and 3 comes from an argument from Taylor expansion?

    Question 3:
    Well, this question is kinda the all over question I guess. When you use the equations in the Schrödinger, we get the equation

    [itex]H^0\psi_n^{(0)} + \lambda(H^0 \psi_n^{(1)} + H' \psi_n^{(0)}) + \lambda^2(H^0 \psi_n^{(2)} + H'\psi_n^{(1)}) + \cdots \\ = E_n^{(0)}\psi_n^{(0)} + \lambda(E_n^{(0)}\psi_n^{(1)} + E_n^{(1)}\psi_n^{(0)}) + \lambda^2(E_n^{(0)}\psi_n^{(2)} + E_n^{(1)}\psi_n^{(1)} + E_n^{(2)}\psi_n^{(0)}) + \cdots [/itex] (Eq. 4)

    Now if what I talked about with regards to preciseness is true, then I understand that this is a perfectly valid form of the schrödinger equation for our hamiltonian, actually precisely describing the solution. And it is from this equation that we choose in which "order" we want to perturbate the system.

    Now comes the funny part. If we choose to perturbate the system to first order, I can see how we easily get to the equation

    [itex] H^0 \psi_n^{(1)} + H' \psi_n^{(0)} = E_n^{(0)}\psi_n^{(1)} + E_n^{(1)}\psi_n^{(0)} [/itex] (Eq. 5)

    And from this we can calculate stuff. However, in the textbook (Griffiths) and other places Im sure, it is then stated that if we do second-order perturbation, we also get the equation

    [itex] H^0 \psi_n^{(2)} + H'\psi_n^{(1)} = E_n^{(0)}\psi_n^{(2)} + E_n^{(1)}\psi_n^{(1)} + E_n^{(2)}\psi_n^{(0)} [/itex] (Eq. 6)

    This makes sense in a way, except that it doesn't to me... Because, if we have equation 4 and then say "lets discard all orders greater than 2", then we're left with the equation

    [itex] \lambda(H^0 \psi_n^{(1)} + H' \psi_n^{(0)}) + \lambda^2(H^0 \psi_n^{(2)} + H'\psi_n^{(1)}) = \lambda(E_n^{(0)}\psi_n^{(1)} + E_n^{(1)}\psi_n^{(0)}) + \lambda^2(E_n^{(0)}\psi_n^{(2)} + E_n^{(1)}\psi_n^{(1)} + E_n^{(2)}\psi_n^{(0)}) [/itex] (Eq. 7)

    In my curious yet unknowing mind, I cannot see how you get from equation 7 to equation 6. In my mind, you cant just say that because of the orders of lambda, the things inside the brackets must be equal. To me its a bit like the following:

    [itex] x(a+b) + x^2(c+d) = x(e+f) + x^2(g+h+i)[/itex]

    In the above example, is it generally true that because of their orders (a+b) must equal (e+f) and (c+d) must equal (g+h+i) ??
  2. jcsd
  3. Mar 11, 2014 #2
    It's an exact equality, by definition. You can write any Hamiltonian ##H## as some simple Hamiltonian ##H_0## plus a difference term ##\lambda H'##: just define ##\lambda H' = H - H_0##. But the perturbation series will converge better if ##\lambda H'## is in some sense a "small" correction.

    Here I think things get a little tricky if you want to be precise. The naive idea is this: ##E_n## is some function of ##\lambda##, and the expression above is the power series of this function (similarly for ##\psi_n##).

    This is too naive, though. The function may not have a power series expansion around ##\lambda = 0##. In general the expressions above are only asymptotic series.

    The idea is that if you do a power series (or asymptotic series) expansion of the left-hand side and the right-hand side, you must get the same series because the left hand side and the right hand side are equal. Therefore the left-hand side and the right-hand side must be equal at each order in ##\lambda##.

    Yes, you should convince yourself that if you require the equation to be true *for all values of ##x##*, then you have ##a + b = e + f## and ##c + d = g + h + i##.
  4. Mar 11, 2014 #3
    Okay, starting from below,

    [itex] x(a+b) + x^2(c+d) = x(e+f) +x^2(g+h+i) \\
    \Longleftrightarrow [(a+b) - (e+f)]x + [(c+d) - (g+h+i)]x^2 = 0[/itex]

    And thus if for all x this equation must be true, then (a+b) = (e+f) and (c+d) = (g+h+i)

    My mind has expanded. Fantastic ;)

    For the rest:
    Thank you for the clarifications. I see the overall idea now and think I can convince myself that it is actually true that one can do this.

    Thank you :)
  5. Mar 16, 2014 #4
    Are you using Griffith, he can be pretty unclear sometimes.
  6. Mar 16, 2014 #5
    Yes, its Griffith's we're using. Yeah sometimes he's a bit too quick and comes with the fantastic "It's easy to see that.." :)
  7. Mar 16, 2014 #6
    I like Shankar much better, his intense rigor can be an annoyance but at least it is always there when you need it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook