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A question about a certain math argument in Griffiths

  1. Jan 26, 2015 #1
    1. In the part of the analytical solution of the harmonic oscillator, p38 in my version of the book a certain mathematical trick is used. The position part of the wave function in function of a dimensionless variable z, is given by

    ##\frac{d^{2}\Psi(z)}{dz^{2}}=(z^{2}-K) \Psi##

    In the limit of z going to +- infinity the solution to this ODE (bearing in mind that we want the function to be normalizable)

    ##\Psi(z)=Ae^{\frac{-z^{2}}{2}}##

    Then he says that for any z now the function can be written as

    ##\Psi(z)=h(z)e^{\frac{-z^{2}}{2}}##

    My question. How certain can I be, only from this reasoning that ##h(z)## does converge? A informal reasoning along the lines:

    Well as z goes to the limit of +- infinity the behaviour of ##\Psi(z)## has to be that of the earlier mentioned exponential. This can only be true if ##h(z)## doesn't diverge because even if ##h(z)=z## for example, it would change the behaviour in the limit to ##\Psi(z)=ze^{\frac{-z^{2}}{2}}##

    This was an informal reasoning that can have gaps in it. Can I be sure that this is 100% true in a formal way as well?
     
    Last edited: Jan 26, 2015
  2. jcsd
  3. Jan 26, 2015 #2

    BvU

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    There is no certainty whatsoever (If you mean ##\displaystyle \lim_{z\rightarrow \infty} h(z) = 0##. None of the h that are found in a later stage do this).

    Usable wave functions must be normalizable, so basically all that is required for that is that the h diverge "less fast" than ##e^{+z^2/2}##.

    And the behaviour of ##\Psi(z)## for ##{z\rightarrow \infty}## doesn't have to be exactly as ##e^{-z^2/2}## (that, by the way would mean ##h\rightarrow 1##), only "like" ##e^{-z^2/2}##. As long as ##\Psi## satisfies the Schroedinger equation, everything is hunky dory.

    Basically what's done here is the factoring out of the asymptotic behaviour and then the treatment continues with looking at what's left over and what equation and conditions it has to fulfil.
     
    Last edited: Jan 26, 2015
  4. Jan 26, 2015 #3
    By converging I meant going to a constant and not zero necessarily.

    Yeah but the constant A is still in front of the exponential so all that requires is that h(z) has an existing limit towards -+ infinity.

    I think you are spot on onto my problem upon underlining exactly. Is it allowed for example in theory that ##h(z)=z## from these facts we have here?

    Thanks for the reply.
     
  5. Jan 26, 2015 #4

    haruspex

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    No, that would not be allowed.
    Let's be clear about what this means: "in the limit, ##\Psi(z)=Ae^{\frac{-z^{2}}{2}}##" (which should really be written ##\Psi(z)##~##Ae^{\frac{-z^{2}}{2}}##).
    It means ##\lim_{z\rightarrow \infty}\Psi(z)e^{\frac{z^{2}}{2}}=A##. From that you can conclude ##\lim_{z\rightarrow \infty}h(z) = A##.
     
  6. Jan 26, 2015 #5

    BvU

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    In that interpretation there is one eigenstate that converges. You can verify that h=const satisfies the Schroedinger equation. The h in all other eigenstates then "don't converge".

    No. All that is requireed is that ##\Psi## satisfies the Schroedinger equation and is normalizable. If the former is satisfied, then all that is required for h is that the h diverge "less fast" than ##e^{ +z^2 /2}##. Any finite power of z is acceptable.

    Yes. In the wavefunction of the first excited state h is proportional to z. Your book top of p 41.
     
  7. Jan 27, 2015 #6
    Thanks a lot, this really helped me. One final small thing so that I reall understand it.

    Consider a function ##f(x)=(x^{2}+x^{3}-2)e^{-x}##

    It is correct to say that this function behaves like ##e^{-x}## in the limit?
     
  8. Jan 27, 2015 #7
    What about what the other poster pointed out below you, that indeed ##h(z)=z## is a form of one of the accepted solutions for ##h(z)## in this problem later on?
     
  9. Jan 27, 2015 #8

    haruspex

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    We seem to be answering different questions. I based my answer on the limiting behaviour condition stated ##\Psi(z)=Ae^{-\frac{z^2}{2}}##, where, presumably, A is a constant. BvU,I believe, is allowing the more permissive condition that ##\Psi(z)## converges to 0 as |z| tends to infinity. But I'm approaching this based purely on the algebra. I have little knowledge of quantum theory.
     
  10. Jan 27, 2015 #9

    BvU

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    I sympathize with the nay sayers, but it doesn't matter: if it satisfies the Schroedinger equation and it is normalizable, then it's a suitable steady state solution.

    When in doubt, make a drawing:

    times_exp-x2.jpg

    Of course a logarithmic y axis mercilessly reveals an ever increasing ratio between the two, but that really doesn't matter.

    A bit later on, when you have the steady state solutions, the probabilities to find the particle outside the classical region drop off exponentially, that's the main thing. It doesn't matter all that much (understatement!) whether the actual probability density is 10-44 or 10-38 (your function, x=100 - where classically x ##\approx# 3) .
     
  11. Jan 27, 2015 #10

    haruspex

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    The usual definition of functions having the same asymptotic behaviour can be found at http://www.math.uiuc.edu/~hildebr/595ama/ama-ch2.pdf. As I posted, f(x)~g(x) as ##x \rightarrow x_0## means ##\lim _{x \rightarrow x_0} f(x)/g(x) =1##. Your f(x) above does not behave like e-x in that sense. But ln(f(x)) ~ -x.
     
  12. Jan 27, 2015 #11

    BvU

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    Indeed we are. And we all agree that the answer to the original question -- with the given meaning of "convergence" -- is that you can not be sure at all that h converges (except for the ground state, which happens to have h = constant :) ). In fact, for all other solutions, you can be quite sure h diverges, in the sense that ##h\uparrow \pm \infty## for ##x\uparrow\pm\infty##. (do I have to draw an updownarrow in these to accomodate the - sign too?)

    What I try to bring across is that the question itself is not relevant in the context of this approach to solve the Schroedinger equation analytically.

    [commercial]:
    The harmonic oscillator example is incredibly important in quantum mechanics and way beyond, in the physics sense (it is the first order approach to any non-trivial potential function) and also in the didactical sense. So, dear coffee and all others: pay attention, invest time and effort -- and ask (e.g. in PF) when in doubt or stuck ;)
     
  13. Jan 27, 2015 #12
    I start to get it a little bit more. Thanks a lot for your replies both of you.

    As you seem to encourage me to post about confusions about the harmonic oscillator there's one more thing that I is troubling me, related to this in some way. That deserves a different thread though, which I'll make when I'll find some time. Thank again.
     
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