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Solving the Shroedinger equation for a harmonic oscillator potential

  1. Feb 13, 2014 #1

    I have been studying Introduction to Quantum Mechanics by Griffith and in a section he solves the Schrodinger equation for a harmonic oscillator potential using the power series method. First he rewrites the shroedinger equation in the form d^2ψ/dε^2 = (ε^2 - K)ψ , where ε= x√(mw/hbar) and K=2E/(ω hbar ) .
    Then he says that at large ε, we can approximate the equation to be d^2ψ/dε^2 ≈ ε^2ψ . So ψ≈Ae^(-ε^2 /2 ) + Be^(ε^2 /2 ) . But at large ε, we have to remove the Be^(ε^2 /2 ) term because otherwise, the wave function wouldn't be normalizable: ψ≈Ae^(-ε^2 /2 )

    and then he does something that I don't understand :

    " ψ(ε) → ( ) e^(-ε^2 /2 ) at large ε

    This suggests we "peel off" the exponential part,

    ψ(ε) = h(ε)e^(-ε^2 /2 )


    and then he goes on to solve the h(ε) rather than the ψ(ε) : h(ε) = ∑aj ε^j

    Why did he do that? Why not from the very beginning assume that ψ(ε)= ∑aj ε^j and find a recursion formula for that ?
  2. jcsd
  3. Feb 13, 2014 #2
    Because he wanted to motivate the decision to use a power series. In my copy Griffith's even makes a footnote stating that this is, as you reasoned, the idea behind solving diff. eq's using power series.
  4. Feb 13, 2014 #3
    but I tried rewriting the Shroedinger equation as d^2ψ/dε^2 - (ε^2 - K)ψ = 0 and then assumed ψ=∑aj ε^j , then substituted in the Shroedinger equation but got a different recursion formula :
    a_(j+2) = (a_(j-2) - a_(j)K ) / (j+2)(j+1) . How can I deduce that K must equal 2j+1 from this recursion relation ?
  5. Feb 13, 2014 #4


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    You can't. That's because it is a three-term recursion relation (i.e. it involves a's with three different subscripts).

    Which is exactly the purpose of factoring out the exponential - to lead to a differential equation that can be solved by a two-term recursion relation and therefore has polynomial solutions.
  6. Feb 14, 2014 #5
    I did not separate out the Gaussian potential once about 5 or 6 years ago. Like Bill_K, I got a three term recursion relation. Best to do as Griffith or other QM texts recommend.
  7. Feb 14, 2014 #6


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    Well the reason he does that is that he needs the exponential to save the normalizability of the function for ε going to infinity...
    Then what else could someone think as the general solution that has exponential damping in the upper limit?
    In general it's the sum of powers of ε, since their growing rate is cancelled out by the exponential's decrease.

    Almost the same approaches you can find in Hydrogen atom solutions...
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