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I have brief questions regarding the attachments, which are notes from the book Introduction to Quantum Mechanics by Griffiths which explains the harmonic oscillator case. Any assistance would be appreciated. The attachments don't look healthy but the questions are quite simple.

How do we get [2.77], isn't $A$ a constant in the general solution [2.75], why does it become a function of $\xi$?

Lastly, why is this method of terminating the power series taken (which involves letting $a_{n + 2} = 0$ for some $n$ and letting either the odd or even terms all be zero) Surely there are other ways to define the power series so that they terminate (maybe for some $n$ let $a_j = 0$ for all $j \geq n$)?.

Thanks a lot for any assistance.

Harmonic Oscillator 1 Harmonic Oscillator 2 Harmonic Oscillator 3 Harmonic Oscillator 4

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    $\begingroup$ Please type out text that you want to quote, since images are not searchable and hard to read. For your question, $A$ is a constant in the solution to an approximation, but 2.77 is an ansatz for the full equation, motivated by knowing the approximate solution, so what irks you about $A$ not being constant anymore? $\endgroup$
    – ACuriousMind ♦
    Commented Jul 12, 2016 at 21:07
  • $\begingroup$ What does he tell you in [2.79], after he says he is going to solve it by a series solution. As far as I remember, the recursion formula is based on Hermite polynomials. $\endgroup$
    – user108787
    Commented Jul 12, 2016 at 21:35

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  1. $A$ is constant in the approximate solution at large $\xi$. So Griffiths makes the ansatz that for general $\xi$, the solution is of the form for some function $A(\xi)$ that becomes "constant" compared to the exponentation at large $\xi$. It's an ansatz, it is not derived.

  2. Look at the condition on $K$ that terminating the sequence starting from a non-zero term imposes: If the sequence terminates at $n$, you have $K=2n+1$. If both odd and even terms were non-zero from the start, then you would have them terminating at $n_\text{even},n_\text{odd}$ with $2n_\text{even}+1=K=2n_\text{odd}+1$, which is impossible.

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  • $\begingroup$ This might be a trivial question. It states we require some $n$, which is the highest $j$ such that $a_{n+2} = 0$. So it seems we are working with a particular $j=n$ but further it gives $$E_{n} = (n + \frac{1}{2})\hbar \omega$$ for any $n = 0,1,2,...$ Should I look at it as if given that we have found one such $j = n$, then any subsequent $n + 1$ would also work and hence satisfy the condition $$K = 2n +1,$$ so this condition then becomes the important part and it holds for all $n = 0,1,2,...$. Is this reasoning fine or am I missing something simpler? Thanks. $\endgroup$
    – user114445
    Commented Jul 14, 2016 at 12:55
  • $\begingroup$ Maybe look at the fact that each $n$ corresponds to a different sequence of $\{a_j\}_j$, so effectually a different solution $\psi_n$... $\endgroup$
    – user100411
    Commented Jul 14, 2016 at 13:10
  • $\begingroup$ @LucioD: I'm not sure I understand the question. The $j$ is fixed, but arbitrary, that is, the argument works for every choice of $j$ (or $n$). $\endgroup$
    – ACuriousMind ♦
    Commented Jul 14, 2016 at 13:18
  • $\begingroup$ @ACuriousMind Yes I think I may have confused myself. Just to confirm, you are saying that the $j$ is fixed but the $a_j$ is arbitrary, hence we would find different solutions $\psi_n$? $\endgroup$
    – user114445
    Commented Jul 14, 2016 at 13:23

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