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The non relativistic Schrodinger equation of the harmonic oscillator in dimensionless variables is $$\frac{d^2 \Psi}{d \xi^2} = (\xi^2 - k)\Psi$$ where $$k \equiv \frac{2E}{\hbar \omega}$$ According to this stack a direct power series solution is a correct way to solve this equation. So I tried substituting $\Psi = \sum{a_n \xi^n}$ into the Schroedinger equation. I get the recurrence relation $$a_{n+2} = \frac{a_{n-2} - ka_n}{(n+1)(n+2)}$$ where $n \ge 2$ and $$a_2 = - \frac{k}{2}a_0$$ $$a_3 = - \frac{k}{6} a_1$$

Now how do I apply the conditions that $\Psi$ goes to zero at plus minus infinity to this solution? Also, how do I find the energy eigenvalues? In the traditional approach, the energy eigenvalues came out naturally when we wanted to terminate the power series upto a certain power. I don't know how and whether I even should be able to do this here, because the general series might be convergent and satisfy the physical properties.

If there is no way to do these without factorizing out the asymptotic solutions, then how do we solve Schrodinger's equations which don't have such nice asymptotic solutions but are suitable candidates for a power series solution?

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  • $\begingroup$ Hint: solving the harmonic oscillators by series is easier, if you first use a substitution to reduce it to the Hermite equation. Expectedly, you will then get the solutions as Hermit polynomials $\endgroup$
    – Roger V.
    Commented Oct 20, 2020 at 17:04
  • $\begingroup$ You mean $\Psi = A(\xi) exp(-\xi^2/2)$ ? I know about that however I want to know if there is a general way to do this because there are potentials for which I can solve the SE using power series but I run into the same problems that I ran into here. And for those potentials, a substitution of $\Psi = A(\xi) exp(-\xi^2/2)$ or anything similar doesn't reduce the equation to an easier one. $\endgroup$ Commented Oct 20, 2020 at 17:11
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    $\begingroup$ In more general case you reduce things to the Hypergeometric equation: Landau&Livshitz have quite a few problems explicitly solved this way (for bound states and for scattering). They are known for hiding some really good stuff in random problems... ;) $\endgroup$
    – Roger V.
    Commented Oct 20, 2020 at 17:17
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    $\begingroup$ And if you can't reduce the equation neither to the hypergeometric one, nor to the Bessel, then you probably shouldn't waste your time. But it is worth trying. $\endgroup$
    – Roger V.
    Commented Oct 20, 2020 at 17:18
  • $\begingroup$ The reason one does the asymptotic analysis in the first place is that it simplifies the algebra a lot. Since you know the answer for the SHO, you could try reverse engineering the conditions you need to impose on the coefficients, but whatever you come up with has to be able to factor the series into an exponentially decaying factor and a Hermite polynomial, which seems like it will be a lot of algebra. I'd advocate the approach Vadim suggested. $\endgroup$
    – Andrew
    Commented Oct 20, 2020 at 17:32

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The reason for the traditional method of pulling out the factor of $e^{-x^2/2}$ and only then seeking a series solution is that, just given a power series in $x$, it is hard to determine how its sum behaves as $|x|\to \infty$. When the series terminates, as it does for the Hermite polynomials, then it is easy to see that the wavefunction goes to zero at infinity. When the series does not terminate then it is one of the few cases in which one can see that the sum of the non-terminating power series behaves as $e^{+x^2}$ at large $|x|$ and so "terminates" is equivalent to square-integrable.

So the bottom line is that just seeking a series solution is not an effective route to the eigenvlaues.

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