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In the coordinate representation solution to the quantum SHO (the solution via differential equations rather than Dirac's "trick") we ultimately work out that our eigenfunction solutions are of the form (up to some constants) $$u(q) = H(q)e^{-\frac{1}{2}q^2}$$ where $H$ are Hermite polynomials with series expansions $H(q) = \sum_0^\infty a_n q^n $ where we can show that the $a_n$ are constrained via the two-term recurrence relation $$a_{n+2} = \frac{2n+1-\lambda}{(n+2)(n+1)}a_n$$ where $\lambda$ is some parameter (dimensionless energy). We can argue that $\lambda$ must be of the form $\lambda = 2k+1$ for $k$ an integer so that we don't admit unbounded solutions, but my problem is as follows:

Both $a_0$ and $a_1$ are some as yet unconstrained coefficients. My textbook (Ballentine) makes no comment on how we constrain them. I referred to Townsend where it is said that we can pick one or the other as 0 in a given eigenfunction. Why is this so? Is this because we must pick $\lambda = 2k+1$ and so, depending on the $\lambda$, one of the recurrence relations will not terminate so that we need to set its original seed value ($a_0$ or $a_2$) to 0? For example, if $\lambda = 2(2)+1 = 5$ Then clearly $a_0$ and $a_2$ are the only nonvanishing coefficients from the "even" side, but $a_1$, $a_3$,$a_5$,... are unconstrained. Do we argue that this implies $a_1 = 0$ for the eigenfunction corresponding to this eigenvalue?

EDIT: It is not unbounded solutions which are the problem; it is solutions which diverge so severely that they are not even in the rigged Hilbert space.

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What the OP is saying sounds essentially correct. The recurrence relation breaks the coefficients into two independent sequences, and when we choose $\lambda$ so that one of the sequences terminates after a finite number of terms (so that the resulting function is a polynomial), the other sequence necessarily does not terminate. Since we know from the structure of this power series that the resulting power series necessarily blows up at infinity, we have to set the first coefficient in the non-terminating sequence to zero. The remaining undetermined coefficient ($a_0$ or $a_1$, depending) can then be fixed via normalization.

Thus, the process of applying the boundary conditions gives us two things: (1) it yields a relationship between the two undetermined coefficients multiplying the two linearly-independent solutions to the differential equation, and (2) it yields the eigenvalues. This is completely analogous to the case of the particle-in-a-box, for instance, where for each $k$, we get two solutions to the differential equation, so that $$ \psi(x) = A\cos(kx)+B\sin(kx)\,. $$ Applying one boundary condition always yields a relationship between $A$ and $B$ (or, sets one of them to zero), and applying the other boundary condition then determines the possible values of $k$.

Here, for the Hermite equation, we can break up the power-series solution to the differential equation as $$ H_{\lambda}(x) = a_0 \sum_{n=0}^{\infty}\frac{a_{2n}}{a_0} x^{2n} + a_1 \sum_{n=0}^{\infty}\frac{a_{2n+1}}{a_1} x^{2n+1}\,, $$ where the ratios $a_{2n}/a_0$ and $a_{2n+1}/a_1$ can be determined as functions of $n$ using the recurrence relation. Then, we apply boundary conditions by stipulating that the function should be finite at infinity, forcing us to choose a $\lambda$ such that one of these series terminates at a finite number of terms. Suppose that it's the even sequence. Then, we are forced to set $a_1=0$ because the function multiplying $a_1$ diverges at infinity. In addition, by choosing such a $\lambda$, we also automatically get an expression for the energy of the state. Finally, we can set the value of $a_0$ by normalizing the resulting function (after undoing the transformations we made to derive the Hermite equation starting from the SHO Schrodinger equation.

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  • $\begingroup$ Thank you very much for this answer; the connection with the boundary conditions and analogy to the practice-in-a-box was very nice too. $\endgroup$
    – EE18
    Commented Feb 6, 2023 at 5:06
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The reason you state for why we must pick one or the other as 0 is completely true. We can only have one value of $\lambda$ (corresponding to our energy level) and for a given $\lambda$ only one of the even or odd series terminates. The other would continue on as you said but we necessitate that our solution must be normalizable. In your question you mention that $a_1,a_3,a_5...$ is unconstrained in your example, however they are constrained by the fact that the solution must be normalizable. That constraint necessitates that the other series (i.e. $a_1,a_3,a_5,...$) must either terminate or be $0$ but we note that it can't terminate as we don't have the correct $\lambda$ value. Therefore, the only way we can have a normalizable solution is if the series is $0$ which happens for our starting coefficient being $0$. In your example for $\lambda=5$ this would mean $a_1=0$ must be true.

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