Imagine a particle obeying Schrodinger's Equation with an harmonic oscillator potential modified with an additional linear potential and cut off with an infinite potential barrier at $x=0$. That is, $$ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}m\omega x^2\Psi-mgx\Psi=E\Psi $$ for $x>0$, and $\Psi(x\le 0)=\Psi(x=\infty)=0$.

Obviously, by completing the square, one could get this equation into the form of the traditional quantum harmonic oscillator. From there, the usual solution can be derived. However, the left-hand boundary condition is impossible to satisfy using the answer given in terms of Hermite polynomials.

Both Griffiths and Shankar claim that the the series solutions to Hermite's Equation diverge except for half-integer values of the energy, which implies that the series must terminate. But if the series terminates, there is no freedom left to set the left-hand boundary to zero.

How do we resolve this issue?


1 Answer 1


First, let's start by completing the square as suggested. The equation will now be of the form $$ \frac{d^2\Psi}{dz^2}+(\nu+\frac{1}{2}-\frac{1}{4}z^2)\Psi=0 $$ with boundary conditions $\Psi(z=L)=\Psi(z=\infty)=0$ and $$ z=\sqrt{\frac{2m\omega}{\hbar}}x+L \\ \nu=\frac{E}{\hbar\omega}+\frac{L^2}{4}-\frac{1}{2} \\ L=-\sqrt{\frac{2mg^2}{\hbar\omega^3}} $$ This is Weber's Equation, which has solutions known as parabolic cylinder functions. The parabolic cylinder functions can be parameterized in several ways, but one way is to separate the series into two parts based on end behaviour: $U(-\frac{1}{2}-\nu,z)$ and $V(-\frac{1}{2}-\nu,z)$.

$U(-\frac{1}{2}-\nu,z)$ always converges to $0$ as $x\rightarrow\infty$ regardless of $\nu$. However, unless $\nu$ is a non-negative integer, it will diverge as $x\rightarrow -\infty$. This means that $U(-\frac{1}{2}-\nu,z)$ is the general solution to this problem since it reproduces the expected behavior for half-integer values of the energy while also having the ability to satisfy other left-hand boundary conditions by tweaking $\nu$.

I'll use $U(-\frac{1}{2}-\nu,z)=D_\nu(z)$ from now on since that is the usual terminology.

Actually solving this problem for a given value of $g$ (or, equivalently, $L$) can be quite tricky. In fact, solutions which are simple, clean, and exact are impossible to come by. Instead, we must solve this problem numerically.

For the sake of illustration, let's suppose $L=-\sqrt{2}$. In this case, we find that $E_\nu=\nu\hbar\omega$, the lowest energy solution is for $\nu\approx 0.234234$, and the normalization constant is $N=\frac{1}{\sqrt{1.66278}}$. Therefore, the groundstate wavefunction would be roughly $$ \Psi_{G.S.}(z)=\frac{1}{\sqrt{1.66278}}D_{0.234234}(z) $$

Here are the next several values of $\nu$ for which energy eigenstates exist for $L=-\sqrt{2}$ and the corresponding normalization constants: $$ \nu\approx 0.234234 \ \ , \ \ N\approx \frac{1}{\sqrt{1.66278}} \\ \nu\approx 1.69746 \ \ , \ \ N\approx \frac{1}{\sqrt{2.51615}} \\ \nu\approx 3.28019 \ \ , \ \ N\approx \frac{1}{\sqrt{13.3306}} \\ \nu\approx 4.92981 \ \ , \ \ N\approx \frac{1}{\sqrt{159.493}} \\ \nu\approx 6.62244 \ \ , \ \ N\approx \frac{1}{\sqrt{3486.3}} \\ \nu\approx 8.34549 \ \ , \ \ N\approx \frac{1}{\sqrt{122853.}} \\ $$

I used Mathematica to derive these results.


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