I'm trying to find solutions to a harmonic oscillator that sits within an infinite square well. I haven't spent too much time yet, and I've had no success so far. I'm wondering how possible or complex an analytic solution would be?
The potential of a simple harmonic oscillator is:
$$V_1(x)=1/2 m \omega_0^2 \, x^2$$
For simplicity, let us set $\omega_0=1$ and work in unitless time.
The potential for the infinite well $V_2(x)$ is infinity outside of the box, $|x|>L/2$ and zero just about everywhere else.
The potential for the modified problem is:
$$V(x)=V_1+V_2 \,=\, 1/2 m x^2 + V_2(x)$$
I would like to find the energy eigenstates of this modified problem, and the eigenvalues. What does your intuition say about how energy eigenvalues of the SHO would be affected by an added infinite well (assume a well width $L$ much larger than the "wavelength" of the ground state), and how does this compare to the actual analytical or a numerical solution?
Best Answer
The wavefunction $\psi(x)$ will satisfy the Schrödinger equation for the harmonic oscillator on an interval $x\in (-\frac L2 , \frac L2)$. We could write it as $$ \psi '' +\left(\frac{2E}{\hbar \omega} - \xi^2\right) \psi =0,\tag{1} $$ where $\xi = \sqrt{\frac{m\omega}\hbar} x$ is the rescaled $x$-coordinate and dash denotes differentiation w.r.t $\xi$.
However, the equation now has new boundary conditions: $$ \psi\left(\pm \sqrt{\frac{m\omega}\hbar} \frac L2\right)=0. \tag{2} $$ So in order to solve it we need to write the general solution of the equation (1). It is done in terms of confluent hypergeometric functions $M$ and $U$: $$ \psi(\xi) = e^{-\frac{\xi^2}2 }\,\left(C_1 \xi\, M\left(\frac34 - \frac{E}{2\hbar \omega},\frac 32,\xi^2\right)+C_2 \xi\, U\left(\frac34 - \frac{E}{2\hbar \omega},\frac 32,\xi^2\right)\right). $$ (For general value of parameter $E$ this wavefunction cannot be reduced to polynomial times exponential).
Since the potential is an even function, we can require that $\psi$ must have a specific parity. This allows us to eliminate one of the coefficients:
$\psi$ odd ($\psi(0)=0$): $C_2=0.$
$\psi$ even ($\psi'(0)=0$): $$ C_2 = \frac{1}{2\sqrt\pi} \Gamma \left(\frac 14 - \frac{E}{2\hbar \omega} \right) C_1$$
Coefficient $C_1$ is thus only determined by the norm of $\psi$ and should not enter the calculations of the energy spectrum. The only parameter left is the energy $E$, which must be determined by imposing the boundary condition (2) (only one of the equations is needed since we already imposed parity). This would gives us the energy spectrum of the system.
Note, that if the value $\xi=\sqrt{\frac{m\omega}\hbar}\frac L2$ happens to coincide with one of the roots of some Hermite polynomial $H_n(\xi)$, then the energy $\hbar\omega(n+\frac12)$ and wavefunction $\psi_n$ of harmonic oscillator would be the solution for this system. But of course, the number $n$ in this case would not mean that this is the $n$-th level.