Making Sense of the Infinite Square Well Solutions

By defining the position and momentum operators, $\hat{x} = x$ and $\hat{p} = -i \hbar \dfrac{d}{dx}$, we have shown that we can recast out energy eigenvalue problem, $\hat{H}\ket{\phi_n} = E_n\ket{\phi_n}$, in terms of a differential equation:

\[\left(-\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}+V(x)\right)\phi(x) = E\phi(x)\]

By doing this, we are using the “position representation” of the state vector, $\langle x | \phi \rangle$. We recognize that this representation is a continuous one, so how do we end up with discrete (quantized) results? Let’s investigate this with the one-dimensional infinite square well.

Consider a 1D potential given by,

\[\displaystyle V(x) = \begin{cases} \infty & \text{if $x < -a/2$} \\ 0 & \text{if $-a/2<x< a/2$} \\ \infty & \text{if $x > a/2$} \end{cases}\]

We defined $k^2 = \dfrac{2mE}{\hbar^2}$ as McIntyre did. Inside the well, we have the differential equation: $\dfrac{d^2}{dx^2}\phi(x) = -k^2\phi(x)$.

We then determined

\[\phi(x) = A \cos(kx) + B\sin(kx)\]

is the general solution. Of course, there’s a complex exponential version of this solution that works as well.

(This thinking is incomplete! I threw out odd solutions!!!) Given the symmetry of the problem ($\phi(x) = \phi(-x)$), we should expect cosine solutions, so that we can limit our solution in this case to: $\phi(x) = A \cos(kx)$

For this problem a slightly better choice for general solution:

\[\phi(x) = C\sin(kx+\delta)\]

We can then use the boundary conditions to determine that $\sin(ka/2+\delta)=0$ and $\sin(-ka/2+\delta)=0$. These give the quantizations for $k$ and $\delta$.

  • Question: With this $k$, what are the allowed energies? Are these energies quantized?
    • Discussion: Does you answer differ substantially from McIntyre’s? Why would we expect it to be relatively similar? How many energies are allowed?
  • Question: Consider two neighboring states ($n$ and $m$), find the energy separation ($\Delta E$) between these states.
    • Discussion: What happens to this energy separation as we tend towards higher level states? Mathematically, what happens as $n,m \rightarrow \infty$?