# College Math Teaching

## August 19, 2011

### Quantum Mechanics and Undergraduate Mathematics XV: sample problem for stationary states

I feel a bit guilty as I haven’t gone over an example of how one might work out a problem. So here goes:

Suppose our potential function is some sort of energy well: $V(x) = 0$ for $0 < x < 1$ and $V(x) = \infty$ elsewhere.
Note: I am too lazy to keep writing $\hbar$ so I am going with $h$ for now.

So, we have the two Schrödinger equations with $\psi$ being the state vector and $\eta_k$ being one of the stationary states:
$-\frac{h^2}{2m} \frac{\partial}{\partial x^2}\eta_k + V(x) \eta_k = ih\frac{\partial}{\partial t} \eta_k$
$-\frac{h^2}{2m} \frac{\partial}{\partial x^2}\eta_k + V(x) \eta_k = e_k \eta_k$

Where $e_k$ are the eigenvalues for $\eta_k$

Now apply the potential for $0 < x < 1$ and the equations become:
$-\frac{h^2}{2m} \frac{\partial}{\partial x^2}\eta_k = ih\frac{\partial}{\partial t} \eta_k$
$-\frac{h^2}{2m} \frac{\partial}{\partial x^2}\eta_k = e_k \eta_k$

Yes, I know that equation II is a consequence of equation I.

Now we use a fact from partial differential equations: the first equation is really a form of the “diffusion” or “heat” equation; it has been shown that once one takes boundary conditions into account, the equation posses a unique solution. Hence if we find a solution by any means necessary, we don’t have to worry about other solutions being out there.

So attempt a solution of the form $\eta_k = X_k T_k$ where the first factor is a function of $x$ alone and the second is of $t$ alone.
Now put into the second equation:

$-\frac{h^2}{2m} X^{\prime\prime}_kT_k = e_k XT$

Now assume $T \ne 0$ and divide both sides by $T$ and do a little algebra to obtain:
$X^{\prime\prime}_k +\frac{2m e_k}{h^2}X_k = 0$
$e_k$ are the eigenvalues for the stationary states; assume that these are positive and we obtain:
$X = a_k cos(\frac{\sqrt{2m e_k}}{h} x) + b_k sin(\frac{\sqrt{2m e_k}}{h} x)$
from our knowledge of elementary differential equations.
Now for $x = 0$ we have $X_k(0) = a_k$. Our particle is in our well and we can’t have values below 0; hence $a_k = 0$. Now $X(x) = b_k sin(\frac{\sqrt{2m e_k}}{h} x)$
We want zero at $x = 1$ so $\frac{\sqrt{2m e_k}}{h} = k\pi$ which means $e_k = \frac{(k \pi h)^2}{2m}$.

Now let’s look at the first Schrödinger equation:
$-\frac{h^2}{2m}X_k^{\prime\prime} T_k = ihT_k^{\prime}X_k$
This gives the equation: $\frac{X_k^{\prime\prime}}{X_k} = -\frac{ 2m i}{h} \frac{T_k^{\prime}}{T_k}$
Note: in partial differential equations, it is customary to note that the left side of the equation is a function of $x$ alone and therefore independent of $t$ and that the right hand side is a function of $T$ alone and therefore independent of $x$; since these sides are equal they must be independent of both $t$ and $x$ and therefore constant. But in our case, we already know that $\frac{X_k^{\prime\prime}}{X_k} = -2m\frac{e_k}{h^2}$. So our equation involving $T$ becomes $\frac{T_k^{\prime}}{T_k} = -2m\frac{e_k}{h^2} i \frac{h}{2m} = i\frac{e_k}{h}$ so our differential equation becomes
$T_k {\prime} = i \frac{e_k}{h} T_k$ which has the solution $T_k = c_k exp(i \frac{e_k}{h} t)$

So our solution is $\eta_k = d_k sin(\frac{\sqrt{2m e_k}}{h} x) exp(i \frac{e_k}{h} t)$ where $e_k = \frac{(k \pi h)^2}{2m}$.

This becomes $\eta_k = d_k sin(k\pi x) exp(i (k \pi)^2 \frac{\hbar}{2m} t)$ which, written in rectangular complex coordinates is $d_k sin(k\pi x) (cos((k \pi)^2 \frac{\hbar}{2m} t) + i sin((k \pi)^2 \frac{\hbar}{2m} t)$

Here are some graphs: we use $m = \frac{\hbar}{2}$ and plot for $k = 1, k = 3$ and $t \in {0, .1, .2, .5}$. The plot is of the real part of the stationary state vector.