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.

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