# College Math Teaching

## August 10, 2011

### Quantum Mechanics and Undergraduate Mathematics XI: an example (potential operator)

Filed under: advanced mathematics, calculus, differential equations, quantum mechanics, science — collegemathteaching @ 8:41 pm

Recall the Schrödinger equations:
$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \eta_k + V(x) \eta_k = e_k \eta_k$ and
$-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \phi + V(x) \phi = i\hbar \frac{\partial}{\partial t}\phi$

The first is the time-independent equation which uses the eigenfuctions for the energy operator (Hamiltonian) and the second is the time-dependent state vector equation.

Now suppose that we have a specific energy potential $V(x)$; say $V(x) = \frac{1}{2}kx^2$. Note: in classical mechanics this follows from Hooke’s law: $F(x) = -kx = -\frac{dV}{dx}$. In classical mechanics this leads to the following differential equation: $ma = m \frac{d^2x}{dt^2} = -kx$ which leads to $\frac{d^2x}{dt^2} + (\frac{k}{m})x = 0$ which has general solution $x = C_1 sin(wt) +C_2cos(wt)$ where $w = \sqrt{\frac{k}{m}}$ The energy of the system is given by $E = \frac{1}{2}mw^2A^2$ where $A$ is the maximum value of $x$ which, of course, is determined by the initial conditions (velocity and displacement at $t = 0$ ).

Note that there are no a priori restrictions on $A$. Notation note: $A$ stands for a real number here, not an operator as it has previously.

So what happens in the quantum world? We can look at the stationary states associated with this operator; that means turning to the first Schrödinger equation and substituting $V(x) = \frac{1}{2}kx^2$ (note $k > 0$ ):

$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \eta_k + \frac{1}{2}kx^2\eta_k = e_k \eta_k$

Now let’s do a little algebra to make things easier to see: divide by the leading coefficient and move the right hand side of the equation to the left side to obtain:

$\frac{d^2}{dx^2} \eta_k + (\frac{2 e_k m}{(\hbar)^2} - \frac{km}{(\hbar)^2}x^2) \eta_k = 0$

Now let’s do a change of variable: let $x = rz$ Now we can use the chain rule to calculate: $\frac{d^2}{dx^2} = \frac{1}{r^2}\frac{d^2}{dz^2}$. Substitution into our equation in $x$ and multiplication on both sides by $r^2$ yields:
$\frac{d^2}{dz^2} \eta_k + (r^2 \frac{2 e_k m}{\hbar^2} - r^4\frac{km}{(\hbar)^2}z^2) \eta_k = 0$
Since $r$ is just a real valued constant, we can choose $r = (\frac{km}{\hbar^2})^{-1/4}$.
This means that $r^2 \frac{2 e_k m}{\hbar^2} = \sqrt{\frac{\hbar^2}{km}}\frac{2 e_k m}{(\hbar)^2} = 2 \frac{e_k}{\hbar}\sqrt{\frac{m}{k}}$

So our differential equation has been transformed to:
$\frac{d^2}{dz^2} \eta_k + (2 \frac{e_k}{\hbar}\sqrt{\frac{m}{k}} - z^2) \eta_k = 0$

We are now going to attempt to solve the eigenvalue problem, which means that we will seek values for $e_k$ that yield solutions to this differential equation; a solution to the differential equation with a set eigenvalue will be an eigenvector.

If we were starting from scratch, this would require quite a bit of effort. But since we have some ready made functions in our toolbox, we note 🙂 that setting $e_k = (2k+1) \frac{\hbar}{2} \sqrt{{k}{m}}$ gives us:
$\frac{d^2}{dz^2} \eta_k + (2K+1 - z^2) \eta_k = 0$

This is the famous Hermite differential equation.

One can use techniques of ordinary differential equations (say, series techniques) to solve this for various values of $n$.
It turns out that the solutions are:

$\eta_k = (-1)^k (2^k k! \sqrt{\pi})^{-1/2} exp(\frac{z^2}{2})\frac{d^k}{dz^k}exp(-z^2) = (2^k k! \sqrt{\pi})^{-1/2} exp(-z^/2)H_k(z)$ where here $H_k(z)$ is the $k'th$ Hermite polynomial. Here are a few of these:

Graphs of the eigenvectors (in $z$) are here:

Of importance is the fact that the allowed eigenvalues are all that can be observed by a measurement and that these form a discrete set.

Ok, what about other operators? We will study both the position and the momentum operators, but these deserve their own post as this is where the fun begins! 🙂