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:

(graphs from 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! 🙂

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