# College Math Teaching

## August 10, 2011

### Quantum Mechanics and Undergraduate Mathematics X: Schrödinger’s Equations

Filed under: advanced mathematics, applied mathematics, calculus, physics, quantum mechanics, science — collegemathteaching @ 1:19 am

Recall from classical mechanics: $E = \frac{1}{2}mv^2 + V(x)$ where $E$ is energy and $V(x)$ is potential energy. We also have position $x$ and momentum $p = mv$ Note that we can then write $E = \frac{p^2}{2m} + V(x)$. Analogues exist in quantum mechanics and this is the subject of:

Postulate 6. Momentum and position (one dimensional motion) are represented by the operators:
$X = x$ and $P = -i\hbar \frac{d}{dx}$ respectively. If $f$ is any “well behaved” function of two variables (say, locally analytic?) then $A = f(X, P) = f(x, -i\hbar \frac{d}{dx} )$.

To see how this works: let $\phi(x) = (2 \pi)^{-\frac{1}{4}}exp(-\frac{x^2}{4})$
Then $X \phi = (2 \pi)^{-\frac{1}{4}}x exp(-\frac{x^2}{4})$ and $P \phi = i\hbar (2 \pi)^{-\frac{1}{4}} 2x exp(-\frac{x^2}{4})$

Associated with these is energy Hamiltonian operator $H = \frac{1}{2m} P^2 + V(X)$ where $P^2$ means “do $P$ twice”. So $H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)$.

Note We are going to show that these two operators are Hermitian…sort of. Why sort of: these operators $A$ might not be “closed” in the sense that $\langle \phi_1, \phi_2 \rangle$ exists but $\langle \phi_1, A \phi_2 \rangle$ might not exist. Here is a simple example: let $\phi_1 = \phi_2 = \sqrt{\frac{2}{\pi} \frac{1}{x^2 + 1}}$. Then $\langle \phi_1, \phi_2 \rangle = 1$ but $\int_{-\infty}^{\infty} x \phi_1 dx$ fails to exist.

So the unstated assumption is that when we are proving that various operators are Hermetian, we mean that they are Hermetian for state vectors which are transformed into functions for which the given inner product is defined.

So, with this caveat in mind, let’s show that these operators are Hermitian.

$X$ clearly is because $\langle \phi_1, x \phi_2 \rangle = \langle x \phi_1, \phi_2 \rangle$. If this statement is confusing, remember that $x$ is a real variable and therefore $\overline{x} = x$. Clearly, any well behaved real valued function of $x$ is also a Hermitian operator. IF we assume that $P$ is a Hermitian operator, then $\langle \phi_1, P^2 \phi_2 \rangle = \langle P\phi_1, P\phi_2 \rangle = \langle P^2 \phi_1, \phi_2 \rangle$. So we must show that $P$ is Hermitian.

This is a nice exercise in integration by parts:
$\langle \phi_1, P\phi_2 \rangle = -i\hbar\langle \phi_1, \frac{d}{dx} \phi_2 \rangle = -i\hbar \int_{-\infty}^{\infty} \overline{\phi_1} \frac{d}{dx} \phi_2 dx$. Now we note that $\overline{\phi_1} \phi_2 |_{-\infty}^{\infty} = 0$ (else the improper integrals would fail to converge this is a property assumed for state vectors; mathematically it is possible that the limit as $x \rightarrow \infty$ doesn’t exist but the integral still converges) and so by the integration by parts formula we get $i\hbar\int_{-\infty}^{\infty} \overline{\frac{d}{dx}\phi_1} \phi_2 dx =\int_{-\infty}^{\infty} \overline{-i\hbar\frac{d}{dx}\phi_1} \phi_2 dx = \langle P\phi_1, \phi_2 \rangle$.

Note that potential energy is a function of $x$ so it too is Hermitian. So our Hamiltonian $H(p,x) = \frac{1}{2m}P^2 + V(X) = -\frac{h^2}{2m}\frac{d^2}{dx^2} + V(x)$ is also Hermitian. That has some consequences:

1. $H \eta_k = e_k \eta_k$
2. $H \psi = i\hbar\frac{\partial}{\partial t} \psi$

Now we substitute for $H$ and obtain:

1. $-\frac{h^2}{2m} \frac{d^2}{dx^2} \eta_k + V(x)\eta_k = e_k \eta_k$

2. $-\frac{h^2}{2m} \frac{\partial^2}{\partial x^2} \psi + V(x)\psi = i\hbar \frac{\partial}{\partial t} \psi$

These are the Schrödinger equations; the first one is the time independent equation. It is about each Hamiltonian energy eigenvector…or you might say each stationary state vector. This holds for each $k$. The second one is the time dependent one and applies to the state vector in general (not just the stationary states). It is called the fundamental time evolution equation for the state vector.

Special note: if one adjusts the Hamiltonian by adding a constant, the eigenvectors remain the same but the eigenvalues are adjusted by adding a constant. So the adjusted time vector gets adjusted by a factor of $exp(-iC \frac{t}{\hbar})$ which has a modulus of 1. So the new state vector describes the same state as the old one.

Next post: we’ll give an example and then derive the eigenvalues and eigenvectors for the position and momentum operators. Yes, this means dusting off the dirac delta distribution.