The quantum harmonic oscillator

As we will see in the next section, the classical forces in chemical bonds can be described to a good approximation as spring-like or Hooke's law type forces. This is true provided the energy is not too high. Of course, at very high energy, the bond reaches its dissociation limit, and the forces deviate considerably from Hooke's law. It is for this reason that it is useful to consider the quantum mechanics of a harmonic oscillator.

We will start in one dimension. Note that this is a gross simplification of a real chemical bond, which exists in three dimensions, but some important insights can be gained from the one-dimensional case. The Hooke's law force is

$$F(x) = -k(x-x_0)$$

where $k$ is the spring constant. This force is derived from a potential energy

$$V(x) = {1 \over 2}k(x-x_0)^2$$

Let us define the origin of coordinates such that $x_0=0$. Then the potential energy

$$V(x) = {1 \over 2}kx^2$$

If a particle of mass $m$ is subject to the Hooke's law force, then its classical energy is

$${p^2 \over 2m} + {1 \over 2}kx^2 = E$$

Thus, we can set up the Schrödinger equation using the prescription from the last lecture. The result is

$$\left[-{\hbar^2 \over 2m}{d^2 \over dx^2} + {1 \over 2}kx^2\right]\psi(x) = E\psi(x)$$

In this case, the Hamiltonian is

$$\hat{H} = -{\hbar^2 \over 2m}{d \over dx^2} + {1 \over 2}kx^2$$

Since $x$ now ranges over the entire real line $x\in(-\infty,\infty)$, the boundary conditions on $\psi(x)$ are conditions at $x=\pm\infty$. At $x=\pm\infty$, the potential energy becomes infinite. Therefore, it must follow that as $x\rightarrow\pm\infty$, $\psi(x)\rightarrow 0$. Hence, we can state the boundary conditions as $\psi(\pm\infty) = 0$.

Solving this differential equation is not an easy task, so we will not attempt to do it. Here, we simply quote the allowed energies and some of the wave functions. The allowed energies are characterized by a single integer $n$, which can be 0,1,2,.... and take the form

$$E_n = \left(n + {1 \over 2}\right)h\nu$$

where $\nu$ is the frequency of the oscillator

$$\nu = {1 \over 2\pi}\sqrt{k \over m}$$

Defining the parameter

$$\alpha = {\sqrt{km} \over \hbar} = {m\omega\over \hbar} = {4\pi^2 m\nu \over h}$$

the first few wave functions are

$$\psi_0(x) = \left({\alpha \over \pi}\right)^{1/4} e^{-\alpha x^2/2}$$

$$\psi_1(x) = \left({4\alpha^3 \over \pi}\right)^{1/4} xe^{-\alpha x^2/2}$$

$$\psi_2(x) = \left({\alpha \over 4\pi}\right)^{1/4} \left(2\alpha x^2-1\right)e^{-\alpha x^2/2}$$

$$\psi_3(x) = \left({\alpha^3 \over 9\pi}\right)^{1/4} \left(2\alpha x^3-3x\right)e^{-\alpha x^2/2}$$

You should verify that these are in fact solutions of the Schrödinger equation by substituting them back into the equation with their corresponding energies.

The figure below shows these wave functions, their associated allowed energies and the corresponding probability densities $p_n(x) = \psi_n^2(x)$:

Figure: Wave functions, allowed energies, and corresponding probability densities for the harmonic oscillator.

Note that since $x\in(-\infty,\infty)$, the normalization condition is

$$\int_{-\infty}^{\infty} \psi_n^2(x)dx=1$$

Despite this, because the potential energy rises very steeply, the wave functions decay very rapidly as $\vert x\vert$ increases from 0 unless $n$ is very large.