Harmonic oscillator (QM)

The harmonic oscillator potential $V(x) = \frac12 m\omega^2 x^2 = \frac 12 Cx^2$ appears everywhere in physics. We will solve for the wave functions $\psi$ that satisfy the Schrodinger equation for this potential.

Recall the solutions to a finite square well potential. Outside the well, $\psi$ decays exponentially like $e^{-\alpha x}$ where

\alpha = \frac{\sqrt{2m(V - E)}}{\hbar}.

In our harmonic oscillator, the potential $V$ is no longer constant, but instead a function of $x$ . If we consider the behavior of $\alpha$ with an $x$ -dependent potential at large $x$ , we see

\begin{align*} \alpha = \frac{\sqrt{2m(\frac12 C x^2-E)}}{\hbar} \to \frac{\sqrt{mCx^2}}{\hbar} \to \text{constant} \cdot x. \end{align*}

Thus for large $x$ , we expect

\psi(x) \sim e^{-\alpha x} \sim e^{-\text{constant} \cdot x^2},

so $\psi$ behaves like a Gaussian. We generally write Gaussians as $e^{-x^2/2a^2}$ where $a$ is a length scale.

From the time-independent Schrodinger equation, we get

\begin{align*} -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} \approx \frac12 Cx^2\psi ~~\text{ for large } x. \end{align*}

TO DO: Finish, from French 164

Operator method

We can avoid the Taylor series method above by doing some clever operator algebra. First notice that our Hamiltonian for the HO is

\HH=\hbar\omega \left( \frac{\hat p^2}{2m\hbar\omega} + \frac{m\omega}{2\hbar}\hat x^2\right).

We will try to factor $\HH$ . Classically $a^2 + b^2 = (a+ib)(a-ib)$ . Since we’re dealing with operators, this isn’t strictly true, but we will try it. Define $p_0 = \sqrt{2m\hbar\omega}$ and $x_0 = \sqrt{2\hbar/m\omega}$ , and write

\begin{align*} \HH &= \hbar \omega [(\hat p/p_0)^2 + (\hat x/x_0)^2] \end{align*}

Then

\begin{align*} \left(\frac{\xx}{x_0} + i\frac{\pp}{p_0}\right)\left(\frac{\xx}{x_0} - i\frac{\pp}{p_0}\right) &= \left(\frac{\xx}{x_0}\right)^2 + \left(\frac{\pp}{p_0}\right)^2 + \frac{i}{x_0p_0} [\xx,\pp]. \end{align*}

We know $[\xx,\pp] = i\hbar$ and $\frac{1}{x_0p_0} = \frac{1}{2\hbar}$ . We define

\begin{align*} \aa := \frac{\xx}{x_0} + i\frac{\pp}{p_0}, \quad\quad \aa\adj = \frac{\xx}{x_0} -i\frac{\pp}{p_0} \end{align*}

Which lets us write the Hamiltonian as

\begin{align*} \HH = \hbar \omega \left(\aa\adj\aa + \frac{1}{2}\right) = \hbar\omega \left(\aa\aa\adj - \frac{1}{2}\right). \end{align*}

We can calculate $[\aa,\aa\adj]=1$ , $[\HH,\aa] = -\hbar\omega\aa$ , and $[\HH,\aa\adj] = \hbar\omega\aa\adj$ (omitted).

Now consider what these operators do to an energy eigenstate. Let $\ket \psi = \aa \ket E$ .

\begin{align*} \HH \ket \psi &= \HH \aa \ket E \\ &= ([\HH,\aa] + \aa\HH) \ket E \\ &= (-\hbar \omega \aa + \aa E)\ket E \\ &= (E - \hbar \omega) \aa \ket E \\ &= (E - \hbar \omega) \ket \psi. \end{align*}

We see that applying $\aa$ lowers the energy by $\hbar \omega$ . We can similarly show that $\aa\adj$ raises the energy by the same amount. We can apply $\aa$ any number of times we want, and it will lower the energy each time. As we know, the expected energy $\braket \HH$ cannot be less than the minimum of the potential, so there must not be a state with negative energy.

Let $\ket 0$ be the lowest energy state. It must be that $\aa \ket 0 = 0$ . Applying $\aa$ again will just result in $0$ each time. The energy of $\ket 0$ is

\begin{align*} \HH \ket0 = \hbar\omega \left(\aa\adj\aa + \frac12\right) \ket0 = \frac12 \hbar\omega\ket0. \end{align*}

So the ground state has eigenenergy $\half \hbar\omega_0$ . This is remarkable: even in a vacuum the harmonic oscillator has some finite energy.

We can find other eigenstates from this ground state using the $\aa$ and $\aa\adj$ operators we found. Consider what happens when $\aa\adj$ is applied:

\begin{align*} \ket{\tilde 1} &= \aa\adj\ket0 \tag{the tilde means not normalized}\\ \HH\ket{\tilde 1} &= \HH \aa\adj \ket 0 \\ &= \left([\HH, \aa\adj] + \aa\adj\HH\right) \ket 0\\ &= \left(\hbar\omega\aa\adj + \aa\adj\frac12\hbar\omega_0\right) \ket0 \\ &= \frac32 \hbar\omega\aa\adj\ket0 \\ &= \frac32 \hbar\omega\ket{\tilde1}. \end{align*}

It follows from this that the energy of the $n$ -th eigenstate $E_n = (n + \half) \hbar\omega$ . $\aa$ and $\aa\adj$ act as “ladder operators” and move one step up/down the eigenstate “ladder”:

The ladder operators give us non-normalized eigenstates $\ket{\tilde n} = \big(\aa\adj\big)^n \tilde 0$ . Assume we have some normalized eigenstate $\ket n$ . We wish to find the magnitude of $|\ket{\widetilde{n+1}}|$ . We can write $|\ket{\widetilde{n+1}}|^2 = \braket{\aa\adj n | \aa\adj n} = \braket{\aa\aa\adj n | n}$ . From the Hamiltonian, we know $\HH = \hbar\omega(\aa\aa\adj - \half) = (n+\half)\hbar\omega$ , so we know that $\aa\aa\adj = n+1$ . Thus we conclude that $|\ket{\widetilde{n+1}}|^2 = (n+1)|\ket{n}|^2$ .

By induction, assuming $\ket 0$ is normalized, the normalized $n$ -th eigenstate is

\begin{align*} \ket n = \frac{1}{\sqrt{n!}} \big(\aa\adj\big)^n \ket 0. \end{align*}

We have found the general eigenstates in terms of the ground state. Let’s now consider the actual spacial representation of $\ket 0$ , which we will call $u_0$ .

\begin{align*} \aa &= \frac{\xx}{x_0} + i\frac{\pp}{p_0} = \sqrt{\frac{m\omega}{2\hbar}} \xx + \frac{i}{\sqrt{2\hbar m\omega}} \pp \\ \aa u_0(x) &= \left( \sqrt{\frac{m\omega}{2\hbar}} x + \frac{i}{\sqrt{2\hbar m\omega}} \frac{\hbar}{i} \frac{\partial}{\partial x} \right) u_0(x) = 0 \\ 0 &= \left( m\omega x + \hbar \frac{\partial}{\partial x} \right) u_0(x) \\ u_0(x) &= \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-m\omega x^2 / 2\hbar}. \end{align*}