Harmonic oscillator wave functions and time dependence

We have the ground state $\psi_0(x') = \langle x'|0\rangle$. Now we build everything else from it with $a^\dagger$.

# The first excited state

Now we can build the first excited state $|1\rangle$ with $|0\rangle$:

$$ a^\dagger|0\rangle = |1\rangle \implies \sqrt{\frac{m\omega}{2\hbar}}\left(\hat x - \frac{i\hat p}{m\omega}\right)|0\rangle = |1\rangle $$

In the position projection, with $\langle x'|$:

$$ \implies \sqrt{\frac{m\omega}{2\hbar}}\left(\langle x'|\hat x|0\rangle - \frac{i}{m\omega}\langle x'|\hat p|0\rangle\right) = \langle x'|1\rangle $$

$$ \implies \sqrt{\frac{m\omega}{2\hbar}}\left(x'\psi_0 - \frac{i}{m\omega}\left(-i\hbar\frac{\partial}{\partial x'}\langle x'|0\rangle\right)\right) = \langle x'|1\rangle $$

$$ \implies \psi_1(x') = \sqrt{\frac{m\omega}{2\hbar}}\left(x'\psi_0 - \frac{\hbar}{m\omega}\frac{\partial\psi_0}{\partial x'}\right) $$

With $x_0^2 = \frac{\hbar}{m\omega}$ (so $\sqrt{\frac{m\omega}{2\hbar}} = \frac{1}{\sqrt 2x_0}$),

$$ \psi_1 = \frac{1}{\sqrt 2x_0}\left(x'\psi_0 - x_0^2\frac{\partial\psi_0}{\partial x'}\right) $$

$$ \boxed{\psi_1(x') = \frac{1}{\sqrt 2x_0}\left(x' - x_0^2\frac{\partial}{\partial x'}\right)\psi_0(x')} $$

Doing the derivative ($\frac{\partial\psi_0}{\partial x'} = -\frac{x'}{x_0^2}\psi_0$) gives $\psi_1 = \frac{\sqrt 2x'}{x_0}\psi_0$: an odd function with one node in the middle.

# The general case

Since $|n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle$, and every $a^\dagger$ turns into $\frac{1}{\sqrt 2x_0}\left(x' - x_0^2\frac{\partial}{\partial x'}\right)$ in the position basis, we can similarly write out $\psi_n$ as

$$ \psi_n(x') = \frac{1}{\sqrt{n!}}\left(\frac{1}{\sqrt 2x_0}\right)^n\left(x' - x_0^2\frac{\partial}{\partial x'}\right)^n\psi_0(x') $$

↳ and if we expand / substitute for $\psi_0 = \langle x'|0\rangle$:

$$ \boxed{\psi_n(x') = \frac{1}{\pi^{1/4}\sqrt{2^nn!}}\left(\frac{1}{x_0^{n+1/2}}\right)\left(x' - x_0^2\frac{\partial}{\partial x'}\right)^n\exp\left(-\frac{1}{2}\left(\frac{x'}{x_0}\right)^2\right)} $$

The polynomials that pop out of $\left(x' - x_0^2\frac{\partial}{\partial x'}\right)^n$ acting on the Gaussian are the Hermite polynomials, $\psi_n \propto H_n\left(\frac{x'}{x_0}\right)e^{-x'^2/2x_0^2}$. More on those in this post.

# Expectation values of x squared and p squared

We may also want to look at the expectation values of $x^2$ and $p^2$. From $\hat x = \sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)$,

$$ \hat x^2 = \left[\sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)\right]^2 = \frac{\hbar}{2m\omega}\left(a^2 + a^{\dagger 2} + a^\dagger a + aa^\dagger\right) $$

In the ground state, only the last term has a non-vanishing expectation value ($a|0\rangle = 0$ kills $a^2$ and $a^\dagger a$, and $\langle 0|a^{\dagger 2}|0\rangle = 0$), with $\langle 0|aa^\dagger|0\rangle = 1$:

$$ \langle x^2\rangle = \frac{\hbar}{2m\omega}, \qquad \langle p^2\rangle = \frac{\hbar m\omega}{2} $$

Recall,

$$ \begin{aligned} \text{KE:} &\quad \left\langle\frac{p^2}{2m}\right\rangle = \frac{\hbar\omega}{4} = \frac{\langle H\rangle}{2} \\ \text{PE:} &\quad \left\langle\frac{m\omega^2x^2}{2}\right\rangle = \frac{\hbar\omega}{4} = \frac{\langle H\rangle}{2} \end{aligned} $$

$\to$ expected from the virial theorem.

And we also note $\langle x\rangle = \langle p\rangle = 0$, from

$$ \begin{aligned} \langle n'|\hat x|n\rangle &= \sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt n\delta_{n',n-1} + \sqrt{n + 1}\delta_{n',n+1}\right) \\ \langle n'|\hat p|n\rangle &= i\sqrt{\frac{m\hbar\omega}{2}}\left(-\sqrt n\delta_{n',n-1} + \sqrt{n + 1}\delta_{n',n+1}\right) \end{aligned} $$

(the diagonal $n' = n$ elements vanish) ↳ this holds for all $n$.

So (normally it’s $\langle x^2\rangle - \langle x\rangle^2$, but $\langle x\rangle = 0$ here)

$$ \langle(\Delta x)^2\rangle = \langle x^2\rangle = \frac{\hbar}{2m\omega}, \qquad \langle(\Delta p)^2\rangle = \langle p^2\rangle = \frac{\hbar m\omega}{2} $$

$$ \implies \langle(\Delta x)^2\rangle\langle(\Delta p)^2\rangle = \frac{\hbar^2}{4} $$

$\to$ for the ground state. The minimum, as promised by the Gaussian.

# For any n

And if we go to higher excited states,

$$ \hat x^2 = \frac{\hbar}{2m\omega}\left(a^2 + a^{\dagger 2} + \underbrace{a^\dagger a}_{N} + aa^\dagger\right) $$

• $a^2$ and $a^{\dagger 2}$ always give 0 expectation value:

$$ \begin{aligned} \langle n|a^2|n\rangle &= \sqrt{n(n - 1)}\langle n|n - 2\rangle = 0 \\ \langle n|a^{\dagger 2}|n\rangle &= \sqrt{(n + 1)(n + 2)}\langle n|n + 2\rangle = 0 \end{aligned} $$

↳ always 0 due to orthogonality.

• $a^\dagger a = N$: $\langle n|N|n\rangle = n$ (which is 0 when $n = 0$).

• For $aa^\dagger$:

$$ \langle n|aa^\dagger|n\rangle = \langle n|a\sqrt{n + 1}|n + 1\rangle = \sqrt{n + 1}\langle n|a|n + 1\rangle = (n + 1)\langle n|n\rangle = n + 1 $$

$$ \therefore \langle aa^\dagger\rangle = n + 1 $$

Hence, in general,

$$ \boxed{\langle x^2\rangle = \frac{\hbar}{2m\omega}(2n + 1)} $$

And $\hat p^2 = -\frac{m\hbar\omega}{2}\left(a^2 + a^{\dagger 2} - aa^\dagger - a^\dagger a\right)$, so the same reasoning gives

$$ \boxed{\langle p^2\rangle = \frac{m\hbar\omega}{2}(2n + 1)} $$

$$ \implies \langle(\Delta x)^2\rangle\langle(\Delta p)^2\rangle = \frac{m\hbar\omega}{2}\cdot\frac{\hbar}{2m\omega}(2n + 1)^2 $$

$$ \boxed{\langle(\Delta x)^2\rangle\langle(\Delta p)^2\rangle = \frac{\hbar^2}{4}(2n + 1)^2 = \hbar^2\left(n + \frac{1}{2}\right)^2} $$

↳ general for any $n$.

# From quantum to classical

The ground state has its highest prob. at the centre, and not at the edges like classical systems. As $n \to \infty$, $\langle x'|n\rangle$ wiggles faster and faster inside the potential, and the probability starts to look more like the classical one.

Classically, a particle oscillating with amplitude $A$ spends more time where it moves slowly, i.e. near the turning points. The time it spends in $dx$ is $\propto \frac{dx}{|v|}$, which gives

$$ P_{\text{cl}}(x) = \frac{1}{\pi\sqrt{A^2 - x^2}}, \qquad A = x_0\sqrt{2n + 1} $$

where $A$ comes from matching the energy, $\frac{1}{2}m\omega^2A^2 = \left(n + \frac{1}{2}\right)\hbar\omega$. Hover (or focus and use the arrow keys) to read off the values:

At $n = 0$ they look nothing alike: the quantum particle sits in the middle, the classical one at the edges, and the quantum one even leaks past the turning points (classically forbidden). At $n = 30$ the quantum density averages out to the classical curve. That’s the correspondence principle: at large quantum numbers (macroscopic energies) we get classical probabilities back.

# Time dependence of the SHO

We know (these are the equations of motion for the operators, and they look exactly like the classical ones)

$$ \frac{d\hat p}{dt} = -m\omega^2\hat x, \qquad \frac{d\hat x}{dt} = \frac{\hat p}{m} $$

With $a = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat x + \frac{i\hat p}{m\omega}\right)$,

$$ \frac{da}{dt} = \sqrt{\frac{m\omega}{2\hbar}}\left(\frac{\hat p}{m} - i\omega\hat x\right) = -i\omega\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x + \frac{i\hat p}{m\omega}\right) $$

$$ \implies \boxed{\frac{da}{dt} = -i\omega a}, \qquad \boxed{\frac{da^\dagger}{dt} = i\omega a^\dagger} $$

$$ \implies a(t) = a(0)\exp(-i\omega t), \qquad a^\dagger(t) = a^\dagger(0)\exp(i\omega t) $$

$$ \begin{aligned} \implies \hat x(t) + \frac{i\hat p(t)}{m\omega} &= \left(\hat x(0) + \frac{i\hat p(0)}{m\omega}\right)\exp(-i\omega t) \\ \hat x(t) - \frac{i\hat p(t)}{m\omega} &= \left(\hat x(0) - \frac{i\hat p(0)}{m\omega}\right)\exp(i\omega t) \end{aligned} $$

Adding and subtracting the two:

$$ \boxed{ \begin{aligned} \hat x(t) &= \hat x(0)\cos\omega t + \frac{\hat p(0)}{m\omega}\sin\omega t \\ \hat p(t) &= -m\omega\hat x(0)\sin\omega t + \hat p(0)\cos\omega t \end{aligned} } $$

Damn. $\therefore$ the $\hat x$ and $\hat p$ operators oscillate, exactly like the classical oscillator!

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