The simple harmonic oscillator
Contents:
# The setup
$\to$ Any potential well can be approximated as a SHO (near its minimum, $V(x) \approx V(x_0) + \frac{1}{2}V''(x_0)(x - x_0)^2$). We’ll see this later as well.
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┃──/\/\/\/\/\/\/\/──│ m │ ──► x
┃ k └─────┘
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For the SHO:
$$ \boxed{\hat H = \frac{\hat p^2}{2m} + \frac{m\omega^2\hat x^2}{2}}, \qquad \omega = \sqrt{\frac{k}{m}} $$
$\to$ $\omega$ is the angular freq. of the classical oscillator.
$\hat x$ and $\hat p$ are Hermitian.
# Annihilation and creation operators
We now define two new operators:
$$ \boxed{a = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat x + \frac{i\hat p}{m\omega}\right)}, \qquad \boxed{a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat x - \frac{i\hat p}{m\omega}\right)} $$
$\to$ the annihilation operator and the creation operator. We can quickly see that
$$ \boxed{[a, a^\dagger] = 1} $$
since only the cross terms survive:
$$ [a, a^\dagger] = \frac{m\omega}{2\hbar}\left(-\frac{i}{m\omega}[\hat x, \hat p] + \frac{i}{m\omega}[\hat p, \hat x]\right) = \frac{m\omega}{2\hbar}\cdot\left(-\frac{2i}{m\omega}\right)(i\hbar) = 1 $$
# The number operator
We also define
$$ \boxed{N = a^\dagger a} $$
$\to$ the number operator.
$$ a^\dagger a = \frac{m\omega}{2\hbar}\left(\hat x^2 + \frac{\hat p^2}{m^2\omega^2}\right) + \frac{i}{2\hbar}[\hat x, \hat p] $$
(the cross terms don’t cancel, as $\hat x\hat p \neq \hat p\hat x$). With $[\hat x, \hat p] = i\hbar$,
$$ \boxed{a^\dagger a = \frac{\hat H}{\hbar\omega} - \frac{1}{2}} \implies N = \frac{\hat H}{\hbar\omega} - \frac{1}{2} $$
or
$$ \boxed{\hat H = \hbar\omega\left(N + \frac{1}{2}\right)} $$
We also denote an energy eigenket of $N$ by its eigenvalue $n$:
$$ \boxed{N|n\rangle = n|n\rangle} $$
$\to$ why? Because $\hat H$ is just a function of $N$, so they share eigenkets. Now,
$$ \hat H|n\rangle = \hbar\omega\left(N + \frac{1}{2}\right)|n\rangle = \hbar\omega\left(n + \frac{1}{2}\right)|n\rangle $$
$$ \implies \boxed{E_n = \left(n + \frac{1}{2}\right)\hbar\omega} $$
$\to$ the energy eigenvalues.
# Ladder relations
We can show that
$$ [N, a] = [a^\dagger a, a] = a^\dagger[a, a] + [a^\dagger, a]a = -a $$
and
$$ [N, a^\dagger] = a^\dagger $$
$$ \implies \boxed{Na^\dagger|n\rangle = \left([N, a^\dagger] + a^\dagger N\right)|n\rangle = (n + 1)a^\dagger|n\rangle} $$
and
$$ \boxed{Na|n\rangle = (n - 1)a|n\rangle} $$
Sit with this.
These relations imply that $a^\dagger|n\rangle$ ($a|n\rangle$) is also an eigenket of $N$, with the eigenvalue increased (or decreased) by one. Same story as the angular momentum ladders.
We can say
$$ \boxed{a|n\rangle = c|n - 1\rangle} $$
because we denote the normalized eigenket corresponding to the eigenvalue $n - 1$ as $|n - 1\rangle$:
$$ N|n - 1\rangle = (n - 1)|n - 1\rangle $$
For a quantum harmonic oscillator, the energy spectrum (and the spectrum of $n$) is non-degenerate $\implies$ there is only one unique quantum state, up to a scalar constant, for each eigenvalue.
Since both $a|n\rangle$ and $|n - 1\rangle$ have the same eigenvalue $(n - 1)$, they must be the same state up to a constant.
To find $c$:
$$ a|n\rangle = c|n - 1\rangle \implies \langle n|a^\dagger = c^{\ast}\langle n - 1| $$
$$ \implies \langle n|a^\dagger a|n\rangle = |c|^2 $$
(assuming $|n - 1\rangle$ is normalized). And $a^\dagger a = N$, with $N|n\rangle = n|n\rangle$:
$$ \implies \langle n|n|n\rangle = |c|^2 \implies \boxed{n = |c|^2} $$
(assuming $|n\rangle$ is normalized). Taking $c$ to be real and positive, we obtain
$$ \boxed{a|n\rangle = \sqrt n|n - 1\rangle} $$
Similarly (using $aa^\dagger = N + 1$),
$$ \boxed{a^\dagger|n\rangle = \sqrt{n + 1}|n + 1\rangle} $$
# n is a non-negative integer
If we keep applying the annihilation operator here,
$$ \begin{aligned} a^2|n\rangle &= \sqrt{n(n - 1)}|n - 2\rangle \\ a^3|n\rangle &= \sqrt{n(n - 1)(n - 2)}|n - 3\rangle \end{aligned} $$
$\to$ the sequence will terminate if $n$ is an integer.
We also have the positivity requirement for the norm of $a|n\rangle$:
$$ n = \langle n|N|n\rangle = \left(\langle n|a^\dagger\right)\left(a|n\rangle\right) \geq 0 $$
$\therefore$ $n$ can never be negative. So, the sequence must terminate with $n = 0$, and $n$ is a non-negative integer.
(If $n$ wasn’t an integer, applying $a$ enough times would land us on a state with negative $n$, which can’t exist.)
So, the smallest value of $n$ is zero, and the ground state energy becomes
$$ \boxed{E_0 = \frac{1}{2}\hbar\omega} $$
# Building every state from the ground state
Now, we can successively apply creation to the ground state:
$$ \begin{aligned} a^\dagger|0\rangle &= |1\rangle \\ a^\dagger|1\rangle &= \sqrt 2|2\rangle \implies |2\rangle = \frac{a^\dagger|1\rangle}{\sqrt 2} = \frac{(a^\dagger)^2}{\sqrt 2}|0\rangle \\ a^\dagger|2\rangle &= \sqrt 3|3\rangle \implies |3\rangle = \frac{(a^\dagger)^3}{\sqrt{2\cdot 3}}|0\rangle \end{aligned} $$
$$ \implies \boxed{|n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle} $$
$\to$ the $n$-th state.
Now we have simultaneous eigenkets of $N$ and $\hat H$, with energy eigenvalues
$$ \boxed{E_n = \left(n + \frac{1}{2}\right)\hbar\omega, \qquad n \in \lbrace 0, 1, 2, \dots\rbrace} $$
# Matrix elements
So, from $a|n\rangle = \sqrt n|n - 1\rangle$ and $a^\dagger|n\rangle = \sqrt{n + 1}|n + 1\rangle$ (and orthogonality):
$$ \boxed{\langle n'|a|n\rangle = \sqrt n\delta_{n',n-1}, \qquad \langle n'|a^\dagger|n\rangle = \sqrt{n + 1}\delta_{n',n+1}} $$
For example,
$$ a \doteq \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & \sqrt 2 & 0 & \cdots \\ 0 & 0 & 0 & \sqrt 3 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} $$
Rewriting $\hat x$ and $\hat p$ with $a$ and $a^\dagger$ (just invert the definitions),
$$ \hat x = \sqrt{\frac{\hbar}{2m\omega}}\left(a + a^\dagger\right), \qquad \hat p = i\sqrt{\frac{m\hbar\omega}{2}}\left(-a + a^\dagger\right) $$
we get the matrix elements for $\hat x$ and $\hat p$:
$$ \boxed{ \begin{aligned} \langle n'|\hat x|n\rangle &= \sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt n\delta_{n',n-1} + \sqrt{n + 1}\delta_{n',n+1}\right) \\ \langle n'|\hat p|n\rangle &= i\sqrt{\frac{m\hbar\omega}{2}}\left(-\sqrt n\delta_{n',n-1} + \sqrt{n + 1}\delta_{n',n+1}\right) \end{aligned} } $$
$\to$ literally just replace $a$ and $a^\dagger$ with their matrix elements and factor out the constants.
Only $N$ and $\hat H$ are diagonal in the $|n\rangle$ basis. $a$ and $a^\dagger$ each live on a single line just off the diagonal, and $\hat x$ and $\hat p$ (mixes of both) live on both of those lines.
# The ground state wave function
Can we obtain the energy eigenfunctions in position space?
So, we know
$$ a|0\rangle = 0 $$
$\to$ the ground state. And $a = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat x + \frac{i\hat p}{m\omega}\right)$
$$ \implies \left(\hat x + \frac{i\hat p}{m\omega}\right)|0\rangle = 0 $$
$\to$ project this onto position space with $\langle x'|$:
$$ \implies \langle x'|\left(\hat x + \frac{i\hat p}{m\omega}\right)|0\rangle = 0 \implies \langle x'|\hat x|0\rangle + \frac{i}{m\omega}\langle x'|\hat p|0\rangle = 0 $$
The first term is $x'\langle x'|0\rangle$ (from the eigenvalue equation), and the second one we know from the wave functions notes: $\langle x'|\hat p|\alpha\rangle = -i\hbar\frac{\partial}{\partial x'}\langle x'|\alpha\rangle$.
$$ \implies x'\langle x'|0\rangle + \frac{i}{m\omega}\left(-i\hbar\frac{\partial}{\partial x'}\langle x'|0\rangle\right) = 0 $$
$$ \implies \boxed{\frac{\hbar}{m\omega}\frac{\partial}{\partial x'}\langle x'|0\rangle + x'\langle x'|0\rangle = 0} $$
$\to$ that’s the equation. Let
$$ x_0 = \sqrt{\frac{\hbar}{m\omega}} $$
$\to$ this is the length scale of the problem / oscillations.
$$ \implies \boxed{\left(x' + x_0^2\frac{\partial}{\partial x'}\right)\langle x'|0\rangle = 0} $$
Now, we know $\langle x'|0\rangle = \psi_0(x')$
$$ \implies \int\frac{d\psi_0}{\psi_0} = -\int\frac{x'}{x_0^2}dx' $$
$\to$ solve it:
$$ \ln\psi_0 = -\frac{x'^2}{2x_0^2} + \text{const} \implies \psi_0(x') = A\exp\left(-\frac{1}{2}\left(\frac{x'}{x_0}\right)^2\right) $$
and normalizing, $\int|\psi_0|^2dx' = A^2\sqrt\pi x_0 = 1$, so $A = \frac{1}{\pi^{1/4}\sqrt{x_0}}$:
$$ \boxed{\langle x'|0\rangle = \frac{1}{\pi^{1/4}\sqrt{x_0}}\exp\left(-\frac{1}{2}\left(\frac{x'}{x_0}\right)^2\right)} $$
This is a Gaussian wave function for the ground state: the Gaussian wave packet with $k = 0$ and $d = x_0$, so the ground state is a minimum uncertainty state.
The highest prob. is at the centre, and not at the edges like classical systems. We can find the excited states as well, which is the next notes.
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