The Harmonic Oscillator

The Hamiltonian of the quantum harmonic oscillator is:

We define the ladder (raising and lowering) operators:

In terms of these operators, the Hamiltonian becomes $H = \hbar\omega(a^\dagger a + 1/2) = \hbar\omega(\hat{N} + 1/2)$, where $\hat{N} = a^\dagger a$ is the number operator. The key commutation relation is $[a, a^\dagger] = 1$.

The number operator $\hat{N}$ has non-negative integer eigenvalues $n = 0, 1, 2, \dots$ The corresponding eigenstates $|n\rangle$ are called number states. The energy eigenvalues are $E_n = \hbar\omega(n + 1/2)$. The ground state energy $E_0 = \hbar\omega/2$ is called the zero-point energy.

The time-independent Schrodinger equation for the harmonic oscillator is:

It is convenient to introduce the dimensionless variable $\zeta = \sqrt{m\omega/\hbar}\,x$ and the dimensionless energy $\epsilon = 2E/(\hbar\omega)$. The equation becomes:

For large $|\zeta|$, the term $\zeta^2$ dominates, so the asymptotic behavior is $\psi \sim e^{\pm\zeta^2/2}$. Only the decaying solution $e^{-\zeta^2/2}$ is normalizable.

We factor out the asymptotic behavior by writing $\psi(\zeta) = h(\zeta)\,e^{-\zeta^2/2}$. Substituting this into the differential equation gives:

This is the Hermite differential equation. Normalizable solutions exist only when $\epsilon - 1 = 2n$ for $n = 0, 1, 2, \dots$, which reproduces the quantization condition $E_n = \hbar\omega(n + 1/2)$. The functions $h_n(\zeta)$ are Hermite polynomials.

Coherent states $|\alpha\rangle$ are eigenstates of the lowering operator: $a|\alpha\rangle = \alpha|\alpha\rangle$. They can be expanded in the number basis as:

The probability of finding $n$ quanta follows a Poisson distribution: $P(n) = |\alpha|^{2n}e^{-|\alpha|^2}/n!$. The mean photon number is $\langle n\rangle = |\alpha|^2$ and the variance is $(\Delta n)^2 = |\alpha|^2$.

Coherent states are minimum-uncertainty states for position and momentum: $\Delta x = \sqrt{\hbar/(2m\omega)}$ and $\Delta p = \sqrt{m\omega\hbar/2}$, giving $\Delta x\,\Delta p = \hbar/2$. Their expectation values $\langle x(t)\rangle$ and $\langle p(t)\rangle$ oscillate exactly like a classical harmonic oscillator.

A coherent state evolves by acquiring a time-dependent phase: $|\alpha(t)\rangle = |\alpha e^{-i\omega t}\rangle$. This is why coherent states are called the most classical quantum states.

A harmonic oscillator is in the state $|\psi\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$. Compute $\langle x\rangle$, $\langle p\rangle$, $\langle x^2\rangle$, $\langle p^2\rangle$, and the uncertainties.

Step 1 — Express $x$ and $p$ in terms of ladder operators: $x = \sqrt{\hbar/(2m\omega)}\,(a + a^\dagger)$, $p = i\sqrt{m\omega\hbar/2}\,(a^\dagger - a)$.

Step 2 — Compute $\langle x\rangle$. $\langle\psi|a + a^\dagger|\psi\rangle/2 = (\langle 0|a|1\rangle + \langle 1|a^\dagger|0\rangle)/2 = (1 + 1)/2 = 1$. Thus $\langle x\rangle = \sqrt{\hbar/(2m\omega)}$.

Step 3 — Compute $\langle p\rangle$. $\langle\psi|a^\dagger - a|\psi\rangle = 0$ because the diagonal matrix elements vanish and the off-diagonal elements cancel. Thus $\langle p\rangle = 0$.

Step 4 — Compute $\langle x^2\rangle$. $x^2 = (\hbar/2m\omega)(a + a^\dagger)^2 = (\hbar/2m\omega)(a^2 + a^{\dagger 2} + a a^\dagger + a^\dagger a)$. Only the number operator terms contribute: $\langle x^2\rangle = (\hbar/2m\omega)(\langle 0|a a^\dagger|0\rangle + \langle 1|a^\dagger a|1\rangle)/2 = (\hbar/2m\omega)(1 + 1)/2 = \hbar/2m\omega$. Wait — $a a^\dagger = N + 1$, so $\langle 0|a a^\dagger|0\rangle = 1$ and $\langle 1|a a^\dagger|1\rangle = 2$. Thus $\langle x^2\rangle = (\hbar/2m\omega)(1 + 2)/2 = 3\hbar/4m\omega$.

Step 5 — Compute $\Delta x$. $\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} = \sqrt{3\hbar/4m\omega - \hbar/2m\omega} = \sqrt{\hbar/4m\omega} = \frac{1}{2}\sqrt{\hbar/m\omega}$.