The Quantum Harmonic Oscillator and Ladder Operators
The Quantum Harmonic Oscillator and Ladder Operators
The quantum harmonic oscillator (QHO) is arguably the single most important exactly solvable system in all of quantum mechanics. Its significance extends far beyond the simple physical system of a particle in a parabolic potential well: the QHO is the universal model for any quantum system near a stable equilibrium, the foundation of quantum field theory (where each field mode is an independent oscillator), the mathematical backbone of quantum optics (where photons are excitations of oscillator modes), and the prototype for coherent and squeezed states. Any smooth potential $V(q)$ with a minimum at $q_0$ can be expanded as $V(q) \approx V(q_0) + \frac{1}{2}V''(q_0)(q-q_0)^2 + \ldots$, reducing the small-oscillation problem to a harmonic oscillator. The range of applicability is, accordingly, enormous.
The Hamiltonian in position representation is $\hat{H} = \hat{p}^2/(2m) + m\omega^2\hat{x}^2/2$. The conventional approach solves the TISE directly as a differential equation, yielding solutions in terms of Hermite polynomials multiplied by a Gaussian. This approach, while complete, obscures the elegant algebraic structure underlying the spectrum. The operator-algebraic approach, using the creation and annihilation (ladder) operators $\hat{a}^\dagger$ and $\hat{a}$, reveals the structure directly: the spectrum is exactly $E_n = (n+1/2)\hbar\omega$ for $n = 0,1,2,\ldots$, with a zero-point energy $E_0 = \hbar\omega/2$ that reflects the uncertainty principle's prohibition on a particle being simultaneously at rest at the equilibrium position.
The ladder operators are defined as $\hat{a} = \sqrt{m\omega/(2\hbar)}\hat{x} + i\hat{p}/\sqrt{2m\hbar\omega}$ and $\hat{a}^\dagger = \sqrt{m\omega/(2\hbar)}\hat{x} - i\hat{p}/\sqrt{2m\hbar\omega}$. They satisfy the canonical commutation relation $[\hat{a},\hat{a}^\dagger] = 1$, which follows immediately from $[\hat{x},\hat{p}] = i\hbar$. The number operator $\hat{N} = \hat{a}^\dagger\hat{a}$ has eigenvalues $n = 0,1,2,\ldots$ — the occupation numbers — and the Hamiltonian is elegantly expressed as $\hat{H} = \hbar\omega(\hat{N}+1/2)$. This formulation is completely algebraic: once the commutation relation $[\hat{a},\hat{a}^\dagger]=1$ is established, the entire spectrum follows without solving any differential equation.
The Fock states (number states) $|n\rangle$ — eigenstates of $\hat{N}$ — are constructed by repeated application of $\hat{a}^\dagger$ to the vacuum $|0\rangle$: $|n\rangle = (\hat{a}^\dagger)^n/\sqrt{n!}|0\rangle$. The ground state $|0\rangle$ is annihilated by $\hat{a}$: $\hat{a}|0\rangle = 0$. In position representation, this condition $\hat{a}\psi_0(x) = 0$ is a first-order ODE with solution $\psi_0(x) \propto e^{-m\omega x^2/(2\hbar)} = e^{-x^2/(2x_0^2)}$ where $x_0 = \sqrt{\hbar/(m\omega)}$ is the natural length scale. Higher states are $\psi_n(x) = (\sqrt{\pi}\,2^n\,n!\,x_0)^{-1/2}H_n(x/x_0)e^{-x^2/(2x_0^2)}$ where $H_n$ are Hermite polynomials. The ladder operators act as $\hat{a}|n\rangle = \sqrt{n}|n-1\rangle$ and $\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$.
Expectation values in Fock states reveal the fundamentally non-classical nature of these states. $\langle n|\hat{x}|n\rangle = 0$ and $\langle n|\hat{p}|n\rangle = 0$ for all $n$ — the particle is on average at the equilibrium position with zero mean momentum. Yet $\langle n|\hat{x}^2|n\rangle = (2n+1)\hbar/(2m\omega)$ and $\langle n|\hat{p}^2|n\rangle = (2n+1)m\hbar\omega/2$. The uncertainty product $\Delta x\Delta p = (n+1/2)\hbar$ grows linearly with $n$, exceeding the minimum $\hbar/2$ for all $n > 0$. This means Fock states are not minimum-uncertainty states except for the vacuum $|0\rangle$.
Coherent states $|\alpha\rangle$ (where $\alpha \in \mathbb{C}$) are defined as eigenstates of the annihilation operator: $\hat{a}|\alpha\rangle = \alpha|\alpha\rangle$. They are not eigenstates of the Hermitian number operator, but rather superpositions of all Fock states with a Poissonian photon-number distribution $P(n) = e^{-|\alpha|^2}|\alpha|^{2n}/n!$ with mean $ar{n} = |\alpha|^2$. Coherent states are minimum-uncertainty states ($\Delta x\Delta p = \hbar/2$) at all times. Under time evolution, a coherent state remains coherent: $|\alpha(t)\rangle = |\alpha e^{-i\omega t}\rangle$. The expectation values $\langle\hat{x}(t)\rangle = x_0\sqrt{2}\text{Re}(\alpha e^{-i\omega t})$ and $\langle\hat{p}(t)\rangle = p_0\sqrt{2}\text{Im}(\alpha e^{-i\omega t})$ trace out the classical elliptical phase-space orbit — coherent states are the closest quantum analog of a classical oscillating particle.
Squeezed states generalize coherent states by allowing the uncertainty in one quadrature to be reduced below the vacuum level at the expense of increased uncertainty in the conjugate quadrature, while still saturating the Heisenberg bound. For a squeezed vacuum with squeezing parameter $r$ along the $x$ quadrature: $\Delta x = x_0 e^{-r}$ and $\Delta p = (\hbar/(2x_0))e^{+r}$, giving $\Delta x\Delta p = \hbar/2$. Squeezed states are generated by the squeezing operator $\hat{S}(\xi) = \exp[(\xi^*(\hat{a})^2 - \xi(\hat{a}^\dagger)^2)/2]$ and are the quantum optical resource enabling sub-shot-noise measurements and gravitational wave detection (LIGO uses squeezed light to improve its sensitivity below the standard quantum limit).
The connection to quantum field theory and condensed matter physics is direct. Each mode of the electromagnetic field is an independent harmonic oscillator; photons are the quanta (Fock state excitations) of these modes. Phonons — quantized lattice vibrations — are the analogous quanta in crystalline solids. The creation operator $\hat{a}_k^\dagger$ creates a photon (or phonon) in mode $k$, and the Hamiltonian of the free field is $\hat{H} = \sum_k \hbar\omega_k(\hat{n}_k + 1/2)$. The infinite zero-point energy $\sum_k \hbar\omega_k/2$ (vacuum energy) is a formal divergence that must be renormalized in quantum field theory but has physical observable consequences such as the Casimir effect and the Lamb shift. This lesson provides the algebraic foundation that makes all of these applications accessible.
The quantum harmonic oscillator also provides a natural framework for studying time evolution and quantum dynamics. The Heisenberg equations of motion for the ladder operators are $d\hat{a}/dt = -i\omega\hat{a}$ and $d\hat{a}^\dagger/dt = +i\omega\hat{a}^\dagger$, giving $\hat{a}(t) = \hat{a}(0)e^{-i\omega t}$ — the operator rotates in phase space exactly as a classical complex amplitude. The time evolution of any observable can be computed from this, giving exact, closed-form results. This integrability is exceptional among quantum systems and is the reason the harmonic oscillator serves as the anchor point for perturbation theory (anharmonic corrections), the basis for thermal field theory (via coherent state path integrals), and the mathematical model for quantum noise in open quantum systems.
For a harmonic oscillator with Hamiltonian $\hat{H} = \hat{p}^2/(2m)+m\omega^2\hat{x}^2/2$, define: $$\hat{a} = \sqrt{\frac{m\omega}{2\hbar}}\hat{x} + \frac{i}{\sqrt{2m\hbar\omega}}\hat{p}, \qquad \hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\hat{x} - \frac{i}{\sqrt{2m\hbar\omega}}\hat{p}.$$ Inversely: $\hat{x} = \sqrt{\hbar/(2m\omega)}(\hat{a}+\hat{a}^\dagger)$ and $\hat{p} = i\sqrt{m\hbar\omega/2}(\hat{a}^\dagger-\hat{a})$. Key properties: $[\hat{a},\hat{a}^\dagger] = 1$, $\hat{H} = \hbar\omega(\hat{a}^\dagger\hat{a}+\frac{1}{2}) = \hbar\omega(\hat{N}+\frac{1}{2})$, $[\hat{N},\hat{a}] = -\hat{a}$, $[\hat{N},\hat{a}^\dagger] = +\hat{a}^\dagger$. Action on Fock states: $\hat{a}|n\rangle = \sqrt{n}|n-1\rangle$, $\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$, $\hat{N}|n\rangle = n|n\rangle$.
A coherent state $|\alpha\rangle$ (where $\alpha \in \mathbb{C}$) is an eigenstate of the annihilation operator: $\hat{a}|\alpha\rangle = \alpha|\alpha\rangle$. Explicitly: $$|\alpha\rangle = e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}|n\rangle = e^{-|\alpha|^2/2}e^{\alpha\hat{a}^\dagger}|0\rangle = \hat{D}(\alpha)|0\rangle,$$ where $\hat{D}(\alpha) = e^{\alpha\hat{a}^\dagger - \alpha^*\hat{a}}$ is the displacement operator. Properties: $\langle\hat{N}\rangle = |\alpha|^2$, photon number distribution $P(n) = e^{-|\alpha|^2}|\alpha|^{2n}/n!$ (Poissonian), $\Delta n = \sqrt{|\alpha|^2} = \sqrt{\bar{n}}$. Overlap: $|\langle\beta|\alpha\rangle|^2 = e^{-|\alpha-\beta|^2}$ — coherent states are non-orthogonal, overcomplete: $\frac{1}{\pi}\int|\alpha\rangle\langle\alpha|d^2\alpha = \hat{\mathbf{1}}$.
The $n$-th energy eigenstate in position representation is: $$\psi_n(x) = \frac{1}{\sqrt{2^n n!}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right)\exp\left(-\frac{m\omega x^2}{2\hbar}\right),$$ where $H_n(\xi)$ is the $n$-th Hermite polynomial: $H_0=1$, $H_1=2\xi$, $H_2=4\xi^2-2$, $H_3=8\xi^3-12\xi$. Recurrence: $H_{n+1}(\xi) = 2\xi H_n(\xi) - 2nH_{n-1}(\xi)$. Classical turning points: $\pm x_{\rm cl}^{(n)} = \pm\sqrt{(2n+1)\hbar/(m\omega)}$ where $E_n = m\omega^2x_{\rm cl}^2/2$. The number of nodes of $\psi_n$ equals $n$. Orthonormality: $\int_{-\infty}^\infty \psi_m^*(x)\psi_n(x)dx = \delta_{mn}$.
From $[\hat{a},\hat{a}^\dagger]=1$ and $\hat{H} = \hbar\omega(\hat{N}+1/2)$, the complete spectrum follows: (1) $\hat{N}$ has non-negative eigenvalues: for eigenvector $\hat{N}|n\rangle=n|n\rangle$, $n = \langle n|\hat{a}^\dagger\hat{a}|n\rangle = \|\hat{a}|n\rangle\|^2 \geq 0$. (2) $\hat{a}|n\rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $n-1$: $\hat{N}(\hat{a}|n\rangle) = [\hat{N},\hat{a}]|n\rangle + \hat{a}\hat{N}|n\rangle = -\hat{a}|n\rangle + n\hat{a}|n\rangle = (n-1)\hat{a}|n\rangle$. Similarly $\hat{N}(\hat{a}^\dagger|n\rangle) = (n+1)(\hat{a}^\dagger|n\rangle)$. (3) Since eigenvalues are non-negative, the chain $n, n-1, n-2, \ldots$ must terminate at 0: $\hat{a}|0\rangle = 0$. This forces $n \in \mathbb{Z}_{\geq 0}$. (4) Normalization: $\|\hat{a}^\dagger|n\rangle\|^2 = \langle n|\hat{a}\hat{a}^\dagger|n\rangle = \langle n|(\hat{N}+1)|n\rangle = n+1$. Hence $\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$ and the spectrum is $E_n = (n+\frac{1}{2})\hbar\omega$, $n = 0,1,2,\ldots$ $\square$
Under Hamiltonian $\hat{H} = \hbar\omega(\hat{N}+1/2)$, a coherent state evolves as: $$e^{-i\hat{H}t/\hbar}|\alpha\rangle = e^{-i\omega t/2}|\alpha e^{-i\omega t}\rangle.$$ The phase $e^{-i\omega t/2}$ is the vacuum phase. The amplitude $\alpha(t) = \alpha e^{-i\omega t}$ rotates in the complex plane at angular frequency $\omega$, so $|\alpha(t)|$ and hence $\bar{n} = |\alpha|^2$ are constant. Expectation values: $\langle\hat{x}(t)\rangle = \sqrt{2\hbar/(m\omega)}\text{Re}(\alpha e^{-i\omega t})$, $\langle\hat{p}(t)\rangle = \sqrt{2m\hbar\omega}\text{Im}(\alpha e^{-i\omega t})$. These are exactly the classical harmonic oscillator equations $x(t) = x_0\cos(\omega t + \phi)$, $p(t) = -m\omega x_0\sin(\omega t+\phi)$ — coherent states are the closest quantum analog of classical oscillation.
Derive the position-space wave function for the ground state of the quantum harmonic oscillator using the condition $\hat{a}|0\rangle = 0$. Then verify that $\psi_1(x) = \hat{a}^\dagger\psi_0(x)/\|\hat{a}^\dagger\psi_0\|$ using the differential form of $\hat{a}^\dagger$. Compute $\langle x^2\rangle_0$ and $\langle p^2\rangle_0$ and verify $E_0 = \langle H\rangle_0 = \hbar\omega/2$.
Ground state condition: $\hat{a}|0\rangle = 0$ in position representation: $\left(\sqrt{\frac{m\omega}{2\hbar}}x + \frac{\hbar}{\sqrt{2m\hbar\omega}}\frac{d}{dx}\right)\psi_0(x) = 0$. This simplifies to $\frac{d\psi_0}{dx} = -\frac{m\omega}{\hbar}x\psi_0$, a separable ODE with solution $\psi_0(x) = Ce^{-m\omega x^2/(2\hbar)}$. Normalization: $|C|^2\sqrt{\pi\hbar/(m\omega)} = 1$, giving $C = (m\omega/(\pi\hbar))^{1/4}$. For $\psi_1$: $\hat{a}^\dagger = \sqrt{m\omega/(2\hbar)}x - \hbar/\sqrt{2m\hbar\omega}\cdot d/dx$. Acting on $\psi_0$: $\hat{a}^\dagger\psi_0 = \sqrt{m\omega/(2\hbar)}x\psi_0 - \hbar/\sqrt{2m\hbar\omega}\cdot(-m\omega x/\hbar)\psi_0 = 2\sqrt{m\omega/(2\hbar)}x\psi_0 = \sqrt{2m\omega/\hbar}x\psi_0$. With $x_0 = \sqrt{\hbar/(m\omega)}$: $\psi_1 \propto (x/x_0)\psi_0 = H_1(x/x_0)\psi_0$. ✓ $\langle x^2\rangle_0 = \int x^2|\psi_0|^2dx = \hbar/(2m\omega)$ (Gaussian integral). $\langle p^2\rangle_0 = m\hbar\omega/2$. $E_0 = \langle p^2\rangle/(2m) + m\omega^2\langle x^2\rangle/2 = \hbar\omega/4 + \hbar\omega/4 = \hbar\omega/2$. ✓
Compute the matrix elements $\langle m|\hat{x}^2|n\rangle$ and $\langle m|\hat{x}^4|n\rangle$ using ladder operators. Use the result to find the expectation value of $\hat{x}^4$ in the $n$-th Fock state (needed for anharmonic perturbation theory).
$\hat{x} = x_0(\hat{a}+\hat{a}^\dagger)$ where $x_0 = \sqrt{\hbar/(2m\omega)}$. Then $\hat{x}^2 = x_0^2(\hat{a}+\hat{a}^\dagger)^2 = x_0^2(\hat{a}^2 + \hat{a}\hat{a}^\dagger + \hat{a}^\dagger\hat{a} + (\hat{a}^\dagger)^2) = x_0^2(\hat{a}^2 + 2\hat{N}+1+(\hat{a}^\dagger)^2)$ (using $\hat{a}\hat{a}^\dagger = \hat{N}+1$). Matrix elements: $\langle m|\hat{x}^2|n\rangle = x_0^2(\sqrt{n(n-1)}\delta_{m,n-2} + (2n+1)\delta_{m,n} + \sqrt{(n+1)(n+2)}\delta_{m,n+2})$. Diagonal: $\langle n|\hat{x}^2|n\rangle = x_0^2(2n+1) = (2n+1)\hbar/(2m\omega)$. For $\hat{x}^4 = x_0^4(\hat{a}+\hat{a}^\dagger)^4$: expand and normal-order. $\langle n|\hat{x}^4|n\rangle = x_0^4(6\hat{N}^2+6\hat{N}+3)|_{\rm diagonal} = x_0^4(6n^2+6n+3) = \frac{\hbar^2}{4m^2\omega^2}(6n^2+6n+3)$. This result is used in first-order perturbation theory for the anharmonic term $\lambda\hat{x}^4$: $E_n^{(1)} = \lambda\langle n|\hat{x}^4|n\rangle = \frac{3\lambda\hbar^2}{4m^2\omega^2}(2n^2+2n+1)$.
Show that a coherent state $|\alpha\rangle$ has Poissonian photon statistics, is a minimum-uncertainty state, and prove the overcomplete resolution of identity $\pi^{-1}\int|\alpha\rangle\langle\alpha|d^2\alpha = \hat{\mathbf{1}}$ (where $d^2\alpha = d\text{Re}(\alpha)\,d ext{Im}(\alpha)$).
Fock expansion: $|\alpha\rangle = e^{-|\alpha|^2/2}\sum_n \alpha^n/\sqrt{n!}|n\rangle$. Photon number probability: $P(n) = |\langle n|\alpha\rangle|^2 = e^{-|\alpha|^2}|\alpha|^{2n}/n!$ — Poissonian with mean $\bar{n} = |\alpha|^2$ and variance $\Delta n^2 = \bar{n}$. ✓ Uncertainty: $\langle\hat{x}\rangle = x_0(\alpha+\alpha^*) = 2x_0\text{Re}(\alpha)$. $\langle\hat{x}^2\rangle = x_0^2\langle(\hat{a}+\hat{a}^\dagger)^2\rangle = x_0^2(\alpha^2+2|\alpha|^2+1+(\alpha^*)^2) = x_0^2(|\alpha+\alpha^*|^2/... )$. More carefully: $\langle\hat{x}^2\rangle = x_0^2(\alpha^2+1+(\alpha^*)^2+2|\alpha|^2-...$ let me use $\hat{a}|\alpha\rangle=\alpha|\alpha\rangle$: $\langle\hat{x}^2\rangle = x_0^2\langle(\hat{a}+\hat{a}^\dagger)^2\rangle = x_0^2(\alpha^2+2|\alpha|^2+1+(\alpha^*)^2)$... $\langle\hat{x}\rangle^2 = x_0^2(\alpha+\alpha^*)^2 = x_0^2(\alpha^2+2|\alpha|^2+(\alpha^*)^2)$. So $\Delta x^2 = x_0^2 = \hbar/(2m\omega)$, independent of $\alpha$. Similarly $\Delta p^2 = m\hbar\omega/2$. $\Delta x\Delta p = \sqrt{\hbar/(2m\omega)}\sqrt{m\hbar\omega/2} = \hbar/2$. ✓ Minimum uncertainty. For the overcomplete identity: $\frac{1}{\pi}\int|\alpha\rangle\langle\alpha|d^2\alpha = \frac{1}{\pi}\int e^{-|\alpha|^2}\sum_{m,n}\frac{\alpha^n(\alpha^*)^m}{\sqrt{n!m!}}d^2\alpha\,|n\rangle\langle m|$. Using polar $\alpha = re^{i\theta}$, $d^2\alpha = r\,dr\,d\theta$: the angular integral $\int_0^{2\pi}e^{i(n-m)\theta}d\theta = 2\pi\delta_{nm}$. Then $\frac{2}{n!}\int_0^\infty e^{-r^2}r^{2n+1}dr = 1$ (by substitution $u=r^2$). So $\frac{1}{\pi}\int|\alpha\rangle\langle\alpha|d^2\alpha = \sum_n|n\rangle\langle n| = \hat{\mathbf{1}}$. ✓
Derive the thermal state density matrix for the quantum harmonic oscillator at temperature $T$, compute the mean photon number $\bar{n}$, and show the photon number distribution is geometric (Bose-Einstein). Find the Wigner function of the thermal state and interpret its Gaussian form.
The thermal (canonical) density matrix is $\rho_{\rm th} = Z^{-1}e^{-\hat{H}/(k_BT)}$ where $Z = \text{Tr}(e^{-\hat{H}/(k_BT)}) = \sum_{n=0}^\infty e^{-\hbar\omega(n+1/2)/(k_BT)} = e^{-\hbar\omega/(2k_BT)}/(1-e^{-\hbar\omega/(k_BT)})$. Define $\beta = \hbar\omega/(k_BT)$. In the Fock basis: $\rho_{\rm th} = (1-e^{-\beta})\sum_n e^{-n\beta}|n\rangle\langle n|$. Mean photon number: $\bar{n} = \text{Tr}(\hat{N}\rho_{\rm th}) = \sum_n n(1-e^{-\beta})e^{-n\beta} = e^{-\beta}/(1-e^{-\beta}) = 1/(e^{\beta}-1)$ — the Bose-Einstein distribution. Probability $P(n) = (1-e^{-\beta})e^{-n\beta} = \bar{n}^n/(1+\bar{n})^{n+1}$ — geometric distribution. Variance: $\Delta n^2 = \bar{n}(\bar{n}+1) > \bar{n}$ (super-Poissonian — thermal noise exceeds shot noise). Wigner function: $W(x,p) = \frac{1}{\pi\hbar}\tanh(\beta/2)\exp\left(-\frac{\tanh(\beta/2)}{\hbar}\left[m\omega x^2+\frac{p^2}{m\omega}\right]\right)$, a Gaussian of width $\sqrt{\coth(\beta/2)/(2m\omega)}$ in $x$ (broader than vacuum for $T>0$). At $T=0$: $\tanh(\beta/2) \to 1$, recovering the vacuum Wigner function. At $T\to\infty$: $\tanh(\beta/2) \approx \beta/2 \to 0$, the Gaussian broadens to reproduce the classical phase-space distribution.
Practice Problems
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1. Key steps: (i) $[\hat{a},\hat{a}^\dagger]=1$ gives $[\hat{N},\hat{a}]=-\hat{a}$, $[\hat{N},\hat{a}^\dagger]=+\hat{a}^\dagger$. (ii) If $|n\rangle$ is an eigenvector with eigenvalue $n$, then $\hat{a}|n\rangle$ has eigenvalue $n-1$ and $\hat{a}^\dagger|n\rangle$ has eigenvalue $n+1$. (iii) Non-negativity: $n = \|\hat{a}|n\rangle\|^2 \geq 0$. (iv) Termination: chain must end at $\hat{a}|0\rangle=0$, forcing $n \in \mathbb{Z}_{\geq 0}$. (v) $\hat{H}|n\rangle = \hbar\omega(n+1/2)|n\rangle$. No differential equations needed — purely algebraic from $[\hat{a},\hat{a}^\dagger]=1$.
2. $\hat{x}^3 = x_0^3(\hat{a}+\hat{a}^\dagger)^3 = x_0^3(\hat{a}^3 + 3\hat{a}^\dagger\hat{a}^2 + 3(\hat{a}^\dagger)^2\hat{a} + (\hat{a}^\dagger)^3 + 3\hat{a} + 3\hat{a}^\dagger)$ (after normal ordering). Diagonal elements $\langle n|\hat{x}^3|n\rangle$: only surviving terms change $n$ by 0 — but the expansion contains $\hat{a}^3$, $\hat{a}(\hat{a}^\dagger)^2\hat{a}$, etc. All terms in $(\hat{a}+\hat{a}^\dagger)^3$ change $n$ by $\pm 1$ or $\pm 3$ — none by 0. Hence $\langle n|\hat{x}^3|n\rangle = 0$, first-order correction vanishes. Second order: $E_n^{(2)} = \sum_{m\neq n}|\langle m|\hat{x}^3|n\rangle|^2/(E_n^{(0)}-E_m^{(0)}) \neq 0$ since off-diagonal matrix elements are non-zero.
3. $\langle\hat{x}(t)\rangle = \langle\psi(t)|x_0(\hat{a}+\hat{a}^\dagger)|\psi(t)\rangle$. $|\psi(t)\rangle = (e^{-i\omega t/2}|0\rangle + e^{-3i\omega t/2}|1\rangle)/\sqrt{2}$. $\langle\hat{x}\rangle = x_0\text{Re}(\langle 0|e^{i\omega t/2}\cdot e^{-3i\omega t/2}\hat{a}^\dagger|0\rangle\cdot 1 + \text{c.c.}) = x_0\cos(\omega t)$. Similarly $\langle\hat{p}(t)\rangle = -m\omega x_0\sin(\omega t) = -p_0\sin(\omega t)$ where $p_0 = m\omega x_0$. $\langle\hat{H}\rangle = (\hbar\omega/2 + 3\hbar\omega/2)/2 = \hbar\omega$.
4. $\hat{D}^\dagger(\alpha) = e^{\alpha^*\hat{a}-\alpha\hat{a}^\dagger} = \hat{D}(-\alpha) = \hat{D}^{-1}(\alpha)$ — unitary. $\hat{D}(\alpha)\hat{a}\hat{D}^\dagger(\alpha)$: use BCH $e^X\hat{A}e^{-X} = \hat{A}+[X,\hat{A}]+...$. With $X = \alpha\hat{a}^\dagger-\alpha^*\hat{a}$: $[X,\hat{a}] = -\alpha[\hat{a}^\dagger,\hat{a}] = \alpha\cdot(-1) = -\alpha$. Higher commutators vanish. So $\hat{D}\hat{a}\hat{D}^\dagger = \hat{a} - \alpha$. Then $\hat{a}|\alpha\rangle = \hat{a}\hat{D}(\alpha)|0\rangle = \hat{D}(\alpha)(\hat{D}^\dagger\hat{a}\hat{D})|0\rangle\cdot... = \hat{D}(\alpha)(\hat{a}+\alpha)|0\rangle = \hat{D}(\alpha)\alpha|0\rangle = \alpha|\alpha\rangle$. ✓
5. 3D states labeled $(n_x,n_y,n_z)$ with $n_x+n_y+n_z = N$. Degeneracy: number of non-negative integer solutions = $\binom{N+2}{2} = (N+1)(N+2)/2$. Ground state ($N=0$): non-degenerate, $E_0 = 3\hbar\omega/2$. Compare 1D: $E_0^{1D} = \hbar\omega/2$. The 3D zero-point energy is three times larger (three independent degrees of freedom).
6. $\hat{a}^\dagger|\alpha\rangle = e^{-|\alpha|^2/2}\sum_n \alpha^n/\sqrt{n!}\cdot\sqrt{n+1}|n+1\rangle$. If $\hat{a}^\dagger$ had eigenvalue $\beta$, all coefficients would satisfy $\sqrt{n+1} = \beta$ for all $n$ — impossible. Hence no eigenstates. $\langle\alpha|\hat{a}^\dagger|\alpha\rangle = \alpha^*$ (complex conjugate of eigenvalue of $\hat{a}$). The non-existence of eigenstates of $\hat{a}^\dagger$ follows from: if $\hat{a}^\dagger|v\rangle = \mu|v\rangle$, taking the adjoint gives $\langle v|\hat{a} = \mu^*\langle v|$, so $\hat{a}$ would have eigenvalue $\mu^*$ with eigenvector $|v\rangle$. Writing in Fock basis: $v_n\sqrt{n} = \mu^* v_{n-1}$ gives $v_n = (\mu^*)^n/\sqrt{n!}\cdot v_0$, so $\|v\|^2 = e^{|\mu|^2}|v_0|^2 = \infty$ — not normalizable.
7. $W_1(x,p) = \frac{1}{\pi\hbar}(4x^2/x_0^2 + 4p^2/(\hbar/x_0)^2 - 2)e^{-x^2/x_0^2-p^2x_0^2/\hbar^2}$ (after calculation using the Wigner function formula). The factor $(4\xi^2-2)$ (where $\xi = x/x_0$) is negative when $|\xi| < 1/\sqrt{2}$. Specifically $W_1 < 0$ in the ellipse $m\omega x^2/\hbar + p^2/(m\hbar\omega) < 1/2$. This negativity of the Wigner function is a non-classical signature — impossible for any classical probability distribution on phase space.
8. Allowed modes between plates: $k_z = n\pi/d$, $n=1,2,...$ Energy $E = \hbar c\sqrt{k_x^2+k_y^2+(n\pi/d)^2}/2$ per mode, summed over all transverse $k_{x,y}$. After regularization (zeta function $\zeta(-3)=-1/120$): $\langle E\rangle/A = -\pi^2\hbar c/(720d^3)$. Force: $F/A = -\partial(E/A)/\partial d = -\pi^2\hbar c/(240d^4)$. Attractive (pressure inward). For $d=1\mu$m: $P \approx 1.3\times10^{-3}$ Pa — measurable and confirmed experimentally.
9. $\bar{n} = 1/(e^{\hbar\omega/(k_BT)}-1)$ (Bose-Einstein). $\langle E\rangle = \hbar\omega(\bar{n}+1/2) = \hbar\omega/2 + \hbar\omega/(e^{\hbar\omega/(k_BT)}-1)$. Classical limit $x = \hbar\omega/(k_BT) \ll 1$: $e^x-1 \approx x$, so $\bar{n} \approx k_BT/\hbar\omega \gg 1$, $\langle E\rangle \approx k_BT$ (equipartition). Quantum limit $x \gg 1$: $\bar{n} \approx e^{-x} \to 0$, $\langle E\rangle \approx \hbar\omega/2$ (zero-point energy dominates).
10. Differentiate $e^{2xt-t^2} = \sum_n H_n(x)t^n/n!$ with respect to $t$: $(2x-2t)e^{2xt-t^2} = \sum_n H_n(x)nt^{n-1}/n! = \sum_n H_n(x)t^{n-1}/(n-1)!$. Substituting the series: $\sum_n(2x H_n - 2H_{n-1})t^n/n! = \sum_n H_{n+1}t^n/n!$. So $H_{n+1} = 2xH_n - 2nH_{n-1}$. ✓ Orthogonality: use the Rodrigues formula $H_n(x) = (-1)^ne^{x^2}d^n(e^{-x^2})/dx^n$ and integrate by parts $n$ times to show $\int H_mH_ne^{-x^2}dx = 0$ for $m\neq n$ and $\sqrt{\pi}2^nn!$ for $m=n$.
11. $P_{\rm forbidden} = 2\int_{x_0}^\infty|\psi_0(x)|^2dx$ where $x_0 = \sqrt{\hbar/(m\omega)}$ is the classical turning point for $n=0$. $\psi_0 = (m\omega/(\pi\hbar))^{1/4}e^{-m\omega x^2/(2\hbar)}$. Let $u = x/x_0$: $P_{\rm forbidden} = \frac{2}{\sqrt{\pi}}\int_1^\infty e^{-u^2}du = \text{erfc}(1) \approx 0.1573$. About 15.7% probability outside classical turning points — a purely quantum effect with no classical analog.
12. BCH with $X = r(\hat{a}^2-(\hat{a}^\dagger)^2)/2$: $[X,\hat{a}] = r\hat{a}^\dagger$, $[X,[X,\hat{a}]] = r^2\hat{a}$, giving series $\hat{S}^\dagger\hat{a}\hat{S} = \hat{a}\cosh r - \hat{a}^\dagger\sinh r$. For squeezed vacuum $|\text{sq}\rangle = \hat{S}(r)|0\rangle$: $\langle\hat{x}^2\rangle = x_0^2\langle(\hat{a}+\hat{a}^\dagger)^2\rangle$. Use $\hat{a} \to \hat{A} = \hat{a}\cosh r - \hat{a}^\dagger\sinh r$ in the squeezed frame (vacuum averages). $\Delta x^2 = x_0^2(\cosh r - \sinh r)^2 = x_0^2 e^{-2r}$. $\Delta p^2 = (\hbar/(2x_0))^2 e^{+2r}$. $\Delta x\Delta p = (\hbar/2)e^{-r}e^{+r} = \hbar/2$. ✓ Minimum uncertainty achieved.