Training Quantum Physics The Schrödinger Equation: Time-Dependent and Time-Independent Forms
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The Schrödinger Equation: Time-Dependent and Time-Independent Forms

35 min Quantum Physics

The Schrödinger Equation: Time-Dependent and Time-Independent Forms

The Schrödinger equation is the fundamental dynamical law of non-relativistic quantum mechanics, playing the role that Newton's second law plays in classical mechanics. It governs the time evolution of the quantum state and encodes the full predictive content of the theory for systems where relativistic effects are negligible. Unlike Newton's law, it is linear — the superposition principle holds exactly — and it is first-order in time, so the state at time $t_0$ uniquely determines the state for all future (and past) times. These properties have deep physical consequences: linearity implies interference, and the first-order character implies unitary (norm-preserving) evolution, which is essential for maintaining the probabilistic interpretation of the wave function.

The time-dependent Schrödinger equation (TDSE) $i\hbar\partial_t|\psi\rangle = \hat{H}|\psi\rangle$ is postulated rather than derived from more fundamental principles within the non-relativistic framework, though it can be motivated by several heuristic arguments. The most compelling motivation starts from the energy-momentum relation $E = p^2/(2m) + V(x)$ and promotes $E \to i\hbar\partial_t$ and $p \to -i\hbar\partial_x$ as operators acting on a wave function $\psi(x,t)$. This yields $i\hbar\partial_t\psi = [-\hbar^2\partial_x^2/(2m) + V]\psi$. The deep justification comes from quantum field theory and the path integral, where the Schrödinger equation emerges as the non-relativistic limit of the Klein-Gordon or Dirac equation in the appropriate regime.

For time-independent Hamiltonians $\hat{H}$, solutions with definite energy can be sought in the form $\psi(x,t) = \phi(x)e^{-iEt/\hbar}$. Substituting into the TDSE yields the time-independent Schrödinger equation (TISE) $\hat{H}\phi = E\phi$ — an eigenvalue equation. The solutions $\phi_n(x)$ are the stationary states: their probability density $|\phi_n(x)|^2$ is independent of time. The time-dependence is contained entirely in the phase factor $e^{-iE_nt/\hbar}$, which cancels in any probability calculation for a single energy eigenstate. This is why such states are called stationary, despite the wave function itself oscillating in time.

The Born interpretation assigns physical meaning to the wave function: $|\psi(x,t)|^2$ is the probability density for finding the particle at position $x$ at time $t$. Normalization $\int|\psi|^2dx = 1$ must hold at all times. The TDSE guarantees that normalization is preserved by the time evolution: using the TDSE and its complex conjugate, one derives the continuity equation $\partial_t\rho + \nabla\cdot\mathbf{j} = 0$ where $\rho = |\psi|^2$ is probability density and $\mathbf{j} = (\hbar/2mi)(\psi^*\nabla\psi - \psi\nabla\psi^*)$ is the probability current density. This is a local conservation law: probability cannot be created or destroyed, only transported.

Boundary conditions imposed on the wave function are physically motivated and mathematically necessary. For a potential that is finite everywhere, $\psi(x)$ and $\psi'(x)$ must be continuous. At an infinite potential wall, $\psi = 0$ (the particle cannot penetrate). For bound states, $\psi(x) \to 0$ as $x \to \pm\infty$ (square integrability). These conditions quantize the allowed energies — a discrete spectrum emerges from what would classically be a continuum of allowed values. This is the origin of atomic energy levels, molecular vibrational levels, and the discreteness of quantum numbers throughout physics.

The particle in an infinite square well (particle in a box) is the paradigmatic exactly solvable system. For $V = 0$ inside $[0,L]$ and $V = \infty$ outside, the boundary conditions $\phi(0) = \phi(L) = 0$ restrict the allowed solutions to $\phi_n(x) = \sqrt{2/L}\sin(n\pi x/L)$ with energies $E_n = n^2\pi^2\hbar^2/(2mL^2) = n^2E_1$ for $n = 1,2,3,...$. The ground state energy $E_1 = \pi^2\hbar^2/(2mL^2) > 0$ is the zero-point energy, a direct consequence of the uncertainty principle: confining a particle to a box of length $L$ gives $\Delta p \geq \hbar/(2L)$, contributing kinetic energy $\sim \hbar^2/(8mL^2)$, of the same order as $E_1$.

The WKB (Wentzel-Kramers-Brillouin) approximation provides an asymptotic solution to the TISE valid when the potential varies slowly on the scale of the de Broglie wavelength: $|d\lambda/dx| \ll 1$ where $\lambda(x) = 2\pi\hbar/p(x) = 2\pi/\sqrt{2m(E-V(x))}/\hbar$. In classically allowed regions ($E > V$), $\psi_{\rm WKB}(x) \approx C/\sqrt{p(x)}\exp(\pm i\int p(x')dx'/\hbar)$. In classically forbidden regions ($E < V$), $\psi_{\rm WKB}(x) \approx C/\sqrt{|p(x)|}\exp(\mp\int|p(x')|dx'/\hbar)$. Connection formulas at turning points (where $E = V$) link the two regions. WKB quantization gives the Bohr-Sommerfeld rule: $\oint p\,dq = (n+\frac{1}{2})\cdot 2\pi\hbar$ for bound states, accurate for large quantum numbers.

A general state can always be expressed as a superposition of stationary states: $\psi(x,t) = \sum_n c_n \phi_n(x)e^{-iE_nt/\hbar}$, where $c_n = \int\phi_n^*(x)\psi(x,0)dx$. The energy of each component is well-defined, but the expectation value $\langle E\rangle = \sum_n|c_n|^2E_n$ is a weighted average. Unlike stationary states, superpositions have time-dependent probability densities: $|\psi(x,t)|^2 = \sum_n|c_n|^2|\phi_n|^2 + \sum_{n\neq m}c_n^*c_me^{i(E_n-E_m)t/\hbar}\phi_n^*\phi_m$, oscillating at the Bohr frequencies $\nu_{nm} = (E_n-E_m)/h$. These oscillations are the quantum-mechanical origin of atomic radiation and Rabi oscillations.

Time-Dependent Schrödinger Equation (TDSE)

For a particle of mass $m$ in potential $V(\mathbf{r},t)$, the state $|\psi(t)\rangle$ evolves according to: $$i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle = \hat{H}(t)|\psi(t)\rangle,$$ where $\hat{H} = \hat{T} + \hat{V} = -\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r},t)$ is the Hamiltonian. In position representation: $$i\hbar\frac{\partial\psi(\mathbf{r},t)}{\partial t} = \left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r},t)\right]\psi(\mathbf{r},t).$$ For time-independent $\hat{H}$, the formal solution is $|\psi(t)\rangle = e^{-i\hat{H}t/\hbar}|\psi(0)\rangle$. The evolution operator $\hat{U}(t) = e^{-i\hat{H}t/\hbar}$ is unitary, guaranteeing conservation of probability.

Probability Current Density

The probability density $\rho(\mathbf{r},t) = |\psi(\mathbf{r},t)|^2$ satisfies the continuity equation: $$\frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf{j} = 0,$$ where the probability current density is: $$\mathbf{j}(\mathbf{r},t) = \frac{\hbar}{2mi}(\psi^*\nabla\psi - \psi\nabla\psi^*) = \frac{1}{m}\text{Re}(\psi^*\hat{\mathbf{p}}\psi).$$ For a plane wave $\psi = Ae^{i\mathbf{k}\cdot\mathbf{r}}$: $\mathbf{j} = |A|^2\hbar\mathbf{k}/m = |A|^2\mathbf{v}$ — probability flows at the classical velocity. In tunneling, $\mathbf{j}$ is non-zero even in classically forbidden regions (evanescent wave), explaining quantum tunneling through barriers.

Infinite Square Well (Particle in a Box)

For $V(x) = 0$, $0 < x < L$ and $V = \infty$ elsewhere, the TISE solutions satisfying boundary conditions $\phi(0) = \phi(L) = 0$ are: $$\phi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right), \quad n = 1,2,3,\ldots$$ with energy eigenvalues $E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} = n^2 E_1$ where $E_1 = \frac{\pi^2\hbar^2}{2mL^2}$ is the zero-point energy. For an electron in $L=1$ nm: $E_1 = 0.376$ eV. The states form a complete orthonormal set: $\int_0^L\phi_m^*(x)\phi_n(x)dx = \delta_{mn}$ and $\sum_n\phi_n^*(x')\phi_n(x) = \delta(x-x')$.

Ehrenfest's Theorem

The expectation values of quantum observables follow classical equations of motion. For position and momentum: $$\frac{d\langle\hat{x}\rangle}{dt} = \frac{\langle\hat{p}\rangle}{m}, \qquad \frac{d\langle\hat{p}\rangle}{dt} = -\left\langle\frac{\partial V}{\partial x}\right\rangle.$$ These follow from the Heisenberg equation of motion $d\hat{A}/dt = [\hat{A},\hat{H}]/(i\hbar) + \partial\hat{A}/\partial t$ applied to $\hat{x}$ and $\hat{p}$, using $[\hat{x},\hat{H}] = i\hbar\hat{p}/m$ and $[\hat{p},V(\hat{x})] = -i\hbar V'(\hat{x})$. Note: $\langle-\partial V/\partial x\rangle \neq -\partial V(\langle x\rangle)/\partial x$ in general — quantum corrections appear. Ehrenfest's theorem shows classical mechanics emerges from quantum mechanics for well-localized wave packets.

WKB Quantization (Bohr-Sommerfeld)

For a potential well with classical turning points $x_1 < x_2$ (where $E = V(x_{1,2})$), the WKB wave function in the allowed region is $\psi_{\rm WKB} \propto [p(x)]^{-1/2}\sin(\int_{x_1}^x p(x')dx'/\hbar + \pi/4)$ where $p(x) = \sqrt{2m(E-V(x))}$. Matching at both turning points requires: $$\int_{x_1}^{x_2} p(x)\,dx = \left(n - \frac{1}{2}\right)\pi\hbar, \quad n = 1,2,3,\ldots$$ equivalently $\oint p\,dq = (n-1/2)\cdot 2\pi\hbar$ (with Maslov index $\mu = 2$ for two soft turning points). For the harmonic oscillator ($V = \frac{1}{2}m\omega^2x^2$): $\oint p\,dq = 2\pi E/\omega$, giving $E_n = (n-1/2)\hbar\omega$ — exactly the correct result up to the zero-point shift, with $n=1$ giving $E = \hbar\omega/2$. The WKB approximation is exact for linear $V$ and becomes exact in the $\hbar\to 0$ limit.

Example 1

Derive the energy eigenvalues and eigenfunctions for the infinite square well of length $L$. Then compute: (a) the normalization constant, (b) $\langle x\rangle_n$, $\langle x^2\rangle_n$, $\langle p\rangle_n$, $\langle p^2\rangle_n$ for the $n$-th state, (c) verify the uncertainty relation $\Delta x\Delta p \geq \hbar/2$.

Inside the well, TISE is $-\hbar^2\phi''/(2m) = E\phi$, giving $\phi'' = -k^2\phi$ with $k = \sqrt{2mE}/\hbar$. General solution: $\phi = A\sin(kx) + B\cos(kx)$. Boundary condition $\phi(0) = 0 \Rightarrow B = 0$. Condition $\phi(L) = 0 \Rightarrow \sin(kL) = 0 \Rightarrow k_n = n\pi/L$, $E_n = \hbar^2k_n^2/(2m) = n^2\pi^2\hbar^2/(2mL^2)$. (a) Normalization: $\int_0^L A^2\sin^2(n\pi x/L)dx = A^2L/2 = 1 \Rightarrow A = \sqrt{2/L}$. (b) $\langle x\rangle_n = \frac{2}{L}\int_0^L x\sin^2(n\pi x/L)dx = L/2$ (midpoint, by symmetry). $\langle x^2\rangle_n = \frac{L^2}{3} - \frac{L^2}{2n^2\pi^2}$. $\langle p\rangle_n = 0$ (real wave function). $\langle p^2\rangle_n = \hbar^2k_n^2 = n^2\pi^2\hbar^2/L^2 = 2mE_n$. (c) $\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} = L\sqrt{1/12 - 1/(2n^2\pi^2)}$. $\Delta p = n\pi\hbar/L$. Product: $\Delta x\Delta p = n\pi\hbar\sqrt{1/12-1/(2n^2\pi^2)} = \hbar\sqrt{n^2\pi^2/12 - 1/2}$. For $n=1$: $\Delta x\Delta p = \hbar\sqrt{\pi^2/12 - 1/2} \approx 0.568\hbar > \hbar/2$. ✓ The bound is tightest for $n=1$ (groundstate).

Example 2

A particle in an infinite square well of length $L$ is initially in the state $\psi(x,0) = \sqrt{30/L^5}\,x(L-x)$ (parabolic profile). Find the expansion coefficients $c_n$, the probability of measuring $E_1$, and the time-evolved probability density $|\psi(x,t)|^2$.

Expand $\psi(x,0) = \sum_n c_n\phi_n(x)$. The coefficients are $c_n = \int_0^L\phi_n^*(x)\psi(x,0)dx = \sqrt{2/L}\sqrt{30/L^5}\int_0^L x(L-x)\sin(n\pi x/L)dx$. Using $\int_0^L x(L-x)\sin(n\pi x/L)dx = 2L^3[1-(-1)^n]/(n\pi)^3$ (odd $n$ only): $$c_n = \frac{4\sqrt{15}}{(n\pi)^3}\cdot[1-(-1)^n] = \begin{cases}\frac{8\sqrt{15}}{(n\pi)^3} & n\text{ odd}\\0 & n\text{ even}\end{cases}.$$ Probability $P(E_1) = |c_1|^2 = 960/\pi^6 \approx 0.9986$ — almost entirely in the ground state! The time-evolved state: $\psi(x,t) = \sum_{n\text{ odd}}c_n\phi_n(x)e^{-iE_nt/\hbar}$. The probability density $|\psi(x,t)|^2$ oscillates at the Bohr frequency $(E_3-E_1)/h = 8E_1/h$, but since $P(E_1) \approx 1$, the oscillations are small. Check: $\sum_{n\text{ odd}}|c_n|^2 = \frac{960}{\pi^6}\sum_{n\text{ odd}}\frac{1}{n^6} = \frac{960}{\pi^6}\cdot\frac{\pi^6}{960} = 1$. ✓

Example 3

Derive the tunneling transmission coefficient for a rectangular barrier of height $V_0 > E$ and width $a$. Find the approximate expression for $T$ when $\kappa a \gg 1$ (opaque barrier), and compute $T$ for an electron ($m = m_e$) with $E = 1$ eV tunneling through a barrier of $V_0 = 2$ eV and $a = 1$ nm.

Define $k = \sqrt{2mE}/\hbar$ (outside), $\kappa = \sqrt{2m(V_0-E)}/\hbar$ (inside barrier). The TISE solutions are: Region I ($x<0$): $\psi = e^{ikx} + Re^{-ikx}$; Region II ($0a$): $\psi = Te^{ikx}$. Matching $\psi$ and $\psi'$ at $x=0,a$ gives 4 equations. The exact result: $$T = \frac{1}{1+\frac{(k^2+\kappa^2)^2}{4k^2\kappa^2}\sinh^2(\kappa a)}.$$ For $\kappa a \gg 1$, $\sinh(\kappa a) \approx e^{\kappa a}/2$: $T \approx \frac{16k^2\kappa^2}{(k^2+\kappa^2)^2}e^{-2\kappa a}$. Numerically: $V_0-E = 1$ eV, $\kappa = \sqrt{2\times9.109\times10^{-31}\times1.6\times10^{-19}}/(1.055\times10^{-34}) = \sqrt{2.918\times10^{-48}}/1.055\times10^{-34} \approx 5.12\times10^9$ m$^{-1}$. $\kappa a = 5.12\times10^9\times10^{-9} \approx 5.12$. Prefactor $\approx 1$ (order unity). $T \approx e^{-2\times5.12} = e^{-10.24} \approx 3.6\times10^{-5}$. About 0.004% transmission — small but measurable, explaining scanning tunneling microscopy and nuclear alpha decay.

Example 4

Derive the continuity equation $\partial_t|\psi|^2 + \nabla\cdot\mathbf{j} = 0$ from the TDSE. Then show that for a stationary state $\psi_n = \phi_n(x)e^{-iE_nt/\hbar}$, the probability current vanishes identically for real $\phi_n$ and compute $j$ for a complex scattering state $\psi = Ae^{ikx} + Be^{-ikx}$.

TDSE: $i\hbar\partial_t\psi = \hat{H}\psi$ and c.c. $-i\hbar\partial_t\psi^* = \hat{H}^*\psi^* = \hat{H}\psi^*$ (real $V$). Then: $\partial_t(\psi^*\psi) = \psi^*\partial_t\psi + (\partial_t\psi^*)\psi = \frac{1}{i\hbar}[\psi^*\hat{H}\psi - \psi\hat{H}\psi^*] = \frac{1}{i\hbar}\left[\psi^*\left(-\frac{\hbar^2}{2m}\nabla^2+V\right)\psi - \psi\left(-\frac{\hbar^2}{2m}\nabla^2+V\right)\psi^*\right]$. The potential terms cancel: $\partial_t\rho = \frac{-\hbar^2}{2mi}(\psi^*\nabla^2\psi - \psi\nabla^2\psi^*) = -\nabla\cdot\left[\frac{\hbar}{2mi}(\psi^*\nabla\psi - \psi\nabla\psi^*)\right] = -\nabla\cdot\mathbf{j}$, using the identity $\psi^*\nabla^2\psi - \psi\nabla^2\psi^* = \nabla\cdot(\psi^*\nabla\psi - \psi\nabla\psi^*)$. ✓ For real $\phi_n$: $j = \hbar(\phi_n\nabla\phi_n - \phi_n\nabla\phi_n)/(2mi) = 0$. For $\psi = Ae^{ikx}+Be^{-ikx}$: $j = \frac{\hbar k}{m}(|A|^2 - |B|^2)$ — incident minus reflected probability flux, conserved in stationary scattering.

Interactive Explorer: Particle in a Box Wave Functions
Energy Eₙ = 0.376 eV
Zero-point energy E₁ = 0.376 eV
de Broglie wavelength = 2.000 nm

Practice Problems

1. Solve the TISE for a finite square well of depth $V_0$ and width $2a$: $V = -V_0$ for $|x| < a$, $V = 0$ for $|x| > a$. Write the matching conditions and the transcendental equation for even-parity bound states.
2. An infinite square well of width $L$ is suddenly expanded to width $2L$ at $t=0$ (the right wall moves instantly). If the particle was in the ground state $\phi_1^{(L)}$ before the expansion, what are the probabilities of measuring each energy level $E_n^{(2L)}$ after expansion?
3. Verify that the probability current for a normalized Gaussian wave packet $\psi = (2\pi\sigma^2)^{-1/4}e^{ik_0 x - x^2/(4\sigma^2)}$ satisfies $\int j\,dx = \hbar k_0/m = v_g$ (total probability flux equals group velocity times probability).
4. Use the WKB approximation to find the energy levels of a particle in the potential $V(x) = C|x|$ (linear potential / "quantum bouncer"). Express in terms of the zeros $a_n$ of the Airy function.
5. Show that for any normalizable solution of the TISE with energy $E$ in a potential $V(x)$, the expectation value of the kinetic energy satisfies $\langle T\rangle = E - \langle V\rangle$ (trivial from $\hat{H} = \hat{T}+\hat{V}$), and use the virial theorem to show $\langle T\rangle = \langle V\rangle$ for the harmonic oscillator.
6. Derive the reflection and transmission coefficients $R$ and $T$ for a step potential $V(x) = V_0\theta(x)$ with $E > V_0$, and verify $R + T = 1$.
7. A particle of mass $m$ is in the 50th energy level of an infinite square well of length $L = 1$ nm. (a) Compute the de Broglie wavelength. (b) Compare the quantum probability density with the classical uniform distribution $P_{\rm cl}(x) = 1/L$. (c) Verify Bohr's correspondence principle.
8. The double well potential: $V(x) = V_0[\delta(x+a)+\delta(x-a)]$ with $V_0 > 0$. Find the allowed energies of bound states (if any) as solutions to a transcendental equation. Discuss the formation of symmetric and antisymmetric states.
9. Derive the time evolution of $\langle x(t)\rangle$ for a particle in the superposition $\psi(x,0) = (\phi_1(x)+\phi_2(x))/\sqrt{2}$ in an infinite square well. Show it oscillates with the Bohr frequency $\omega_{21} = (E_2-E_1)/\hbar$.
10. Show that for the particle in a box, the operator $\hat{p} = -i\hbar\partial_x$ is Hermitian on the full real line but NOT Hermitian on $[0,L]$ with Dirichlet boundary conditions. What does this tell us about whether momentum is an observable in a box?
11. Apply WKB to the harmonic oscillator $V = \frac{1}{2}m\omega^2x^2$. Compute the action integral $\oint p\,dq$ and recover the exact energy levels $E_n = (n+1/2)\hbar\omega$, noting the Maslov index correction.
12. Estimate the tunneling rate for an alpha particle (mass $4m_p$, charge $2e$) inside a nucleus of radius $R = 7$ fm, escaping a Coulomb barrier with $Q = 5$ MeV. Use Gamow's WKB formula $\Gamma \propto e^{-2G}$ where $G = \int_R^{r_c}\kappa(r)dr$ and $r_c$ is the classical turning point.
Show Answer Key

1. Even parity: $\phi = A\cos(kx)$ inside ($k=\sqrt{2m(E+V_0)}/\hbar$), $\phi = Be^{-\kappa|x|}$ outside ($\kappa = \sqrt{-2mE}/\hbar$, with $E<0$). Matching at $x=a$: $A\cos(ka) = Be^{-\kappa a}$ and $-Ak\sin(ka) = -B\kappa e^{-\kappa a}$. Dividing: $k\tan(ka) = \kappa$. With $z = ka$, $z_0 = a\sqrt{2mV_0}/\hbar$: $z\tan z = \sqrt{z_0^2 - z^2}$ — transcendental equation; always at least one even-parity bound state.

2. New eigenstates: $\phi_n^{(2L)}(x) = \sqrt{1/L}\sin(n\pi x/(2L))$. $c_n = \int_0^L\phi_1^{(L)}\phi_n^{(2L)}dx = \frac{\sqrt{2}}{L}\int_0^L\sin(\pi x/L)\sin(n\pi x/(2L))dx$. Non-zero only for certain $n$. $P_1 = |c_1|^2 = (8/\pi)^2\cdot(1/(1^2-4))^2\cdot$... detailed computation gives $P(E_1^{(2L)}) = 32/(3\pi)^2 \approx 0.360$.

3. $j(x) = \frac{\hbar}{2mi}(\psi^*\partial_x\psi - \psi\partial_x\psi^*) = \frac{\hbar k_0}{m}|\psi|^2 + \text{(real part terms that integrate to zero for Gaussian)}$. $\int j\,dx = \frac{\hbar k_0}{m}\int|\psi|^2dx = \hbar k_0/m = v_g$. ✓ The cross terms vanish since $\psi^*\partial_x\psi - \psi\partial_x\psi^* = 2ik_0|\psi|^2 - x|\psi|^2/(2\sigma^2) + \text{c.c.}$ and the latter integrates to zero.

4. Classically, $p(x) = \sqrt{2m(E-C|x|)}$. Turning points at $\pm x_0 = \pm E/C$. WKB: $\int_0^{x_0}\sqrt{2m(E-Cx)}dx = \frac{2}{3}\sqrt{2m}\cdot E^{3/2}/C = (n-1/4)\pi\hbar$ (one hard wall at $x=0$, Maslov index $\mu=1$). Solving: $E_n = (9/8)^{1/3}(C\hbar)^{2/3}(4n-1)^{2/3}(\pi^2/2m)^{1/3}/C$... The exact answer involves zeros $a_n$ of Ai$(-z)$: $E_n = a_n(C^2\hbar^2/(2m))^{1/3}$.

5. $\langle T\rangle + \langle V\rangle = \langle\hat{H}\rangle = E$, so $\langle T\rangle = E - \langle V\rangle$. Virial theorem for QHO: $\langle\hat{H}\rangle = \langle T\rangle + \langle V\rangle$ and by the virial theorem (valid when $\langle d\hat{r}\cdot\hat{p}/dt\rangle = 0$ in stationary states): $2\langle T\rangle = \langle \mathbf{r}\cdot\nabla V\rangle$. For $V = \frac{1}{2}m\omega^2x^2$: $x\partial_xV = 2V$, so $2\langle T\rangle = 2\langle V\rangle$, giving $\langle T\rangle = \langle V\rangle = E_n/2 = (n+1/2)\hbar\omega/2$.

6. $k_1 = \sqrt{2mE}/\hbar$ (left), $k_2 = \sqrt{2m(E-V_0)}/\hbar$ (right). $\psi_I = e^{ik_1x}+Re^{-ik_1x}$, $\psi_{II} = Te^{ik_2x}$. Matching: $1+R = T$, $ik_1(1-R) = ik_2T$. Solving: $R = (k_1-k_2)/(k_1+k_2)$, $T = 2k_1/(k_1+k_2)$. Reflection coeff $|R|^2$, transmission $T_{ m coeff} = (k_2/k_1)|T|^2 = 4k_1k_2/(k_1+k_2)^2$. Verify: $|R|^2 + T_{ m coeff} = [(k_1-k_2)^2+4k_1k_2]/(k_1+k_2)^2 = 1$. ✓

7. (a) $\lambda_{50} = 2L/50 = 0.04$ nm. (b) $|\phi_{50}|^2 = (2/L)\sin^2(50\pi x/L)$ oscillates 50 times in $[0,L]$; average value is $1/L = P_{\rm cl}$. (c) For large $n$, the rapid oscillations of $\sin^2(n\pi x/L)$ average to $1/2$, giving $|\phi_n|^2 \to 1/L$ (classical uniform distribution) — the correspondence principle.

8. For the delta function double well, $V = V_0[\delta(x+a)+\delta(x-a)]$ with $V_0 > 0$ (repulsive) — actually for bound states need $V_0 < 0$. For $V_0 < 0$ (attractive), symmetric state: $\psi = A\cosh(\kappa x)$ inside, $Be^{-\kappa|x|}$ outside (for $|x|>a$). Matching: $2m|V_0|/\hbar^2 = \kappa(1+\tanh(\kappa a))$ (symmetric), $\kappa(1-\tanh(\kappa a))$ (antisymmetric). Energy splitting $\Delta E \propto e^{-2\kappa a}$ — basis of covalent bonding.

9. $\psi(x,t) = (\phi_1 e^{-iE_1t/\hbar}+\phi_2 e^{-iE_2t/\hbar})/\sqrt{2}$. $\langle x\rangle = \frac{1}{2}[\langle\phi_1|x|\phi_1\rangle+\langle\phi_2|x|\phi_2\rangle + 2\text{Re}(e^{i(E_1-E_2)t/\hbar}\langle\phi_1|x|\phi_2\rangle)]$. First two terms = $L/2$. Cross term $\langle\phi_1|x|\phi_2\rangle = -16L/(9\pi)^2$ (by integration). Thus $\langle x(t)\rangle = L/2 - \frac{32L}{9\pi^2}\cos(\omega_{21}t)$ with $\omega_{21} = (E_2-E_1)/\hbar = 3\pi^2\hbar/(2mL^2)$.

10. For $\hat{p}$ to be Hermitian on $[0,L]$: $\langle\phi|\hat{p}\psi\rangle = \langle\hat{p}\phi|\psi\rangle$ requires boundary term $[\phi^*\psi]_0^L = 0$. With Dirichlet BCs $\psi(0)=\psi(L)=0$ but $\phi$ arbitrary in the domain of $\hat{p}^\dagger$, this fails. Momentum is NOT an observable in the box — there is no complete orthonormal set of $\hat{p}$ eigenstates satisfying the BCs. Instead, $\hat{p}^2$ (proportional to $\hat{H}$) is the relevant observable.

11. Turning points at $\pm x_0 = \pm\sqrt{2E/(m\omega^2)}$. $\oint p\,dq = 2\int_{-x_0}^{x_0}\sqrt{2m(E-\frac{1}{2}m\omega^2x^2)}dx$. Substituting $x = x_0\sin\theta$: $= 2\sqrt{2m}\cdot x_0\cdot\frac{\sqrt{2E/(m)}}{2}\cdot\pi = \pi\cdot 2E/\omega = 2\pi E/\omega$. Setting equal to $(n-1/2)\cdot 2\pi\hbar$ (Maslov correction for two soft turning points, $\mu = 2$, subtract $\mu\pi\hbar/2 = \pi\hbar$): $2\pi E/\omega = 2\pi(n+1/2)\hbar$ iff accounting correctly: $E_n = (n+1/2)\hbar\omega$ with $n = 0,1,2,...$ ✓

12. Coulomb barrier: $V(r) = 2e^2/(4\pi\epsilon_0 r) = k_e(2e)(2e)/r$. Turning point: $r_c = 4k_ee^2/Q \approx 4\times1.44\text{ MeV·fm}/5\text{ MeV} \approx 57.6$ fm. Gamow factor: $G = \int_R^{r_c}\sqrt{2\mu(V(r)-Q)/\hbar^2}dr$ where $\mu = 4m_p$. Result: $G = \pi Z_1Z_2e^2/\hbar v - 2Z_1Z_2e^2\sqrt{R/Q}\cdot(2/\hbar)\/\sqrt{...}$. For $Z_1=Z_2=2$, $Q=5$ MeV: $G \approx \pi\times2\times2\times1.44/(\hbar v) - \text{correction} \approx 16$ (estimate). Decay rate $\Gamma \propto e^{-2G} \approx e^{-32}$ — qualitatively explaining long alpha-decay half-lives.