Appendix G: Answers to Selected Exercises

"You do not really understand something unless you can explain it to your grandmother." -- Attributed (probably apocryphally) to Albert Einstein

This appendix provides complete worked solutions to selected exercises from Chapters 1-10. For calculation problems, every significant step is shown. For conceptual problems, rubric criteria indicate what a complete answer should contain. Problems are referenced by their numbering in the corresponding chapter's exercise set.

Use these solutions to check your work after making a serious attempt. Reading solutions without first struggling with the problem provides almost no learning benefit (see the metacognitive research in the Preface).

Chapter 1: The Quantum Revolution

Problem B.1: Photoelectric Effect Calculations

Given: Work function of platinum, $\phi = 5.65$ eV.

(a) Threshold wavelength:

At threshold, the photon energy equals the work function exactly:

$$E_{\text{photon}} = h\nu_{\text{threshold}} = \phi$$

$$\frac{hc}{\lambda_{\text{threshold}}} = \phi$$

$$\lambda_{\text{threshold}} = \frac{hc}{\phi} = \frac{(6.626 \times 10^{-34}\,\text{J}\cdot\text{s})(3.00 \times 10^8\,\text{m/s})}{5.65\,\text{eV} \times 1.602 \times 10^{-19}\,\text{J/eV}}$$

$$\lambda_{\text{threshold}} = \frac{1.988 \times 10^{-25}}{9.051 \times 10^{-19}} = 2.196 \times 10^{-7}\,\text{m}$$

$$\boxed{\lambda_{\text{threshold}} \approx 220\,\text{nm}}$$

This is in the ultraviolet, which is why visible light cannot eject electrons from platinum.

(b) Maximum kinetic energy for $\lambda = 150$ nm:

$$K_{\max} = h\nu - \phi = \frac{hc}{\lambda} - \phi$$

$$K_{\max} = \frac{1240\,\text{eV}\cdot\text{nm}}{150\,\text{nm}} - 5.65\,\text{eV} = 8.27\,\text{eV} - 5.65\,\text{eV}$$

$$\boxed{K_{\max} = 2.62\,\text{eV}}$$

(Here we used the convenient shortcut $hc = 1240$ eV$\cdot$nm.)

(c) Maximum speed:

$$K_{\max} = \frac{1}{2}m_e v_{\max}^2$$

$$v_{\max} = \sqrt{\frac{2K_{\max}}{m_e}} = \sqrt{\frac{2 \times 2.62 \times 1.602 \times 10^{-19}\,\text{J}}{9.109 \times 10^{-31}\,\text{kg}}}$$

$$v_{\max} = \sqrt{\frac{8.394 \times 10^{-19}}{9.109 \times 10^{-31}}} = \sqrt{9.215 \times 10^{11}} = 9.60 \times 10^5\,\text{m/s}$$

$$\boxed{v_{\max} \approx 9.60 \times 10^5\,\text{m/s} \approx 0.0032c}$$

This is non-relativistic, so our classical kinetic energy formula is justified.

(d) Stopping potential:

The stopping potential $V_0$ is the voltage needed to bring the fastest electrons to rest:

$$eV_0 = K_{\max}$$

$$\boxed{V_0 = \frac{K_{\max}}{e} = 2.62\,\text{V}}$$

Problem B.3: De Broglie Wavelengths

(a) Electron accelerated through $V = 54$ V (Davisson-Germer):

The kinetic energy is $K = eV = 54$ eV $= 54 \times 1.602 \times 10^{-19} = 8.651 \times 10^{-18}$ J.

The momentum is $p = \sqrt{2m_e K}$:

$$p = \sqrt{2(9.109 \times 10^{-31})(8.651 \times 10^{-18})} = \sqrt{1.576 \times 10^{-47}} = 3.970 \times 10^{-24}\,\text{kg}\cdot\text{m/s}$$

$$\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{3.970 \times 10^{-24}} = 1.669 \times 10^{-10}\,\text{m}$$

$$\boxed{\lambda \approx 0.167\,\text{nm} = 1.67\,\text{\AA}}$$

This is comparable to the lattice spacing of nickel (2.15 \AA), which is why Davisson and Germer observed diffraction.

(b) Proton with $K = 1.0$ MeV:

$$p = \sqrt{2m_p K} = \sqrt{2(1.673 \times 10^{-27})(1.0 \times 10^6 \times 1.602 \times 10^{-19})}$$

$$p = \sqrt{2(1.673 \times 10^{-27})(1.602 \times 10^{-13})} = \sqrt{5.360 \times 10^{-40}} = 2.315 \times 10^{-20}\,\text{kg}\cdot\text{m/s}$$

$$\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{2.315 \times 10^{-20}} = 2.863 \times 10^{-14}\,\text{m}$$

$$\boxed{\lambda \approx 28.6\,\text{fm}}$$

This is comparable to nuclear sizes (~1-10 fm), explaining why MeV protons are used to probe nuclear structure.

Problem A.2: Wave-Particle Duality (Conceptual)

Rubric criteria for a complete answer:

1. Acknowledges the friend's observations are correct (1 point): Yes, the photoelectric effect demonstrates particle-like behavior, and the double-slit experiment demonstrates wave-like behavior.

2. Explains why this is not a contradiction (2 points): Physics does not contradict itself because the two experiments probe different aspects of the same entity. No single experiment simultaneously demonstrates both behaviors. The photon is neither a classical particle nor a classical wave -- it is a quantum object that exhibits particle-like behavior in some experimental contexts and wave-like behavior in others.

3. Introduces complementarity (2 points): Bohr's complementarity principle states that wave and particle descriptions are complementary, not contradictory. The experimental setup determines which aspect is manifested. Attempting to determine "which slit" the photon goes through (particle information) destroys the interference pattern (wave behavior). You cannot observe both simultaneously.

4. Avoids the common error (1 point): Does NOT say "light is sometimes a wave and sometimes a particle." The correct statement is that light is always a quantum object; the wave and particle descriptions are approximations that capture different aspects of its behavior.

Chapter 2: The Wave Function and the Schrodinger Equation

Problem B.1: Normalization

(a) $\psi(x) = Ax(L-x)$ for $0 \leq x \leq L$:

$$\int_0^L |A|^2 x^2(L-x)^2\,dx = 1$$

Expand: $x^2(L-x)^2 = x^2(L^2 - 2Lx + x^2) = L^2 x^2 - 2Lx^3 + x^4$.

$$|A|^2 \left[\frac{L^2 x^3}{3} - \frac{2Lx^4}{4} + \frac{x^5}{5}\right]_0^L = |A|^2 \left[\frac{L^5}{3} - \frac{L^5}{2} + \frac{L^5}{5}\right]$$

$$= |A|^2 L^5 \left[\frac{1}{3} - \frac{1}{2} + \frac{1}{5}\right] = |A|^2 L^5 \left[\frac{10 - 15 + 6}{30}\right] = |A|^2 \frac{L^5}{30}$$

Setting this equal to 1:

$$\boxed{A = \sqrt{\frac{30}{L^5}}}$$

(b) $\psi(x) = Ae^{-|x|/a}$ for $-\infty < x < \infty$:

By symmetry:

$$|A|^2 \int_{-\infty}^{\infty} e^{-2|x|/a}\,dx = 2|A|^2 \int_0^{\infty} e^{-2x/a}\,dx = 2|A|^2 \cdot \frac{a}{2} = |A|^2 a = 1$$

$$\boxed{A = \frac{1}{\sqrt{a}}}$$

Problem B.2: Probability Calculations

Given: $\psi(x) = \sqrt{2/L}\sin(\pi x/L)$ on $[0, L]$.

(a) Probability in $[0, L/4]$:

$$P = \frac{2}{L}\int_0^{L/4} \sin^2\left(\frac{\pi x}{L}\right)dx$$

Use the identity $\sin^2\theta = \frac{1}{2}(1 - \cos 2\theta)$:

$$P = \frac{2}{L} \cdot \frac{1}{2}\int_0^{L/4}\left[1 - \cos\left(\frac{2\pi x}{L}\right)\right]dx$$

$$= \frac{1}{L}\left[x - \frac{L}{2\pi}\sin\left(\frac{2\pi x}{L}\right)\right]_0^{L/4}$$

$$= \frac{1}{L}\left[\frac{L}{4} - \frac{L}{2\pi}\sin\left(\frac{\pi}{2}\right)\right] = \frac{1}{4} - \frac{1}{2\pi}$$

$$\boxed{P\left(0 \leq x \leq \frac{L}{4}\right) = \frac{1}{4} - \frac{1}{2\pi} \approx 0.0908}$$

This is less than 1/4, which makes physical sense: the ground state wave function peaks at $x = L/2$, so the particle is less likely to be found near the walls.

(b) Probability in $[L/4, 3L/4]$:

By the same method:

$$P = \frac{1}{L}\left[x - \frac{L}{2\pi}\sin\left(\frac{2\pi x}{L}\right)\right]_{L/4}^{3L/4}$$

$$= \frac{1}{L}\left[\left(\frac{3L}{4} - \frac{L}{2\pi}\sin\frac{3\pi}{2}\right) - \left(\frac{L}{4} - \frac{L}{2\pi}\sin\frac{\pi}{2}\right)\right]$$

$$= \frac{1}{L}\left[\frac{3L}{4} + \frac{L}{2\pi} - \frac{L}{4} + \frac{L}{2\pi}\right] = \frac{1}{2} + \frac{1}{\pi}$$

$$\boxed{P\left(\frac{L}{4} \leq x \leq \frac{3L}{4}\right) = \frac{1}{2} + \frac{1}{\pi} \approx 0.818}$$

Check: The total probability over $[0, L]$ must be 1. By symmetry, $P(3L/4 \leq x \leq L) = P(0 \leq x \leq L/4) \approx 0.0908$. Sum: $0.0908 + 0.818 + 0.0908 \approx 1.00$. $\checkmark$

Problem A.3: Why Must $\psi$ Be Continuous? (Conceptual)

Rubric criteria:

1. Identifies the kinetic energy issue (2 points): The kinetic energy operator involves the second derivative: $\hat{T} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$. If $\psi$ has a discontinuity, $d\psi/dx$ contains a delta function and $d^2\psi/dx^2$ contains a derivative of a delta function, which makes the kinetic energy expectation value $\langle \hat{T}\rangle$ infinite.

2. Connects to probability current (1 point): A discontinuous $\psi$ would produce a discontinuous probability current $j = \frac{\hbar}{2mi}(\psi^*\nabla\psi - \psi\nabla\psi^*)$, violating conservation of probability at the point of discontinuity.

3. Notes the exception (1 point): An infinite potential (like the walls of the infinite square well) can force $\psi = 0$ at a boundary, but $\psi$ itself is still continuous -- it is $d\psi/dx$ that is discontinuous at infinite potential boundaries.

Chapter 3: Solving the Schrodinger Equation

Problem 3.1: Infinite Square Well Energy Levels

Given: Electron in an infinite square well of width $a = 0.5$ nm.

(a) Ground state energy $E_1$:

$$E_n = \frac{n^2 \pi^2 \hbar^2}{2m_e a^2}$$

$$E_1 = \frac{\pi^2 (1.055 \times 10^{-34})^2}{2(9.109 \times 10^{-31})(0.5 \times 10^{-9})^2}$$

$$= \frac{\pi^2 \times 1.113 \times 10^{-68}}{2 \times 9.109 \times 10^{-31} \times 2.5 \times 10^{-19}}$$

$$= \frac{1.098 \times 10^{-67}}{4.555 \times 10^{-49}} = 2.411 \times 10^{-19}\,\text{J}$$

$$\boxed{E_1 = 1.506\,\text{eV} \approx 1.51\,\text{eV}}$$

(b) First excited state $E_2$:

$$E_2 = 4E_1 = \boxed{6.02\,\text{eV}}$$

(c) Wavelength of the $n = 2 \to n = 1$ photon:

$$\Delta E = E_2 - E_1 = 3E_1 = 4.53\,\text{eV}$$

$$\lambda = \frac{hc}{\Delta E} = \frac{1240\,\text{eV}\cdot\text{nm}}{4.53\,\text{eV}} = \boxed{274\,\text{nm}}$$

(d) This is in the ultraviolet region of the electromagnetic spectrum.

Problem 3.3: Expectation Values for the Infinite Well

Given: $\psi_n(x) = \sqrt{2/a}\sin(n\pi x/a)$ on $[0, a]$.

(a) $\langle x \rangle$:

$$\langle x \rangle = \frac{2}{a}\int_0^a x\sin^2\left(\frac{n\pi x}{a}\right)dx$$

Using $\sin^2\theta = \frac{1}{2}(1 - \cos 2\theta)$:

$$= \frac{2}{a}\left[\frac{1}{2}\int_0^a x\,dx - \frac{1}{2}\int_0^a x\cos\left(\frac{2n\pi x}{a}\right)dx\right]$$

The first integral is $a^2/4$. The second integral, by integration by parts, gives:

$$\int_0^a x\cos\left(\frac{2n\pi x}{a}\right)dx = \left[\frac{ax}{2n\pi}\sin\left(\frac{2n\pi x}{a}\right)\right]_0^a + \frac{a^2}{4n^2\pi^2}\left[\cos\left(\frac{2n\pi x}{a}\right)\right]_0^a = 0$$

since $\sin(2n\pi) = 0$ and $\cos(2n\pi) - \cos(0) = 0$. Therefore:

$$\boxed{\langle x \rangle = \frac{a}{2}}$$

for all $n$. This makes sense by symmetry: the probability density is symmetric about $x = a/2$.

(e) $\sigma_x = \sqrt{\langle x^2\rangle - \langle x\rangle^2}$:

From the standard result (derived via integration by parts):

$$\langle x^2\rangle = a^2\left(\frac{1}{3} - \frac{1}{2n^2\pi^2}\right)$$

$$\sigma_x = a\sqrt{\frac{1}{12} - \frac{1}{2n^2\pi^2}}$$

(f) From $\langle p^2\rangle = n^2\pi^2\hbar^2/a^2$ and $\langle p\rangle = 0$:

$$\sigma_p = \frac{n\pi\hbar}{a}$$

(g) The uncertainty product is:

$$\sigma_x\sigma_p = n\pi\hbar\sqrt{\frac{1}{12} - \frac{1}{2n^2\pi^2}}$$

For $n = 1$: $\sigma_x\sigma_p = \pi\hbar\sqrt{\frac{1}{12} - \frac{1}{2\pi^2}} = \hbar\sqrt{\frac{\pi^2}{12} - \frac{1}{2}} \approx \hbar\sqrt{0.822 - 0.5} = \hbar\sqrt{0.322} \approx 0.568\hbar$.

Since $\hbar/2 = 0.5\hbar$, we have $\sigma_x\sigma_p \approx 0.568\hbar > \hbar/2$. $\checkmark$

The product is smallest (closest to the Heisenberg limit $\hbar/2$) for $\boxed{n = 1}$, the ground state. As $n \to \infty$, the product grows without bound.

Problem 3.4: Fourier Expansion of $\Psi(x,0) = Ax(a-x)$

(a) Normalization: From the same integral as Ch 2 Problem B.1(a) with $L \to a$:

$$\boxed{A = \sqrt{\frac{30}{a^5}}}$$

(b) Expansion coefficients:

$$c_n = \sqrt{\frac{2}{a}}\int_0^a \sin\left(\frac{n\pi x}{a}\right) \cdot \sqrt{\frac{30}{a^5}}\,x(a-x)\,dx = \sqrt{\frac{60}{a^6}}\int_0^a x(a-x)\sin\left(\frac{n\pi x}{a}\right)dx$$

Evaluating (using integration by parts twice or a table):

$$\int_0^a x(a-x)\sin\left(\frac{n\pi x}{a}\right)dx = \frac{2a^3}{n^3\pi^3}[1 - \cos(n\pi)] = \begin{cases} \frac{4a^3}{n^3\pi^3} & n\text{ odd} \\ 0 & n\text{ even}\end{cases}$$

Therefore:

$$c_n = \begin{cases} \sqrt{\frac{60}{a^6}} \cdot \frac{4a^3}{n^3\pi^3} = \frac{4\sqrt{60}}{n^3\pi^3} = \frac{8\sqrt{15}}{n^3\pi^3} & n\text{ odd} \\ 0 & n\text{ even}\end{cases}$$

(c) Probabilities:

$$P(E_1) = |c_1|^2 = \frac{960}{1 \cdot \pi^6} = \frac{960}{\pi^6} \approx \boxed{0.9986}$$

$$P(E_2) = |c_2|^2 = \boxed{0}$$

The astonishing result is that $\Psi(x,0) = Ax(a-x)$ is 99.86% in the ground state! This makes sense: the parabola $x(a-x)$ has the same shape as $\sin(\pi x/a)$ -- symmetric, single-humped, zero at the boundaries.

Chapter 4: The Quantum Harmonic Oscillator

Problem 4.4: Direct Verification of the Ground State

Claim: $\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^2/2\hbar}$ satisfies $-\frac{\hbar^2}{2m}\frac{d^2\psi_0}{dx^2} + \frac{1}{2}m\omega^2 x^2\psi_0 = \frac{1}{2}\hbar\omega\psi_0$.

Step 1: Compute $d\psi_0/dx$:

$$\frac{d\psi_0}{dx} = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(-\frac{m\omega}{\hbar}x\right)e^{-m\omega x^2/2\hbar} = -\frac{m\omega}{\hbar}x\,\psi_0$$

Step 2: Compute $d^2\psi_0/dx^2$:

$$\frac{d^2\psi_0}{dx^2} = -\frac{m\omega}{\hbar}\psi_0 + \left(-\frac{m\omega}{\hbar}x\right)\left(-\frac{m\omega}{\hbar}x\right)\psi_0 = -\frac{m\omega}{\hbar}\psi_0 + \frac{m^2\omega^2}{\hbar^2}x^2\psi_0$$

Step 3: Substitute into the TISE:

$$-\frac{\hbar^2}{2m}\left[-\frac{m\omega}{\hbar}\psi_0 + \frac{m^2\omega^2}{\hbar^2}x^2\psi_0\right] + \frac{1}{2}m\omega^2 x^2\psi_0$$

$$= \frac{\hbar\omega}{2}\psi_0 - \frac{m\omega^2}{2}x^2\psi_0 + \frac{m\omega^2}{2}x^2\psi_0 = \frac{1}{2}\hbar\omega\,\psi_0 \quad \checkmark$$

The potential energy and kinetic energy terms involving $x^2$ cancel exactly, leaving $E_0 = \frac{1}{2}\hbar\omega$. $\square$

Problem 4.6: Hermite Polynomials via the Rodrigues Formula

(a) Using $H_n(\xi) = (-1)^n e^{\xi^2}\frac{d^n}{d\xi^n}e^{-\xi^2}$:

$n = 0$: $H_0 = (-1)^0 e^{\xi^2} \cdot e^{-\xi^2} = \boxed{1}$

$n = 1$: $H_1 = (-1)^1 e^{\xi^2}\frac{d}{d\xi}e^{-\xi^2} = -e^{\xi^2}(-2\xi)e^{-\xi^2} = \boxed{2\xi}$

$n = 2$: $\frac{d^2}{d\xi^2}e^{-\xi^2} = \frac{d}{d\xi}(-2\xi e^{-\xi^2}) = -2e^{-\xi^2} + 4\xi^2 e^{-\xi^2} = (4\xi^2 - 2)e^{-\xi^2}$

$H_2 = (+1)e^{\xi^2}(4\xi^2 - 2)e^{-\xi^2} = \boxed{4\xi^2 - 2}$

$n = 3$: $\frac{d^3}{d\xi^3}e^{-\xi^2} = \frac{d}{d\xi}[(4\xi^2 - 2)e^{-\xi^2}] = (8\xi)e^{-\xi^2} + (4\xi^2-2)(-2\xi)e^{-\xi^2} = (-8\xi^3 + 12\xi)e^{-\xi^2}$

$H_3 = (-1)^3 e^{\xi^2}(-8\xi^3 + 12\xi)e^{-\xi^2} = 8\xi^3 - 12\xi = \boxed{8\xi^3 - 12\xi}$

(b) Verify $H_{n+1} = 2\xi H_n - 2nH_{n-1}$:

For $n = 1$: $H_2 \stackrel{?}{=} 2\xi H_1 - 2H_0 = 2\xi(2\xi) - 2(1) = 4\xi^2 - 2$. $\checkmark$

For $n = 2$: $H_3 \stackrel{?}{=} 2\xi H_2 - 4H_1 = 2\xi(4\xi^2 - 2) - 4(2\xi) = 8\xi^3 - 4\xi - 8\xi = 8\xi^3 - 12\xi$. $\checkmark$

Problem 4.9: Classical Turning Points at Large $n$

At the classical turning point, all energy is potential: $E_n = \frac{1}{2}m\omega^2 x_{\text{tp}}^2$.

$$\left(n + \frac{1}{2}\right)\hbar\omega = \frac{1}{2}m\omega^2 x_{\text{tp}}^2$$

$$x_{\text{tp}}^2 = \frac{(2n+1)\hbar}{m\omega}$$

$$\boxed{x_{\text{tp}} = \pm\sqrt{\frac{(2n+1)\hbar}{m\omega}}}$$

For large $n$, $x_{\text{tp}} \approx \sqrt{2n\hbar/(m\omega)}$, which grows as $\sqrt{n}$. The probability of finding the quantum particle beyond the classical turning point decreases as $n$ increases -- the correspondence principle at work.

Chapter 5: The Hydrogen Atom

Problem 1: Coordinate Conversion

(a) $(x, y, z) = (1, 1, \sqrt{2})$:

$$r = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = 2$$

$$\theta = \arccos(z/r) = \arccos(\sqrt{2}/2) = \pi/4$$

$$\phi = \arctan(y/x) = \arctan(1) = \pi/4$$

$$\boxed{(r, \theta, \phi) = (2, \pi/4, \pi/4)}$$

Verification: $x = r\sin\theta\cos\phi = 2\sin(\pi/4)\cos(\pi/4) = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 1$ $\checkmark$

$y = r\sin\theta\sin\phi = 1$ $\checkmark$; $z = r\cos\theta = 2\cos(\pi/4) = \sqrt{2}$ $\checkmark$

Problem 7: Spherical Harmonics Normalization and Orthogonality

(a) $Y_0^0 = 1/\sqrt{4\pi}$:

$$\int_0^{2\pi}\int_0^{\pi}\frac{1}{4\pi}\sin\theta\,d\theta\,d\phi = \frac{1}{4\pi}\cdot 2\pi \cdot \int_0^{\pi}\sin\theta\,d\theta = \frac{1}{2}[-\cos\theta]_0^{\pi} = \frac{1}{2}(1+1) = \boxed{1}\;\checkmark$$

(b) With $Y_1^0 = \sqrt{3/(4\pi)}\cos\theta$ and $Y_1^1 = -\sqrt{3/(8\pi)}\sin\theta\,e^{i\phi}$:

$$\int_0^{2\pi}\int_0^{\pi}(Y_1^0)^* Y_1^1\sin\theta\,d\theta\,d\phi$$

The $\phi$ integral gives $\int_0^{2\pi}e^{i\phi}\,d\phi = 0$. Therefore the entire integral vanishes:

$$\boxed{\langle Y_1^0 | Y_1^1 \rangle = 0}\;\checkmark$$

The orthogonality follows from the different $m$ values: any two spherical harmonics with $m \neq m'$ are orthogonal because the $\phi$ integral $\int_0^{2\pi}e^{i(m'-m)\phi}\,d\phi$ vanishes for $m \neq m'$.

Chapter 6: Operators, Commutators, and the Uncertainty Principle

Problem 2: Operator Products

(a) $\hat{x}\hat{p}\psi = \hat{x}\left(-i\hbar\frac{d\psi}{dx}\right) = -i\hbar x\frac{d\psi}{dx}$

(b) $\hat{p}\hat{x}\psi = -i\hbar\frac{d}{dx}(x\psi) = -i\hbar\left(\psi + x\frac{d\psi}{dx}\right) = -i\hbar\psi - i\hbar x\frac{d\psi}{dx}$

Comparing (a) and (b):

$$[\hat{x}, \hat{p}]\psi = \hat{x}\hat{p}\psi - \hat{p}\hat{x}\psi = -i\hbar x\frac{d\psi}{dx} - \left(-i\hbar\psi - i\hbar x\frac{d\psi}{dx}\right) = i\hbar\psi$$

$$\boxed{[\hat{x}, \hat{p}] = i\hbar}$$

This is the canonical commutation relation -- the cornerstone of quantum mechanics.

(c) $\hat{x}^2\hat{p}\psi = -i\hbar x^2\frac{d\psi}{dx}$

(d) $\hat{p}\hat{x}^2\psi = -i\hbar\frac{d}{dx}(x^2\psi) = -i\hbar\left(2x\psi + x^2\frac{d\psi}{dx}\right)$

Therefore: $[\hat{x}^2, \hat{p}]\psi = -i\hbar x^2\psi' + i\hbar(2x\psi + x^2\psi') = 2i\hbar x\psi$, giving $[\hat{x}^2, \hat{p}] = 2i\hbar\hat{x}$.

Problem 5: Parity Operator

(a) Linearity: $\hat{\Pi}[c_1\psi_1 + c_2\psi_2](x) = [c_1\psi_1 + c_2\psi_2](-x) = c_1\psi_1(-x) + c_2\psi_2(-x) = c_1\hat{\Pi}\psi_1 + c_2\hat{\Pi}\psi_2$. $\checkmark$

(b) $\hat{\Pi}^2\psi(x) = \hat{\Pi}[\psi(-x)] = \psi(-(-x)) = \psi(x)$. So $\hat{\Pi}^2 = \hat{I}$. $\checkmark$

(c) If $\hat{\Pi}|\pi\rangle = \pi|\pi\rangle$, then $\hat{\Pi}^2|\pi\rangle = \pi^2|\pi\rangle = |\pi\rangle$, so $\pi^2 = 1$, giving $\boxed{\pi = \pm 1}$.

Even functions ($\psi(-x) = \psi(x)$) have eigenvalue $+1$; odd functions ($\psi(-x) = -\psi(x)$) have eigenvalue $-1$.

(d) Hermiticity: We need $\langle\phi|\hat{\Pi}|\psi\rangle = \langle\psi|\hat{\Pi}|\phi\rangle^*$.

$$\langle\phi|\hat{\Pi}|\psi\rangle = \int_{-\infty}^{\infty}\phi^*(x)\psi(-x)\,dx$$

Substitute $u = -x$: $= \int_{\infty}^{-\infty}\phi^*(-u)\psi(u)(-du) = \int_{-\infty}^{\infty}\phi^*(-u)\psi(u)\,du = \langle\hat{\Pi}\phi|\psi\rangle = \langle\psi|\hat{\Pi}|\phi\rangle^*$. $\checkmark$

Chapter 7: Time Evolution

Problem 7.1: Time-Dependent State

(a) $\Psi_3(x,t) = \psi_3(x)e^{-iE_3 t/\hbar}$ where $E_3 = \frac{9\pi^2\hbar^2}{2ma^2}$ and $\psi_3(x) = \sqrt{2/a}\sin(3\pi x/a)$.

(b) $|\Psi_3(x,t)|^2 = |\psi_3(x)|^2|e^{-iE_3 t/\hbar}|^2 = |\psi_3(x)|^2 \cdot 1 = \frac{2}{a}\sin^2(3\pi x/a)$.

This is time-independent. $\checkmark$ The complex phase cancels in the modulus squared -- this is why it is called a "stationary state."

(c) Since $|\Psi_3|^2$ is time-independent and symmetric about $x = a/2$:

$\langle\hat{x}\rangle = a/2$ (time-independent). $\langle\hat{p}\rangle = m\frac{d\langle\hat{x}\rangle}{dt} = 0$ (by Ehrenfest's theorem, since $\langle\hat{x}\rangle$ is constant).

Both are time-independent, as expected for a stationary state. Ehrenfest's theorem guarantees that for any stationary state, $\langle\hat{x}\rangle$ and $\langle\hat{p}\rangle$ are constant in time.

Problem 7.6: Gaussian Wave Packet Spreading

Given: $\sigma_0 = 1$ nm, $k_0 = 0$.

The width at time $t$ is:

$$\sigma(t) = \sigma_0\sqrt{1 + \left(\frac{\hbar t}{2m\sigma_0^2}\right)^2}$$

The spreading timescale is $\tau = 2m\sigma_0^2/\hbar$.

For an electron ($m_e = 9.109 \times 10^{-31}$ kg):

$$\tau_e = \frac{2(9.109 \times 10^{-31})(10^{-9})^2}{1.055 \times 10^{-34}} = \frac{1.822 \times 10^{-48}}{1.055 \times 10^{-34}} = 1.727 \times 10^{-14}\,\text{s} \approx 17.3\,\text{fs}$$

(a) At $t = 1$ fs: $t/\tau_e = 0.058$, so $\sigma \approx \sigma_0\sqrt{1 + 0.003} \approx 1.002$ nm.

(b) At $t = 1$ ps = 1000 fs: $t/\tau_e = 57.9$, so $\sigma \approx \sigma_0 \cdot 57.9 \approx 57.9$ nm.

(c) At $t = 1$ ns: $t/\tau_e = 57,900$, so $\sigma \approx 57.9\,\mu$m.

Doubling time: $\sigma = 2\sigma_0$ when $t/\tau_e = \sqrt{3}$, so $t_{\text{double}} = \sqrt{3}\,\tau_e \approx \boxed{30\,\text{fs}}$.

For a proton ($m_p = 1.673 \times 10^{-27}$ kg):

$$\tau_p = \frac{2(1.673 \times 10^{-27})(10^{-9})^2}{1.055 \times 10^{-34}} = 3.171 \times 10^{-11}\,\text{s} \approx 31.7\,\text{ps}$$

The proton packet spreads roughly 1,836 times slower than the electron, because the spreading timescale $\tau \propto m$.

Doubling time for proton: $t_{\text{double}} = \sqrt{3}\,\tau_p \approx \boxed{54.9\,\text{ps}}$.

Chapter 8: Linear Algebra and Dirac Notation

Problem 1: Spin-1/2 State

Given: $|\psi\rangle = \frac{3}{5}|\uparrow\rangle + \frac{4i}{5}|\downarrow\rangle$.

(a) $\langle\psi| = \frac{3}{5}\langle\uparrow| + \frac{-4i}{5}\langle\downarrow| = \frac{3}{5}\langle\uparrow| - \frac{4i}{5}\langle\downarrow|$

(Remember: the bra is the conjugate transpose, so $4i \to -4i$.)

(b) $\langle\psi|\psi\rangle = \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9}{25} + \frac{16}{25} = 1$. $\checkmark$

(c) $|\langle\uparrow|\psi\rangle|^2 = |3/5|^2 = 9/25 = 0.36$. This is the probability of measuring spin-up.

$|\langle\downarrow|\psi\rangle|^2 = |4i/5|^2 = 16/25 = 0.64$. This is the probability of measuring spin-down.

(d) $0.36 + 0.64 = 1$. $\checkmark$

Problem 2: Inner Products

Given: $|\alpha\rangle = 2i|1\rangle + 3|2\rangle$, $|\beta\rangle = |1\rangle - i|2\rangle$, $\{|1\rangle, |2\rangle\}$ orthonormal.

(a) $\langle\alpha|\beta\rangle = (-2i)(1) + (3)(-i) = -2i - 3i = \boxed{-5i}$

(The bra $\langle\alpha| = -2i\langle 1| + 3\langle 2|$.)

(b) $\langle\beta|\alpha\rangle = (1)(2i) + (i)(3) = 2i + 3i = \boxed{5i}$

(c) $\langle\alpha|\beta\rangle^* = (-5i)^* = 5i = \langle\beta|\alpha\rangle$. $\checkmark$

(d) $\langle\alpha|\alpha\rangle = |2i|^2 + |3|^2 = 4 + 9 = 13$; $\langle\beta|\beta\rangle = |1|^2 + |-i|^2 = 1 + 1 = 2$.

Chapter 9: Eigenvalue Problems and Spectral Theory

Problem 3: Eigenvalue Problem for $\hat{S}_y$

(a) $\det(S_y - \lambda I) = 0$:

$$\det\begin{pmatrix} -\lambda & -i\hbar/2 \\ i\hbar/2 & -\lambda\end{pmatrix} = \lambda^2 - \frac{\hbar^2}{4} = 0$$

$$\boxed{\lambda = \pm\frac{\hbar}{2}}$$

(b) For $\lambda = +\hbar/2$: $(S_y - \frac{\hbar}{2}I)\mathbf{v} = 0$:

$$\begin{pmatrix} -\hbar/2 & -i\hbar/2 \\ i\hbar/2 & -\hbar/2\end{pmatrix}\begin{pmatrix}a \\ b\end{pmatrix} = 0$$

First row: $-a - ib = 0$, so $b = ia$. Normalizing: $|a|^2 + |b|^2 = |a|^2(1+1) = 1$, so $a = 1/\sqrt{2}$.

$$\boxed{|+\rangle_y = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ i\end{pmatrix} = \frac{1}{\sqrt{2}}(|\uparrow\rangle + i|\downarrow\rangle)}$$

For $\lambda = -\hbar/2$: first row gives $a - ib = 0$, so $b = -ia$.

$$\boxed{|-\rangle_y = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -i\end{pmatrix} = \frac{1}{\sqrt{2}}(|\uparrow\rangle - i|\downarrow\rangle)}$$

Chapter 10: Symmetry and Conservation Laws

Problem 1: Translation Operator

(a) $\hat{T}^\dagger(a) = e^{+i\hat{p}a/\hbar}$ (since $\hat{p}^\dagger = \hat{p}$).

$$\hat{T}^\dagger(a)\hat{T}(a) = e^{i\hat{p}a/\hbar}e^{-i\hat{p}a/\hbar} = e^0 = \hat{I}\;\checkmark$$

(b) $\hat{T}(a)\hat{T}(b) = e^{-i\hat{p}a/\hbar}e^{-i\hat{p}b/\hbar} = e^{-i\hat{p}(a+b)/\hbar} = \hat{T}(a+b)$.

(The exponentials combine because $\hat{p}$ commutes with itself: $[\hat{p}, \hat{p}] = 0$.)

(c) $\hat{T}^{-1}(a) = \hat{T}^\dagger(a) = e^{i\hat{p}a/\hbar} = e^{-i\hat{p}(-a)/\hbar} = \hat{T}(-a)$. $\checkmark$

(d) $\langle x|\hat{T}(a)|\psi\rangle = \int\langle x|\hat{T}(a)|x'\rangle\langle x'|\psi\rangle\,dx'$.

Since $\hat{T}(a)|x'\rangle = |x'+a\rangle$, we get $\langle x|x'+a\rangle = \delta(x - x' - a)$. Therefore:

$$\langle x|\hat{T}(a)|\psi\rangle = \int\delta(x-x'-a)\psi(x')\,dx' = \psi(x-a)$$

$$\boxed{\hat{T}(a)\text{ shifts the wave function to the right by }a.}$$

Problem 4: Operator Transformation Under Translation (BCH Lemma)

(a) Using $e^{\hat{A}}\hat{B}e^{-\hat{A}} = \hat{B} + [\hat{A}, \hat{B}] + \frac{1}{2!}[\hat{A}, [\hat{A}, \hat{B}]] + \cdots$ with $\hat{A} = i\hat{p}a/\hbar$ and $\hat{B} = \hat{x}$:

$[\hat{A}, \hat{B}] = \frac{ia}{\hbar}[\hat{p}, \hat{x}] = \frac{ia}{\hbar}(-i\hbar) = a$

Since $[\hat{A}, a] = 0$ (a constant), all higher commutators vanish:

$$\hat{T}^\dagger(a)\hat{x}\hat{T}(a) = \hat{x} + a\hat{I}\;\checkmark$$

(b) With $\hat{B} = \hat{p}$: $[\hat{A}, \hat{p}] = \frac{ia}{\hbar}[\hat{p}, \hat{p}] = 0$. All terms beyond $\hat{B}$ vanish:

$$\hat{T}^\dagger(a)\hat{p}\hat{T}(a) = \hat{p}\;\checkmark$$

(c) Translation shifts where you are ($x \to x + a$) but not how fast you are moving ($p$ unchanged). This is exactly what happens classically when you change your coordinate origin. The momentum of a particle does not depend on where you put the origin.

Additional solutions for Chapters 11-40 are available in the online supplement at the textbook's companion website. Solutions to all problems (not just selected ones) are available in the Instructor's Manual.