Multiplying and subtracting, the $V\psi_n^*\psi_n$ terms cancel exactly since $V$ is real, leaving

$$-\frac{\hbar^2}{2m}\left[\psi_n^*\frac{\mathrm d^2\psi_n}{\mathrm dx^2} - \psi_n\frac{\mathrm d^2\psi_n^*}{\mathrm dx^2}\right] = (E_n - E_n^*)\psi_n^*\psi_n$$

The bracket is a perfect derivative, giving the stated form. Integrating over all $x$:

$$-\frac{\hbar^2}{2m}\left[\psi_n^*\frac{\mathrm d\psi_n}{\mathrm dx} - \psi_n\frac{\mathrm d\psi_n^*}{\mathrm dx}\right]_{-\infty}^{+\infty} = (E_n - E_n^*)\int_{-\infty}^{+\infty}|\psi_n|^2\,\mathrm dx$$

The left side vanishes because $\psi_n \to 0$ at infinity. The integral on the right is 1 for a normalized state — in any case strictly positive. Therefore

$$E_n - E_n^* = 0 \quad\Rightarrow\quad E_n \text{ is real}$$

Why it matters. An energy eigenvalue is something you can measure, and measurements return real numbers. This shows the formalism delivers that automatically, provided $V$ is real. Make $V$ complex and the argument fails — which is exactly how absorption is modelled, with the imaginary part giving states a finite lifetime.

The $V$ terms cancel as before, the bracket is again a perfect derivative, and integrating kills the boundary term:

$$0 = (E_n - E_m)\int_{-\infty}^{+\infty}\psi_m^*\psi_n\,\mathrm dx$$

If $E_m \ne E_n$ the prefactor is non-zero, so the integral must vanish. That is Eq. (4.49).

What it does not cover. If $E_m = E_n$ — a degenerate level — the prefactor is zero and the equation says nothing at all. That is precisely the gap §4.5 acknowledges: eigenfunctions of a degenerate eigenvalue are not uniquely determined, and you must choose an orthogonal set (Gram–Schmidt). The proof does not fail there; it simply has no content.

Why it matters. Orthogonality is what makes the expansion coefficients extractable at all — every step of Eq. (4.51) depends on the cross terms vanishing.

$$E_2 - E_1 = \frac{3\pi^2\hbar^2}{2ma^2}$$

(a) With $a = a_0$ the bracket is exactly the Rydberg energy:

$$E_2 - E_1 = 3\pi^2 \times 13.6\ \mathrm{eV} \approx 403\ \mathrm{eV}$$

The right order for atomic physics — soft X-ray. The crude box overestimates real atomic transitions (a few eV) because a real atom’s potential is a smooth Coulomb well rather than hard walls, but the scale is right.

(b) With $a = 1$ fm and $m = m_n$:

$$E_2 - E_1 \approx 614\ \mathrm{MeV}$$

The point of the pair. Same formula, no new physics, and six orders of magnitude between them. The mass rises by 1839 (lowering the levels) but the box shrinks by a factor of $5\times10^5$, and that enters squared. Confinement scale alone explains why atomic physics is measured in eV and nuclear physics in MeV.

$$\Psi_n(x,t) = \frac{N}{2i}\left[e^{+i(k_nx - E_nt/\hbar)} - e^{-i(k_nx + E_nt/\hbar)}\right]$$

The first term is a plane wave travelling to the right with momentum $+\hbar k_n$; the second travels to the left with $-\hbar k_n$. Equal amplitudes, so the standing wave is an equal superposition of the two.

This is §3.4’s momentum result, arrived at from the other direction. There, transforming $\sin k_nx$ gave two momentum peaks near $\pm n\hbar\pi/a$; here the same fact is visible in one line of algebra. And it explains why $\langle p\rangle = 0$ while $\Delta p \ne 0$: the two travelling waves cancel in the mean and add in the spread.

A caution about the picture. It is tempting to say the particle “bounces back and forth”. Nothing in the formalism supports that: $|\Psi|^2$ is completely static, so there is no bouncing to observe. The superposition is of two momentum possibilities, not two phases of a journey.

(a) The two-dimensional versions of Eqs. (4.43) and (4.44):

$$\psi_{n_x,n_y}(x,y) = N\sin\left(\frac{n_x\pi x}{a}\right)\sin\left(\frac{n_y\pi y}{b}\right),\qquad E_{n_x,n_y} = \frac{\pi^2\hbar^2}{2m}\left[\frac{n_x^2}{a^2} + \frac{n_y^2}{b^2}\right]$$

(b) In units of $\pi^2\hbar^2/2ma^2$:

	first four levels
$a = b$	2 (×1, 11) · 5 (×2, 21) · 8 (×1, 22) · 10 (×2, 31)
$a = 2b$	5 (×1) · 8 (×1) · 13 (×1) · 17 (×1) — all non-degenerate

The square box has degeneracies for the reason §4.4 gives: swapping $x$ and $y$ is a symmetry, so $(1,2)$ and $(2,1)$ must share an energy. Stretch it to $a = 2b$ and that symmetry is gone, and so are the degeneracies.

(a) $|c_1|^2 = \tfrac14$ and $|c_2|^2 = \tfrac34$ give

$$\Psi = \tfrac12\,\psi_1e^{-iE_1t/\hbar} + \tfrac{\sqrt3}{2}\,\psi_2e^{-iE_2t/\hbar}$$

which is normalized since $\tfrac14 + \tfrac34 = 1$.

(b) $\langle E^2\rangle = \tfrac14E_1^2 + \tfrac34E_2^2$, so

$$(\Delta E)^2 = \tfrac14E_1^2 + \tfrac34E_2^2 - \left(\tfrac14E_1 + \tfrac34E_2\right)^2 = \tfrac{3}{16}(E_1-E_2)^2$$$$\Delta E = \frac{\sqrt3}{4}|E_1 - E_2|$$

Compare Eq. (4.57)‘s equal mix, which gave $\tfrac12|E_1-E_2|$. An uneven split has a smaller spread — sensible, since pushing all the weight onto one level would give $\Delta E = 0$.

(c)

$$|\Psi|^2 = \tfrac14|\psi_1|^2 + \tfrac34|\psi_2|^2 + \tfrac{\sqrt3}{4}\left[\psi_1^*\psi_2e^{+i(E_1-E_2)t/\hbar} + \text{c.c.}\right]$$

The cross term oscillates at $|E_1-E_2|/\hbar$, so the period is $T = 2\pi\hbar/|E_1-E_2|$. Substituting $|E_1-E_2| = 4\Delta E/\sqrt3$:

$$T = \frac{\sqrt3\,\pi\hbar}{2\Delta E}$$

which is $\delta t\,\Delta E \approx \hbar$ again, with the numerical factor depending on the mixing. The period is set by the level separation, not by the mixing; only the depth of the oscillation depends on the coefficients.

(a) $\displaystyle\int_0^a\sin^2k_nx\,\mathrm dx = \frac{a}{2}$ (the mean of $\sin^2$ is $\tfrac12$ over a whole number of half-periods, which $k_n = n\pi/a$ guarantees). So $N^2a/2 = 1$ and $N = \sqrt{2/a}$.

(b) With the product-to-sum identity,

$$\int_0^a\sin k_mx\sin k_nx\,\mathrm dx = \frac12\int_0^a\left[\cos\frac{(m-n)\pi x}{a} - \cos\frac{(m+n)\pi x}{a}\right]\mathrm dx$$

Each cosine has an integer number of full periods in $[0,a]$ and integrates to zero — provided $m \ne n$. When $m = n$ the first cosine becomes $\cos 0 = 1$ and contributes $a/2$, which is exactly part (a).

(c) Multiply the series by $\psi_n$ and integrate; orthogonality leaves one term:

$$\int_0^a\psi_nf\,\mathrm dx = c_n \quad\Rightarrow\quad c_n = \sqrt{\frac2a}\int_0^a\sin k_nx\,f(x)\,\mathrm dx$$

With the book’s convention of absorbing $N$ into $\psi_n$ so that $f = \sum c_n\sin k_nx$, this becomes $c_n = \frac2a\int_0^a\sin k_nx\,f\,\mathrm dx$ — the standard Fourier sine coefficient.

$$c_1 = \int_0^{a/2}\sqrt{\frac2a}\sin\frac{\pi x}{a}\cdot\sqrt{\frac4a}\sin\frac{2\pi x}{a}\,\mathrm dx = \frac{2\sqrt2}{a}\int_0^{a/2}\sin\frac{\pi x}{a}\sin\frac{2\pi x}{a}\,\mathrm dx$$

Using $\sin A\sin B = \tfrac12[\cos(A-B) - \cos(A+B)]$:

$$= \frac{\sqrt2}{a}\left[\frac{a}{\pi}\sin\frac{\pi x}{a} - \frac{a}{3\pi}\sin\frac{3\pi x}{a}\right]_0^{a/2} = \frac{\sqrt2}{\pi}\left[1 + \frac13\right] = \frac{4\sqrt2}{3\pi}$$$$P(E_1) = |c_1|^2 = \frac{32}{9\pi^2} \approx 0.360$$

Worth computing the rest. $|c_2|^2 = \tfrac12$ exactly, and $|c_4|^2 = 0$ exactly — as does every even $n$ except 2.

$$\frac{E_nT}{\hbar} = \frac{n^2\pi^2\hbar}{2ma^2}\cdot\frac{4ma^2}{\pi\hbar} = 2\pi n^2$$

Since $n$ is an integer, $n^2$ is an integer, and $e^{-i2\pi n^2} = 1$ for every $n$. Every term of the superposition therefore returns to its $t = 0$ value at the same instant, so $\Psi(x, T) = \Psi(x, 0)$ exactly — and the motion is periodic with period $T$.

The essential ingredient is that $E_n \propto n^2$. The ratio of any two energies is a ratio of integers, so all the phases are commensurate. A potential whose levels are not in rational ratios has no exact revival — the state comes arbitrarily close to its initial form but never returns to it.

The widget on §4.6 traces $\langle x\rangle$ over exactly this period, which is why that curve closes.

From $E_2 - E_1 = hc/\lambda$,

$$|\Delta E| = \frac{hc}{\lambda^2}\Delta\lambda \quad\Rightarrow\quad \Delta\lambda = \frac{\lambda^2}{hc}\Delta E$$

Substituting $\Delta E = \hbar/\tau$ and $\hbar = h/2\pi$:

$$\Delta\lambda = \frac{\lambda^2}{hc}\cdot\frac{\hbar}{\tau} = \frac{\lambda^2}{2\pi c\tau}$$

Note what it does not contain. $\Delta\lambda$ depends on $\lambda$ and $\tau$ but not on the atom, the transition, or the intensity — it is fixed entirely by how long the state lives.

$$\tau = \frac{\hbar}{\Delta E} = \frac{6.582\times10^{-16}\ \mathrm{eV\,s}}{2.5\times10^{9}\ \mathrm{eV}} \approx 2.6\times10^{-25}\ \mathrm{s}$$

Read the relation backwards. No clock can time $10^{-25}$ s, and no detector can see a particle whose decay length is $c\tau \approx 8\times10^{-17}$ m — a thousandth of a proton radius. The Z is never observed to travel; it is created and decays at the same point.

So its lifetime is not measured. Its width is: experiments scan the collision energy and record the resonance peak’s shape, and $\Delta E = 2.5$ GeV is that measured width. Equation (4.59) then converts it into a lifetime that could not be observed any other way.

Note the scale: 2.5 GeV against a mass energy of 91.2 GeV means the Z’s mass is uncertain by 2.7% of itself. A “particle” too short-lived to have a well-defined mass.