(a) Outside the well $E < V = 0$, so $\mathrm d^2\psi/\mathrm dx^2 = \alpha^2\psi$ and the solutions are $e^{\pm\alpha x}$. Finiteness picks $e^{+\alpha x}$ on the left and $e^{-\alpha x}$ on the right. Inside, $E + V_0 > 0$ gives $\mathrm d^2\psi/\mathrm dx^2 = -k_0^2\psi$, so $\psi$ is a combination of $\cos k_0x$ and $\sin k_0x$ — and positive parity kills the sine, because $\sin$ is odd. The same coefficient $A$ appears on both outside pieces for the same reason.

That is the whole content of parity here: it is a symmetry of the Hamiltonian, and it halves the work by forbidding one of the two solutions in each region before any boundary condition is applied.

(b) Match at $x = a$:

Dividing the second by the first removes $A$ and $C$ together:

Matching at $x = -a$ gives the identical condition — which is the payoff of parity. One interface does the work of two.

(c) See the correction below.

(d) For negative parity the inside solution is $C\sin k_0x$, and the same division gives $k_0\cot k_0a = -\alpha$ — which is Eq. (5.11) exactly. The reason is worth stating: an antisymmetric function must vanish at the origin, and $\psi(0) = 0$ is precisely the condition the infinite wall imposed in Fig. 5.1. A symmetric well of half-width $a$ therefore contains a perfect copy of Fig. 5.1’s spectrum, hidden in its odd states.

(a) For a function of $r$ alone the radial Laplacian is $\nabla^2\psi = \frac{1}{r}\frac{\mathrm d^2(r\psi)}{\mathrm dr^2}$. With $\psi = u/r$ we have $r\psi = u$, so $\nabla^2\psi = \frac{1}{r}\frac{\mathrm d^2u}{\mathrm dr^2}$. Substituting and multiplying through by $r$ gives the stated equation — a one-dimensional eigenvalue problem in $r$.

This is the same substitution §5.2b used for the two-proton problem, and the reason it is worth doing once carefully: it converts every spherically symmetric 3-D problem in this book into a 1-D one.

(b) Inside, $u'' = -k_0^2u$ so $u = C\sin k_0r + C'\cos k_0r$. Here is the new ingredient: $\psi = u/r$ must stay finite at the origin, so $u(0) = 0$, which kills the cosine. Outside, $u'' = \alpha^2u$ and finiteness at infinity leaves $u = Ae^{-\alpha r}$. Dividing by $r$ gives the stated $\psi$.

(c) Matching $u$ and $u'$ at $r = R$ and dividing:

which is Eq. (5.11) with $a \to R$ — identical to Fig. 5.1’s condition, and for exactly the reason problem 1(d) gave: $u(0) = 0$ is an infinite wall at the origin. So the counting carries over unchanged: the first bound state needs $w > \pi/2R$, the second $w > 3\pi/2R$, where $w = \sqrt{2mV_0/\hbar^2}$. Squaring those thresholds,

as required.

With $\hbar = m = a = 1$ the numbers are clean. From $E = \pi^2/2$ and $V_0 = 2\pi^2$:

Eq. (5.19) reads $k_0\cot k_0a = k\cot(ka+\delta)$. Since $\cot$ has period $\pi$ and $ka = \pi$, the left side collapses:

so $\delta = \operatorname{arccot}(2.4431) = 0.3888$, and the family of valid values is $\dots,\ -2.7528,\ 0.3888,\ 3.5304,\ \dots$ spaced by $\pi$. The book quotes the third of these.

Why the ambiguity is real and not sloppiness. Eq. (5.19) constrains $\sin(kx+\delta)$ only through a cotangent, and shifting $\delta$ by $\pi$ flips the sign of the whole outside wave function — which is not an observable change, since $\psi$ and $-\psi$ describe the same state. Only differences of $\delta$ between energies matter, which is why §5.1b’s time delay $\tau = 2\hbar\,\mathrm d\delta/\mathrm dE$ is well defined even though $\delta$ itself is not.

(a) This problem runs the eigenvalue equation backwards. Rearranged, it is not a differential equation at all but a formula:

Differentiate twice:

so $\psi''/\psi = \alpha^2 - 2\alpha/x$. Substituting, with $E = -\hbar^2\alpha^2/2m$:

The $\alpha^2$ terms cancel exactly — which is what makes this $\psi$ an eigenfunction rather than merely a function. And $V = \infty$ for $x < 0$ is what forces $\psi(0) = 0$, which the factor of $x$ in $Nxe^{-\alpha x}$ delivers.

(b) The first excited state has one node at some $x > 0$: it rises, comes back through zero, dips to a negative minimum, and decays to zero from below. It is also more spread out, because it is less tightly bound.

(a) The time factor $e^{-iEt/\hbar}$ cancels between $\Psi$ and $\Psi^*$, so $j$ may be computed from $\psi_E$ alone. On the right this is immediate: with $\psi = A_Te^{ikx}$, $\partial\psi/\partial x = ikA_Te^{ikx}$, and

On the left the algebra produces four terms. Two of them are the incident and reflected currents, $+|A_I|^2\hbar k/m$ and $-|A_R|^2\hbar k/m$ — note the sign, which is the reflected wave carrying probability the other way. The two cross terms, proportional to $A_I^*A_R e^{-2ikx}$ and its conjugate, cancel exactly, because they enter as a quantity minus its own complex conjugate multiplied by $i$ — leaving something real that turns out to be zero. That cancellation is the crux: without it $j$ would depend on $x$, and no conservation statement would be possible.

(b) For a stationary state $|\Psi(x,t)|^2 = |\psi_E(x)|^2$ has no time dependence at all, so for any $x_1, x_2$,

The current is therefore the same everywhere. Equating its value on the left to its value on the right:

The common factor $\hbar k/m$ cancels — and it can only cancel because $k$ is the same on both sides, which is true here because $V = 0$ on both sides and would not be true for a step. Dividing by $|A_I|^2$ gives $1 - R = T$.

Write $q \equiv k_B$ and apply the four matching conditions. With $A_I = 1$ the result is the same closed form §5.2a derived, since the algebra never cared whether $q$ was real or imaginary:

Taking $|A_T|^2$ and simplifying the denominator with $|e^{\pm iqa}|=1$ gives

which is the stated result with $S = (k^2-k_B^2)/2kk_B$. Then $R = 1 - T$ gives the second formula immediately — so problem 5’s conservation law does half of problem 6’s work.

Complete transparency. $T = 1$ requires $\sin k_Ba = 0$, i.e.

The barrier width is then a whole number of half wavelengths, so the waves reflected from its two faces are exactly out of phase and cancel. Nothing is absorbed and nothing is reflected: the barrier becomes invisible.

The turning point. Eq. (5.44) with $e^2/4\pi\epsilon_0 = 1.44\ \mathrm{MeV\,fm}$:

360 times the $2\ \mathrm{fm}$ range of the nuclear force.

Two protons. $E_G = 493\ \mathrm{keV}$ from §5.2b, so

Two helium nuclei. Eq. (5.51) must be generalized, because it was written for unit charges. Restoring them, $e^2 \to Z_1Z_2e^2$:

For $^4\mathrm{He}$, $Z_1Z_2 = 4$ and $\mu = m_{\mathrm{He}}/2 \approx 2m_p$, so $E_G$ grows by $4^2 = 16$ from the charge and by about 4 from the mass — a factor of 64, giving $E_G = 31.3\ \mathrm{MeV}$. The exponent grows by $\sqrt{64} = 8$:

48 orders of magnitude below the proton case, at the same energy.

The expression.

The estimate. Both decays are an alpha particle leaving a heavy nucleus of almost the same charge ($Z = 90$ and $92$), so $E_G$ and $\nu$ are very nearly common to the two and cancel in a ratio:

With $Z_1Z_2 = 2\times90$ and $\mu \approx 3.93\,u$, Eq. (5.51) gives $E_G \approx 1.25\times10^5\ \mathrm{MeV}$, so $\sqrt{E_G/E}$ falls from $163.3$ to $154.3$ — a drop of $8.97$ — and

The measured value is $2.4\times10^4$ years. A factor of four, from a model with one adjustable input, across a range of four and a half orders of magnitude in half-life.

What the answer is really showing. A $12\%$ increase in the energy released shortens the half-life by a factor of nearly $8000$. This is the Geiger–Nuttall law, discovered empirically in 1911 and unexplained until Gamow derived exactly this exponent in 1928 — the first successful application of quantum mechanics to the nucleus. It is why measured alpha half-lives run from microseconds to longer than the age of the universe while the energies behind them vary by less than a factor of three.