(a) $m_l$ takes $2l+1 = 7$ values and $m_s$ takes $2s+1 = 3$, independently:

(b) Eq. (8.6) runs from $l+s$ down to $|l-s|$ in steps of 1:

(c) Each $j$ carries $2j+1$ values of $m_j$:

The counts agree, and the note is telling you this is not a coincidence. The labels $(m_l, m_s)$ and $(j, m_j)$ describe the same 21 states — coupling is a change of basis, and a change of basis cannot alter a dimension. That is also why Eq. (8.6) stops at $|l-s|$: sort the uncoupled states by $m_j = m_l + m_s$ into a staircase and peel multiplets off the top, and the columns run out exactly there.

(a) Setting $\sqrt{l(l+1)}\,\hbar = 9.1\times10^{-33}$ J s gives $\sqrt{l(l+1)} = 86.5$, so

(b) $2l+1 = \mathbf{173}$ discrete values of $L_z$.

(c) Adjacent values differ by $\hbar$, against a magnitude $\sqrt{86\cdot87}\,\hbar = 86.5\hbar$:

And that is the answer the problem is really after. A 1.16% step is already hard to resolve; for a spinning object of everyday size $l$ is around $10^{30}$ and the steps are $10^{-30}$ of the total. The quantization never switches off — there are always exactly $2l+1$ values — it just becomes unmeasurably fine. §8.1’s widget makes the same point with the cone angle, which at $l = 86$ is already down to $6.15°$.

(a) With $l = 0$ the only angular momentum is the electron’s spin, $s = \tfrac12$, so $m_s = \pm\tfrac12$ — two states. By Eq. (8.8) their magnetic energies are $\mp 2\mu_B B m_s$, i.e. $\pm\mu_B B$, so the gap is

(b) The proton is also spin-½, so it has its own two orientations, and Eq. (8.12) gives its moment as $2.79\,\mu_N m_j$. Each electron level therefore splits again, by

Four levels in total, in two widely separated pairs — the electron splitting is 658 times the proton splitting, because $\mu_B/\mu_N = m_p/m_e = 1836$ and the proton’s 2.79 recovers only part of that.

(a) Each mass sits at distance $a/2$ from the centre, so

Classically $E = L^2/2I$. This system has no potential energy at all — nothing but rotation — so the Hamiltonian is $\hat L^2/2I$ and its eigenvalues follow immediately from Eq. (8.23):

(b) $2l+1$, one for each $m_l$ — the field-free degeneracy that §8.2’s magnetic field lifts.

(c) With $m = m_p$ and $a = 0.075$ nm,

Now compare with $kT$ at room temperature, which is $0.0259$ eV. The rotational excitation costs less than the thermal energy available, so H₂ molecules in a room are rotating — the states are thermally populated.

(a) By the chain rule, $\dfrac{\partial\psi}{\partial\phi} = \dfrac{\partial\psi}{\partial x}\dfrac{\partial x}{\partial\phi} + \dfrac{\partial\psi}{\partial y}\dfrac{\partial y}{\partial\phi} + \dfrac{\partial\psi}{\partial z}\dfrac{\partial z}{\partial\phi}$. From the coordinate relations, $\partial x/\partial\phi = -r\sin\theta\sin\phi = -y$, $\partial y/\partial\phi = +r\sin\theta\cos\phi = +x$, and $\partial z/\partial\phi = 0$. Substituting,

which is exactly the bracket in $\hat L_z$.

(b) $-i\hbar\,\dfrac{\partial}{\partial\phi}\dfrac{e^{im_l\phi}}{\sqrt{2\pi}} = -i\hbar(im_l)\dfrac{e^{im_l\phi}}{\sqrt{2\pi}} = m_l\hbar Z_{m_l}$.

(c) $(r,\theta,\phi)$ and $(r,\theta,\phi+2\pi)$ are the same point in space, so a wave function that assigns two different values to them would not be a function of position. Requiring $e^{im_l(\phi+2\pi)} = e^{im_l\phi}$ gives $e^{2\pi im_l} = 1$, and hence

(d) $\displaystyle\int_0^{2\pi}\frac{e^{-im_l'\phi}}{\sqrt{2\pi}}\frac{e^{im_l\phi}}{\sqrt{2\pi}}\mathrm d\phi = \frac{1}{2\pi}\int_0^{2\pi}e^{i(m_l-m_l')\phi}\,\mathrm d\phi$. For $m_l = m_l'$ the integrand is 1 and the result is 1; otherwise the exponential completes a whole number of cycles and integrates to zero.

(e) Multiply the expansion by $Z^*_{m_l}$ and integrate. By (d) every term vanishes except the one with matching index, which returns $c_{m_l}$.

Multiplying the two exponential forms:

Four terms, all of the same magnitude. Comparing with Eq. (8.27), the outcomes are $m_l = +3, +1, -1, -3$, each with probability $\tfrac14$.

Note which values are absent: $m_l = 0$ and $m_l = \pm2$ never occur. A product of trigonometric functions does not spread over every harmonic — it picks out a specific few, exactly as multiplying two signals in DSP produces sum and difference frequencies and nothing else.

Use the same trick:

So $m_l = 0, +2, -2$ with amplitudes in the ratio $\tfrac12 : \tfrac14 : \tfrac14$. Squaring gives $\tfrac14 : \tfrac1{16} : \tfrac1{16}$, and normalizing by their sum $\tfrac38$:

Compare with problem 6. There, four equal amplitudes gave four equal probabilities. Here the amplitudes are unequal, and squaring exaggerates the imbalance: a 2:1 amplitude ratio becomes 4:1 in probability. Confirmed numerically above.

Note also that $\cos^2\phi$ is real and even, so its decomposition is symmetric in $\pm m_l$ — the state has $\langle L_z\rangle = 0$ even though it is not an eigenstate.

Separation. $\hat L_z$ acts only on $\phi$, so its eigenfunctions fix the $\phi$ dependence as $e^{im_l\phi}$ (problem 5). Substituting $Y = F(\theta)e^{im_l\phi}$ into $\hat L^2Y = l(l+1)\hbar^2Y$, the $\partial^2/\partial\phi^2$ term returns $-m_l^2$, the exponential cancels throughout, and the stated ODE remains.

The three substitutions.

$F_a = A$: both derivatives vanish, leaving $\left(l(l+1) - m_l^2/\sin^2\theta\right)A = 0$ for all $\theta$. The $\theta$-dependent term must vanish on its own, so $m_l = 0$, and then $l(l+1) = 0$ gives $l = 0$. $(l, m_l) = (0,0)$.

$F_b = B\cos\theta$: $F'' = -B\cos\theta$ and $(\cos\theta/\sin\theta)F' = -B\cos\theta$, so the equation becomes $B\cos\theta\left(-2 + l(l+1) - m_l^2/\sin^2\theta\right) = 0$. Again $m_l = 0$, and $l(l+1) = 2$ gives $(l, m_l) = (1,0)$.

$F_c = C\sin\theta$: substituting and multiplying through by $\sin\theta/C$ leaves $1 - m_l^2 + \sin^2\theta\left(l(l+1) - 2\right) = 0$. Both pieces must vanish separately, giving $m_l^2 = 1$ and $l(l+1) = 2$: $(l, m_l) = (1,\pm1)$.

These are exactly Table 8.1’s $l \le 1$ entries, recovered from the differential equation rather than quoted.

The 3-D oscillator’s level $n$ has energy $(n + \tfrac32)\hbar\omega$ and degeneracy $(n+1)(n+2)/2$, with eigenfunctions being polynomials of degree $n$ in $x, y, z$ times the Gaussian $e^{-r^2/2a^2}$.

(a) $n = 0$: one state, the bare Gaussian. It depends only on $r$, so it is spherically symmetric — and §8.3a proved $\hat{\mathbf L}R(r) = 0$. Hence $l = 0$, $m_l = 0$.

(b) $n = 1$: three states, spanned by $x$, $y$, $z$ times the Gaussian. Table 8.1 says the combinations with definite $L_z$ are

all with $l = 1$, since each is $r$ times a $Y_{1,m_l}$.

(c) $n = 2$: six states, spanned by the quadratics $x^2, y^2, z^2, xy, yz, zx$ times the Gaussian. Five combinations match $Y_{2,m_l}$ — for instance $(x\pm iy)^2$ gives $m_l = \pm2$, $(x\pm iy)z$ gives $m_l = \pm1$, and $3z^2 - r^2$ gives $m_l = 0$. That accounts for five.

The sixth is $r^2 = x^2+y^2+z^2$ itself, which is spherically symmetric and therefore has $l = 0$, $m_l = 0$.

So $6 = 5 + 1$: this level carries two different values of $l$ at the same energy.