Exercises — Spin — intrinsic angular momentum

What the rule says: $m_s$ runs from $-s$ to $+s$ in whole-number steps. Apply: from $-\frac{1}{2}$ up by $1$ gives $-\frac{1}{2},+\frac{1}{2}$. Stop (the next step $+\frac{3}{2}$ exceeds $+s$). Answer: $m_s=-\frac{1}{2}$ and $m_s=+\frac{1}{2}$ — exactly two values. Why exactly two? Because $2s+1=2(\frac{1}{2})+1=2$. This is the reason Stern–Gerlach gives two spots.

Rule: number of orientations $=2s+1$. Apply: $2(\frac{3}{2})+1=4$. Answer: 4 spots (from $m_s=-\frac{3}{2},-\frac{1}{2},+\frac{1}{2},+\frac{3}{2}$).

Rule: $|\vec{S}|=\hslash \sqrt{s(s+1)}$. Apply: $\sqrt{\frac{1}{2}\cdot \frac{3}{2}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$. Answer: $|\vec{S}|=\frac{\sqrt{3}}{2}\hslash \approx 0.866\hslash$.

Why use cosine? The $z$-axis is one side of a right triangle whose hypotenuse is $\vec{S}$. The side along $z$ is the shadow $S_z$. "Adjacent over hypotenuse" is exactly $\cos \theta$ — that is the tool that turns a shadow-length into an angle. $\cos \theta =\frac{S_z}{|\vec{S}|}=\frac{\frac{1}{2}\hslash }{\frac{\sqrt{3}}{2}\hslash }=\frac{1}{\sqrt{3}}.$ $\theta =\arccos \frac{1}{\sqrt{3}}\approx 54.7^{\circ }.$ Answer: $\theta \approx 54.7^{\circ }$. It is not $0^{\circ }$ — the arrow tilts, always.

$\cos \theta =\frac{-\frac{1}{2}\hslash }{\frac{\sqrt{3}}{2}\hslash }=-\frac{1}{\sqrt{3}}\Rightarrow \theta =\arccos \left(-\frac{1}{\sqrt{3}}\right)\approx 125.3^{\circ }.$ Why the negative sign matters: a negative cosine means the arrow points into the lower hemisphere (angle past $90^{\circ }$). $125.3^{\circ }=180^{\circ }-54.7^{\circ }$, the exact mirror of spin-up. Both cases covered: up tilts $54.7^{\circ }$ from $+z$, down tilts $54.7^{\circ }$ from $-z$.

Rule: ${\mu }_z=-g_s\frac{e}{2m_e}{m}_s\hslash$. With $g_s=2$ and $|m_s|=\frac{1}{2}$: $|{\mu }_z|=2\cdot \frac{e}{2m_e}\cdot \frac{1}{2}\hslash =\frac{e\hslash }{2m_e}={\mu }_B\approx 9.274\times 10^{-24}\text{J/T}.$ Answer: $|{\mu }_z|={\mu }_B\approx 9.27\times 10^{-24}\text{J/T}$. The factor $g_s=2$ is what makes spin twice as magnetic as the naive current-loop guess would give.

Set up: the most-aligned state has $S_z=s\hslash$ and $|\vec{S}|=\hslash \sqrt{s(s+1)}$. So $\cos \theta =\frac{s\hslash }{\hslash \sqrt{s(s+1)}}=\frac{s}{\sqrt{s(s+1)}}=\sqrt{\frac{s}{s+1}}.$ Limit: as $s\to \infty$, $\frac{s}{s+1}\to 1$, so $\cos \theta \to 1$ and $\theta \to 0^{\circ }$. Interpret: for huge $s$ (large systems), the arrow can line up almost perfectly with $z$ — quantum tilt becomes negligible and we recover the classical picture of a vector pointing straight up. For small $s$ (like the electron's $\frac{1}{2}$), the tilt is large ($54.7^{\circ }$) and irreducibly quantum. This is the correspondence principle in action. Sanity check the edge: $s=\frac{1}{2}\Rightarrow \sqrt{\frac{1/2}{3/2}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}$ ✔ matches L2.1.

Two energies: ${\mu }_z=\mp {\mu }_B$ (up gives $-{\mu }_B\cdot (-1)$... let's be careful). For $m_s=+\frac{1}{2}$, ${\mu }_z=-{\mu }_B$, so $U_{\uparrow }=-(-{\mu }_B)B_0=+{\mu }_B{B}_0$. For $m_s=-\frac{1}{2}$, ${\mu }_z=+{\mu }_B$, so $U_{\downarrow }=-{\mu }_B{B}_0$. Gap: $\Delta U=U_{\uparrow }-U_{\downarrow }={\mu }_B{B}_0-(-{\mu }_B{B}_0)=2{\mu }_B{B}_0.$ Why it splits lines: one energy level becomes two (up and down) separated by $2{\mu }_B{B}_0$. Photons emitted from these now come in slightly different energies → a single spectral line splits into a close pair. That splitting scales with $B_0$, which is the experimental signature of the Zeeman effect.

(a) Force. $F_z={\mu }_z\frac{\partial B_z}{\partial z}$, with ${\mu }_z=\pm {\mu }_B$: $|F_z|=(9.274\times 10^{-24})(1.8\times 10^3)=1.669\times 10^{-20}\text{N}.$ So the two forces are $+1.67\times 10^{-20}$ N and $-1.67\times 10^{-20}$ N — equal size, opposite sign → beam splits symmetrically into two. (b) Velocity kick. Newton: $a=F_z/m$, then $v_z=at$: $a=\frac{1.669\times 10^{-20}}{1.8\times 10^{-25}}=9.27\times 10^4{\text{m/s}}^2,$ $v_z=at=(9.27\times 10^4)(1.0\times 10^{-4})=9.27\text{m/s}.$ Answer: each atom is pushed with $\approx 1.67\times 10^{-20}$ N and leaves with $|v_z|\approx 9.3$ m/s (one beam up, one down). The discreteness of $m_s$ shows up as two clean beams, not a smear.

Pauli: two electrons sharing all of $n,\ell ,m_{\ell }$ must differ in $m_s$ (see Quantum Numbers). So one has $m_s=+\frac{1}{2}$, the other $m_s=-\frac{1}{2}$. Total shadow: $S_z^{\text{total}}=(+\frac{1}{2}+(-\frac{1}{2}))\hslash =0.$ Total moment: ${\mu }_z^{\text{total}}=(-{\mu }_B)+(+{\mu }_B)=0$. Answer: a filled orbital is spin-neutral: $S_z^{\text{total}}=0$ and ${\mu }_z^{\text{total}}=0$. This is why full shells contribute no net spin magnetism — the two opposite arrows cancel.

Set up: treat $L\approx m_{e}v{r}_e$ and set $L=\frac{1}{2}\hslash$. Solve for $v$: $v\approx \frac{\hslash /2}{m_e{r}_e}=\frac{(1.055\times 10^{-34})/2}{(9.11\times 10^{-31})(2.82\times 10^{-15})}.$ Compute the denominator: $9.11\times 10^{-31}\times 2.82\times 10^{-15}=2.569\times 10^{-45}$. Compute: $v\approx \frac{5.275\times 10^{-35}}{2.569\times 10^{-45}}\approx 2.05\times 10^{10}\text{m/s}$. Compare: $v/c\approx \frac{2.05\times 10^{10}}{3.0\times 10^8}\approx 68$. Conclusion: the surface would move about 68 times the speed of light — physically impossible. Therefore spin is not literal spatial rotation. It is an intrinsic quantum label, exactly as the parent note warns. (The rough $\sim 10^1$–$10^2$ overshoot is the standard textbook result; the point is the enormous violation, not the last digit.)

Smallest angle comes from the largest shadow, $m_s=+s$ (already found in L3.1): $\cos {\theta }_{\min }=\sqrt{\frac{s}{s+1}}.$ Never $0^{\circ }$: $\cos \theta =1$ would need $\frac{s}{s+1}=1$, i.e. $s+1=s$, impossible for any finite $s$. So $\cos {\theta }_{\min }<1$ always ⇒ ${\theta }_{\min }>0$ always. Limit: as $s\to \infty$, ${\theta }_{\min }\to 0^{\circ }$ (classical limit); for the electron $s=\frac{1}{2}$ it is the maximal $54.7^{\circ }$. Why (deep reason): if the arrow lay perfectly along $z$, all of its length would be in $S_z$ and none in $S_x,S_y$ — meaning $S_x=S_y=0$ exactly and known simultaneously. The commutator $[S_x,S_y]=i\hslash S_z\ne 0$ forbids that. The leftover length in figure s01 (the part sticking out sideways) is precisely the uncertainty that must remain. So the tilt is not a quirk — it is the geometry of the uncertainty principle.

Projection uses $m$, size uses $\sqrt{s(s+1)}$ ::: $S_z=m_s\hslash$ (shadow); $|\vec{S}|=\hslash \sqrt{s(s+1)}$ (full length) — the second is always larger.