Question bank — Spin — intrinsic angular momentum

False. A field only sets the energy of each orientation via $U=-{\mu }_{z}B$ (where ${\mu }_z$ is the moment's $z$-projection and $B$ the field magnitude); the length $|\vec{S}|=\hslash \sqrt{s(s+1)}$ is fixed by the particle's identity, like its mass or charge.

False. Pointing along $z$ means $S_z=|\vec{S}|$ and hence $S_x=S_y=0$ known exactly — but $[S_x,S_y]=i\hslash S_z$ forbids both from being sharp once $S_z$ is fixed. The compromise is a fixed tilt: $\cos \theta =\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}$, so $\theta \approx 54.7^{\circ }$ (see the spin-cone figure).

False. $S_z=m_s\hslash$ is only the projection (shadow) onto $z$; the full length is $|\vec{S}|=\frac{\sqrt{3}}{2}\hslash$, larger because $S_x,S_y$ also contribute to ${\vec{S}}^2$.

True. The count is $2s+1=2(\frac{1}{2})+1=2$, matching $m_s=\pm \frac{1}{2}$. Each value of $m_s$ is one discrete deflection, so the beam splits into exactly two.

False. $L=0$ kills the orbital moment, but the lone valence electron's spin moment remains — that residual moment is precisely what split the beam and forced us to invent spin.

False. Spin is intrinsic: it exists even for a point particle with no spatial extent and no orbit. A classical spinning sphere giving $\frac{\hslash }{2}$ would need its surface moving faster than light.

False. Orbital motion gives $g_{\ell }=1$, but the electron spin has the anomalous $g_s\approx 2$. This "extra factor of two" is a genuinely relativistic/quantum result, explained by the Dirac Equation.

False (trap). A massive spin-1 particle would give $2s+1=3$. But a photon is massless, so it can be seen in only two helicity states ($m=\pm 1$, the two circular polarizations); the $m=0$ state is forbidden (see the "why" item below). Massless particles break the naive count.

The writer used the projection formula for the magnitude. Magnitude uses $\sqrt{s(s+1)}$, giving $\frac{\sqrt{3}}{2}\hslash$; only $S_z$ uses $m_s$.

$m_s$ runs from $-s$ to $+s$ in integer steps, i.e. $-\frac{1}{2},+\frac{1}{2}$ only. Negative values are mandatory and there are exactly $2s+1=2$ of them, never three.

The sign is wrong (electron charge is negative, so ${\vec{\mu }}_s$ points opposite to $\vec{S}$: a minus sign), and the factor $g_s\approx 2$ is missing: ${\vec{\mu }}_s=-g_s\frac{e}{2m_e}\vec{S}$.

The three components can't all be sharp at once, so $S_x^2+S_y^2$ contributes even when $S_z$ is maximal. That extra piece turns $s^2$ into $s(s+1)$.

"Up/down" is specific to spin-$\frac{1}{2}$. For a massive $s=1$, $m_s=-1,0,+1$ gives $2s+1=3$ spots. The number of orientations grows with $s$.

Splitting needs a gradient: $F_z={\mu }_z\partial B_z/\partial z$. A uniform field ($\partial B_z/\partial z=0$) exerts zero net force — it only makes the moment precess (Larmor), not deflect (see the precession figure).

A classical moment could point any direction, smearing deflections continuously; instead $m_s$ takes only discrete values, so only discrete forces occur — discreteness of quantization made visible.

It is a permanent, unchangeable property of the particle itself, not caused by motion through space — you cannot spin it up, slow it down, or remove it.

If $\vec{S}$ aligned with $z$, all three components would be simultaneously sharp ($S_x=S_y=0$, $S_z=|\vec{S}|$), contradicting $[S_x,S_y]=i\hslash S_z$. Nature keeps $S_z$ and $|\vec{S}|$ sharp and blurs $S_x,S_y$ around a cone whose half-angle is the fixed $\theta \approx 54.7^{\circ }$.

A uniform field gives a torque $\vec{\tau }=\vec{\mu }\times \vec{B}$ that makes $\vec{S}$ precess (Larmor) without changing $S_z$, so no net translational force acts. But each orientation still has energy $U=-{\mu }_{z}B$, and the two ${\mu }_z$ values give two energies — an energy split, which is the Zeeman Effect.

The classical current-loop picture predicts $g=1$; the observed doubling of the spin magnetism has no classical explanation and only emerges from the relativistic Dirac Equation.

Both obey the ladder rule that forces $2s$ to be a non-negative integer, allowing $s=\frac{1}{2}$. Orbital $\ell$ has an extra constraint — its wavefunction must return to itself after a full $360^{\circ }$ spatial turn, forbidding half-integers. Spin, tied to the double-cover group $SU(2)$ of rotations (where a $360^{\circ }$ turn flips the sign and only $720^{\circ }$ restores it), carries no such single-valuedness demand, so half-integers survive.

A massless particle moves at $c$, so you can never jump to a frame where it is at rest and turn its spin around; only its projection along its own motion (the helicity) is physical, and gauge invariance removes the longitudinal ($m=0$) mode. Hence a photon shows only $m=\pm 1$, the two transverse (circular) polarizations.

Half-integer spin makes electrons fermions, whose total wavefunction must be antisymmetric under exchange; that antisymmetry is exactly what forbids two electrons from sharing all quantum numbers.

The magnitude depends only on $s$, never on $m_s$; both $m_s=\pm \frac{1}{2}$ share $s=\frac{1}{2}$, so both have $|\vec{S}|=\frac{\sqrt{3}}{2}\hslash$ and differ only in orientation.

$2s+1=1$: a single undeflected spot. No magnetic moment from spin means no splitting — this is the "expected" classical-looking result.

$2s+1=4$ spots, with $m_s=-\frac{3}{2},-\frac{1}{2},+\frac{1}{2},+\frac{3}{2}$. Note none of these is $0$, so the pattern is symmetric about the centre with no central spot.

No — with $s=0$ the only $m_s$ is $0$ and $\sqrt{s(s+1)}=0$, giving $0/0$. There is no vector to tilt; the "angle" is undefined because the spin is simply zero.

$\cos \theta =\frac{s}{\sqrt{s(s+1)}}=\sqrt{\frac{s}{s+1}}\to 1$, so $\theta \to 0$. Large spins behave more classically, aligning almost fully with the axis — the quantum "tilt" fades.

The two deflections swap sides: the spin-up and spin-down beams exchange which way they bend, since $F_z={\mu }_z\partial B_z/\partial z$ flips sign for each. The number of spots (two) is unchanged.

Zero net translational force, so no deflection; the moment only precesses (Larmor) and its orientations acquire different energies $U=-{\mu }_{z}B$ (the basis of the Zeeman Effect).