Spin

At last: spin—the word that appears everywhere in quantum mechanics and still refuses to behave like an ordinary spinning object.

The short version is this: spin is intrinsic angular momentum. It is not the orbital angular momentum $\mathbf L=\mathbf r\times\mathbf p$, and an electron is not a tiny rigid sphere literally rotating about its own axis. Spin is a quantum degree of freedom with the same angular-momentum algebra as orbital angular momentum.

Historically, atomic spectra—including the anomalous Zeeman effect—helped reveal that orbital angular momentum alone was not enough. The modern theory does not introduce spin merely as an ad hoc classical rotation; it treats spin as an intrinsic quantum observable whose predictions have been tested extensively.

This English edition preserves the article’s informal learning-journal voice while correcting several points transparently. In particular, it corrects the spelling of boson, distinguishes intrinsic spin from a classical moment of inertia, and states measurement and uncertainty claims in operator language. Historical formula rasters are retained in source order when accurate. Ten unambiguous formula screenshots are replaced by accessible TeX: two simple notation screenshots and eight screenshots whose visible equations contain substantive errors. Each scientific replacement is identified where it occurs.

Spin uses the angular-momentum algebra

For orbital angular momentum, the simultaneous eigenstates of $L^2$ and $L_z$ obey

$$ L^2|\ell,m\rangle=\hbar^2\ell(\ell+1)|\ell,m\rangle, \qquad L_z|\ell,m\rangle=\hbar m|\ell,m\rangle, $$

with $m=-\ell,-\ell+1,\ldots,\ell$.

Spin follows the same algebra after replacing $\ell$ by $s$ and $m$ by $m_s$:

$$ S^2|s,m_s\rangle=\hbar^2s(s+1)|s,m_s\rangle, \qquad S_z|s,m_s\rangle=\hbar m_s|s,m_s\rangle. $$

The source next writes the eigenfunction as $f_{s,m_s}$ and then switches to Dirac notation. Those two simple formula screenshots are safely represented here as semantic TeX:

$$ f_{s,m_s}\quad\longleftrightarrow\quad |s,m_s\rangle. $$

This is a change of representation, not a new physical state. A wavefunction and a ket are two ways to represent the same abstract state, once a basis or coordinate representation has been chosen.

The older notation is still useful:

Integer and half-integer spin

Particles with integer spin are bosons; particles with half-integer spin are fermions. Examples include spin-0 pions and spin-1 photons among bosons, and spin-$\tfrac12$ electrons, protons, and neutrons among fermions. Protons and neutrons are composite particles, but their total spin is still $\tfrac12$.

The spin-statistics theorem connects integer spin with Bose-Einstein statistics and half-integer spin with Fermi-Dirac statistics in relativistic quantum field theory. Fermions obey the Pauli exclusion principle; bosons do not. That distinction is central to atomic, molecular, and solid-state physics.

The source points to a later article for this connection. See The Pauli Exclusion Principle (Quantum Mechanics I Studied #44) for the repository’s canonical follow-up to the old platform log 220651906098.

For the electron and proton discussed here, $s=\tfrac12$. Therefore

$$ m_s=+\frac12\quad\text{or}\quad m_s=-\frac12. $$

There are two basis states, so every spin operator is represented by a $2\times2$ matrix in this basis. The source first repeats the function-style label,

and then sketches a generic $2\times2$ matrix eigenvalue equation acting on a two-component column:

The function-style notation can now give way to a two-component spinor:

Choose the normalized basis vectors

so a general normalized spinor is

$$ |\chi\rangle=a|+\rangle+b|-\rangle =\begin{pmatrix}a\\b\end{pmatrix}, \qquad |a|^2+|b|^2=1. $$

Up and down along the z-axis

The original post first asks whether an electron should be pictured as a tiny object moving along a line:

That picture is not a literal model of spin. Instead, a spin measurement has discrete outcomes, and the state encodes their probabilities:

The historical probability raster at this point wrote $a^2$ and $b^2$. It is omitted because $a$ and $b$ may be complex. The Born probabilities are

$$ P(+) = |a|^2,\qquad P(-) = |b|^2,\qquad |a|^2+|b|^2=1. $$

Let

$$ \chi_+=\left|\frac12,\frac12\right\rangle=|\uparrow\rangle, \qquad \chi_-=\left|\frac12,-\frac12\right\rangle=|\downarrow\rangle. $$

These symbols all refer to the same two basis states in different notations:

The important point is that “up” and “down” refer to the outcomes of $S_z$, not to a tiny arrow whose direction exists independently of measurement in the classical sense.

Matrix representation for spin one-half

The angular-momentum eigenvalue equations are

and for $s=\tfrac12$ they become

In the $S_z$ basis,

Start with an unknown $2\times2$ operator matrix:

Because the chosen columns are eigenvectors, the diagonal matrices follow immediately:

$$ S^2=\frac34\hbar^2 I, \qquad S_z=\frac\hbar2 \begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$

Ladder operators

The ladder action is

Writing those four actions as column-vector equations gives

and therefore

$$ S_+=\hbar\begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad S_-=\hbar\begin{pmatrix}0&0\\1&0\end{pmatrix}. $$

Using

we obtain

Thus

$$ S_x=\frac\hbar2\begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad S_y=\frac\hbar2\begin{pmatrix}0&-i\\i&0\end{pmatrix}, \quad S_z=\frac\hbar2\begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$

Equivalently, $S_i=(\hbar/2)\sigma_i$, where $\sigma_i$ are the Pauli matrices.

The historical raster that followed is omitted because it is not a summary of spin matrices: it shows orbital-angular-momentum differential operators and prints the wrong sign for $L_z$. In the standard position representation, the corrected operator is

$$ L_z=-i\hbar\frac{\partial}{\partial\phi}. $$

Orbital angular momentum can also be represented by matrices after choosing a basis. The special convenience here is simply that spin one-half has a two-dimensional state space, so the matrices are tiny.

Eigenstates of S x and S y

The source works through the ordinary $2\times2$ eigenvalue problem:

For $S_x$, the normalized eigenstates are

$$ \chi^{(x)}_+=\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}, \qquad \chi^{(x)}_-=\frac1{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}, $$

and for $S_y$ they are

$$ \chi^{(y)}_+=\frac1{\sqrt2}\begin{pmatrix}1\\i\end{pmatrix}, \qquad \chi^{(y)}_-=\frac1{\sqrt2}\begin{pmatrix}1\\-i\end{pmatrix}. $$

Begin with the general state in the $S_z$ basis:

The two $S_x$ basis columns are

and solving for the new expansion coefficients gives

The historical probability raster that followed used ordinary squares. It is omitted because the amplitudes may be complex. The correct probabilities for measuring $S_x$ are

$$ P_x(+) = \left|\frac{a+b}{\sqrt2}\right|^2,\qquad P_x(-) = \left|\frac{a-b}{\sqrt2}\right|^2. $$

For a normalized state, the two probabilities add to one.

Incompatible components and measurement

Because

$$ [S_x,S_z]=-i\hbar S_y\neq0, $$

$S_x$ and $S_z$ are not generally simultaneously sharp. Suppose the initial state is

$$ |\chi\rangle=a|+z\rangle+b|-z\rangle,\qquad |a|^2+|b|^2=1. $$

The historical raster here also used ordinary $a^2$ and $b^2$ for probabilities, so it is omitted. The correct $S_z$ probabilities are $P_z(+)=|a|^2$ and $P_z(-)=|b|^2$.

If an ideal projective measurement of $S_z$ returns $+\hbar/2$, the immediate post-measurement state is $\chi_+$:

The two possible $S_x$ outcome states are

For the prepared state $|+z\rangle$,

$$ |+z\rangle=\frac1{\sqrt2} \left(|+x\rangle+|-x\rangle\right). $$

so each $S_x$ outcome has probability one-half:

This does not mean that no state can ever have information about both components. It means there is no common eigenbasis that makes both components perfectly definite. Their uncertainties satisfy

$$ \Delta S_x\,\Delta S_z\geq\frac\hbar2|\langle S_y\rangle|. $$

Exercise 4.28: expectation values, second moments, and variances

For the normalized spinor

the exercise asks for

and then asks us to verify

The source calculation proceeds component by component:

The historical $S_z$ calculation is omitted because it mislabeled the expectation value as $\langle S_x\rangle$ and used a plus sign between the probabilities. The corrected result is

$$ \langle S_z\rangle=\frac{\hbar}{2}\left(|a|^2-|b|^2\right). $$

The two preceding source figures concern second moments, not variances:

$$ \langle S_x^2\rangle+\langle S_y^2\rangle+\langle S_z^2\rangle =\langle S^2\rangle=\frac{3\hbar^2}{4}. $$

Since $S_i^2=(\hbar^2/4)I$ for all three components,

$$ \langle S_x^2\rangle=\langle S_y^2\rangle=\langle S_z^2\rangle=\frac{\hbar^2}{4}. $$

For a pure spin-one-half state, $|\langle\mathbf S\rangle|=\hbar/2$, so

$$ (\Delta S_x)^2+(\Delta S_y)^2+(\Delta S_z)^2 =\langle S^2\rangle-|\langle\mathbf S\rangle|^2 =\frac{\hbar^2}{2}. $$

Exercise 4.30: spin along an arbitrary direction

Let

and define the component of spin along that direction:

Using spherical coordinates,

we seek the eigenvalues and eigenvectors of

The matrix is

$$ S_{\hat r}=\mathbf S\cdot\hat{\mathbf r} =\frac\hbar2 \begin{pmatrix} \cos\theta & e^{-i\phi}\sin\theta\\ e^{i\phi}\sin\theta & -\cos\theta \end{pmatrix}. $$

The historical negative-eigenvalue derivation is omitted because it used $\tan(\theta/2)$ where the displayed component ratio requires $\cot(\theta/2)$. With the first component set to one, a correct unnormalized negative branch is

$$ \begin{pmatrix}1\\-e^{i\phi}\cot(\theta/2)\end{pmatrix}. $$

The eigenvalues are again $\pm\hbar/2$. One conventional normalized choice of eigenvectors is

$$ \chi_+(\theta,\phi)= \begin{pmatrix} \cos(\theta/2)\\e^{i\phi}\sin(\theta/2) \end{pmatrix}, \qquad \chi_-(\theta,\phi)= \begin{pmatrix} -e^{-i\phi}\sin(\theta/2)\\\cos(\theta/2) \end{pmatrix}. $$

An overall phase multiplying either eigenvector has no physical effect.

Exercise 4.31: the spin-1 representation

For $s=1$, the task is to construct

The hint first asks how many eigenstates there are for

It then asks how each of the following operators acts on those states:

For $s=1$, the allowed magnetic quantum numbers are $m_s=1,0,-1$, so the representation is three-dimensional:

The historical $S^2$ raster is omitted because its middle line incorrectly maps $f_{1,0}$ to $f_{1,1}$. Total spin does not change $m_s$:

$$ S^2 f_{1,m_s}=2\hbar^2 f_{1,m_s},\qquad m_s=1,0,-1. $$

In particular, $S^2f_{1,0}=2\hbar^2f_{1,0}$. The $S_z$ eigenvalue equations are

and therefore

The ladder action is

which produces

The historical final raster is omitted because it prints $S_y=(S_++S_-)/(2i)$ even though its matrix uses the required difference. The corrected identities are

$$ S_x=\frac{S_++S_-}{2},\qquad S_y=\frac{S_+-S_-}{2i}. $$

In the ordered basis $\{|1,1\rangle,|1,0\rangle,|1,-1\rangle\}$,

$$ S_x=\frac{\hbar}{\sqrt2} \begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}, \quad S_y=\frac{\hbar}{\sqrt2} \begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix}, \quad S_z=\hbar \begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}. $$

That is the main pattern to remember: choose a spin quantum number $s$, build the $2s+1$ basis states labeled by $m_s$, and represent every spin operator on that finite-dimensional space.

What to carry forward

• Spin is intrinsic angular momentum, not literal classical self-rotation.
• A spin-$\tfrac12$ state is a normalized two-component spinor.
• $S_i=(\hbar/2)\sigma_i$ in the standard basis.
• The squared magnitudes of the expansion coefficients in an operator’s eigenbasis give the probabilities of its measurement outcomes.
• Different spin components do not commute, so they are not generally simultaneously sharp.
• A spin-$s$ representation has dimension $2s+1$.

The notation can feel like a pile of arbitrary conventions at first. Once the basis, operators, and measurement rule are kept separate, however, the whole subject becomes a compact piece of linear algebra rather than a mysterious picture of a tiny spinning ball.