Spin as Intrinsic Angular Momentum

Spin is angular momentum carried by an internal quantum degree of freedom. It is not built from $\mathbf R\times\mathbf P$, but its components satisfy the same angular momentum algebra:

$$[S_i,S_j] = i\hbar\sum_k\epsilon_{ijk}S_k.$$

The algebra organizes unitary rotation representations into multiplets labeled by $s$ and $m_s$ and determines how spin adds to orbital angular momentum. A magnetic coupling requires additional physical information about the magnetic moment. The introductory conceptual page is What Spin Is and Is Not; here we build the operators on the internal factor explicitly.

Internal Hilbert Space

For a spin-$s$ degree of freedom, the spin Hilbert space is finite-dimensional:

$$\mathcal H_s\simeq\mathbb C^{2s+1}.$$

A particle with spatial motion and spin has a tensor-product structure of the form

$$\mathcal H = L^2(\mathbb R^3)\otimes\mathcal H_s$$

in the simplest nonrelativistic one-particle model. Orbital operators act on the spatial factor, while spin operators act on the internal factor:

$$\mathbf L = \mathbf R\times\mathbf P, \qquad \mathbf S \text{ acts on }\mathcal H_s.$$

In that product model,

$$[S_i,R_j]=0, \qquad [S_i,P_j]=0, \qquad [S_i,L_j]=0.$$

This is the formal reason spin is not another name for orbital angular momentum. The two can be added because they are both angular momenta, but they act on different degrees of freedom.

Spin Algebra

The spin components obey

$$[S_x,S_y]=i\hbar S_z, \qquad [S_y,S_z]=i\hbar S_x, \qquad [S_z,S_x]=i\hbar S_y.$$

Define the spin Casimir

$$S^2=S_x^2+S_y^2+S_z^2.$$

It commutes with all components:

$$[S^2,S_i]=0.$$

As with any angular momentum, one usually diagonalizes $S^2$ and one component, conventionally $S_z$. The other components cannot generally be sharp at the same time because they do not commute with $S_z$.

S Squared and z Projection

The standard spin basis is

$$\lvert s,m_s\rangle,$$

with

$$S^2\lvert s,m_s\rangle = \hbar^2s(s+1)\lvert s,m_s\rangle,$$

and

$$S_z\lvert s,m_s\rangle = \hbar m_s\lvert s,m_s\rangle.$$

For fixed $s$,

$$m_s=-s,-s+1,\ldots,s.$$

The number of orthogonal spin basis states is

$$2s+1.$$

The allowed spin values are

$$s=0,\frac12,1,\frac32,\ldots.$$

The algebraic derivation of these labels is the same as for general angular momentum and is given in Eigenvalues of J Squared and Jz. What is special about spin is the physical realization: the finite-dimensional space is internal, not a space of scalar functions on the sphere.

Ladder Operators

Define

$$S_\pm=S_x\pm iS_y.$$

They raise and lower the spin projection:

$$S_\pm\lvert s,m_s\rangle = \hbar \sqrt{s(s+1)-m_s(m_s\pm1)} \, \lvert s,m_s\pm1\rangle.$$

The ladder stops at the endpoints:

$$S_+\lvert s,s\rangle=0, \qquad S_-\lvert s,-s\rangle=0.$$

For spin-$1/2$, this ladder has only two states. For spin-$1$, it has three states. For spin-$3/2$, it has four states.

Spin $s$	Allowed $m_s$ values	Dimension
$0$	$0$	$1$
$1/2$	$-1/2,1/2$	$2$
$1$	$-1,0,1$	$3$
$3/2$	$-3/2,-1/2,1/2,3/2$	$4$

Rotations of Spin States

A rotation by angle $\theta$ about a unit axis $\hat{\mathbf n}$ acts on spin states as

$$U_s(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf S \right).$$

For spin-$1/2$,

$$\mathbf S=\frac{\hbar}{2}\boldsymbol\sigma,$$

so

$$U_{1/2}(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{2}\theta\,\hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

This is the half-angle spinor rotation law developed in Spin Rotations. For higher spin, the same exponential is used, but the matrices representing $S_x,S_y,S_z$ are $(2s+1)\times(2s+1)$ matrices; see Higher Spin Systems.

Spin Versus Orbital Angular Momentum

Spin and orbital angular momentum obey the same commutation relations but have different physical realizations. An integer-spin irreducible representation is mathematically equivalent to the matching fixed-$\ell$ orbital representation; what differs is the degree of freedom on which it acts.

Feature	Orbital angular momentum	Spin angular momentum
Operator	$\mathbf L=\mathbf R\times\mathbf P$	$\mathbf S$ on an internal space
Acts on	spatial wavefunctions	spin indices or spinors
Possible labels	integer orbital $\ell$ for scalar wavefunctions	integer or half-integer $s$
Rotation representation	functions on space	finite-dimensional $SU(2)$ representations

For an ordinary scalar wavefunction, orbital angular momentum is integer-valued because single-valuedness around the azimuthal angle forces integer $m$. Spin is not constrained by being a scalar function on space. Half-integer spin belongs to spinor representations of $SU(2)$, the double cover of $SO(3)$.

This distinction is why a spin-$1/2$ electron can have intrinsic angular momentum even in an $s$ orbital with $\ell=0$.

Total Angular Momentum

When a system has both orbital and spin angular momentum, the total generator of rotations is

$$\mathbf J=\mathbf L+\mathbf S.$$

Because $\mathbf L$ and $\mathbf S$ commute with each other in the simple product model, $\mathbf J$ also satisfies the angular momentum algebra:

$$[J_i,J_j] = i\hbar\sum_k\epsilon_{ijk}J_k.$$

The total generator is what rotates the full state, including both spatial and spin degrees of freedom. This is the starting point for total angular momentum, the one-particle addition of orbital and spin angular momentum, and interactions such as spin–orbit coupling.

Measurements and Interpretation

A measurement of $S_z$ has possible outcomes

$$\hbar m_s, \qquad m_s=-s,\ldots,s.$$

For spin-$1/2$, the two outcomes are

$$\pm\frac{\hbar}{2}.$$

For a different measurement axis $\hat{\mathbf n}$, the measured operator is

$$S_{\hat n} = \hat{\mathbf n}\cdot\mathbf S.$$

The spectrum is the same set of projections, but the eigenbasis is rotated. This is the formal core behind Stern–Gerlach analyzers and spin-component measurements.

The magnitude associated with $S^2$ is

$$\sqrt{s(s+1)}\,\hbar,$$

not $s\hbar$. The projection along a chosen axis is $\hbar m_s$. The chosen axis is part of the measurement or labeling convention unless a Hamiltonian, field, or apparatus selects it physically.

Magnetic Moment Preview

Spin often carries a magnetic moment proportional to $\mathbf S$:

$$\boldsymbol\mu_S=\gamma\mathbf S.$$

For an electron, the relation is conventionally written

$$\boldsymbol\mu_S = -g_s\mu_B\frac{\mathbf S}{\hbar}.$$

The magnetic coupling

$$H=-\boldsymbol\mu_S\cdot\mathbf B$$

turns spin projections into energy splittings. The sign and $g$ factor are physical input beyond the abstract spin algebra; see Spin in Magnetic Fields and Magnetic Moments and g-Factors.

Common Mistakes

• Treating spin as literal rotation of an extended charged object.
• Thinking half-integer spin contradicts ordinary three-dimensional rotations; it reflects the $SU(2)$ double cover of $SO(3)$.
• Forgetting that $\mathbf S$ acts on an internal Hilbert-space factor, not on position coordinates.
• Writing the $S^2$ eigenvalue as $\hbar^2s^2$ instead of $\hbar^2s(s+1)$.
• Assuming spin-$1/2$ intuition automatically describes spin-$1$ or spin-$3/2$ systems.
• Confusing the spin label $s$ with the projection label $m_s$.
• Treating a magnetic moment as part of the spin algebra; it is an additional physical relation.

Exercises

1. For the state $|s,s\rangle$, find $\langle S_x\rangle$, $\langle S_y\rangle$, and their variances. Why does a maximally polarized state still have transverse uncertainty?

Solution

The ladder operators change $m_s$, so their expectation values in $|s,s\rangle$ vanish; hence both transverse means vanish. Symmetry about $z$ gives equal transverse second moments. Since

$$\langle S_x^2+S_y^2\rangle =\hbar^2s(s+1)-\hbar^2s^2=\hbar^2s,$$

we obtain $(\Delta S_x)^2=(\Delta S_y)^2=\hbar^2s/2$. Their uncertainty product is $\hbar^2s/2=\hbar|\langle S_z\rangle|/2$, saturating the Robertson bound. A sharp maximal projection does not assign sharp values to incompatible transverse components.

2. Show that $\mathbf J=\mathbf L+\mathbf S$ satisfies the angular momentum algebra when $[L_i,S_j]=0$.

Solution

Compute

$$[J_i,J_j] = [L_i+S_i,L_j+S_j].$$

Using $[L_i,S_j]=0$ and $[S_i,L_j]=0$,

$$[J_i,J_j] = [L_i,L_j]+[S_i,S_j].$$

The two algebras give

$$[L_i,L_j] = i\hbar\sum_k\epsilon_{ijk}L_k,$$

and

$$[S_i,S_j] = i\hbar\sum_k\epsilon_{ijk}S_k.$$

Therefore

$$[J_i,J_j] = i\hbar\sum_k\epsilon_{ijk}(L_k+S_k) = i\hbar\sum_k\epsilon_{ijk}J_k.$$

3. For a spin-$1/2$ state, what are the possible outcomes of measuring $S_{\hat n}$ along any unit axis $\hat{\mathbf n}$?

Solution

Changing the measurement axis rotates the spin basis but does not change the spectrum of a spin component. For $s=1/2$, the allowed projection labels are

$$m_s=-\frac12,\frac12.$$

Therefore the possible outcomes are

$$-\frac{\hbar}{2}, \qquad \frac{\hbar}{2}.$$

4. Why can an electron in an orbital state with $\ell=0$ still have angular momentum?

Solution

The label $\ell=0$ says the orbital angular momentum is zero. It does not remove the electron’s intrinsic spin. The full one-electron Hilbert space includes a spin-$1/2$ internal factor, so the electron can still have

$$S^2=\hbar^2\frac12\left(\frac12+1\right) = \frac34\hbar^2.$$

Thus a spatial $s$ orbital can have zero orbital angular momentum while the electron still carries spin angular momentum.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• B. C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, 2nd ed., Springer, 2015.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.