Spin Coherent States

Spin coherent states are the closest quantum states to classical spin arrows. For spin $s>0$, they are obtained by rotating the highest-weight state $\lvert s,s\rangle$ to point along a direction $\hat{\mathbf n}$ on the sphere. They generalize the spin-$1/2$ Bloch-sphere states, but with an important warning: for $s>1/2$, the sphere labels only a special family of states, not all pure states in the $(2s+1)$-dimensional Hilbert space.

They are useful because they connect three languages:

• the geometry of rotations and the sphere;
• semiclassical spin dynamics in the large-$s$ limit;
• spin path integrals and many-body spin models.

This page introduces the construction and the formulas most often used in quantum mechanics. The bridge from the spin Berry action to Wess–Zumino-type terms is From Berry Phase to Topological Terms.

Highest-Weight Starting Point

Fix an irreducible spin-$s$ representation with basis

$$\lvert s,m\rangle, \qquad m=-s,-s+1,\ldots,s.$$

The highest-weight state satisfies

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

It is the quantum state most sharply aligned with $+z$. A spin coherent state pointing along a unit vector $\hat{\mathbf n}$ is a rotated version of this state:

$$\lvert\hat{\mathbf n};s\rangle = R(\hat{\mathbf n}) \lvert s,s\rangle.$$

A common rotation convention is

$$R(\hat{\mathbf n}) = e^{-i\phi S_z/\hbar} e^{-i\theta S_y/\hbar},$$

where

$$\hat{\mathbf n} = (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta).$$

Changing the rotation convention can multiply the state by a direction-dependent phase. The ray and the expectation-value direction are unchanged.

Stereographic Parameter

An algebraically convenient form uses the complex coordinate

$$\zeta = e^{i\phi}\tan\frac{\theta}{2}.$$

Then

$$\lvert\zeta;s\rangle = \frac{1}{(1+\lvert\zeta\rvert^2)^s} \exp \left( \frac{\zeta S_-}{\hbar} \right) \lvert s,s\rangle.$$

Expanding the exponential gives

$$\lvert\zeta;s\rangle = \frac{1}{(1+\lvert\zeta\rvert^2)^s} \sum_{k=0}^{2s} \binom{2s}{k}^{1/2} \zeta^k \lvert s,s-k\rangle.$$

For $s=1/2$, this becomes

$$\lvert\zeta;1/2\rangle = \frac{ \lvert\uparrow\rangle+\zeta\lvert\downarrow\rangle }{ \sqrt{1+\lvert\zeta\rvert^2} },$$

which is the usual Bloch-sphere spinor in stereographic coordinates. This coordinate patch excludes the south pole, where $\zeta$ diverges; the coherent ray remains regular and is described using a second patch. Relative to the rotation convention above, $|\zeta;s\rangle=e^{is\phi}|\hat{\mathbf n};s\rangle$ away from the patch boundary.

Spin Expectation Values

The defining classical-looking property is

$$\langle\hat{\mathbf n};s| \mathbf S |\hat{\mathbf n};s\rangle = \hbar s\,\hat{\mathbf n}.$$

Thus the coherent state points along $\hat{\mathbf n}$ with maximal expectation-value length $\hbar s$. The invariant magnitude obtained from $S^2$ is instead $\hbar\sqrt{s(s+1)}$. It is not an eigenstate of all spin components. It is an eigenstate of the component along its own direction:

$$\hat{\mathbf n}\cdot\mathbf S \lvert\hat{\mathbf n};s\rangle = \hbar s \lvert\hat{\mathbf n};s\rangle.$$

Components perpendicular to $\hat{\mathbf n}$ fluctuate. In a frame where $\hat{\mathbf n}=\hat z$,

$$\Delta S_x^2 = \Delta S_y^2 = \frac{\hbar^2s}{2}, \qquad \Delta S_z^2 = 0.$$

The relative transverse fluctuation scales as

$$\frac{\Delta S_\perp}{\lvert\langle\mathbf S\rangle\rvert} \sim \frac{1}{\sqrt{s}}.$$

This is why spin coherent states become classical-looking for large $s$.

Overlap and Localization on the Sphere

Spin coherent states are normalized but not orthogonal. If $\gamma$ is the angle between $\hat{\mathbf n}$ and $\hat{\mathbf m}$, then

$$\left| \langle\hat{\mathbf n};s|\hat{\mathbf m};s\rangle \right|^2 = \left( \frac{1+\hat{\mathbf n}\cdot\hat{\mathbf m}}{2} \right)^{2s} = \left( \cos^2\frac{\gamma}{2} \right)^{2s}.$$

For spin $1/2$, nearby points on the Bloch sphere have substantial overlap. For large $s$, the overlap becomes sharply peaked near $\hat{\mathbf n}=\hat{\mathbf m}$. This is the sphere version of classical localization.

The complex phase of the overlap depends on gauge choices for the coherent states. Around closed loops, that phase is related to the solid-angle Berry phase. The spin-$1/2$ case is worked out in Berry Phase for Spin-1/2; for spin $s$, the geometric phase is multiplied by $2s$ relative to the spin-$1/2$ aligned-state convention.

Resolution of Identity

Although spin coherent states are not orthogonal, they form an overcomplete set. The identity on the spin-$s$ Hilbert space can be resolved as

$$\frac{2s+1}{4\pi} \int_{S^2} d\Omega\, \lvert\hat{\mathbf n};s\rangle \langle\hat{\mathbf n};s\rvert = I_{2s+1}.$$

This formula is the key reason coherent states are useful in calculations. It lets one insert a continuous set of spin directions instead of a discrete $m$ basis.

Overcomplete does not mean linearly independent. There are infinitely many coherent states on the sphere but only $2s+1$ dimensions in the Hilbert space.

Bloch Sphere Versus General Spin

For spin $1/2$, every pure state is a coherent state. The Bloch sphere is the full projective pure-state space:

$$\mathbb{CP}^1 \cong S^2.$$

For spin $s>1/2$, the pure-state space has complex dimension $2s$, while the coherent-state sphere has real dimension $2$. Therefore most spin-$s$ pure states are not coherent states.

Examples of noncoherent spin states include:

• the spin-one state $|1,0\rangle$, whose spin expectation vanishes;
• spin-squeezed states;
• states with nonzero quadrupole or higher multipole structure not captured by a single direction.

The coherent-state sphere is still extremely useful. It is the classical phase space for one spin degree of freedom, not the full quantum state space for higher spin.

Spin Path Integral Preview

The resolution of identity allows a spin transition amplitude to be written as an integral over paths $\hat{\mathbf n}(t)$ on the sphere. With the stereographic state defined above and the amplitude convention $e^{iS/\hbar}$, the kinetic term is $i\hbar\langle\zeta;s|\partial_t|\zeta;s\rangle$. Direct differentiation gives $i\langle\zeta;s|\partial_\phi|\zeta;s\rangle=-s(1-\cos\theta)$, so the real-time action contains

$$S_{\mathrm{Berry}} = -\hbar s \int dt\, (1-\cos\theta)\dot\phi.$$

The remaining term is the classical coherent-state expectation of the Hamiltonian,

$$H_{\mathrm{cl}}(\hat{\mathbf n}) = \langle\hat{\mathbf n};s| H |\hat{\mathbf n};s\rangle.$$

Schematically,

$$S[\hat{\mathbf n}] = \int dt\, \left[ -\hbar s(1-\cos\theta)\dot\phi - H_{\mathrm{cl}}(\hat{\mathbf n}) \right].$$

In the rotation gauge $e^{-i\phi S_z/\hbar}e^{-i\theta S_y/\hbar}|s,s\rangle$, the kinetic term is instead $+\hbar s\cos\theta\dot\phi$; the two expressions differ by a total derivative. Gauge changes cannot reverse the Berry curvature. For the aligned-state convention here, the closed-loop geometric phase is $-s\Omega$ modulo $2\pi$, with oriented solid angle $\Omega$. Physical phases for closed paths are defined modulo $2\pi$ and are tied to the solid angle enclosed by the path.

Common Mistakes

• Thinking spin coherent states are the same as harmonic-oscillator coherent states. They share overcompleteness and semiclassical behavior, but their phase spaces and algebra are different.
• Treating the coherent-state sphere as the full pure-state space for every spin.
• Forgetting that coherent states are nonorthogonal.
• Dropping the phase convention when comparing coherent-state overlaps.
• Interpreting $\langle\mathbf S\rangle=\hbar s\hat{\mathbf n}$ as zero uncertainty in all spin components.
• Using the spin-$1/2$ Bloch-sphere intuition for higher-spin squeezing or multipole structure without checking the extra degrees of freedom.

Exercises

1. Recover the spin-$1/2$ Bloch state.

Starting from

$$\lvert\zeta;1/2\rangle = \frac{ \lvert\uparrow\rangle+\zeta\lvert\downarrow\rangle }{ \sqrt{1+\lvert\zeta\rvert^2} }, \qquad \zeta=e^{i\phi}\tan\frac{\theta}{2},$$

show that this is equivalent to the usual Bloch-sphere state.

Solution

Use

$$1+\tan^2\frac{\theta}{2} = \frac{1}{\cos^2(\theta/2)}.$$

Then

$$\frac{1}{\sqrt{1+\lvert\zeta\rvert^2}} = \cos\frac{\theta}{2},$$

and

$$\frac{\zeta}{\sqrt{1+\lvert\zeta\rvert^2}} = e^{i\phi} \sin\frac{\theta}{2}.$$

Therefore

$$\lvert\zeta;1/2\rangle = \cos\frac{\theta}{2}\lvert\uparrow\rangle + e^{i\phi} \sin\frac{\theta}{2}\lvert\downarrow\rangle,$$

which is the standard Bloch-sphere parameterization.

2. Derive the resolution of identity from rotational invariance.

For one irreducible spin-$s$ space, define

$$Q=\int_{S^2}d\Omega\, \lvert\hat{\mathbf n};s\rangle \langle\hat{\mathbf n};s\rvert.$$

Show that $Q$ commutes with every spin rotation, then determine it using irreducibility and its trace. Why would the trace alone not prove the operator identity?

Solution

A rotation maps a coherent ray at $\hat{\mathbf n}$ to the ray at $R\hat{\mathbf n}$. Any phase from the chosen representatives cancels in the projector. Therefore

$$U(R)QU(R)^\dagger =\int_{S^2}d\Omega\, |R\hat{\mathbf n};s\rangle\langle R\hat{\mathbf n};s| =Q,$$

where the last equality uses the rotation-invariant sphere measure. On this single irreducible spin space, the Schur-lemma argument gives $Q=cI_{2s+1}$. Each projector has unit trace, so

$$c(2s+1)=\operatorname{Tr}Q=\int_{S^2}d\Omega=4\pi.$$

Thus $Q=4\pi I_{2s+1}/(2s+1)$, which proves the stated resolution. The trace fixes the coefficient only after symmetry and irreducibility establish that the operator is scalar; many nonscalar operators have the same trace. If several spin sectors or repeated copies are included, this argument must instead be applied to the appropriate irreducible subspace.

3. Large-spin localization.

If two coherent states are separated by a small angle $\gamma$, use

$$\left| \langle\hat{\mathbf n};s|\hat{\mathbf m};s\rangle \right|^2 = \left( \cos^2\frac{\gamma}{2} \right)^{2s}$$

to estimate the overlap for small $\gamma$ and large $s$.

Solution

For small $\gamma$,

$$\cos\frac{\gamma}{2} \approx 1-\frac{\gamma^2}{8}.$$

Thus

$$\cos^2\frac{\gamma}{2} \approx 1-\frac{\gamma^2}{4}.$$

For large $s$,

$$\left( 1-\frac{\gamma^2}{4} \right)^{2s} \approx e^{-s\gamma^2/2}.$$

The angular width therefore scales like $1/\sqrt{s}$.

4. Why are higher-spin coherent states not all pure states?

Give a dimension-counting argument.

Solution

The spin-$s$ Hilbert space has complex dimension $2s+1$, so its projective pure-state space has complex dimension $2s$ and real dimension $4s$.

The coherent states are labeled by a point on $S^2$, which has real dimension $2$. These dimensions agree only when $s=1/2$. For $s>1/2$, the coherent-state sphere is a lower-dimensional submanifold of the full pure-state space.

5. Check the Berry-connection sign directly for spin $1/2$ using $|\psi\rangle=(\cos(\theta/2),e^{i\phi}\sin(\theta/2))^{\mathsf T}$.

Solution

At fixed $\theta$, only the second component depends on $\phi$, so

$$i\langle\psi|\partial_\phi\psi\rangle =i\bigl[i\sin^2(\theta/2)\bigr] =-\frac{1-\cos\theta}{2}.$$

Thus a positively oriented constant-latitude loop gives $\gamma_B=-\pi(1-\cos\theta)=-\Omega/2$. Multiplying the spinor by $e^{-i\phi/2}$ changes the local connection to $\cos\theta/2$, but this representative is not periodic after $\phi\to\phi+2\pi$; the endpoint phase must also be included. The physical closed-loop phase agrees.

References

• F. T. Arecchi, E. Courtens, R. Gilmore, and H. Thomas, “Atomic Coherent States in Quantum Optics”, Physical Review A 6, 2211–2237 (1972).
• A. Auerbach, Interacting Electrons and Quantum Magnetism, Springer, 1994.
• J. R. Klauder and B.-S. Skagerstam, eds., Coherent States, World Scientific, 1985.
• A. M. Perelomov, Generalized Coherent States and Their Applications, Springer, 1986.
• J. M. Radcliffe, “Some properties of coherent spin states”, Journal of Physics A: General Physics 4, 313–323 (1971).