Spinors and 2π Rotations

A spinor can acquire a minus sign under a physical $2\pi$ rotation. For spin-$1/2$,

$$U(\hat{\mathbf n},2\pi)=-I, \qquad U(\hat{\mathbf n},4\pi)=I.$$

The statement is both simple and easy to misread. A single isolated ray is unchanged by the replacement $\lvert\psi\rangle\mapsto-\lvert\psi\rangle$, because quantum pure states are rays. The sign becomes physically meaningful only when it is compared with another phase reference: another path in an interferometer, another internal component, or another branch of a coherent superposition.

This page is the spinor-specific explanation of the sign. The group-theoretic double cover is developed in SU(2) versus SO(3), the rotation operator is derived in Spin Rotations, and the ray language is introduced in Rays and Global Phase.

The Calculation

For a spin-$1/2$ state, a rotation by angle $\theta$ about the unit vector $\hat{\mathbf n}$ is represented by

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

Because

$$(\hat{\mathbf n}\cdot\boldsymbol\sigma)^2=I,$$

the exponential reduces to

$$U(\hat{\mathbf n},\theta) = \cos\frac{\theta}{2}\,I - i\sin\frac{\theta}{2}\, \hat{\mathbf n}\cdot\boldsymbol\sigma.$$

At one full turn,

$$\cos\pi=-1, \qquad \sin\pi=0,$$

so

$$U(\hat{\mathbf n},2\pi)=-I.$$

At two full turns,

$$\cos 2\pi=1, \qquad \sin 2\pi=0,$$

so

$$U(\hat{\mathbf n},4\pi)=I.$$

Thus the state vector has a $4\pi$ period, while the corresponding ordinary spatial rotation has a $2\pi$ period.

Same Ray, Different Representative

The sign change is not a contradiction with the rule that global phase is physically irrelevant. A pure state is represented by a ray,

$$\lvert\psi\rangle \sim e^{i\alpha}\lvert\psi\rangle,$$

so

$$\lvert\psi\rangle \sim -\lvert\psi\rangle.$$

For a single isolated spin state, the transition probability to any other ray is unchanged:

$$\left| \langle\phi|(-\psi)\rangle \right|^2 = \left| -\langle\phi|\psi\rangle \right|^2 = \left| \langle\phi|\psi\rangle \right|^2.$$

The careful statement is therefore:

A $2\pi$ spinor rotation changes a Hilbert-space representative by $-1$, but not the isolated physical ray.

The representative still matters as soon as the phase is compared with another coherent amplitude. Interference is exactly such a comparison.

Relative Phase Can Be Observed

Consider a coherent path degree of freedom with two orthogonal alternatives $\lvert a\rangle$ and $\lvert b\rangle$. Let the spin state be the same in both paths initially:

$$\lvert\Psi_{\mathrm{in}}\rangle = \frac{1}{\sqrt2} (\lvert a\rangle+\lvert b\rangle) \otimes \lvert\chi\rangle.$$

Now rotate the spinor in path $b$ by $2\pi$, while leaving path $a$ untouched. Since $U(2\pi)=-I$ on the spinor,

$$\lvert\Psi_{\mathrm{out}}\rangle = \frac{1}{\sqrt2} (\lvert a\rangle-\lvert b\rangle) \otimes \lvert\chi\rangle.$$

The spin factor itself is still the same ray in each path. The observable change is the relative phase between the path amplitudes. If the two paths are recombined, this relative minus sign swaps which output port is bright and which output port is dark.

A simple beam-splitter model makes this explicit. Suppose the recombination basis is

$$\lvert +\rangle = \frac{\lvert a\rangle+\lvert b\rangle}{\sqrt2}, \qquad \lvert -\rangle = \frac{\lvert a\rangle-\lvert b\rangle}{\sqrt2}.$$

Before the $2\pi$ spin rotation, the path state is $\lvert +\rangle$. After the spin rotation in one arm, the path state is $\lvert -\rangle$. The measurement did not detect an absolute phase of one isolated vector; it detected a relative phase in a coherent superposition.

The SU(2) Path Picture

The double-cover language gives the same result geometrically. Physical rotations of ordinary vectors form $SO(3)$. Spinors transform under $SU(2)$, and the covering map

$$SU(2)\to SO(3)$$

identifies $U$ and $-U$ as the same ordinary rotation.

A path in $SO(3)$ that represents one full turn closes in $SO(3)$:

$$R(\hat{\mathbf n},2\pi)=I_3.$$

But the corresponding lifted path in $SU(2)$ begins at $I$ and ends at $-I$:

$$I \longrightarrow -I.$$

Running the loop a second time lifts to a path from $-I$ back to $I$. This is the origin of the phrase ”$4\pi$ periodicity” of spinors. The local angular-momentum algebra is the same, but the global topology of the rotation group is different.

Bloch Vector Caution

The Bloch vector of a spin-$1/2$ state is

$$\mathbf r = \langle\boldsymbol\sigma\rangle.$$

Under a spinor rotation $U(\hat{\mathbf n},\theta)$, the Bloch vector rotates by the ordinary three-dimensional angle $\theta$. It returns to itself after $2\pi$.

This does not mean the spinor representative has returned. The density matrix

$$\rho = \lvert\psi\rangle\langle\psi\rvert$$

is unchanged by $\lvert\psi\rangle\mapsto-\lvert\psi\rangle$:

$$(-\lvert\psi\rangle)(-\langle\psi\rvert) = \lvert\psi\rangle\langle\psi\rvert.$$

So the Bloch sphere is an excellent picture of spin directions and spin measurement probabilities, but it deliberately forgets the global spinor phase. The missing phase can reappear when the spinor is placed in an interferometric comparison.

Interferometry Preview

The clean experimental idea is to give one coherent branch an extra spin rotation while keeping another branch as a phase reference. A spin-$1/2$ branch rotated through $2\pi$ acquires a relative phase $\pi$ compared with the unrotated branch; after a $4\pi$ rotation the relative phase returns to zero modulo $2\pi$.

Neutron interferometry is especially natural because neutrons carry spin-$1/2$ and can be split and recombined coherently. In a schematic interferometer,

$$\frac{1}{\sqrt2} (\lvert a\rangle+\lvert b\rangle) \otimes \lvert\chi\rangle \quad \longrightarrow \quad \frac{1}{\sqrt2} (\lvert a\rangle+e^{i\varphi}\lvert b\rangle) \otimes \lvert\chi\rangle,$$

where the spin rotation in one arm contributes

$$\varphi=\pi \quad \text{for a }2\pi\text{ spinor rotation},$$

and

$$\varphi=0 \quad \text{mod }2\pi \quad \text{for a }4\pi\text{ spinor rotation}.$$

The observed quantity is a shift of interference fringes or output intensities. This is why the phrase “global phase is unobservable” must be used with care: a phase that is global within one branch can become relative between branches. For the broader matter-wave setting, see Interference with Matter and Interferometry.

What Is and Is Not Being Rotated

Several objects are present at once:

Object	Transformation under a $2\pi$ physical rotation
Ordinary spatial vector	Returns to itself
Bloch vector	Returns to itself
Spin-$1/2$ state vector representative	Multiplied by $-1$
Spin-$1/2$ ray	Same ray
Interferometer branch phase relative to a reference branch	Can shift by $\pi$

The first two belong to the ordinary $SO(3)$ picture. The spinor representative belongs to the $SU(2)$ picture. The ray is the physical pure state when no external phase reference is present. Interferometry supplies such a reference.

Relation to Berry Phase

The $2\pi$ sign is closely related to geometric phase, but it should not be conflated with every Berry-phase effect. A spin-$1/2$ state adiabatically carried around a closed circuit on the Bloch sphere acquires a geometric phase proportional to the enclosed solid angle:

$$\gamma_{\mathrm B} = -\frac{\Omega}{2}.$$

For a loop corresponding to one full great-circle sweep of the spin direction, this formula gives a phase modulo $2\pi$ consistent with the spinor sign. The detailed adiabatic setting, gauge choices, and solid-angle convention belong in Berry Phase for Spin-1/2. The present page uses only the rotation-group fact $U(2\pi)=-I$.

Common Mistakes

• Saying the particle must be “physically rotated by $720^\circ$” in the sense of a tiny classical object. Spin is not literal mechanical spinning.
• Saying the $2\pi$ sign is directly observable for a single isolated ray. The observable effect requires comparison with another coherent amplitude.
• Saying the sign is meaningless because global phase is unobservable. The sign is unobservable as an isolated global phase, but observable as a relative phase.
• Confusing the Bloch vector with the spinor. The Bloch vector returns after $2\pi$ even when the representative spinor changes sign.
• Treating $SU(2)$ and $SO(3)$ as the same group because their Lie algebras are locally isomorphic.
• Forgetting that integer-spin representations return after $2\pi$, while half-integer spin representations acquire a minus sign.

Cross-Links

• Spin Rotations
• What Spin Is and Is Not
• Spin-1/2 Hilbert Space
• Pauli Matrices
• Bloch Sphere
• Spin Measurements
• Higher Spin Systems
• SO(3) and SU(2) Preview
• Projective Representations
• SU(2)
• SU(2) versus SO(3)
• Rays and Global Phase
• Projective Hilbert Space
• Berry Phase for Spin-1/2
• Interference with Matter
• Interferometry

References

• Y. Aharonov and L. Susskind, “Observability of the Sign Change of Spinors under 2π Rotations,” Physical Review 158, 1237-1238, 1967.
• H. Rauch, A. Zeilinger, G. Badurek, A. Wilfing, and W. Bauspiess, “Verification of coherent spinor rotation of fermions,” Physics Letters A 54, 425-427, 1975.
• H. Rauch and S. A. Werner, Neutron Interferometry: Lessons in Experimental Quantum Mechanics, 2nd ed., Oxford University Press, 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.
• B. C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, 2nd ed., Springer, 2015.

Exercises

1. Compute $U(\hat{\mathbf n},2\pi)$ and $U(\hat{\mathbf n},4\pi)$ from the half-angle formula.

Solution

The spin-$1/2$ rotation operator is

$$U(\hat{\mathbf n},\theta) = \cos\frac{\theta}{2}\,I - i\sin\frac{\theta}{2}\, \hat{\mathbf n}\cdot\boldsymbol\sigma.$$

At $\theta=2\pi$,

$$\cos\pi=-1, \qquad \sin\pi=0,$$

so $U(\hat{\mathbf n},2\pi)=-I$. At $\theta=4\pi$,

$$\cos2\pi=1, \qquad \sin2\pi=0,$$

so $U(\hat{\mathbf n},4\pi)=I$.

2. Show that a $2\pi$ rotation in only one branch of an interferometer can change the output port.

Solution

Start with the path state

$$\lvert\psi_{\mathrm{path}}\rangle = \frac{\lvert a\rangle+\lvert b\rangle}{\sqrt2} = \lvert +\rangle.$$

If the spinor in branch $b$ is rotated by $2\pi$, the branch amplitude acquires a minus sign:

$$\lvert\psi_{\mathrm{path}}\rangle \longmapsto \frac{\lvert a\rangle-\lvert b\rangle}{\sqrt2} = \lvert -\rangle.$$

Thus a recombiner that sends $\lvert+\rangle$ and $\lvert-\rangle$ to different output ports will register a change. The detected change is a relative phase between branches, not an absolute phase of a single isolated state.

3. Why does the Bloch vector not show the $2\pi$ spinor sign?

Solution

The Bloch vector is computed from expectation values or from the density matrix

$$\rho=\lvert\psi\rangle\langle\psi\rvert.$$

Changing $\lvert\psi\rangle$ to $-\lvert\psi\rangle$ leaves $\rho$ unchanged:

$$(-\lvert\psi\rangle)(-\langle\psi\rvert) = \lvert\psi\rangle\langle\psi\rvert.$$

Therefore all Bloch-vector components are unchanged. The Bloch sphere represents rays, not a chosen phase representative of each ray.

4. Explain in one sentence why $4\pi$ periodicity does not violate the fact that ordinary space has $2\pi$ rotational symmetry.

Solution

Ordinary spatial rotations live in $SO(3)$ and close after $2\pi$, while spin-$1/2$ state-vector representatives live in the double cover $SU(2)$, where the lifted path closes only after $4\pi$.