Wigner Rotations

Two noncollinear boosts generally produce a boost and a rotation. For a particle state, the Wigner rotation is the part left after comparing the transformed momentum with its chosen standard boost. Its angle and sign depend on the ordered transformations and on that comparison. A worked perpendicular-boost calculation makes these choices explicit and provides a check on spin transport.

Required background. Wigner Classification Preview defines $\mathcal W(\Lambda,p)$ and the canonical standard boosts; SL(2, ℂ) and the Lorentz Group fixes their matrix lifts.

Canonical rotation after two perpendicular boosts

Use natural units for this calculation, start at $k=(m,0,0,0)$ with $m>0$, and apply the passive $x$ boost of rapidity $a$ followed by the passive $y$ boost of rapidity $b$:

$$\Lambda=B_y(b)B_x(a).$$

The rightmost factor acts first. Both boosts use the negative time–space off-diagonal signs of the toolkit. The final momentum is

$$\Lambda k =m(\cosh a\cosh b,\,-\sinh a,\, -\cosh a\sinh b,\,0).$$

Choose the rotationless standard boost $L(p)$ satisfying $L(p)k=p$. Then

$$\mathcal W(\Lambda,k)=L(\Lambda k)^{-1}\Lambda =R_z(\Omega).$$

The right-handed rotation angle in this convention obeys

$$\tan\frac{\Omega}{2} =\tanh\frac a2\,\tanh\frac b2.$$

For finite real rapidities, choose the continuous branch $-\pi/2<\Omega<\pi/2$. The angle is positive when $a,b>0$, and it vanishes when either boost vanishes. Here $a,b$ are rapidities, not velocities.

Extracting the rotation by polar decomposition

The two boost lifts multiply to

$$A=A_y(b)A_x(a) =C I-X\sigma_x-Y\sigma_y-iZ\sigma_z,$$

where

$$\begin{aligned} C&=\cosh(a/2)\cosh(b/2),\\ X&=\sinh(a/2)\cosh(b/2),\\ Y&=\cosh(a/2)\sinh(b/2),\\ Z&=\sinh(a/2)\sinh(b/2). \end{aligned}$$

The imaginary term has its sign because $\sigma_y\sigma_x=-i\sigma_z$. In the polar decomposition $A=H U$, $H$ is positive Hermitian with determinant one and $U$ is unitary. The matrix $H$ is the lift of the canonical boost to $\Lambda k$: indeed $AA^\dagger=H^2$ determines the transformed rest momentum.

Set $U=e^{-i\Omega\sigma_z/2}$. In $H=AU^\dagger$, the coefficient of $i\sigma_z$ is

$$C\sin(\Omega/2)-Z\cos(\Omega/2).$$

Hermiticity forces it to vanish, yielding the half-angle formula. With the stated branch the resulting Hermitian factor is positive, so it is the polar boost rather than its negative.

An equivalent expression useful for a numerical check is

$$\tan\Omega =\frac{\sinh a\,\sinh b}{\cosh a+\cosh b}.$$

This calculation identifies the order of factors: $\Lambda=L(\Lambda k)R_z(\Omega)$. Writing a rotation on the other side of a boost changes the boost factor and must not be done without transforming its direction.

Checks on signs and the spin-half action

For $a=0.7$ and $b=0.4$, the formula gives $\Omega\approx0.1325898745$ radians, about $7.60^\circ$. Reversing the ordered boosts gives the opposite rotation angle for the corresponding canonical factorization. Collinear boosts, by contrast, have no such rotation.

At small rapidities,

$$\Omega=\frac{ab}{2} +O(a^3b+ab^3).$$

The rotation begins at second order, consistent with the noncommuting boost generators. A group commutator of four infinitesimal boosts has a different ordered product from this two-boost comparison and should not be assigned the same factor one half automatically.

For spin one half, the rest-spin matrix is

$$D^{(1/2)}(R_z(\Omega)) =e^{-i\Omega\sigma_z/2}.$$

An initially $+x$ spin column $(1,1)^T/\sqrt2$ becomes $(e^{-i\Omega/2},e^{i\Omega/2})^T/\sqrt2$. Its Pauli expectation is $(\cos\Omega,\sin\Omega,0)$. A $z$-polarized state instead acquires a spin-dependent phase without changing that axis.

These refer to canonical spin labels at the new momentum. A detector’s spin observable must be transformed consistently when comparing descriptions in different frames.

Momentum-dependent rotations of packets

For an initial momentum other than rest, use the full remainder

$$\mathcal W(\Lambda,p) =L(\Lambda p)^{-1}\Lambda L(p).$$

Even a single boost generally gives a nontrivial rotation when its direction is not collinear with the particle momentum. Different momenta in a packet can therefore rotate through different spin matrices.

For example, a packet whose spin coefficients initially factor as $f_\sigma(p)=g(p)\zeta_\sigma$ generally transforms into coefficients involving $D(\mathcal W(\Lambda,p))\zeta$. They need not retain that factorization. The complete transformation is unitary; a reduced spin density matrix can nevertheless change after momentum is traced out. The result depends on the specified spin basis and measurement, and is not a loss of unitarity of the full state.

Changing the standard boosts changes the spin basis and the Wigner matrices by momentum-dependent basis factors, as described by the induced construction. An angle quoted without a standard-boost convention is incomplete.

Thomas precession requires a transport convention

Restore $c$. Consider a torque-free gyroscope transported along an accelerated worldline, with its spin Fermi–Walker transported. Compare its rest-frame spin direction with laboratory axes using instantaneous rotationless boosts. For laboratory velocity $\mathbf v(t)$ and acceleration $\mathbf a=d\mathbf v/dt$, the kinematic Thomas angular velocity with respect to laboratory time is

$$\boldsymbol\omega_T =\frac{\gamma^2}{\gamma+1}\, \frac{\mathbf a\times\mathbf v}{c^2}.$$

This is a statement about successive frame transport, not a direct substitution into the ordered two-boost angle above. Comparing transported axes in the inverse direction reverses the written rotation sign. Physical torques, such as electromagnetic spin precession, must be added separately in the relevant dynamics.

For uniform circular motion with orbital angular velocity $\boldsymbol\omega_{\rm orb}$,

$$\boldsymbol\omega_T =-(\gamma-1)\boldsymbol\omega_{\rm orb} \approx-\frac{v^2}{2c^2}\boldsymbol\omega_{\rm orb}.$$

The precession is opposite to the orbital rotation under this convention. The factor one half is the low-speed kinematic correction that appears when rest-frame and laboratory spin couplings are compared. It is not, by itself, a derivation of a complete atomic spin–orbit Hamiltonian.

Exercises

1. Derive the formula for $\tan\Omega$ from the half-angle expression.

Solution

Set $t=\tanh(a/2)\tanh(b/2)$ and use $\tan\Omega=2t/(1-t^2)$. Multiplying numerator and denominator by $\cosh^2(a/2)\cosh^2(b/2)$ and applying the double-angle identities gives $\sinh a\,\sinh b/(\cosh a+\cosh b)$. The denominator is positive, which fixes the continuous branch stated above.

2. Why does a $+z$ spin eigenstate fail to reveal this rotation through its mean spin vector alone?

Solution

A rotation about $z$ changes that spinor by $e^{-i\Omega/2}$, leaving its mean spin vector unchanged. A transverse spin preparation or a suitable relative-phase experiment can reveal the rotation.

3. Verify the exact circular-motion relation from the displayed Thomas angular velocity.

Solution

For circular motion, $\mathbf a\times\mathbf v=-v^2\boldsymbol\omega_{\rm orb}$. Using $\gamma^2v^2/[c^2(\gamma+1)]=\gamma-1$ gives the result. Expanding $\gamma-1=v^2/(2c^2)+O(v^4/c^4)$ gives the low-speed limit.

References

• J. D. Jackson, Classical Electrodynamics, 3rd ed., Wiley, 1999 — Thomas precession and spin transport.
• W. N. Polyzou, W. Glöckle, and H. Witała, “Spin in Relativistic Quantum Theory,” Few-Body Systems 54, 1667–1704, 2013, doi:10.1007/s00601-012-0526-8, arXiv:1208.5840 — standard boosts, Wigner rotations, and spin observables.
• N. Straumann, “Unitary Representations of the inhomogeneous Lorentz Group and their Significance in Quantum Physics,” 2008, arXiv:0809.4942 — the little-group construction.