Time Reversal for Spin-1/2 Particles

For a spin-$1/2$ degree of freedom, time reversal is not just complex conjugation. It must also flip the spin vector. In the standard $S_z$ basis, a convenient convention is

$$\Theta = -i\sigma_yK,$$

where $K$ complex-conjugates spinor components in that basis. The antiunitary operator $\Theta$ satisfies

$$\Theta S_i\Theta^{-1} = -S_i, \qquad S_i=\frac{\hbar}{2}\sigma_i,$$

and, most importantly,

$$\Theta^2=-I$$

on a single spin-$1/2$ Hilbert space. This minus sign is the local seed of Kramers degeneracy and of many later distinctions between integer-spin and half-integer-spin systems.

Convention and Basis

Use the ordered basis

$$\lvert\uparrow\rangle = \begin{pmatrix}1\\0\end{pmatrix}, \qquad \lvert\downarrow\rangle = \begin{pmatrix}0\\1\end{pmatrix},$$

where these are eigenstates of $S_z$. In this basis,

$$\sigma_y = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}, \qquad -i\sigma_y = \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}.$$

Thus $\Theta=-i\sigma_yK$ means: first complex-conjugate the components, then multiply by $-i\sigma_y$. The symbol $K$ is basis-dependent; the antiunitary operator $\Theta$ is the object with physical meaning.

Equivalently,

$$\Theta = e^{-i\pi S_y/\hbar}K$$

in the same $S_z$ basis. The unitary factor is a $\pi$ rotation about the $y$ axis in spin space. The complex conjugation supplies antiunitarity.

Action on Basis Spinors

Applying $\Theta$ to the standard basis gives

$$\Theta\lvert\uparrow\rangle = \lvert\downarrow\rangle, \qquad \Theta\lvert\downarrow\rangle = -\lvert\uparrow\rangle.$$

For a general spinor

$$\lvert\psi\rangle = \alpha\lvert\uparrow\rangle + \beta\lvert\downarrow\rangle,$$

antilinearity gives

$$\Theta\lvert\psi\rangle = -\beta^*\lvert\uparrow\rangle + \alpha^*\lvert\downarrow\rangle.$$

The complex conjugation of $\alpha$ and $\beta$ is not optional. It is exactly where many sign and phase mistakes enter calculations with time reversal.

Spin Reversal

Time reversal should reverse angular momentum. For spin-$1/2$, this means

$$\Theta\boldsymbol\sigma\Theta^{-1} = -\boldsymbol\sigma,$$

or componentwise,

$$\Theta\sigma_x\Theta^{-1}=-\sigma_x, \qquad \Theta\sigma_y\Theta^{-1}=-\sigma_y, \qquad \Theta\sigma_z\Theta^{-1}=-\sigma_z.$$

The $\sigma_y$ component is the one most likely to trip up a calculation, because $K\sigma_yK^{-1}=-\sigma_y$. Both the unitary factor and the complex conjugation must be included.

Consequently,

$$\Theta\mathbf S\Theta^{-1} = -\mathbf S.$$

This is the spin analog of the orbital rule

$$\Theta\mathbf L\Theta^{-1} = -\mathbf L,$$

discussed on the overview page for Time Reversal.

Why the Square Is Minus One

Let $U=-i\sigma_y$, so $\Theta=UK$. Since $U$ is real in the displayed basis,

$$\Theta^2 = UKUK = UU^* = U^2.$$

But

$$U^2 = (-i\sigma_y)^2 = -I.$$

Therefore

$$\Theta^2=-I.$$

This result does not depend on the arbitrary overall phase of $\Theta$. If

$$\Theta' = e^{i\chi}\Theta,$$

then antiunitarity gives

$$(\Theta')^2 = e^{i\chi}\Theta e^{i\chi}\Theta = e^{i\chi}e^{-i\chi}\Theta^2 = \Theta^2.$$

So the sign of $\Theta^2$ is not removable by a phase redefinition. It is a projective representation feature of half-integer spin.

Hamiltonian Constraint

Any Hermitian Hamiltonian on a single spin-$1/2$ space can be written

$$H = aI+\mathbf b\cdot\boldsymbol\sigma,$$

with real $a$ and real $\mathbf b$. Under spin-$1/2$ time reversal,

$$\Theta H\Theta^{-1} = aI-\mathbf b\cdot\boldsymbol\sigma.$$

If no external time-reversal-odd parameter is being transformed along with the system, time-reversal invariance requires

$$\mathbf b=0.$$

This is why an isolated single spin cannot have a preferred Zeeman axis without some time-reversal-breaking environment or field. In contrast, a magnetic-field Hamiltonian

$$H(\mathbf B) = -\gamma\mathbf S\cdot\mathbf B$$

obeys

$$\Theta H(\mathbf B)\Theta^{-1} = H(-\mathbf B).$$

A fixed background $\mathbf B$ therefore breaks time-reversal symmetry, while the family of Hamiltonians is covariant if the magnetic field is also reversed. This distinction is essential in Spin in Magnetic Fields.

Kramers Pair Preview

Suppose a Hamiltonian is time-reversal invariant:

$$\Theta H\Theta^{-1}=H.$$

If

$$H\lvert\psi\rangle = E\lvert\psi\rangle,$$

then $\Theta\lvert\psi\rangle$ is also an energy eigenstate with the same energy. The antiunitary sign $\Theta^2=-I$ also forces orthogonality:

$$\langle\psi|\Theta\psi\rangle=0.$$

This is the algebraic core of Kramers Degeneracy. The theorem has its own hypotheses and applications, but the spin-$1/2$ calculation here explains why the degeneracy is tied to half-integer spin and antiunitary time reversal rather than to ordinary unitary symmetry alone.

Common Mistakes

• Writing $\Theta=-i\sigma_y$ and forgetting the complex conjugation operator $K$.
• Treating $K$ as basis-independent. It is complex conjugation in a specified basis.
• Assuming $\Theta^2=-I$ can be changed to $+I$ by multiplying $\Theta$ by a phase.
• Forgetting that antiunitarity conjugates scalar coefficients.
• Testing time reversal of a magnetic-field Hamiltonian without saying whether the external field is transformed.
• Assuming every spin system has $\Theta^2=-I$. Two spin-$1/2$ particles have a product time-reversal operator with square $+I$.

Cross-Links

• Time Reversal
• Antiunitary Time Reversal
• Time Reversal for Spinless Particles
• Kramers Degeneracy
• Symmetry Classification Preview
• Topological Insulators
• From Discrete Symmetries to CPT
• Antiunitary Symmetries
• Spin-1/2 Hilbert Space
• Pauli Matrices
• Spin in Magnetic Fields
• Symmetry Constraints on Hamiltonians
• Discrete Symmetries in Hamiltonians
• Pauli Matrix Table

References

• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• A. Messiah, Quantum Mechanics, Dover, 1999.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003.

Exercises

1. Use $\Theta=-i\sigma_yK$ to compute $\Theta\lvert\uparrow\rangle$ and $\Theta\lvert\downarrow\rangle$.

Solution

In the $S_z$ basis, both basis spinors are real, so $K$ leaves their components unchanged. Then

$$\Theta\lvert\uparrow\rangle = \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix} = \lvert\downarrow\rangle.$$

Similarly,

$$\Theta\lvert\downarrow\rangle = \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix} \begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}-1\\0\end{pmatrix} = -\lvert\uparrow\rangle.$$

2. Show that $\Theta^2=-I$ by applying $\Theta$ twice to an arbitrary spinor.

Solution

Let

$$\lvert\psi\rangle = \alpha\lvert\uparrow\rangle + \beta\lvert\downarrow\rangle.$$

The first application gives

$$\Theta\lvert\psi\rangle = -\beta^*\lvert\uparrow\rangle + \alpha^*\lvert\downarrow\rangle.$$

Apply $\Theta$ again, remembering antilinearity:

$$\Theta^2\lvert\psi\rangle = -(\alpha^*)^*\lvert\uparrow\rangle + (-\beta^*)^*\lvert\downarrow\rangle = -\alpha\lvert\uparrow\rangle -\beta\lvert\downarrow\rangle.$$

Therefore

$$\Theta^2\lvert\psi\rangle = -\lvert\psi\rangle.$$

3. For $H=aI+\mathbf b\cdot\boldsymbol\sigma$, impose $\Theta H\Theta^{-1}=H$.

Solution

Since $\Theta I\Theta^{-1}=I$ and $\Theta\sigma_i\Theta^{-1}=-\sigma_i$,

$$\Theta H\Theta^{-1} = aI-\mathbf b\cdot\boldsymbol\sigma.$$

Equating this with

$$H = aI+\mathbf b\cdot\boldsymbol\sigma$$

requires

$$\mathbf b=0.$$

Thus a single spin-$1/2$ Hamiltonian with a fixed preferred axis is not time-reversal invariant unless that axis is supplied by an external parameter that is also transformed.

4. Prove the orthogonality statement $\langle\psi|\Theta\psi\rangle=0$ when $\Theta^2=-I$.

Solution

Let

$$c = \langle\psi|\Theta\psi\rangle.$$

Antiunitarity gives

$$\langle\Theta\psi|\Theta^2\psi\rangle = \langle\psi|\Theta\psi\rangle^* = c^*.$$

But $\Theta^2=-I$, so the left side is

$$\langle\Theta\psi|-\psi\rangle = -\langle\Theta\psi|\psi\rangle = -c^*.$$

Hence $c^*=-c^*$, so $c=0$.