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

$$\Theta^2=-I$$

on a single spin-$1/2$ Hilbert space. This minus sign forces each pure spin state to be orthogonal to its time reverse. If the Hamiltonian also preserves time reversal, that partner has the same energy.

Time Reversal in the Spin-z 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

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

For example, $(|\uparrow\rangle+i|\downarrow\rangle)/\sqrt2$ has spin along $+y$. Its transformed spinor is $(i|\uparrow\rangle+|\downarrow\rangle)/\sqrt2$, which points along $-y$. Dropping conjugation would give the wrong direction.

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 statement concerns a Hamiltonian on this two-dimensional spin space alone. With orbital or momentum degrees of freedom, a term such as $\mathbf b(\mathbf k)\cdot\boldsymbol\sigma$ can preserve time reversal if $\mathbf b(-\mathbf k)=-\mathbf b(\mathbf k)$. Time reversal then relates different momenta rather than forbidding every spin-dependent term. In contrast, a magnetic-field Hamiltonian

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

obeys

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

For nonzero magnetic coupling, a fixed background $\mathbf B\ne0$ therefore breaks ordinary time reversal. The field-reversed family remains covariant. 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 $+I$.

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 nonzero fixed splitting in this two-dimensional space breaks time reversal. Transforming an external parameter compares a different Hamiltonian and does not establish invariance at the original parameter value.

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 $c=0$.

References

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