Time Reversal

Time reversal of a Dirac solution reverses the time argument, conjugates amplitudes, and rotates the spinor components so that both spin and momentum reverse. The positive-frequency sector remains positive-frequency. In an electromagnetic problem the background must also transform: electric field is even under time reversal, whereas magnetic field is odd. A transformation between the two backgrounds is not automatically a symmetry of either background held fixed.

Required background. Antiunitary Time Reversal owns the general reason for antilinearity; Gamma-Matrix Conventions and Minimal Coupling fix the relativistic operators. Helpful background. Time Reversal for Spin-Half Particles sets the two-component phase convention; Kramers Degeneracy gives the corresponding spectral theorem.

Antiunitary map in the Dirac basis

Let $K$ denote componentwise complex conjugation in the position representation and the chosen Dirac spinor basis. On initial data define

$$\Theta=U_TK,\qquad U_T=-\gamma^1\gamma^3 =\begin{pmatrix} -i\sigma^2&0\\ 0&-i\sigma^2 \end{pmatrix}.$$

This phase agrees with the Pauli operator $-i\sigma^2K$ used on the spin-half owner page. The matrix is unitary, and the full map is antiunitary: $\Theta(z\psi)=z^*\Theta\psi$. For a time-dependent solution, the reversed solution is

$$\psi_T(t,\mathbf x) =U_T\psi^*(-t,\mathbf x).$$

The matrix identities needed for the Hamiltonian calculation are

$$\begin{aligned} U_T\beta^*U_T^{-1}&=\beta,\\ U_T\alpha_i^*U_T^{-1}&=-\alpha_i,\\ U_T\Sigma_i^*U_T^{-1}&=-\Sigma_i. \end{aligned}$$

Complex conjugation also acts on the differential momentum: $K\mathbf pK^{-1}=-\mathbf p$. It is not enough to conjugate the finite spinor matrices while treating $\mathbf p=-i\hbar\nabla$ as a real number.

For this choice $U_T$ is real and

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

A common alternative matrix is $U_{T,0}=i\gamma^1\gamma^3$. Our convention is $U_T=iU_{T,0}$. The alternative has $U_{T,0}^2=I$, but its antiunitary square is still $U_{T,0}U_{T,0}^*=-I$. More generally multiplying $\Theta$ by a unit-modulus phase does not change its square. Matrix squares and antiunitary-operator squares must be kept distinct.

Reversed electromagnetic dynamics

Let the potentials be real, with

$$H_q(t) =c\boldsymbol\alpha\cdot (\mathbf p-q\mathbf A(t,\mathbf x)) +\beta mc^2+q\Phi(t,\mathbf x).$$

Complex conjugating the Schrödinger equation and reversing its time argument gives

$$i\hbar\partial_t\psi_T =U_TH_q^*(-t)U_T^{-1}\psi_T.$$

The two minus signs, from conjugating $i$ and from differentiating $-t$, cancel. Using the matrix and momentum identities yields the target Hamiltonian

$$H_{q,T}(t) =H_q[\Phi_T,\mathbf A_T](t),$$

with

$$\begin{aligned} \Phi_T(t,\mathbf x)&=\Phi(-t,\mathbf x),\\ \mathbf A_T(t,\mathbf x)&=-\mathbf A(-t,\mathbf x). \end{aligned}$$

The charge $q$ and mass are unchanged. Differentiating these potentials gives

$$\mathbf E_T(t,\mathbf x)=\mathbf E(-t,\mathbf x), \qquad \mathbf B_T(t,\mathbf x)=-\mathbf B(-t,\mathbf x).$$

These are active maps of the complete solution and prescribed background. They apply also to a time-dependent driving protocol: the protocol must be reversed along with the state. Beisert’s spinor-field notes provide the corresponding relativistic transformation framework.

For a complex Klein–Gordon amplitude, the counterpart is

$$\phi_T(t,\mathbf x)=\phi^*(-t,\mathbf x)$$

with the same transformed potentials and the same charge. The two temporal sign changes again preserve the covariant second-order equation. For a spinless scalar this map squares to $+I$. Complex conjugation at the same time argument would be a different operation and would instead reverse the charge in the scalar equation.

Positive frequency, momentum, and spin

Consider a free positive-energy mode

$$\psi(t,\mathbf x) =u(\mathbf p,\chi) e^{-iE_{\mathbf p}t/\hbar+i\mathbf p\cdot\mathbf x/\hbar}.$$

Time reversal gives

$$\psi_T(t,\mathbf x) =U_Tu(\mathbf p,\chi)^* e^{-iE_{\mathbf p}t/\hbar-i\mathbf p\cdot\mathbf x/\hbar}.$$

Thus energy remains $+E_{\mathbf p}$ and momentum becomes $-\mathbf p$. With the normalized free spinors of Free Dirac Spinors, their two-component label transforms as

$$\chi_T=-i\sigma^2\chi^*,\qquad \begin{pmatrix}a\\b\end{pmatrix} \longmapsto \begin{pmatrix}-b^*\\a^*\end{pmatrix}.$$

The spin expectation reverses. This map does not send the particle to the negative-energy branch or change its charge.

Both spin and momentum reverse, so helicity is even:

$$\Theta\frac{\boldsymbol\Sigma\cdot\mathbf p}{2|\mathbf p|} \Theta^{-1} =\frac{\boldsymbol\Sigma\cdot\mathbf p}{2|\mathbf p|}.$$

The component identity $U_T(\gamma^5)^*U_T^{-1}=\gamma^5$ also preserves chirality. By contrast, parity reverses momentum while leaving spin axial, and exchanges chirality. These statements concern the specified free component maps; they do not establish time-reversal invariance of arbitrary chiral interactions.

For the probability current,

$$\begin{aligned} \rho_T(t,\mathbf x)&=\rho(-t,\mathbf x),\\ \mathbf j_T(t,\mathbf x)&=-\mathbf j(-t,\mathbf x). \end{aligned}$$

This is the expected reversal of flux while preserving the positive Dirac probability density.

A fixed magnetic field is a different test

For a static Hamiltonian to possess time-reversal symmetry, the transformed operator and its domain must describe the same physical background, possibly after a compensating gauge transformation. A static central electrostatic potential with $\mathbf A=0$ satisfies this condition. A fixed nonzero uniform magnetic field does not: it is mapped to $-\mathbf B$.

This distinction is visible in Relativistic Landau Levels. Time reversal maps an eigenmode at $\mathbf B$ to one at $-\mathbf B$ with reversed spin and longitudinal momentum. It is not a proof that every mode has a partner in the same fixed magnetic background.

For a self-adjoint, time-reversal-invariant Dirac operator with a preserved domain, $\Theta^2=-I$ permits the usual Kramers theorem for its normalizable discrete eigenstates. The theorem’s antiunitary proof belongs to that owner. It should not be invoked when the external field or a boundary condition breaks the required symmetry. Nor should a claim about normalizable bound-state pairs be transferred unqualified to generalized continuum eigenfunctions.

Exercises

An arbitrary phase. Let $\Theta'=e^{i\theta}\Theta$. Show that $(\Theta')^2=-I$, and identify where treating the map as linear would fail.

Solution

Antilinearity gives $\Theta e^{i\theta}=e^{-i\theta}\Theta$. Therefore

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

Replacing the second phase by $e^{i\theta}$ would incorrectly produce a phase-dependent square.

A driven uniform electric field. In temporal gauge take $\mathbf A(t)=-f(t)\hat{\mathbf z}$, $\Phi=0$. Find the reversed potential and electric field.

Solution

The original field is $\mathbf E(t)=f'(t)\hat{\mathbf z}$. The transformed potential is $\mathbf A_T(t)=f(-t)\hat{\mathbf z}$, and hence

$$\mathbf E_T(t) =-\partial_t\mathbf A_T(t) =f'(-t)\hat{\mathbf z} =\mathbf E(-t).$$

The sign in the vector potential is needed to obtain the even electric-field transformation.

Reversing a spinor twice. Apply $(a,b)^{\mathsf T}\mapsto(-b^*,a^*)^{\mathsf T}$ twice. Explain why the resulting minus sign is compatible with unchanged probability density.

Solution

The second application gives $(-a,-b)^{\mathsf T}$. The density is unchanged by that overall sign, but the operator identity $\Theta^2=-I$ still has consequences for inner products and Kramers pairing. It cannot be removed by rephasing $\Theta$.

References

• Beisert, Niklas. Quantum Field Theory I. ETH Zurich lecture notes, autumn semester 2012, chapter 5, section 5.3. Free Spinor Field.
• Sakurai, Jun John, and Jim Napolitano. Modern Quantum Mechanics. 3rd ed. Cambridge University Press (2020). doi:10.1017/9781108587280.
• Thaller, Bernd. The Dirac Equation. Springer (1992). doi:10.1007/978-3-662-02753-0.