Charge Conjugation

Charge conjugation at the wave-equation level is an antilinear map from a charged solution to a solution with the opposite charge. For a Dirac spinor it requires a matrix as well as componentwise conjugation. The map reverses the Fourier frequency of a mode, so identifying it with a positive-energy antiparticle requires the quantum-field interpretation. The unitary charge-conjugation operator on quantum states is a distinct object from this antilinear map of commuting solution columns.

Required background. Minimal Coupling, the Covariant Dirac Equation, and the Klein–Gordon Equation supply the equations and their charge conventions. Helpful background. Majorana Spinors owns the matrix real structure; Dirac Negative-Energy Solutions explains the field interpretation; Charge Conjugation Preview owns the general charge-sector overview.

Charge reversal in the scalar equation

Take real electromagnetic potentials and the signed-charge derivative

$$D_\mu^{(q)} =\partial_\mu+\frac{iq}{\hbar}A_\mu.$$

Complex conjugation gives

$$\left(D_\mu^{(q)}\phi\right)^* =D_\mu^{(-q)}\phi^*.$$

Therefore if

$$\left(D_\mu^{(q)}D^{(q)\mu} +\frac{m^2c^2}{\hbar^2}\right)\phi=0,$$

then $\phi_C=\phi^*$ satisfies the same form of equation with charge $-q$ in the same background. There is no reversed time argument in this transformation.

Equivalently, keep the symbol $q$ fixed and transform the background as $A_{\mu,C}=-A_\mu$. Both descriptions reverse the product $qA_\mu$ entering the equation. They are alternative descriptions of the same target differential operator. Reversing both $q$ and $A_\mu$ would leave their product unchanged and would not reproduce this derivation.

The gauge phase is consistent with the charge change. Under $\phi\mapsto e^{iq\chi/\hbar}\phi$,

$$\phi_C\mapsto e^{-iq\chi/\hbar}\phi_C.$$

It transforms in the conjugate charge representation. In the fixed-$q$, reversed-background description, the corresponding gauge parameter is $-\chi$.

Dirac matrix and the conjugate equation

Use the Dirac basis and define

$$C_D=i\gamma^2\gamma^0,\qquad B=C_D(\gamma^0)^{\mathsf T}=i\gamma^2.$$

The charge-conjugate column is

$$\psi_C=C_D\bar\psi^{\,\mathsf T} =B\psi^*.$$

The real-structure derivation on Majorana Spinors establishes

$$B(\gamma^\mu)^*B^{-1}=-\gamma^\mu, \qquad BB^*=I.$$

Complex conjugate the original equation:

$$\left(-i\hbar c(\gamma^\mu)^* D_\mu^{(-q)}-mc^2\right)\psi^*=0.$$

Multiplication by $B$ then gives

$$\left(i\hbar c\gamma^\mu D_\mu^{(-q)} -mc^2\right)\psi_C=0.$$

The sign from the gamma identity compensates the sign from conjugating $i$. Multiplication by $C_D$ alone, without the adjoint and transpose, would not implement this operation.

Let $\mathscr C=BK$ on commuting initial-data columns. Then

$$\mathscr C^2=BB^*=I.$$

The matrix $C_D$ itself instead satisfies $C_D^2=-I$ in this convention. There is no contradiction: it is not the full antilinear map. A unit-modulus phase multiplying $\mathscr C$ cancels in its square.

Hamiltonian and frequency bookkeeping

The instantaneous Hamiltonian identity is

$$\mathscr C H_q[\Phi,\mathbf A](t)\mathscr C^{-1} =-H_{-q}[\Phi,\mathbf A](t).$$

To verify it, use

$$B\beta^*B^{-1}=-\beta,\qquad B\alpha_i^*B^{-1}=\alpha_i,$$

as well as $K\mathbf pK^{-1}=-\mathbf p$. The extra overall minus sign is necessary. Since $\mathscr C$ is antilinear, $\mathscr C i=-i\mathscr C$, it follows that

$$i\hbar\partial_t\psi_C =H_{-q}(t)\psi_C.$$

For a static problem, $H_q u=Eu$ implies

$$H_{-q}\mathscr C u=-E\,\mathscr C u.$$

For a free positive-frequency mode, the Fourier factor changes as

$$e^{-iEt/\hbar+i\mathbf p\cdot\mathbf x/\hbar} \longmapsto e^{+iEt/\hbar-i\mathbf p\cdot\mathbf x/\hbar}.$$

Thus the commuting solution map reverses the frequency and canonical Fourier momentum signs. It does not directly map one positive-energy wavefunction to another. For a charged bound-state problem, the relation likewise compares an energy $E$ in one charge equation with $-E$ in the conjugate-charge equation. Interpreting the latter as a physical antiparticle state requires the field mode expansion.

The distinction from time reversal is concrete: time reversal conjugates the mode and reverses $t$, thereby preserving the frequency sign. Charge conjugation here leaves the time argument unchanged.

Commuting currents and fermionic currents

For an ordinary commuting Dirac wavefunction, the positive probability density is unchanged:

$$\psi_C^\dagger\psi_C=\psi^\dagger\psi.$$

The spatial probability current is also unchanged at the same spacetime point,

$$c\psi_C^\dagger\boldsymbol\alpha\psi_C =c\psi^\dagger\boldsymbol\alpha\psi.$$

The second identity follows by transposing the scalar expression after using $B^\dagger\alpha_iB=\alpha_i^*$. In the description that reverses the charge parameter, multiplying this probability current by the new charge changes the electric current sign.

The scalar Klein–Gordon current has a different algebraic behavior. With the normalization of Conserved Current,

$$j^\mu_{-q}[\phi^*]=-j^\mu_q[\phi].$$

Its density is a signed frequency-sector quantity, not the positive Dirac norm. If the defining charge is also relabeled as $-q$, the product instead satisfies

$$(-q)\,j^\mu_{-q}[\phi^*] =q\,j^\mu_q[\phi].$$

Keeping $q$ fixed while reversing the background gives a C-odd scalar electric current, because only the signed current then reverses. The two descriptions agree on the target differential operator; their current assignments must still be tracked with their charge labels. A negative-frequency scalar solution cannot be read directly as a positive-norm, charge-reversed particle. Neither identity should be replaced by an intuition that every relativistic wavefunction has the same kind of probability current.

For an anticommuting quantum Dirac field, the properly defined electric current is C-odd. Reordering fermionic fields contributes the extra sign absent from commuting columns. For example, a field-theory table assigning a minus sign to $\bar\Psi\gamma^\mu\Psi$ under C cannot be applied to the positive c-number density $\psi^\dagger\psi$. The statistics distinction and Majorana consequences are developed by Dreiner, Haber, and Martin (2010).

The quantum-state operation and its boundary

In a charge-conjugation-invariant quantum field theory, the implementing operator $\widehat C$ is unitary. It exchanges particle and antiparticle creation operators and preserves positive excitation energy. Its action on a field may be written schematically as

$$\widehat C\,\widehat\Psi(x)\,\widehat C^{-1} =\eta_C C_D\, \overline{\widehat\Psi}(x)^{\mathsf T}.$$

The adjoint on the operator-valued field does not make $\widehat C$ antiunitary on quantum states. For scalar numbers $z$, $\widehat C(z|\Psi\rangle)= z\widehat C|\Psi\rangle$, whereas the classical column map satisfies $\mathscr C(z\psi)=z^*\mathscr C\psi$. These are different mathematical operations with related roles.

The positive-energy reinterpretation of negative-frequency coefficients belongs to Dirac Negative-Energy Solutions. A background held fixed also requires a separate symmetry test: reversing the external source or charge sector does not prove degeneracy within an electron-only Hamiltonian.

Finally, the condition $\psi=\mathscr C\psi$ is a real structure on a neutral spinor. A single such structure is incompatible with an ordinary nonzero U(1) charge under arbitrary gauge phases. The component construction and its scope belong to Majorana Spinors.

Exercises

Two target descriptions. For a static electrostatic potential $\Phi(\mathbf x)$, write the target scalar interaction after charge conjugation in both descriptions. What goes wrong if both labels are reversed?

Solution

Keeping the background fixed gives $(-q)\Phi=-q\Phi$. Keeping $q$ fixed gives $q(-\Phi)=-q\Phi$. Reversing both gives $(-q)(-\Phi)=q\Phi$, the original coupling instead of the required conjugate one.

A spectral sign. Suppose a static $H_q$ has an eigenvector of energy $E$. Use antilinearity and the Hamiltonian identity to derive the energy in the $-q$ equation.

Solution

For a real eigenvalue, $\mathscr C H_q u=E\mathscr C u$. But $\mathscr C H_q=-H_{-q}\mathscr C$. Therefore $H_{-q}\mathscr C u=-E\mathscr C u$. The conclusion concerns solution energies of the two first-quantized operators, not the sign of a physical antiparticle excitation energy after quantization.

Why the density cannot be C-odd. Show directly that $B$ preserves the Dirac norm. Identify the assumption that changes when using a fermionic field-current table.

Solution

Since $B^\dagger B=I$, $\psi_C^\dagger\psi_C= \psi^{\mathsf T}\psi^*= \psi^\dagger\psi$ for commuting components. For fermionic fields, exchanging the order of components is not a sign-free operation. Defined local field bilinears therefore have different C transformation signs.

References

• Beisert, Niklas. Quantum Field Theory I. ETH Zurich lecture notes, autumn semester 2012, chapter 5, section 5.3. Free Spinor Field.
• Dreiner, Herbi K., Howard E. Haber, and Stephen P. Martin. “Two-component spinor techniques and Feynman rules for quantum field theory and supersymmetry.” Physics Reports 494, 1–196 (2010). doi:10.1016/j.physrep.2010.05.002. Corrected manuscript.
• Thaller, Bernd. The Dirac Equation. Springer (1992). doi:10.1007/978-3-662-02753-0.