Past exam of the quantum field theory course of the University of Oxford / 2023 / 3

In terms of the Pauli matrices, ${\gamma }^0={\sigma }_3$, ${\gamma }^1=i{\sigma }_1$, ${\gamma }^2=i{\sigma }_2$. The Pauli matrix multiplication law implies ${\sigma }_j^2=I_2$ and ${\sigma }_j{\sigma }_k=-{\sigma }_k{\sigma }_j$ for $j\ne k$. ConsequentlyThese diagonal and off-diagonal cases are exactly the Clifford algebra relationThe identity matrix is implicit on the right-hand side in the source. Useful conventions for the remaining parts are $({\gamma }^0{)}^{†}={\gamma }^0$, $({\gamma }^j{)}^{†}=-{\gamma }^j$ for $j=1,2$, and ${\gamma }^{0\ast }={\gamma }^0$, ${\gamma }^{1\ast }=-{\gamma }^1$, ${\gamma }^{2\ast }={\gamma }^2$.

A parity symmetry in quantum field theory reverses spatial orientation, so its spatial matrix must lie in the determinant-$-1$ component of the orthogonal group. In two spatial dimensions, reversing both coordinates gives $-I_2$, which has determinant $+1$ and is a rotation through $\pi$ in the special orthogonal group in two dimensions. It cannot test a discrete orientation-reversing symmetry independent of ordinary rotations. Reflecting one coordinate, for example $\mathrm{diag}(-1,1)$, has determinant $-1$ and supplies the required parity symmetry in quantum field theory.

In three spatial dimensions, $-I_3$ has determinant $-1$, so simultaneous inversion of all three coordinates is the standard choice. A single-coordinate reflection also reverses orientation there, and differs from $-I_3$ by a proper rotation through $\pi$. Thus, in a rotationally invariant theory, it tests the same discrete symmetry. The word is in the source specifies conventional representatives: the essential distinction is orientation reversal, rather than a unique choice of axis. Both transformations preserve the ordinary spatial integration measure through the absolute value of their Jacobian determinants.

Put $x_P=(t,-x,y)$ and $U_P={\gamma }^1$. Since $U_P^{†}=-{\gamma }^1$ and ${\gamma }^0{\gamma }^1{\gamma }^0=-{\gamma }^1$, the Dirac adjoint transforms asThe chain rule gives ${\partial }_{\mu }\psi (x_P)=r_{\mu }({\partial }_{\mu }\psi )(x_P)$, with $(r_0,r_1,r_2)=(1,-1,1)$ and no sum in this formula. The Clifford algebra givesMultiplication by $r_{\mu }$ therefore restores each kinetic term:However, the Dirac scalar bilinear changes sign because $({\gamma }^1{)}^2=-I_2$:Changing variables in the action has unit absolute Jacobian determinant. HenceA nonzero Dirac mass term for this single two-component field breaks parity symmetry in quantum field theory with the specified transformation.

The specified time reversal includes complex conjugating $i{\gamma }^{\mu }{\partial }_{\mu }$; omitting that operation changes the answer. Put $x_T=(-t,x,y)$, $U_T={\gamma }^2$, and $(r_0,r_1,r_2)=(-1,1,1)$. As in c, the Dirac adjoint in the source's transformed bilinear is ${\bar{\psi }}^{\prime }(x)=\bar{\psi }(x_T){\gamma }^2$. The conjugated Dirac operator has coefficient matricesCombining these coefficients with the chain rule signs and the Clifford algebra gives, without summing over $\mu$,For example the time coefficient is ${\gamma }^2(-i{\gamma }^0){\gamma }^2(-1)=i{\gamma }^0$, and the second spatial coefficient is ${\gamma }^2(-i{\gamma }^2){\gamma }^2=i{\gamma }^2$. Thus the kinetic term is invariant. The real Dirac mass term again reverses sign since ${\bar{\psi }}^{\prime }{\psi }^{\prime }=\bar{\psi }({\gamma }^2{)}^2\psi =-\bar{\psi }\psi$. The spacetime integration has unit absolute Jacobian determinant, soEach operation reverses the sign of the mass, while their composition reverses it twice and preserves the kinetic term. In an operator formulation the complex conjugation of coefficients expresses the antiunitary operator character of time reversal; the transformation must not be treated as an ordinary unitary substitution with all $i$'s held fixed. These statements concern the stated free action.

First conjugate the Dirac equation: $(-i{\gamma }^{\mu \ast }{\partial }_{\mu }-m){\psi }^{\ast }=0$. A matrix $\Gamma$ taking ${\psi }^{\ast }$ into another solution of the same Dirac equation must obeyBecause ${\gamma }^1$ anticommutes with the real ${\gamma }^0,{\gamma }^2$ and commutes with the purely imaginary ${\gamma }^1$, a unitary choice isAny constant overall phase is equally valid. To verify invariance of the action, rather than only of its equation, keep the fermion exchange signs. Define the charge-conjugation matrix $C=\Gamma {\gamma }^0={\sigma }_2$. It obeysFor a classical Grassmann field, interchanging two components contributes a minus sign; equivalently use normal ordering for quantum fermionic operators. ThusandThe last term integrates to a boundary term, vanishing with the usual decay or compatible boundary conditions. Both the kinetic term and the Dirac mass term therefore leave $S$ invariant under charge conjugation. Treating the components as commuting numbers would give the wrong mass-bilinear sign; the fermion statistics is essential.