Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 48 / 1

Write $I=I_2$. The Pauli matrix multiplication law gives ${\sigma }^i{\sigma }^j+{\sigma }^j{\sigma }^i=2{\delta }^{ij}I$. Block multiplication of the Weyl representation of the gamma matrices givesandThus the mixed time-space anticommutators vanish, the time-time anticommutator is $2I_4$, and the spatial anticommutators are $-2{\delta }^{ij}{I}_4$. Together these prove the Clifford algebra identityThe negative spatial sign comes from the lower-left block of each spatial gamma matrix.

Construct the Lorentz-spinor generators from a Clifford algebra asThey are antisymmetric in $\rho ,\sigma$. Reordering a third gamma matrix with the Clifford algebra yieldsUse the commutator derivation identity on the two factors in $M^{\tau \nu }=\frac{1}{4}[{\gamma }^{\tau },{\gamma }^{\nu }]$:This derives every Lorentz algebra commutation relation, so the matrices $M^{\rho \sigma }$ form its spinor representation. Their finite exponentials describe the connected Spinor representation of the Lorentz group, more precisely the lift to its double cover.

For $x^{\prime }=\Lambda x$, choose the corresponding spin liftThe spinor field transforms asThe infinitesimal Clifford algebra identity from part (b) exponentiates to $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$, establishing Lorentz covariance of the Dirac operator. At fixed coordinates the infinitesimal variation also includes the orbital term $-{\Omega }^{\mu }{}_{\nu }{x}^{\nu }{\partial }_{\mu }\psi$.

A spatial rotation generator $M^{ij}$ is anti-Hermitian, but a Lorentz boost generator is Hermitian. For example,Its eigenvalues are $\pm \frac{1}{2}$. A nonzero rapidity $\zeta$ gives boost eigenvalues $e^{\pm \zeta /2}$, which are not all on the unit circle. This proves that the finite-dimensional component Spinor representation of the Lorentz group cannot be unitary for a positive-definite component inner product, even after a change of basis. It does not contradict a unitary action on the physical space of quantum states.

Consequently ${\psi }^{\prime †}(x^{\prime }){\psi }^{\prime }(x^{\prime })={\psi }^{†}{S}^{†}S\psi$ need not equal ${\psi }^{†}\psi$: it is the time component of the Dirac current, not a Lorentz scalar. The appropriate invariant form is indefinite. From $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$ one obtainsThis Dirac spinor pseudo-unitarity gives ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$ and hence the Dirac scalar bilinear

Use the PDF's convention in which $C$ multiplies ${\psi }^{\ast }$, rather than ${\bar{\psi }}^T$. Its unitarity converts the given gamma matrix identity intoApplying it twice givesFor real infinitesimal Lorentz transformation parameters,Therefore ${\psi }^c=C{\psi }^{\ast }$ transforms in exactly the same spinor representation as $\psi$. The coordinate pullback is real and transforms identically as well. This is the antilinear charge-conjugation intertwiner relation; exponentiating also gives $C{S}^{\ast }=SC$ for connected transformations.

For real mass, complex conjugation of the free Dirac equation gives $(-i{\gamma }^{\mu \ast }{\partial }_{\mu }-m){\psi }^{\ast }=0$. Multiply by $C$ and move it through the gamma matrices:ThusAs an explicit check, in the given Weyl representation of the gamma matrices one may take $C=i{\gamma }^2$. It is unitary and obeys the required conjugation identity. The free-equation conclusion does not assert that a charge-conjugated field retains the same charge in a fixed electromagnetic background.