Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 301 / 2

The gamma matrices represent the spacetime Clifford algebra. With the Minkowski metric convention above,The braces denote the anticommutator. Consequently $({\gamma }^0{)}^2=I_4$, $({\gamma }^i{)}^2=-I_4$, and distinct gamma matrices anticommute. The identity matrix is essential: the right side is a matrix equality, not a scalar equality. These signs agree with the explicit Dirac representation of the gamma matrices in the next part.

Write every displayed block identity as $I_2$. Direct block multiplication in the Dirac representation of the gamma matrices givesThus their mixed anticommutator is zero. For two spatial indices,Using the Pauli matrix multiplication law givesThese exhaust all time-time, time-space and space-space cases of the Clifford algebra. No special choice of individual Pauli matrices beyond their stated anticommutator is needed.

A Dirac-to-Weyl gamma-matrix similarity transform with the required signs isIt is a unitary matrix, with $U^{-1}=U^{†}=2^{-1/2}\left(\begin{array}{cc}{I}_2& I_2\\ -I_2& I_2\end{array}\right)$. Multiplication givesThe matrix $U$ commutes with every ${\tilde{\gamma }}^i$: their common block structure uses the same antisymmetric $2\times 2$ block matrix, and the scalar blocks of $U$ commute with each Pauli matrix. Hence $U{\tilde{\gamma }}^i{U}^{-1}={\tilde{\gamma }}^i$, exactly the spatial matrices specified in the Weyl representation of the gamma matrices. A similarity transformation preserves every Clifford algebra relation, so no new consistency check is needed for the transformed representation.

Use the Clifford algebra to write ${\gamma }^{\nu }{\gamma }^{\mu }=2{\eta }^{\mu \nu }{I}_4-{\gamma }^{\mu }{\gamma }^{\nu }$. The Lorentz generator from gamma-matrix commutators is thereforeThis also checks the diagonal case $\mu =\nu$, where $S^{\mu \mu }=0$ because $({\gamma }^{\mu }{)}^2={\eta }^{\mu \mu }{I}_4$. The metric term in the printed expression is understood to multiply the identity matrix. The generators are antisymmetric in their two spacetime indices.

The commutator with the scalar metric term vanishes. Reorder the third gamma matrix using the Clifford algebra twice:Subtract the last term and multiply by $1/2$ to obtainThis is the infinitesimal vector transformation law implemented by the Spinor representation of the Lorentz group. The first contraction is ${\eta }^{\nu \rho }$ as printed in the PDF; the TeX conversion's ${\eta }^{\rho \rho }$ is erroneous and would fail, for example, when $\mu =0,\nu =1,\rho =2$: the actual commutator is zero.

Apply the commutator derivation identity $[S,AB]=[S,A]B+A[S,B]$ and the preceding gamma matrix transformation identity. Keeping the matrix order unchanged givesThusThe order of the gamma matrices matters, since they generally do not commute. The PDF has the four ordinary metric contractions shown here; the conversion's stray powers such as ${\eta }^{\nu }{\rho }^{\sigma }$ are not mathematical expressions to retain.

Since $S^{\rho \sigma }=\frac{1}{4}[{\gamma }^{\rho },{\gamma }^{\sigma }]$, apply the commutator derivation identity to this commutator and insert part (ii):This is exactly the printed Lorentz algebra relation, including its final two signs; their alternative appearance in other formulas comes from $S^{\rho \mu }=-S^{\mu \rho }$. Therefore the matrices S form a representation of the Lorentz algebra. Their definition contains no extra factor of $i$, so the spatial rotation generators below are represented by skew-Hermitian matrices. Adding a factor of $i$ would change the convention for the Lie algebra structure constants and cannot be done silently.

For the stated matrix exponential parametrization in the connected proper orthochronous Lorentz group, use the corresponding Spinor representation of the Lorentz group:The spacetime argument is inverse-transformed because this is an active transformation of the field at a fixed coordinate $x$. Equivalently, with $x^{\prime }=\Lambda x$, ${\psi }^{\prime }(x^{\prime })=D(\Lambda )\psi (x)$. The identity $[S^{\mu \nu },{\gamma }^{\rho }]={\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu }$ integrates to $D^{-1}{\gamma }^{\mu }D={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$, which ensures covariance of the Dirac equation.

Strictly, $D$ depends on a lift to the Spin group, not just on the final Lorentz transformation: two lifts differ by sign. The specified generator exponential chooses a lift along its continuous path from the identity. This qualification is essential for the full-rotation comparison in part (g).

In either the displayed Dirac representation of the gamma matrices or the displayed Weyl representation of the gamma matrices, the spatial blocks giveThe Pauli matrix commutator identity $[{\sigma }^j,{\sigma }^k]=2i{ϵ}^{jkl}{\sigma }^l$ therefore yieldsThe Levi-Civita symbol is normalized by ${ϵ}^{123}=1$. The constant is $-i/2$, consistent with rotation generators represented by skew-Hermitian matrices under the chosen Lorentz algebra convention. For $j=k$ both sides vanish automatically.

Insert ${\Omega }_{jk}=-{ϵ}_{jkl}{\varphi }^l$ and the result for $S^{jk}$ into the Spinor representation of the Lorentz group. The contraction of two Levi-Civita symbols ${\sum }_{j,k}{ϵ}_{jkl}{ϵ}^{jkm}=2{\delta }_l{}^m$ givesThe matrix exponential of this block diagonal matrix is the stated $O$, so ${\psi }^{\prime }(x)=O\psi ({\Lambda }^{-1}x)$. The plus sign in its exponent follows from both minus signs above; it is tied to the rotation-parameter convention in the PDF.

For a rotation through $2\pi$ about $z$, the Pauli matrix has eigenvalues $\pm 1$, hence $e^{i\pi {\sigma }^3}=-I_2$. The spacetime rotation is the identity, whileThis is the spinor sign under a full spatial rotation. A vector returns to itself already at $2\pi$. The comparison concerns the exponentiated Lorentz group representations: a Lie algebra by itself has no distinguished element called a $2\pi$ rotation. The sign records the nontrivial lift of this rotation loop to the Spin group.