# Quantum field theory 1, lecture 19

Spinor representations of the Lorentz group. As a final building block for the construction of quantum electrodynamics (QED) or quantum chromodynamics (QCD) for the strong interactions, or the standard model (SM) of particle physics which unifies the weak and electromagnetic interactions, we need a field for fermions and its transformation with respect to the Lorentz group. Electrons have half-integer spin. With respect to the rotation group electrons transform according to a two-dimensional representation, the spinor representation. Before generalising to the Lorentz group, we first investigate the spinor representation of the rotation group $$SO(3)$$, which is a subgroup of the Lorentz group. For nonrelativistic electrons this subgroup is all that matters.

The two-dimensional spinor representation of the rotation group involves two complex fields $$\chi _1(x)$$ and $$\chi _2(x)$$, that we order in a two-component complex vector field: \begin{equation*} \chi (x) = \begin{pmatrix} \chi _1(x) \\ \chi _2(x) \end{pmatrix} =\begin{pmatrix} \chi _1 \\ \chi _2 \end{pmatrix}. \end{equation*} The $$SO(3)$$-rotations act on this field as \begin{equation*} \delta \chi = i\epsilon _z T_z \chi + \delta '\chi , \quad \delta '\chi (x) = x^\rho \delta \Lambda _\rho ^{~\sigma }\partial _\sigma \chi (x). \end{equation*} We will omit $$\delta '$$ in the notation from now on since this universal contribution from the change of coordinates is the same for all fields. It is implicitly added if we transform fields. The spinor representation of $$SO(3)$$ is two-dimensional. The three $$2\times 2$$ matrices $$T_z$$ are given by the Pauli matrices, \begin{equation*} T_z = \tfrac{1}{2} \tau _z, \quad z= 1,2,3. \end{equation*} The generators $$T_z$$ of the two-dimensional representations are identical to the spin-matrices introduced earlier. We use here a common symbol $$T$$ for generators in arbitrary representations. Since the Pauli matrix $$\tau _2$$ is purely imaginary, the two component field $$\chi (x)$$ has to be complex. No real two-component representation of $$SO(3)$$ exists. The fermion fields are Grassmann variables. This is not relevant for symmetry transformations.

For a relativistic quantum field theory for electrons or protons, neutrons or neutrinos we have to answer two questions

• What are the spinor representations of the Lorentz group, i.e. what are the generators $$T_z$$ for $$z=1,\ldots ,6 ?$$
• Are there two-dimensional representations, i.e. are there six $$2\times 2$$ matrices that obey \begin{equation*} [T_x,T_y] = if_{xyz}T_z? \end{equation*}

Two-dimensional representations are the minimal setting, since already the rotation-subgroup requires a two-component complex field. We will see that neutrinos can be described by such a two-dimensional representation. For charged fermions we will find a four-dimensional complex representation, the so called ”Dirac spinors“.

A systematic construction of the spinor representations of the Lorentz group belongs to the mathematical field of representation theory. We do not attempt here to follow a systematic construction principle. We rather follow the results of Dirac, Pauli and Weyl, indicate the generators in the appropriate representations, and verify that they obey the commutation relations of the Lorentz group.

Dirac Spinors. Dirac spinors are four-dimensional complex fields \begin{equation*} \psi = \begin{pmatrix} \psi _1 \\ \psi _2 \\ \psi _3 \\ \psi _4 \end{pmatrix}. \end{equation*} It is convenient to label the six generators in the corresponding four-dimensional spinor representation by \begin{equation*} i\epsilon _z T_z = \tfrac{i}{2}\varepsilon _{\hat{\mu }\hat{\nu }}T^{\hat{\mu }\hat{\nu }}, \quad T^{\hat{\mu }\hat{\nu }} = -T^{\hat{\nu }\hat{\mu }},\quad \varepsilon _{\hat{\mu }\hat{\nu }} = -\varepsilon _{\hat{\nu }\hat{\mu }}. \end{equation*} With $$\hat{\mu }, \hat{\nu }=0,1,2,3$$ the six generators $$T^{\hat{\mu }\hat{\nu }}$$ are now labelled by $$\hat{\mu }\hat{\nu }$$ instead of $$z$$. The factor $$\tfrac{1}{2}$$ accounts for $$\tfrac{1}{2}(\varepsilon _{12} T^{12} + \varepsilon _{21} T^{21}) = \varepsilon _{12} T^{12}$$ etc. The pairs $$(\hat{\mu } \hat{\nu })$$ are just labels of the six generators and we have put the hats on $$\hat{\mu }$$ and $$\hat{\nu }$$ in order to avoid confusion: the matrices $$T^{\hat{\mu }\hat{\nu }}$$ are fixed $$4\times 4$$ matrices and Lorentz transformations do not act on $$(\hat{\mu } \hat{\nu })$$ as they do on fields. As an example $$T^{12}$$ is itself a $$4 \times 4$$ matrix with elements $$(T^{12})^\mu _{\;\nu }$$ or $$(T^{12})_\mu ^{\;\nu }$$. Thus $$\hat{\mu }\hat{\nu } =12$$ is just a convenient label for this matrix, which we could also have labelled equivalently by $$z=3$$.

Dirac matrices. The matrices $$T^{\hat{\mu }\hat{\nu }}$$ are obtained as the commutators of the Dirac matrices $$\gamma ^{\hat{\mu }}$$ $$T^{\hat{\mu }\hat{\nu }} = -\tfrac{i}{4}\left [\gamma ^{\hat{\mu }},\gamma ^{\hat{\nu }}\right ]. \label{eq:Tmunuhat}$$ The Dirac matrices $$\gamma ^{\hat{\mu }}$$ are four complex $$4\times 4$$ matrices, given explicitly by $$\gamma ^0 = \begin{pmatrix} 0 & -i\mathbf{1} \\ -i\mathbf{1} & 0 \end{pmatrix}, \quad \gamma ^k = \begin{pmatrix} 0 & -i\tau _k \\ i\tau _k & 0 \end{pmatrix}, \quad k=1,2,3 \quad , \label{eq:Diracmatrices}$$ with $$\tau _k,\; k=1,2,3$$ the Pauli matrices. In the following, we often omit the hat for $$\gamma ^\mu$$, or $$T^{\mu \nu }$$. We should always recall, however, that Lorentz transformations act only on fields, whereas the matrices $$\gamma ^\mu$$ are kept fixed.
An explicit computation of the generators \eqref{eq:Tmunuhat} from the Dirac matrices \eqref{eq:Diracmatrices} is a good exercise. One finds that they are of the block-diagonal form $$T^{\mu \nu }= \begin{pmatrix} T^{\mu \nu }_+ & 0 \\ 0 & T^{\mu \nu }_- \end{pmatrix} \label{eq:TmunuForm}$$ where the $$T^{\mu \nu }_{\pm }$$ are $$2\;\times \;2$$ matrices. The $$ij$$-components are rotations, \begin{equation*} T^{ij}_{+}= T^{ij}_{-} = \tfrac{1}{2}\varepsilon ^{ijk}\tau _k, \quad i,j,k \in \{1,2,3\}. \end{equation*} For a rotation around the $$z$$-axis $$(\varepsilon _{12} = -\varepsilon _{21} \equiv \varepsilon _3)$$, one has \begin{equation*} \begin{split} \varepsilon _3 T_3 &\equiv \tfrac{1}{2}(\varepsilon _{12}T^{12} + \varepsilon _{21}T^{21}),\\ &= \varepsilon _3 T^{12} = \varepsilon _3(\tfrac{1}{2} \varepsilon ^{123} \tau _3) = \varepsilon _3 \tfrac{\tau _3}{2}, \end{split} \end{equation*} confirming $$T_3 = \tfrac{\tau _3}{2}$$. If we denote \begin{equation*} \begin{pmatrix} \psi _1 \\ \psi _2 \end{pmatrix} = \psi _L, \quad \begin{pmatrix} \psi _3 \\ \psi _4 \end{pmatrix} = \psi _R, \quad \begin{pmatrix} \psi _L \\ \psi _R \end{pmatrix} = \psi , \end{equation*} then $$\psi _L$$ and $$\psi _R$$ transform as 2-component spinors with respect to rotations.

The generators $$T^{0k}$$ are boosts, \begin{equation*} T_{+}^{0k} = -T^{0k}_{-} = -\tfrac{i}{2} \tau _k. \end{equation*} The boost generators are not hermitian. They act on $$\psi _L$$ and $$\psi _R$$ with different signs.

Commutation relation of generators. The commutation relations can be computed as $$\left [T^{\mu \nu }, T^{\rho \sigma }\right ] = i\left (\eta ^{\mu \rho }T^{\nu \sigma } - \eta ^{\mu \sigma }T^{\nu \rho }+ \eta ^{\nu \sigma }T^{\mu \rho } - \eta ^{\nu \rho }T^{\mu \nu }\right ). \label{eq:commutationRelationTmunu}$$ These are indeed the commutation relations of the Lorentz group.
We can compare with the defining vector representation by the identification \begin{equation*} \begin{split} T_1 &= T^{23},\quad T_2 = T^{31}, \quad T_3 = T^{12},\\ T_4 &= T^{01},\quad T_5 = T^{02}, \quad T_6 = T^{03}. \end{split} \end{equation*} The contravariant vector representation obtains from the covariant vector representation by raising the first index and lowering the second, which changes the sign of $$T_4$$, $$T_5$$ and $$T_6$$. In this representation the generators are given explicitly by \begin{equation*} \left (T^{\hat{\mu }\hat{\nu }}\right )^{\mu }_{~ ~\nu } = -i\left (\eta ^{\hat{\mu }\mu }\delta ^{\hat{\nu }}_\nu - \eta ^{\hat{\nu }\mu }\delta ^{\hat{\mu }}_{\nu }\right ). \end{equation*} Examples are \begin{equation*} \begin{split} (T_1)^\mu _{~ ~\nu } &= (T^{23})^\mu _{~ ~\nu } = -i\left (\delta ^{2\mu }\delta ^3_\nu - \delta ^{3\mu }\delta ^2_\nu \right ) = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0& 0& i& 0 \end{pmatrix},\\ \text{or} \\ (T_4)^\mu _{~ ~\nu } &= (T^{01})^\mu _{~ ~\nu } = i\left (\delta ^{0\mu }\delta ^1_\nu + \delta ^{1\mu }\delta ^0_\nu \right ). \end{split} \end{equation*} In this representation the commutation relation \eqref{eq:commutationRelationTmunu} is easily established.

Weyl Spinors. The matrices $$T^{\mu \nu }$$ in eq. \eqref{eq:TmunuForm} are block-diagonal. This implies that Dirac spinors are reducible representations of the continuous Lorentz group. The irreducible representations are the two-dimensional representations $$\psi _L$$ and $$\psi _R$$, which do not mix under Lorentz transformations. Mathematically, there are two invariant subspaces, and the Dirac representation is therefore reducible. The Weyl representation is the two-dimensional irreducible representation (irrep). The decomposition of the Dirac representation can be formulated in a four-component notation \begin{equation*} \psi _L = \begin{pmatrix} \psi _1 \\ \psi _2 \\ 0 \\ 0 \end{pmatrix}, \quad \quad \quad \psi _R = \begin{pmatrix} 0 \\ 0 \\ \psi _3 \\ \psi _4 \end{pmatrix}. \end{equation*} With only a slight abuse of notation we employ the same notation also for the two-component Weyl spinors \begin{equation*} \psi _L = \begin{pmatrix} \psi _1 \\ \psi _2 \end{pmatrix}, \quad \quad \quad \psi _R = \begin{pmatrix} \psi _3 \\ \psi _4 \end{pmatrix}. \end{equation*} Weyl spinors describe neutrinos. The naming of left-handed Weyl spinors $$\psi _L$$ and right-handed Weyl spinors $$\psi _R$$ will be understood later. Neutrinos are left-handed Weyl spinors, while the complex conjugated field describes right-handed antineutrinos. Electrons, quarks and other charged fermions are described by Dirac spinors. This is related to the fact that the symmetries related to conserved charges cannot be implemented for Weyl spinors. Dirac spinors describe particles and their antiparticles, e.g. electrons and positrons. Also the parity transformation maps between $$\psi_L$$ and $$\psi_R$$.

Parity Transformation. The parity transformation is defined by \begin{equation*} \psi (x) \to \gamma ^0 \psi (Px),\quad Px=(x^0,-\vec{x}). \end{equation*} For the action on Weyl spinors we observe \begin{equation*} \gamma ^0 \begin{pmatrix} \psi _L \\ \psi _R \end{pmatrix} =-i \begin{pmatrix} \psi _R\\ \psi _L \end{pmatrix}, \end{equation*} and therefore \begin{equation*} (\psi ^{'})_L = -i \psi _R, \quad (\psi ^{'})_R = -i\psi _L. \end{equation*} Parity exchanges left-handed and right-handed Weyl spinors. This is indeed one of the reasons why one needs a left-handed and a right-handed Weyl spinor to describe electrons. Since neutrinos are described only by a left-handed Weyl spinor, they necessarily violate parity!

Projection Matrix. A projection from the Dirac to the Weyl representation can be defined in terms of the matrix $$\gamma ^5$$ by \begin{equation*} \psi _L = \tfrac{1}{2}(1+\gamma ^5)\psi , \end{equation*} \begin{equation*} \psi _R = \tfrac{1}{2}(1-\gamma ^5)\psi . \end{equation*} In our representation of Dirac matrices $$\gamma ^5$$ has the simple form $$\gamma ^5 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \label{eq:gamma}$$ where the $$1$$ represents a $$2\times 2$$-unit-matrix. One can check easily the relations $$[\gamma ^5, T^{\mu \nu }]=0,\qquad (\gamma ^5)^2 = 1. \label{eq:gammaRelations}$$ For the Dirac matrices \eqref{eq:Diracmatrices} one verifies by explicit computation the anticommutation relation $$\{\gamma ^\mu ,\gamma ^5\} = 0. \label{eq:gammaAnticommutationRelation}$$ This anticommutation can be used for a definition of $$\gamma ^5$$ in an arbitrary representation of the Dirac matrices. It implies the relations \eqref{eq:gammaRelations} . For a proof we use that $$\gamma ^5$$ commutes with a product of two Dirac matrices, \begin{equation*} \gamma ^5 T^{\mu \nu } = -\tfrac{i}{4} \gamma ^5(\gamma ^\mu \gamma ^\nu - \gamma ^\nu \gamma ^\mu ) = \tfrac{i}{4}(\gamma ^\mu \gamma ^5 \gamma ^\nu -\gamma ^\nu \gamma ^5 \gamma ^\mu )\\ =-\tfrac{i}{4}(\gamma ^\mu \gamma ^\nu - \gamma ^\nu \gamma ^\mu )\gamma ^5 = T^{\mu \nu }\gamma ^5. \end{equation*} The anticommutation relation \eqref{eq:gammaAnticommutationRelation} is obeyed by the definition \begin{equation*} \gamma ^5 = -i\gamma ^0\gamma ^1\gamma ^2\gamma ^3. \end{equation*} which yields eq. \eqref{eq:gamma} for our particular representation of the Dirac matrices. In our particular representation the projectors are very simple \begin{equation*} \frac{1+\gamma ^5}{2} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad \frac{1-\gamma ^5}{2}= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \end{equation*} or \begin{equation*} \gamma ^5 \psi _L = \psi _L, \quad \gamma ^5 \psi _R = -\psi _R. \end{equation*}

Dirac Matrices. The defining property for Dirac matrices is given by anticommutation relation \begin{equation*} \{\gamma ^\mu , \gamma ^\nu \} = 2\eta ^{\mu \nu }. \end{equation*} This is known as the Clifford algebra. From this relation one can derive all the commutator relations for the $$T^{\mu \nu }$$ and $$\gamma ^5$$. In particular, one has $$(\gamma ^k)^2$$ =1, k= 1,2,3 and $$(\gamma ^0)^2 = -1$$. Different books on quantum field theory will use different representations of the Clifford algebra. Different representations are related by a similarity transformation \begin{equation*} \gamma ^\mu \to \gamma ^{\prime \mu } = A \gamma ^\mu A^{-1}. \end{equation*} For any regular matrix $$A$$ this transformation does not change the anticommutator relations: \begin{equation*} \{\gamma '^{\mu }, \gamma '^{\nu }\} = A\{\gamma ^\mu , \gamma ^\nu \} A^{-1} = 2A\eta ^{\mu \nu } A^{-1} = \eta ^{\mu \nu }. \end{equation*}

### 11 Quantum electrodynamics

We are now ready to construct the action for quantum electrodynamics (QED). Charged fermions as electrons, muons or quarks are described by Grassmann variables in the Dirac representation of the Lorentz group. We start with free electrons, and add the interactions with photons subsequently.

#### 11.1 Invariant action for free electrons.

Kinetic term. We want to use the spinor representation discussed in the previous section to establish a Lorentz invariant action for fermions. We can construct a kinetic term with only one derivative: $$S= \int d^4x ~ \mathscr{L},\quad \quad \quad \mathscr{L}=i\bar{\psi }\gamma ^\mu \partial _\mu \psi =i \bar{\psi }_\alpha (\gamma ^\mu )_{\alpha \beta } \partial _\mu \psi _\beta . \label{eq:DiracAction}$$ As usual, $$\psi$$ denotes a column vector and $$\bar{\psi }$$ is a line vector, \begin{equation*} \psi = \begin{pmatrix} \psi _1 \\ \psi _2 \\ \psi _3 \\ \psi _4 \end{pmatrix}, \quad \quad \quad \bar{\psi }= \left (\bar{\psi }_1, \bar{\psi }_2, \bar{\psi }_3, \bar{\psi }_4 \right ). \end{equation*} The Dirac indices or spinor indices $$\alpha , \beta =1,2,3,4$$ should not be confused with Lorentz-indices $$\mu =0,1,2,3$$. For Weyl spinors we use only two spinor indices, and in other dimensions the dimension of the Dirac representation differs from $$d$$. The kinetic term for fermions involves only one derivative.

A priori, $$\psi _\alpha$$ and $$\bar{\psi }_\alpha$$ are independent Grassmann variables, and Grassmann variables are neither real nor complex numbers. In contrast to the kinetic term for scalars, which requires two time-derivatives, the formulation with only one time-derivative is closer to the formulation for non-relativistic particles for which we have derived the functional integral from the operator formalism. Under an infinitesimal Lorentz transformation, $$\psi$$ and $$\bar{\psi }$$ transform as $$\begin{split} \delta \psi &= \tfrac{i}{2} \epsilon _{\mu \nu } T^{\mu \nu }\psi , \\ \delta \bar{\psi } &= -\tfrac{i}{2}\epsilon _{\mu \nu }\bar{\psi }T^{\mu \nu }. \end{split} \label{eq:LorentzTrafoPsi}$$ One can introduce a complex structure in the Grassmann algebra by defining $$\psi ^{*}$$ through \begin{equation*} \bar{\psi } = \psi ^{\dagger }\gamma ^{0} = (\psi ^{*})^T \gamma ^0. \end{equation*}

This is the defining relation for $$\psi ^*$$ in terms of $$\bar{\psi }$$. One can check the consistency of complex conjugation with Lorentz transformations, \begin{equation*} \delta \psi ^* = -\tfrac{i}{2} \epsilon _{\mu \nu } T^{\mu \nu } \psi ^*,\quad \quad \quad \delta \bar{\psi } = (\delta \psi )^{\dagger } \gamma ^0 . \end{equation*} Having defined $$\psi ^*$$, one could define real and imaginary parts $$\psi _{\text{R}} = \tfrac{1}{2}(\psi + \psi ^*)$$ and $$\psi _{\text{I}} = -\tfrac{i}{2}(\psi -\psi ^{*})$$ and use those as independent Grassmann variables.

Transformation of Spinor Bilinears. We next have to verify the Lorentz-invariance of the Dirac action \eqref{eq:DiracAction}. For this purpose we compute the behavior of general bilinear forms of spinors under Lorentz transformations. It is sufficient to consider infinitesimal Lorentz transformations.

As a first relation we proof the invariance of $$\bar{\psi }\psi$$, \begin{equation*} \delta (\bar{\psi } \psi ) = 0. \end{equation*} Insertion of eq. \eqref{eq:LorentzTrafoPsi} yields directly \begin{equation*} \delta (\bar{\psi }\psi ) = \delta \bar{\psi }\psi + \bar{\psi }\delta \psi =-\tfrac{i}{2}(\bar{\psi } T^{\mu \nu }\psi - \bar{\psi } T^{\mu \nu } \psi ) = 0. \end{equation*} We recall that there is an additional contribution from the transformation of coordinates that we do not display explicitly. Thus $$\bar{\psi }\psi$$ transforms as a scalar field under Lorentz transformations.

Next we show that $$\bar{\psi } \gamma ^\mu \psi$$ transforms as a contravariant vector under Lorentz transformations. \begin{equation*} \delta (\bar{\psi } \gamma ^\mu \psi ) = \delta \Lambda ^\mu _{\;\nu }(\bar{\psi } \gamma ^\nu \psi ) = \epsilon ^\mu _{\;\nu }\bar{\psi }\gamma ^\nu \psi . \end{equation*}

This can be seen in three steps. First we note that \begin{equation*} \delta (\bar{\psi }\gamma ^\rho \psi ) = \delta \bar{\psi }\gamma ^\rho \psi + \bar{\psi }\gamma ^\rho \delta \psi = -\tfrac{i}{2} \epsilon _{\mu \nu }(\bar{\psi }T^{\mu \nu }\gamma ^\rho \psi - \bar{\psi }\gamma ^\rho T^{\mu \nu }\bar{\psi }) = -\tfrac{i}{2}\epsilon _{\mu \nu }\bar{\psi }[T^{\mu \nu },\gamma ^\rho ]\psi . \end{equation*} Second, we employ the identity \begin{equation*} \gamma ^\mu \gamma ^\nu \gamma ^\rho = \gamma ^\mu \{\gamma ^\nu , \gamma ^\rho \} - \gamma ^\mu \gamma ^\rho \gamma ^\nu = 2\eta ^{\nu \rho }\gamma ^\mu -\gamma ^\mu \gamma ^\rho \gamma ^\nu . \end{equation*} in order to establish the commutator \begin{equation*} \begin{split} [T^{\mu \nu },\gamma ^\rho ] &= -\tfrac{i}{4}(\gamma ^\mu \gamma ^\nu \gamma ^\rho - \gamma ^\nu \gamma ^\mu \gamma ^\rho -\gamma ^\rho \gamma ^\mu \gamma ^\nu +\gamma ^\rho \gamma ^\nu \gamma ^\mu )\\ &=-\tfrac{i}{4}(2\eta ^{\nu \rho }\gamma ^\mu -\gamma ^\mu \gamma ^\rho \gamma ^\nu -2\eta ^{\mu \rho }\gamma ^\nu + \gamma ^\nu \gamma ^\rho \gamma ^\mu -2\eta ^{\mu \rho }\gamma ^\nu \\ &\qquad + \gamma ^\mu \gamma ^\rho \gamma ^\nu + 2\eta ^{\nu \rho }\gamma ^\mu -\gamma ^\nu \gamma ^\rho \gamma ^\mu )\\ &=-i(\eta ^{\nu \rho }\gamma ^\mu -\eta ^{\mu \rho }\gamma ^\nu ). \end{split} \end{equation*}

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