Quantum field theory 1, lecture 19
Classification of representaions
Representations of SU(2) are characterized by spin \(n\) of half integer or integer value. Accordingly, the representations of the Lorentz group can be classified as \((2n+1,2\tilde n+1)\). For example \[\begin{split} (1,1) & = \text{scalar or singlet},\\ (2,1) & = \text{left-handed spinor},\\ (1,2) & = \text{right-handed spinor},\\ (2,2) & = \text{vector}. \end{split}\] One can also construct tensor product representations as done for a single copy of SU\((2)\) and decompose the latter again in terms of irreducible representations.
Pauli spinor representation
In the non-relativistic description of spin-1/2 particles due to Pauli, the generators of rotation are \[J_i = \frac{1}{2} \sigma_i,\] where the hermitian Pauli matrices are given by \[\sigma_1 = \begin{pmatrix} && 1 \\ 1 && \end{pmatrix}, \quad\quad\quad \sigma_2 = \begin{pmatrix} && - i \\ i && \end{pmatrix}, \quad\quad\quad \sigma_3 = \begin{pmatrix} 1 && \\ && -1 \end{pmatrix},\] and fulfill the algebraic relation \[\sigma_i \sigma_j = \delta_{ij} \; \mathbb{1} + i\, \epsilon_{ijk}\; \sigma_k.\] In other words, the Pauli matrices provide a mapping between the space of rotations SO(3) and the space of unitary matrices SU(2). More concrete, an infinitesimal rotation \[\Lambda^i_{\;\;j} = \delta^i_{\;\; j} + \delta\omega^i_{\;\;j},\] corresponds to \[L(\Lambda)=\mathbb{1} + \frac{i}{4} \delta\omega_{ij} \, \epsilon_{ijk} \, \sigma_k.\] By exponentiating this one obtains the mapping. Note, however, that the group SU(2) covers SO(3) twice in the sense that a rotation by 360 degrees corresponds to \(L(\Lambda) = - \mathbb{1}\).
Left handed spinor representations
We now construct first the left handed spinor representation of the Lorentz group by using that it agrees with the Pauli representation for normal (spatial) rotations. When acting on the left-handed representation (2,1), the generator \(\tilde N_j\) vanishes. Since \(J_j = N_j + \tilde N_j\) and \(K_j = i (N_j - \tilde N_j)\) one has \[N_j = J_j = - i K_j = \frac{1}{2} \sigma_j, \quad \quad \tilde N_j=0.\] Using \(\eqref{eq:defJ}\) and \(\eqref{eq:defK}\) this yields for the left handed spinor representation \[\begin{split} (M_L^{jk}) & = \epsilon_{jkl} N_l = \frac{1}{2} \epsilon_{jkl} \, \sigma_l,\\ (M_L^{j0}) & = i N_j = i \frac{1}{2} \sigma_j. \end{split} \label{eq:LeftLorentzRep}\] As the name suggests, this representation acts in the space of left-handed spinors which are two-components entities, for example \[\psi_L=\begin{pmatrix} \psi_{1} \\ \psi_{2} \end{pmatrix}.\] We also use a notation with explicit indices \(\psi_a\) with \(a=1,2\). The infinitesimal transformation in \(\eqref{eq:LorentzTransformInfinitesimal}\) reads with the matrices \(\eqref{eq:LeftLorentzRep}\) \[\delta \psi_a = \frac{i}{2} \delta \omega_{\mu\nu} (M_L^{\mu\nu})_a^{\;\;b} \; \psi_b. \label{eq:LorentzTransforLeftDownSpinor}\]
Map to special linear group SL\((2, \mathbb{C})\)
From the concrete matrix representation we find that \[\frac{i}{2} \delta\omega_{\mu\nu} (M_L^{\mu\nu})_a^{\;\;b} = \frac{i}{2} (\delta a_k + i \delta b_k) (\sigma_k)_a^{\;\;b}\] with some coefficients \(\delta a_l\) and \(\delta b_l\). The right hand side is a general, complex but traceless \(2\times 2\) matrix and therefore an element of the Lie algebra of SL\((2,\mathbb{C})\). Accordingly there is a map from the Lie algebra of the Lorentz group SO\((1,3,\mathbb{R})\) to the Lie algebra of SL\((2, \mathbb{C})\). In fact, there is an isomorphism between groups SL\((2,\mathbb{C})/Z_2 = \text{SO}(1,3,\mathbb{R})\). (Excercise: Construct the corresponding map between group elements explicitely.)
Tensor representations
One can also construct spinors in tensor product representations. They have several indices and transform accordingly. For example, a spinor with two left-handed indices transforms according to \[\delta \chi_{ab} = \frac{i}{2} \delta\omega_{\mu\nu} \left[ (M^{\mu\nu}_L)_a^{\;\;c} \delta_b^{\;\;d} + \delta_a^{\;\;c} (M^{\mu\nu}_L)_b^{\;\;d} \right] \chi_{cd}.\] One can decompose such a spinor with two indices into an anti-symmetric and a symmetric part, corresponding to the decomposition \[(2,1) \otimes (2,1) = (1,1)_A \oplus (3,1)_S. \label{eq:LetfSpinorTensorProduct}\] This is a decomposition into a spin singlet and a spin triplet or spin one representation.
Invariant symbol in left-handed singlet
From \(\eqref{eq:LetfSpinorTensorProduct}\) it follows that there must be a Lorentz-singlet with two left-handed spinor indices and it has to be anti-symmetric. The corresponding invariant symbol can be taken as \(\varepsilon_{ab}\) with components \(\varepsilon_{21}=1\), \(\varepsilon_{12}=-1\) and \(\varepsilon_{11}=\varepsilon_{22}=0\). Indeed one finds from the concrete presentation that \[(M_L^{\mu\nu})_a^{\;\;c} \varepsilon_{cb} + (M_L^{\mu\nu})_b^{\;\;c} \varepsilon_{ac} = 0. \label{eq:MLEpsilon}\] This is essentialy due to \(\sigma_j \sigma_2 + \sigma_2 \sigma_j^T = 0\) for \(j=1,2,3\) as can be checked easily. For clarity the non-vanishing components are \[\varepsilon^{12} = - \varepsilon^{21} = \varepsilon_{21} = - \varepsilon_{12} = 1. \label{eq:varepsilonComponents}\] As an invariant symbol, \(\varepsilon_{ab}\) plays for SU\((2)\) a similar role as the metric \(\eta_{\mu\nu}\) for the Lorentz group. It is natural to use \(\varepsilon_{ab}\) and its inverse \(\varepsilon^{ab}\) to pull the indices \(a,b,c\) up and down. We can write for example a left-handed spinor with upper spinor index \[\psi^a = \varepsilon^{ab}\psi_b = - \varepsilon^{ba}\psi_b = - \psi_b \varepsilon^{ba} = \psi_b \varepsilon^{ab}.\] (We wrote different equivalent expressions to show that care is needed here with minus signs.) From eq. \(\eqref{eq:MLEpsilon}\) it follows also that \[(M_L^{\mu\nu})_{a}^{\;\;b} = (M_L^{\mu\nu})^{b}_{\;\;a},\] as well as \[(M_L^{\mu\nu})_{ab} = (M_L^{\mu\nu})_{ba},\] so that \[\varepsilon^{ab} (M_L^{\mu\nu})_{ab} = (M_L^{\mu\nu})_a^{\;\;a}=0.\] The symbol \(\delta^a_b\) is also invariant when spinors with upper left-handed indices have the Lorentz-transformation behavior \[\delta \psi^a = - \frac{i}{2} \delta \omega_{\mu\nu} \psi^b (M_L^{\mu\nu})_b^{\;\;a} = - \frac{i}{2} \delta \omega_{\mu\nu} (M_L^{\mu\nu})^a_{\;\;b} \psi^b.\] This implies also that upper and lower indices can be contracted, for example \[\psi^a \chi_a = \varepsilon^{ab}\psi_b \chi_a = - \varepsilon^{ba} \psi_b \chi_a = - \psi_b \chi^b,\] is invariant. (Again we wrote several equivalent expressions to show that care is needed with signs.)
Right handed spinor representation
Similarly one finds for the right-handed spinor representation (1,2) using \[N_j = 0, \quad \quad \tilde N_j = J_j = i K_j = \frac{1}{2} \sigma_j,\] the relations \[\begin{split} (M_R^{jk}) & = \epsilon_{jkl} \tilde N_l = \frac{1}{2} \epsilon_{jkl} \, \sigma_l\\ (M_R^{j0}) & = - i \tilde N_j = - i \frac{1}{2} \sigma_j. \end{split} \label{eq:RightLorentzRep}\] The representation \(\eqref{eq:RightLorentzRep}\) acts in the space of right handed spinors, for example \[\bar\psi = \begin{pmatrix} \bar\psi^{1} \\ \bar\psi^{2} \end{pmatrix}.\] For right handed spinors we will also use a notation with an explicit index that has a dot in order to distinguish it from a left-handed index, i. e. \(\bar\psi^{\dot a}\) with \(\dot a = 1,2\) denotes a right-handed spinor. The infinitesimal transformation in \(\eqref{eq:LorentzTransformInfinitesimal}\) reads with the matrices in \(\eqref{eq:RightLorentzRep}\) \[\delta \bar\psi^{\dot a} = \frac{i}{2} \delta \omega_{\mu\nu} (M_R^{\mu\nu})^{\dot a}_{\;\;\dot b} \; \bar\psi^{\dot b}.\]
Invariant symbol in right-handed singlet
In a completely analogous way the relation \[(1,2) \times (1,2) = (1,1)_A + (1,3)_S\] implies that there is a Lorentz singlet with two right-handed spinor indices. The corresponding symbol can be taken as \(\varepsilon^{\dot a \dot b}\), with inverse \(\varepsilon_{\dot a \dot b}\), with components as in \(\eqref{eq:varepsilonComponents}\). This symbol is used to lower and raise right-handed indices. Spinors with lower right handed index transform under Lorentz-transformations as \[\delta \bar\psi_{\dot a} = - \frac{i}{2} \delta\omega_{\mu\nu}\bar\psi_{\dot b} (M_R^{\mu\nu})^{\dot b}_{\;\;\dot a} = - \frac{i}{2} \delta\omega_{\mu\nu} (M_R^{\mu\nu})^{\;\;\dot b}_{\dot a} \bar\psi_{\dot b}. \label{eq:LorentzTransfoRightLower}\]
Invariant symbols for vectors
Consider now an object with a left-handed and a right-handed index. It is in the representation \((2,2)\) which should also contain the vector. There is therefore an invariant symbol which can be chosen as \[(\sigma^\mu)_{a\dot a} = (\mathbb{1}, \boldsymbol{\sigma}),\] and similarly \[(\bar \sigma^\mu)^{\dot a a} = (\mathbb{1}, - \boldsymbol{\sigma}).\] These symbols are invariant in the sense that they get mapped to themselves when all indices are transformed appropriately, e. g. \[\Lambda^\mu_{\;\;\nu} L_a^{\;\;b} R_{\dot a}^{\;\;\dot b} (\sigma^\nu)_{b \dot b} = (\sigma^\mu)_{a\dot a}.\] Here \(L_a^{\;\;b}\) and \(R_{\dot a}^{\;\;\dot b}\) are finite Lorentz transformation matrices in appropriate representations for left handed and right handed spinors respectively.
It turns out that the matrices for infinitesimal Lorentz transformations can be written as \[\begin{split} (M_L^{\mu\nu})_a^{\;\;b} & = \frac{i}{4} (\sigma^\mu \bar \sigma^\nu - \sigma^\nu \bar \sigma^\mu)_a^{\;\;b},\\ (M_R^{\mu\nu})^{\dot a}_{\;\;\dot b} & = \frac{i}{4} (\bar \sigma^\mu \sigma^\nu - \bar \sigma^\nu \sigma^\mu)^{\dot a}_{\;\;\dot b}. \end{split}\] Some useful identities are \[\begin{split} (\sigma^\mu)_{a \dot a} (\sigma_\mu)_{b \dot b} & = -2 \, \varepsilon_{ab} \varepsilon_{\dot a \dot b}\, ,\\ % (\bar \sigma^\mu)^{\dot a a} (\bar\sigma_\mu)^{\dot b b} & = -2 \varepsilon^{ab} \varepsilon^{\dot a \dot b},\\ % \varepsilon^{ab} \varepsilon^{\dot a \dot b} (\sigma^\mu)_{a\dot a} (\sigma^\nu)_{b\dot b} & = - 2 \,\eta^{\mu\nu},\\ % (\bar \sigma^\mu)^{\dot a a} & = \varepsilon^{ab} \varepsilon^{\dot a \dot b} (\sigma^\mu)_{b\dot b},\\ % (\sigma^\mu\bar \sigma^\nu+\sigma^\nu \bar \sigma^\mu)_a^{\;\;b} & = - 2 \; \eta^{\mu\nu} \delta_a^b,\\ % \text{Tr} (\sigma^\mu \bar \sigma^\nu) & = \text{Tr} ( \bar \sigma^\mu \sigma^\nu) = -2\, \eta^{\mu\nu},\\ % \bar \sigma^\mu \sigma^\nu \bar \sigma_\mu & = 2\, \bar \sigma^\nu,\\ % \sigma^\mu \bar \sigma^\nu \sigma_\mu & = 2 \, \sigma^\nu. \end{split}\] We leave it as an excercise to prove these.
Complex conjugation
The matrices \(\eqref{eq:LeftLorentzRep}\) and \(\eqref{eq:RightLorentzRep}\) are hermitian conjugate of each other, i. e. \[(M_L^{\mu\nu})^\dagger = M_R^{\mu\nu}, \quad\quad (M_R^{\mu\nu})^\dagger = M_L^{\mu\nu}. \label{eq:hermitianConjugationMinkowski}\] The hermitian conjugate of the Lorentz transformation \(\eqref{eq:LorentzTransforLeftDownSpinor}\) is given by \[\left[ \delta \psi_a \right]^\dagger = -\frac{i}{2} \delta\omega_{\mu\nu}^* \left[ \psi_b \right]^\dagger \underbrace{\left[ (M_L^{\mu\nu})_a^{\;\;b} \right]^\dagger}_{=(M_R^{\mu\nu})^{\dot b}_{\;\;\dot a}}. \label{eq:hermConj}\] For \(\delta \omega_{\mu\nu}\in \mathbb{R}\) this is of the same form as eq. \(\eqref{eq:LorentzTransfoRightLower}\). In Minkowski space it is therefore consistent to take \(\psi^\dagger\) to be a right-handed spinor with lower dotted index, we write \[\left[ \psi_a \right]^\dagger = (\psi^\dagger)_{\dot a}, \quad\quad \quad \text{(Minkowski)}\] and in an analogous way one finds that it is consistent to write \[\left[ \bar\psi^{\dot a} \right]^\dagger = (\bar\psi^\dagger)^{a}, \quad\quad \quad \text{(Minkowski)}\] which is a left-handed spinor. So far we have considered Minkowski space only. In Euclidean space or for more general complex \(\delta \omega_{\mu\nu}\) the hermitian conjugation is more complicated. For complex \(\delta \omega_{\mu\nu}\) eq. \(\eqref{eq:hermConj}\) constitutes a transformation behavior that is not of any already discussed type. For a consistent analytic continuation it is actually necessay to have all fields transforming such that the infinitesimal transformation law involves only \(\delta \omega_{\mu\nu}\) (and not \(\delta\omega_{\mu\nu}^*\)).
Dirac spinors in chiral basis
Dirac spinors are composed of a left handed and a right handed spinor. In the chiral basis they read \[\Psi = \begin{pmatrix} \psi_a \\ \bar\xi^{\dot a} \end{pmatrix}, \quad\quad\quad \bar \Psi = \begin{pmatrix} \xi^a, \bar \psi_{\dot a} \end{pmatrix}. \label{eq:DiracSpinor}\] Note that, as the notation suggests, \(\xi^a\) transforms as a left handed spinor and \(\bar \psi_{\dot a}\) as a right-handed one. One should see \(\Psi\) and the Dirac conjugate \(\bar \Psi\) as independent fields and they are in any case represented as independent Grassmann variables. In Minkowski space one can identify \[\bar \Psi = \Psi^\dagger \beta, \quad\quad\quad \Psi = \beta^{-1}\bar\Psi^\dagger \quad\quad\quad \text{(Minkowski space)}.\] with \[\beta = \begin{pmatrix} && \delta^{\dot a}_{\;\;\dot b} \\ \delta_a^{\;\;b} && \end{pmatrix}, \quad \quad\quad \beta^{-1} = \begin{pmatrix} && \delta_a^{\;\;b} \\ \delta^{\dot a}_{\;\;\dot b} && \end{pmatrix}.\] This is useful to ckeck that actions are hermitean or real and lead to unitary time evolution.
Clifford algebra
The gamma matrices are in this representation given by1 \[\gamma^\mu = \begin{pmatrix} && -i (\sigma^\mu)_{a\dot b} \\ -i (\bar \sigma^\mu)^{\dot a b} && \end{pmatrix}. \label{eq:GammaMatrices}\] They fulfill an anti-commutation relation \[\{\gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu= 2 \eta^{\mu\nu}. \label{eq:CliffordRel}\] An anti-commutation relation like \(\eqref{eq:CliffordRel}\) defines a Clifford algebra and can also be taken as the starting point for the coonstruction of spinor representations of the Lorentz group. Note that one may redefine the gamma matrices and spinors \[\gamma^\mu \to U \gamma^\mu U^\dagger, \quad\quad\quad \Psi \to U \Psi,\] to obtain another representation of the Clifford algebra that works equally well.
Antisymmetric matrices
Define the commutator of gamma-matrices as2 \[\sigma^{\mu\nu} = - \frac{i}{2} \left[ \gamma^\mu, \gamma^\nu \right] = \begin{pmatrix} \frac{i}{2}(\sigma^\mu\bar\sigma^\nu- \sigma^\nu\bar\sigma^\mu)_a^{\;\;b} && \\ && \frac{i}{2} (\bar \sigma^\mu \sigma^\nu - \bar \sigma^\nu \sigma^\mu)^{\dot a}_{\;\;\dot b} \end{pmatrix}.\] It is useful to define also \[\begin{split} (\tau^{\mu\nu})_a^{\;\;b} = \frac{i}{2}(\sigma^\mu \bar \sigma^\nu - \sigma^\nu \bar \sigma^\mu)_a^{\;\;b},\\ (\bar \tau^{\mu\nu})^{\dot a}_{\;\;\dot b} = \frac{i}{2}(\bar \sigma^\mu \sigma^\nu - \bar \sigma^\nu \sigma^\mu)^{\dot a}_{\;\;\dot b}, \end{split}\] such that \[\sigma^{\mu\nu} = \begin{pmatrix} \tau^{\mu\nu} && \\ && \bar \tau^{\mu\nu} \end{pmatrix}.\] As matrix, \(\sigma^{\mu\nu}\) can be written in terms of Pauli matrices, \[\begin{split} \sigma^{ij} = & \epsilon^{ijk} \begin{pmatrix} \sigma^k && \\ && \sigma^k \end{pmatrix},\\ \sigma^{j0} = & - \sigma^{0j} = \begin{pmatrix} i \sigma^j && \\ && -i \sigma^j \end{pmatrix}. \end{split}\] Commutation relations with gamma matrices are \[\begin{split} \frac{1}{2}\left[ \sigma^{\mu\nu}, \gamma^\rho \right] & = - \frac{i}{4} \left[ \left[ \gamma^\mu, \gamma^\nu \right], \gamma^\rho \right]\\ & = - \frac{i}{4} \left( \left\{ \gamma^\mu, \left\{ \gamma^\nu, \gamma^\rho \right\} \right\} - \left\{ \gamma^\nu, \left\{ \gamma^\mu, \gamma^\rho \right\} \right\} \right)\\ & = - i \left( \gamma^\mu \eta^{\nu \rho} -\gamma^\nu \eta^{\mu\rho} \right). \end{split}\] This can be understood as a kind of adjoint representation of the Lie algebra of the Lorentz group.
One can write the Lie algebra generators acting on Dirac spinors directly as \[M^{\mu\nu} = \frac{1}{2} \sigma^{\mu\nu}.\] This definition works also in other representations of the Clifford algebra because only the defining anticommutation relation \(\eqref{eq:CliffordRel}\) has been used. In other words, we could have started with some representation of \(\eqref{eq:CliffordRel}\) and would have obtained automatically a spinor representation of the Lorentz group. This would have been a reducible representation though, because it involves left-handed and right-handed spinors simultaneously. The representation can be decmposed again into its parts with gamma five, which we introduce now.
Gamma five
The matrix \(\gamma_5\) is defined as3 \[\begin{split} \gamma_5 & = - i \gamma^0 \gamma^1 \gamma^2 \gamma^3\\ & = \frac{i}{4!} \epsilon_{\mu\nu\rho\sigma} \gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma. \end{split}\] The four-dimensional Levi-Civita symbol is completely anti-symmetric with \(\epsilon_{0123} = -1\). In chiral representation one has \[\gamma^5 = \begin{pmatrix} \mathbb{1} && \\ && - \mathbb{1} \end{pmatrix} = \begin{pmatrix} \delta_a^{\;\;b} && \\ && - \delta^{\dot a}_{\;\;\dot b} \end{pmatrix} .\] The projectors to the left- and right-handed spinors are \[P_L = \frac{1}{2} (\mathbb{1} + \gamma_5) = \begin{pmatrix} \delta_a^{\;\;b} && 0 \\ 0 && 0 \end{pmatrix},\] and \[P_R = \frac{1}{2} (\mathbb{1} - \gamma_5) = \begin{pmatrix} 0 && 0 \\ 0 && \delta^{\dot a}_{\;\;\dot b} \end{pmatrix}.\] They can be used to project Dirac spinors with four components to Weyl spinors with only two nonvanishing components.
Due to the definitions one has \[\begin{split} \left\{ \gamma_5, \gamma^\mu \right\} & = 0\\ \left[\gamma_5, \sigma^{\mu\nu}\right] & = 0\\ (\gamma_5)^2 & = \mathbb{1}. \end{split}\]
This is the convention of Weinberg and Wetterich. There is an alternative definition of the gamma matrices often used in the literature (for example Peskin & Schroeder, Srednicki, Elvang & Huang) where one has \[\gamma^\mu = \begin{pmatrix} && (\sigma^\mu)_{a\dot a} \\ (\bar \sigma^\mu)^{\dot a a} && \end{pmatrix}, \quad \quad \{\gamma^\mu, \gamma^\nu \} = - 2 \, \eta^{\mu\nu}.\]↩︎
These are the conventions of Peskin & Schroeder. Weinberg uses \({\cal J}^{\mu\nu} = \frac{1}{2} \sigma^{\mu\nu}\). Wetterich uses the opposite sign↩︎
These is the convention of Weinberg, Wetterich. Peskin & Schroeder, Srednicki define \(\gamma_5\) with opposite sign↩︎