Lorentz Transformations and Field Theory

SR and LT (Read the above for a recap of Special Relativity and Lorentz Transformations)

• Scalars $\varphi (x)\to {\varphi }^{\prime }(x^{\prime })\equiv \varphi (x)$ Similarly, ${\partial }_{\mu }\varphi (x)\to {\partial }_{\mu }^{\prime }{\varphi }^{\prime }(x^{\prime })={\Lambda }_{\mu }^{\nu }{\partial }_{\nu }{\varphi }^{\prime }(x^{\prime })\equiv {\Lambda }_{\mu }^{\nu }{\partial }_{\nu }\varphi (x)$ and therefore, \partial_\mu\phi(x)\partial^\mu\phi(x)=\partial_\mu\phi(x)\partial_\nu\phi(x)g^{\mu\nu}\rightarrow \Lambda^\rho_\mu\partial_\rho\phi'(x')\Lambda^\sigma_\nu\partial_\sigma\phi'(x')g^{\mu\nu}$$$$=\partial_\rho\phi(x)\partial_\sigma\phi(x)\Lambda^\rho_\mu\Lambda^\sigma_\nu g^{\mu\nu}=\partial_\rho\phi(x)\partial_\sigma\phi(x)g^{\rho\sigma} so that the Lagrangian transforms simply as a scalar, $\mathcal{L}(x)\to {\mathcal{L}}^{\prime }(x^{\prime })\equiv \mathcal{L}(x)$ and theKlein-Gorden equation is also invariant.

• Vectors A^\mu(x)\rightarrow A'^\mu(x')=\Lambda^\mu_\nu A^\nu(x)$$$$F^{\mu\nu}(x)\rightarrow F'^{\mu\nu}(x')=\Lambda^\mu_\rho\Lambda^\nu_\sigma F^{\rho\sigma}(x)

• Fermions ${\Psi }_a\to {\Psi }_a^{\prime }=\mathcal{D}(\Lambda {)}_{ab}{\Psi }_b\equiv ({\Lambda }_{\frac{1}{2}}{)}_{ab}{\Psi }_b$ where a,b = 1,…,4 denote the components of the Dirac Spinor and D(Λ) is 4x4 matrix in Dirac space. $({\Lambda }_{\frac{1}{2}}{)}_{ab}=e^{1-\frac{i}{2}{ϵ}_{\rho \sigma }{\hat{S}}^{\rho \sigma }}$ is the spinor representation of the LT D(Λ). How does the Dirac matrices transform? ${\gamma }_{ab}^{\mu }\to {\gamma }_{ab}^{\prime \mu }={\Lambda }_{\nu }^{\mu }{\gamma }_{ab}^{\nu }$ Now, from Dirac algebra, $\{{\gamma }^{\prime \mu },{\gamma }^{\prime \nu }\}=\{{\Lambda }_{\rho }^{\mu }{\gamma }^{\rho },{\Lambda }_{\sigma }^{\nu }{\gamma }^{\sigma }\}={\Lambda }_{\rho }^{\mu }{\Lambda }_{\sigma }^{\nu }\{{\gamma }^{\rho },{\gamma }^{\sigma }\}=2g^{\rho \sigma }{\Lambda }_{\rho }^{\mu }{\Lambda }_{\sigma }^{\nu }{\delta }_{ab}=2g^{\mu \nu }{\delta }_{ab}$

• Invariance of the Dirac equation $(i{\gamma }^{\mu }{\partial }_{\mu }-m)\Psi (x)=0$ Under a LT, (i\gamma^\mu\partial_\mu-m)\Psi(x)\rightarrow(i\gamma^\mu\Lambda^\nu_\mu\partial_\nu-m)\Psi'(x')=(i\gamma^\mu\Lambda^\nu_\mu\partial_\nu-m)\Lambda_\tfrac{1}{2}\Psi(x')$$$$=\Lambda_\tfrac{1}{2}\Lambda_\tfrac{1}{2}^{-1}(i\gamma^\mu\Lambda^\nu_\mu\partial_\nu-m)\Lambda_\tfrac{1}{2}\Psi(x')=\Lambda_\tfrac{1}{2}(i\underbrace{\Lambda_\tfrac{1}{2}^{-1}\gamma^\mu\Lambda_\tfrac{1}{2}}_{\Lambda^\mu_\sigma\gamma^\sigma}\Lambda^\nu_\mu\partial_\nu-m)\Psi(x')$$$$=\Lambda_\tfrac{1}{2}(i\Lambda_\sigma^\mu\gamma^\sigma\Lambda^\nu_\mu\partial_\nu-m)\Psi(x')=\Lambda_\tfrac{1}{2}(i\gamma^\sigma\partial_\sigma-m)\Psi(x')=0 In other words, the Dirac equation is Lorentz Invariant.

• What is Ŝ? $({\hat{S}}_{\rho \sigma }{)}_{ab}=\frac{i}{4}([{\gamma }^{\rho },{\gamma }^{\sigma }]{)}_{ab}$ Weyl basis for Dirac matrices

  In the Weyl Representation, \hat{S}^{0i}=\frac{i}{4}[\gamma^0,\gamma^i]=-\frac{i}{2}\begin{pmatrix}\sigma_i&0\\0&-\sigma_i\end{pmatrix}$$$$\hat{S}^{ij}=\frac{i}{4}[\gamma^i,\gamma^j]=\frac{1}{2}\epsilon^{ijk}\begin{pmatrix}\sigma_k&0\\0&\sigma_k\end{pmatrix} The LT is not unitary because although the rotation generators are Hermitian, the boost generators are not. Hence, we define $\bar{\Psi }={\Psi }^{†}{\gamma }^0$ where ${\hat{S}}^{ij}{\gamma }^0={\gamma }^0{\hat{S}}^{ij},{\hat{S}}^{0i}{\gamma }^0=-{\gamma }^0{\hat{S}}^{0i}$ such that $\bar{\Psi }\to {\Psi }^{†}(1+\frac{i}{2}{ϵ}_{\rho \sigma }({\hat{S}}^{\rho \sigma }{)}^{†}){\gamma }^0={\Psi }^{†}{\gamma }^0(1+\frac{i}{2}{ϵ}_{\rho \sigma }{\hat{S}}^{\rho \sigma })\equiv \bar{\Psi }{\Lambda }_{\frac{1}{2}}^{-1}$