%%============================================================================== \begin{lecture} \item Construction of the Dirac equation \item Free-particle solutions of the Dirac equation \end{lecture} %%============================================================================== %%============================================================================== \begin{problemset} \item Fock states and coherent states \item Free-particle solutions of the Dirac equation \item Review of the Lorentz group \end{problemset} %%============================================================================== %%============================================================================== \reference{PS:35--49} %%============================================================================== \chapter{The Dirac Field} \section{The Dirac Equation} \label{sec:dirac_equation} \imp{So far: Simplest relativistic field equation \therefore Klein-Gordon equation}\\ % \imp{Now: Second simplest relativistic field equation \therefore Dirac equation} \begin{lot} \item \Emph{Observation 1:} Lorentz symmetry of the KG equation:\\ % \imp{We view Lorentz transformations as \emph{active} transformations, mapping solutions to different solutions! This is equivalent to the \emph{passive} viewpoint where the coordinate system is transformed instead.} % \begin{lot} \item \consider Coordinate transformation: $x'=\Lambda x$ \& Field transformation: $\phi'(x')=\phi(x)$ \item \consider $\phi$ with $(\partial^2+m^2)\phi(x)=0$ for all $x$ \item \therefore $\phi'(x)=\phi(\Lambda^{-1}x)$ is a new solution:\\ % \imp{Use the chain rule in the first step twice:} % \begin{subalign} (g^{\mu\nu}\partial_\mu\partial_\nu+m^2)\phi'(x) &=[g^{\mu\nu}\tensor{(\Lambda^{-1})}{^\sigma_\mu}\partial_\sigma\, \tensor{(\Lambda^{-1})}{^\rho_\nu}\partial_\rho+m^2]\phi(\Lambda^{-1}x) \\ &\mimp{Use invariance of the metric} \\ &=(g^{\sigma\rho}\partial_\sigma\partial_\rho+m^2)\phi(\Lambda^{-1}x) \\ &=(\partial^2+m^2)\phi(\Lambda^{-1}x)\stackrel{\text{$\phi$ solution}}{=}0 \end{subalign} % \imp{Here $\partial_\sigma\phi(\Lambda^{-1}x)$ must be read as $\left.\partial_\sigma\phi(y)\right|_{y=\Lambda^{-1}x}$, i.e., we compute the derivative of the function $\phi$ with respect to its argument $y$ and then plug in the value $\Lambda^{-1}x$.} \end{lot} \item \Emph{Observation 2:} \consider Vector fields under rotations: $\vec \phi'(\vec x)=R\vec\phi(R^{-1}\vec x)$\\ % \therefore In general\note{, a field $\phi(x)\in\complex^n$ can transform under $n$-dimensional Lorentz transformations as} % \begin{align} \phi'_a(x)=M_{ab}(\Lambda)\phi_b(\Lambda^{-1}x) \qquad a=1,\dots,n \end{align} % where % \begin{align} M(\Lambda')M(\Lambda)\phi(\Lambda^{-1}\Lambda'^{-1}x)\stackrel{!}{=}M(\Lambda'\Lambda)\phi((\Lambda'\Lambda)^{-1}x) \end{align} % is a $n$-dimensional \Emph{representation} of the Lorentz group $\POLG$. \item We want a \Emph{first-order} relativistic field equation: % \begin{align} (\partial^\mu\partial_\mu+\const)\phi=0 \quad\Rightarrow\quad (i\blacksquare^\mu\partial_\mu+\const)\phi=0 \end{align} % \imp{The $i$ anticipates wave-like solutions for real $\blacksquare$.} \item Then \imp{(combine 1 \& 2)} % \begin{lot} \item \consider Coordinate transformation: $x'=\Lambda x$ \& Field transformation: $\phi'(x')=M(\Lambda)\phi(x)$ \item \consider $\phi$ with $(i\blacksquare^\mu\partial_\mu+\const)\phi(x)=0$ for all $x$ \item When is $\phi'(x)=M(\Lambda)\phi(\Lambda^{-1}x)$ is a new solution? % \begin{align} (i\blacksquare^\mu\partial_\mu+\const)\phi'(x) =[i\blacksquare^\mu\tensor{(\Lambda^{-1})}{^\nu_\mu}\partial_\nu+\const]\,M(\Lambda)\phi(\Lambda^{-1}x) \stackrel{!}{=}0 \end{align} % \imp{Multiply with $M^{-1}(\Lambda)$:} % \begin{align} \Leftrightarrow\quad [i\underbrace{M^{-1}(\Lambda)\blacksquare^\mu M(\Lambda)\tensor{(\Lambda^{-1})}{^\nu_\mu}}_{% \stackrel{!}{=}\blacksquare^\nu}\partial_\nu+\const]\, \phi(\Lambda^{-1}x) \stackrel{!}{=}0 \end{align} % \therefore $\blacksquare^\mu\equiv\gamma^\mu$ must be $n\times n$-matrices with % \begin{emphalign} M^{-1}(\Lambda)\gamma^\mu M(\Lambda)=\tensor{\Lambda}{^\mu_\nu}\gamma^\nu \label{eq:g_M} \end{emphalign} % \note{The $\gamma$-matrices ``translate'' the ``spinor''-representation $M(\Lambda)$ into the ``vector''-representation $\Lambda$ and vice versa.} \end{lot} \item \imp{How to find $\gamma^\mu$ and $M(\Lambda)$?} $\POLG$ is a \Emph{Lie group} \imp{(\seepset{3})}: % \begin{subalign} \Lambda &= \exp\left[-\frac{i}{2}\omega_{\alpha\beta}\J^{\alpha\beta}\right] \stackrel{\omega\ll 1}{\approx} \Id-\frac{i}{2}\omega_{\alpha\beta}\J^{\alpha\beta}\label{eq:vec_rep}\\ M(\Lambda) &= \exp\left[-\frac{i}{2}\omega_{\alpha\beta}S^{\alpha\beta}\right] \stackrel{\omega\ll 1}{\approx} \Id-\frac{i}{2}\omega_{\alpha\beta}S^{\alpha\beta}\label{eq:spin_rep} \end{subalign} % $\omega_{\alpha\beta}$ antisymmetric tensor \therefore 3 rotations (angles) + 3 boosts (rapidities) \note{It is $(\J^{\alpha\beta})_{\mu\nu}=i(\delta^\alpha_\mu\delta^\beta_\nu-\delta^\alpha_\nu\delta^\beta_\mu)$.} \imp{The $4\times 4$ matrices $\J^{\alpha\beta}$ generate the vector-representation $\Lambda$, $(\hlf,\hlf)$, the $n\times n$-matrices $S^{\alpha\beta}$ the spinor-representation $M(\Lambda)$, $(\hlf,0)\oplus(0,\hlf)$. The generators are antisymmetric in the spacetime indices.} \begin{itemize} \item Infinitesimal form of \cref{eq:g_M}: % \begin{align} \com{\gamma^\mu}{S^{\alpha\beta}}\nte\tensor{(\J^{\alpha\beta})}{^\mu_\nu}\gamma^\nu \nte i(g^{\alpha\mu}\gamma^\beta-g^{\beta\mu}\gamma^\alpha) \label{eq:g_M_L} \end{align} \item $\J^{\alpha\beta}$ \therefore \Emph{Lie-algebra} of Lorentz group \imp{($J=S,\J$, \seepset{3})} % \begin{align} \com{J^{\mu\nu}}{J^{\rho\sigma}}\nte i(g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma} -g^{\nu\sigma}J^{\mu\rho}+g^{\mu\sigma}J^{\nu\rho}) \label{eq:LA} \end{align} % \imp{The Lie algebra defines the structure of the Lie group by integration and is therefore the same for all representations.} \end{itemize} \item \Emph{Solution:} Dirac's trick: \consider $\gamma^\mu$ such that % \begin{emphalign} \acom{\gamma^\mu}{\gamma^\nu}=2 g^{\mu\nu}\,\Id_{n\times n} \quad\text{\termdef{Dirac algebra}} \end{emphalign} % \note{This is the 16-dimensional \emph{Clifford algebra} $\mathrm{C}\ell_{1,3}(\complex)$.}\\ % Then % \begin{emphalign} S^{\mu\nu}\equiv \frac{i}{4}\com{\gamma^\mu}{\gamma^\nu} \label{eq:Smunu} \end{emphalign} % satisfies the Lorentz algebra \cref{eq:LA} and \cref{eq:g_M_L}. \item \Emph{Representations:} \begin{itemize} \item At least $4$-dimensional\\ % \note{(think of the $\gamma^\mu$ as Majorana modes and construct ladder operators \therefore 2 modes)} \item All 4-dimensional representations are unitarily equivalent\\ % \note{(actually, they constitute the \emph{unique} irrep of the Dirac algebra which is 4-dimensional)} \item We use the \Emph{Weyl representation} \note{(sometimes called \emph{chiral representation})}: % \begin{emphalign} \gamma^0=\begin{pmatrix}0 & \Id \\ \Id & 0\end{pmatrix} \quad\text{and}\quad \gamma^i=\begin{pmatrix}0 & \sigma^i \\ -\sigma^i & 0\end{pmatrix} \end{emphalign} % \item Henceforth: $\Lambda_\hlf\equiv M(\Lambda)$\\ % \imp{Two ``copies'' of a spin-$\hlf$ projective representation.} \end{itemize} \item Setting $\const=-m$, we find: % \begin{emphalign} (i\g^\mu\pmu-m)\Psi=0 \quad\text{\termdef{Dirac equation}} \label{eq:DE} \end{emphalign} % \imp{$\Psi(x)$ is a \emph{bispinor}-field with values in $\complex^4=\complex^2\oplus\complex^2$.} \item The components of the Dirac spinor field satisfy the KG equation: % \begin{align} 0={\color{marked}(-i\gmu\pmu-m)}(i\gamma^\nu\partial_\nu-m)\Psi\nte (\partial^2+m^2)\Psi \label{eq:Dirac_KG} \end{align} % \imp{The Dirac differential operator is the ``square root'' of the Klein-Gordon differential operator. On the right hand side of \cref{eq:Dirac_KG} there is an identity $\Id_{4\times 4}$ that we omit.} \item \Emph{Dirac adjoint:} \imp{Goal: Lagrangian (which must be a Lorentz scalar).}\\ % \imp{\therefore How to form Lorentz scalars from spinors?} % \begin{lot} \item First try: $\Psi^\dag\Psi$ % \begin{align} \Psi'^\dag\Psi' =\Psi^\dag \underbrace{\Lambda_\hlf^\dag\Lambda_\hlf}_{\neq\Id}\Psi\neq \Psi^\dag\Psi \end{align} % \imp{$\Lambda_\hlf$ is not unitary because $S^{\mu\nu}$ is not Hermitian for boosts ($\mu=0$ and $\nu=1,2,3$).\\ % This is a consequence of the non-compactness of the Lorentz group due to boosts!} \item Define % \begin{emphalign} \ol \Psi=\Psi^\dag\gamma^0 \quad\text{\termdef{Dirac adjoint}} \end{emphalign} % \easytoshow $\ol\Psi'\Psi'=\ol\Psi\Lambda_\hlf^{-1}\Lambda_\hlf\Psi=\ol\Psi\Psi$ \limplies Lorentz scalar\\ % \imp{Use \cref{eq:Smunu} and \cref{eq:spin_rep} and the Dirac algebra to show this!} \end{lot} \item \Emph{Lagrangian:} % \begin{emphalign} \L_\mathrm{Dirac}=\ol\Psi(i\gmu\pmu-m)\Psi \end{emphalign} % \imp{\easytoshow Euler-Lagrange equations yield Dirac equation.} \end{lot} \begin{blocknote} \begin{itemize} \item Let $\sigma^\mu\equiv (1,\vec \sigma)^T$ and $\ol\sigma^\mu\equiv (1,-\vec \sigma)^T$ and $\g^\mu = \begin{pmatrix}0 & \sigma^\mu \\ \ol\sigma^\mu & 0 \end{pmatrix}$\\ % \therefore Dirac equation: % \begin{align} \begin{pmatrix}-m & i\sigma\partial \\ i\ol\sigma\partial & -m\end{pmatrix}\bispi{\psi_L}{\psi_R}=0 \end{align} \item $\psi_{L}$ and $\psi_{R}$ are called left- and right-handed \termdef{Weyl spinors} \item They do not mix under Lorentz transformations\\ % \note{They form the $(\hlf,0)$ and $(0,\hlf)$ projective irreps of the Lorentz group. Note that the reducibility of the $(\hlf,0)\oplus(0,\hlf)$ bispinor representation is manifest in the Weyl basis:} % \note{% \begin{subequations} \label{eq:generators_weyl} \begin{align} S^{0i}&=\frac{i}{4}\com{\gamma^0}{\gamma^i}= -\frac{i}{2}\begin{pmatrix}\sigma^i & 0 \\ 0 & -\sigma^i\end{pmatrix} \quad\text{(Boosts, anti-Hermitian)} \\ S^{ij}&=\frac{i}{4}\com{\gamma^i}{\gamma^j} =\frac{1}{2}\vep^{ijk}\begin{pmatrix}\sigma^k & 0 \\ 0 & \sigma^k\end{pmatrix} \quad\text{(Rotations, Hermitian)} \end{align} \end{subequations} } \item For $m=0$, the Dirac equation decouples into the \termdef{Weyl equations}: % \begin{align} i\ol\sigma\partial\,\psi_L=0 \quad\text{and}\quad i\sigma\partial\,\psi_R=0 \end{align} % \note{Solutions $\psi_R$ and $\psi_L$ are eigenstates of the helicity operator $h=\hat p\frac{\vec\sigma}{2}$ with $h=+\hlf$ called \emph{right-handed} and $h=-\hlf$ \emph{left-handed}. Here, $\hat p =\vec p/E$ is the normalized 3-momentum for a massless particle.} \end{itemize} \end{blocknote} \section{Free-Particle Solutions of the Dirac Equation} \note{Here we consider the Dirac equation as a wave equation for a \emph{single particle}, or, equivalently, the classical field equation of a complex bispinor field; what follows is therefore ``first-quantized'' quantum mechanics. We do this because we need the eigenfunctions of the Dirac differential operator to construct the field operators when we quantize the Dirac field (``second quantization'').} \imp{Detailed calculations: \seepset{3}.} \begin{lot} \item \cref{eq:DE} \limplies $(\partial^2+m^2)\Psi=0$ \imp{(Klein-Gordon equation)}, therefore % \begin{align} \Psi^\pm(x)=\psi^\pm(p)e^{\mp ipx} \quad\text{with}\; p^2=m^2\; \text{and}\; p^0>0 \label{eq:DE_ansatz} \end{align} % \imp{Here $\psi^{\pm}(p)\in\complex^4$ is a complex-valued four-component bispinor.}\\ % \imp{We set $p^0>0$ for both positive ($+$) and negative ($-$) frequency solutions and change the sign of $p$ in the exponent (to simplify the discussion below).} \item \cref{eq:DE_ansatz} in \cref{eq:DE} yields % \begin{align} (\pm\g^\mu p_\mu-m)\psi^\pm(p) =\begin{pmatrix}-m & \pm p\sigma \\ \pm p\ol\sigma & -m\end{pmatrix}\bispi{\psi^\pm_L}{\psi^\pm_R} =0 \label{eq:DE_1} \end{align} \item Note \imp{(\seepset{3})}: % \begin{itemize} \item $(p\sigma)(p\ol\sigma)=p^2=m^2$ \item Eigenvalues of $p\sigma$ and $p\ol\sigma$: $p^0\pm|\vec p|$ \therefore for $p^0>0$ and $m>0$ positive spectrum\\ % \imp{In particular, $p\sigma$ and $p\ol\sigma$ are invertible and the positive square roots $\sqrt{p\sigma}$ and $\sqrt{p\ol\sigma}$ are Hermitian.} \end{itemize} \item \consider $\psi^\pm_L=\sqrt{p\sigma}\,\xi^\pm$ with arbitrary, normalized [$(\xi^\pm)^\dag\xi^\pm=1$] \Emph{spinor} $\xi^\pm\in\complex^2$: % \begin{align} &\text{\cref{eq:DE_1}}\;\Rightarrow -m\sqrt{p\sigma}\,\xi^\pm \pm p\sigma\,\psi^\pm_R=0 \\ &\stackrel{\sqrt{p\sigma}\sqrt{p\ol\sigma}=m}{\Leftrightarrow}\quad \psi^\pm_R=\pm\frac{m}{\sqrt{p\sigma}}\xi^\pm=\pm\sqrt{p\ol\sigma}\xi^\pm \end{align} % \imp{The second equation in \cref{eq:DE_1} yields the same solution.} \item \Emph{Solutions:} \imp{Conventional notation: $\xi^+\mapsto \xi$, $\xi^-\mapsto \eta$ and $\psi^+\mapsto u$, $\psi^-\mapsto v$}\\ % \imp{Basis states: $\xi^s$ with $\xi^1=\bispi{1}{0}$ and $\xi^2=\bispi{0}{1}$ (same for $\eta^s$)} % \begin{emphalign} \Psi^+(x)&=\underbrace{\bispi{\sqrt{p\sigma}\xi^s}{\sqrt{p\ol\sigma}\xi^s}}_{u^s(p)}e^{-ipx} \quad\text{(positive frequency solutions)} \\ \Psi^-(x)&=\underbrace{\bispi{\sqrt{p\sigma}\eta^s}{-\sqrt{p\ol\sigma}\eta^s}}_{v^s(p)}e^{+ipx} \quad\text{(negative frequency solutions)} \end{emphalign} % with $p^2=m^2$, $p^0>0$ and $s=1,2$ \imp{\therefore \emph{Four} linearly independent solutions for each 3-momentum $\vec p$ ($\pm$ and $s=1,2$).} \item Some relations \imp{(\seepset{3})}: % \begin{itemize} \item \Emph{Orthonormality:} Let $\ol u^s\equiv (u^s)^\dag\gamma^0$ and $\ol v^s\equiv (v^s)^\dag\gamma^0$, then % \begin{eqaligned}[eq:uv] \ol u^r u^s&=2m\,\delta^{rs}\quad\text{and}\quad(u^r)^\dag u^s=2\Ep{p}\,\delta^{rs}\\ \ol v^r v^s&=-2m\,\delta^{rs}\quad\text{and}\quad(v^r)^\dag v^s=2\Ep{p}\,\delta^{rs}\\ \ol v^r u^s&=\ol u^r v^s=0\\ u^{r\dag}(\vec p)v^s(-\vec p)&=v^{r\dag}(-\vec p)u^s(\vec p)=0 \end{eqaligned} % \imp{Note that $\ol u u$ is Lorentz invariant whereas $u^\dag u\propto \Ep{p}$ is not!\\ % Note that $(u^r)^\dag v^s\neq 0$ and $(v^r)^\dag u^s\neq 0$!\\ % For massless particles, the normalization condition is given by $(u^r)^\dag u^s=2\Ep{p}\,\delta^{rs}$.} \item \Emph{Spin sums:} Let $\slashed p\equiv \gamma^\mu p_\mu$ (\termdef{Feynman slash notation}), then % \begin{eqaligned}[eq:uvsum] \sum_s u^s(p)\ol u^s(p)&=\slashed p + m\Id\\ \sum_s v^s(p)\ol v^s(p)&=\slashed p - m\Id \end{eqaligned} % \note{Useful if one wants to sum over spin-polarizations of fermions (\fref later).} \end{itemize} \end{lot} %%============================================================================== \begin{lecture} \item Dirac field bilinears \item Fermi statistics and the quantization of the Dirac field \end{lecture} %%============================================================================== %%============================================================================== \reference{PS:49--62} %%============================================================================== \section{Dirac Field Bilinears} \begin{lot} \item \Emph{Definition:} % \begin{emphalign} \gamma^5\equiv i\gamma^0\gamma^1\gamma^2\gamma^3 \note{= -\frac{i}{4!}\vep^{\mu\nu\rho\sigma}\gamma_\mu\gamma_\nu\gamma_\rho\gamma_\sigma} \stackrel{\text{Weyl basis}}{=} \begin{pmatrix}-\mathds{1} & 0 \\ 0 & \mathds{1} \end{pmatrix} \end{emphalign} % with % \begin{align} (\gamma^5)^\dag=\gamma^5, \quad (\gamma^5)^2=\mathds{1}, \quad \acom{\gamma^5}{\gamma^\mu}=0 \end{align} % \imp{The $\gamma^5$ matrix is labeled with ``5'' instead of ``4'' for historical reasons ($\gamma^0$ was sometimes called $\gamma^4$). Important: $\gamma^5$ does \emph{not} belong to $\gamma^\mu=(\gamma^0,\gamma^1,\gamma^2,\gamma^3)$ but is just a name. Sums like $\gamma^\mu \partial_\mu$ only run over $\mu=0,1,2,3$.} \note{The last relation implies $\com{\gamma^5}{S^{\mu\nu}}=0$, i.e., the Dirac bispinor representation must be reducible according to Schur's lemma: $(\hlf,0)\oplus(0,\hlf)$} \item The following \Emph{bilinears} $\ol\Psi\Gamma\Psi$ transform under the Lorentz group as \ldots % \begin{emphalign} \begin{array}{rlll} \Gamma = & 1 & \text{scalar} & \times 1\\ & \gamma^\mu & \text{vector} & \times 4\\ & \sigma^{\mu\nu}\equiv\frac{i}{2}\com{\gamma^\mu}{\gamma^\nu}=i\gamma^{[\mu}\gamma^{\nu]} & \text{tensor} & \times 6\\ & \gamma^\mu\gamma^5& \text{pseudo-vector} & \times 4\\ & \gamma^5& \text{pseudo-scalar} & \times 1 \end{array} \label{eq:bilinears} \end{emphalign} % \imp{The notation $\gamma^{[\mu}\dots\gamma^{\nu]}$ denotes the completely antisymmetrized product.} \imp{Any $4\times 4$ matrix $\Gamma$ can be decomposed into these 16 matrices with definite transformation properties under Lorentz transformations.} \imp{The prefix \emph{pseudo-} marks quantities that transform under continuous Lorentz transformations $\Lambda\in\POLG$ as usual but pick up an additional sign under parity transformations.} \note{This is similar to the cross product $a\times b$ in three dimensions which produces a \emph{pseudo-vector} from the two vectors $a$ and $b$ with respect to the Euclidean group (= isometries of Euclidean space). E.g., angular momentum $\vec L=\vec r\times\vec p$ is not a vector but a pseudo-vector.} For example, % \begin{align} (j^\mu)'=\ol\Psi'\gamma^\mu\Psi' =\ol\Psi \Lambda_{\hlf}^{-1}\gamma^\mu \Lambda_{\hlf}\Psi =\tensor{\Lambda}{^\mu_\nu}\ol\Psi \gamma^\nu \Psi =\tensor{\Lambda}{^\mu_\nu}j^\nu \label{eq:current_transform} \end{align} % transforms as a Lorentz 4-vector. \note{\easytoshow $j^\mu$ is the conserved Noether current corresponding to the continuous symmetry $\Psi\to e^{i\alpha}\Psi$ of the Dirac Lagrangian.} \imp{Note that $\gamma^\mu$ does \emph{not} transform under Lorentz transformations [it has no prime in \cref{eq:current_transform}, recall our derivation of the gamma matrices in \cref{sec:dirac_equation}]. The tuple of gamma matrices $\gamma^\mu$ is not a Lorentz 4-vector (despite its Greek upper index!) but a fixed set of basis vectors in the Dirac algebra; $\gamma_\mu\equiv g_{\mu\nu}\gamma^\nu$ is defined as usual.} \end{lot} \section{Quantization of the Dirac Field} \begin{lot} \item Lagrangian: $\L=\ol\Psi(i\gmu\pmu-m)\Psi$ \item Canonical momentum: $\Pi_a=\pdiff{\L}{\dot\Psi_a}=i\Psi_a^*$ \item Hamiltonian: $H=\int\dd{3}{x}\Psi^\dag\underbrace{[-i\vec\alpha\nabla+m\beta]}_{=H_D}\Psi$ with $\vec\alpha=\gamma^0\vec\gamma$ and $\beta=\gamma^0$\\ % \imp{$H_D$ is the Dirac Hamiltonian of single-particle quantum mechanics.}\\ % \therefore Expand $\Psi$ in eigenmodes of $H_D$ to diagonalize $H$ \item Eigenmodes: $H_Du^s(\vec p)e^{i\vec p \vec x}=\Ep{p}\rep$ and $H_Dv^s(\vec p)e^{-i\vec p \vec x}=-\Ep{p}\rep$\\ % \note{This can be seen from $[i\gamma^0\partial_0+i\vec\gamma\nabla-m]\Psi=0$ (\bref last lecture).} \item Mode expansion: % \begin{align} \Psi(\vec x)=\sum_s\int\ddp{3}{p}\frac{1}{\sqrt{2\Ep{p}}} \,\left[a_{\vec p}^su^s(\vec p)e^{i\vec p\vec x}+b_{\vec p}^sv^s(\vec p)e^{-i\vec p\vec x}\right] \label{eq:D_me} \end{align} % \imp{$a^s_{\vec p}$ and $b^s_{\vec p}$ are operator-valued expansion coefficients. We do not yet fix their algebra!} \item Use % \begin{align} H_D\Psi(\vec x)=\sum_s\int\ddp{3}{p}\sqrt{\frac{\Ep{p}}{2}} \,\left[a_{\vec p}^su^s(\vec p)e^{i\vec p\vec x}-b_{\vec p}^sv^s(\vec p)e^{-i\vec p\vec x}\right] \end{align} % then \imp{(using the orthonormality relations \cref{eq:uv})} % \begin{align} H=\int\dd{3}{x}\Psi^\dag H_D\Psi\nte\sum_s\int\ddp{3}{p}\,\Ep{p}( a_{\vec p}^{s\dag}a_{\vec p}^{s}-b_{\vec p}^{s\dag}b_{\vec p}^{s}) \label{eq:D_H} \end{align} % \imp{You do not need reordering of operators to show this. The algebra is still undefined!} \end{lot} \subsubsection{First try: Commutator} \begin{lot} \setcounter{enumi}{6} \item Canonical quantization with equal-time \Emph{commutators}: % \begin{eqaligned} \com{\Psi_a(\vec x)}{\Pi_b(\vec y)} &= i\delta_{ab}\df{3}{\vec x-\vec y} \quad\Leftrightarrow\quad\com{\Psi_a(\vec x)}{\Psi_b^\dag(\vec y)} = \delta_{ab}\df{3}{\vec x-\vec y}\\ \com{\Psi_a(\vec x)}{\Psi_b(\vec y)} &= 0 \label{eq:cq_boson} \end{eqaligned} \item \easytoshow Mode algebra % \begin{align} \com{a^r_{\vec p}}{a^{s\dag}_{\vec q}} &= \com{b^r_{\vec p}}{b^{s\dag}_{\vec q}} = (2\pi)^3\delta^{rs}\df{3}{\vec p-\vec q}\\ \com{a^r_{\vec p}}{b^{s(\dag)}_{\vec q}} &= 0 \end{align} % \imp{Show (using the mode expansion \cref{eq:D_me} and the spin sums \cref{eq:uvsum}) that this is equivalent to the commutators of the fields.}\\ % \note{Beware: Eq. (3.89) of P\&S is mathematically ill-defined since $\Psi\Psi^\dag$ is a matrix but $\Psi^\dag\Psi$ is not (it's just sloppy math that doesn't belong in a textbook for students). Do it right, i.e., componentwise: $\com{\Psi_a(\vec x)}{\Psi^\dag_b(\vec y)}=\delta_{ab}\df{3}{\vec x-\vec y}$.} \therefore Irreducible Representation = \Emph{Bosonic} Fock space \item \Emph{Problem:} $\left(b^{s\dag}_{\vec p}\right)^n\ket{0}$ has energy $-n\Ep{p}\xrightarrow{n\to\infty} -\infty$\\ % \therefore No stable vacuum state \note{(the spectrum of $H$ is unbounded below)} \item Fix (?): $b\leftrightarrow b^\dag$ \aside{(Use colors to modify the previous derivation.)} % \begin{lot} \item $\Psi(\vec x)=\dots [a_{\vec p}^s\dots+b_{\vec p}^{s\dag}\dots]$ \item $H=\dots (a_{\vec p}^{s\dag}a_{\vec p}^{s}-b_{\vec p}^{s}b_{\vec p}^{s\dag})$ \item $\com{b^r_{\vec p}}{b^{s\dag}_{\vec q}} = -(2\pi)^3\delta^{rs}\df{3}{\vec p-\vec q}$ \item $H=\dots (a_{\vec p}^{s\dag}a_{\vec p}^{s}-b_{\vec p}^{s\dag}b_{\vec p}^{s})+\const$ \item $\com{H}{b^{s\dag}_{\vec p}}=\Ep{p}b^{s\dag}_{\vec p}$ \therefore $b^{s\dag}_{\vec p}$ creates a particle with \emph{positive} energy! \therefore $H\geq 0$\\ % \imp{It \Emph{seems} that we solved the problem: The spectrum of the Hamiltonian is now bounded from below.} \item \Emph{But:} % \begin{align} \|b^{s\dag}_{\vec p}\ket{0}\|^2 =\bra{0}\com{b^s_{\vec p}}{b^{s\dag}_{\vec p}}\ket{0} =-(2\pi)^3\df{3}{0}<0 \end{align} % \therefore Negative norm states \imp{(i.e., the constructed representation is not a Hilbert space)} \end{lot} \item \Emph{Conclusion:} \cref{eq:cq_boson} implies % \begin{itemize} \item either an instability of the vacuum \item or a loss of unitarity \end{itemize} % \therefore No consistent quantization possible! \end{lot} \subsubsection{Second try: Anticommutator} \begin{lot} \setcounter{enumi}{6} \item Canonical quantization with equal-time \Emph{anti}commutators: % \begin{emphalign} \acom{\Psi_a(\vec x)}{\Psi_b^\dag(\vec y)} = \delta_{ab}\df{3}{\vec x-\vec y} \quad\text{and}\quad \acom{\Psi_a(\vec x)}{\Psi_b(\vec y)} = 0 \label{eq:cq_fermion} \end{emphalign} % \imp{Note that these are \emph{equal-time} anticommutators!} \item \easytoshow Mode algebra % \begin{emphalign} \acom{a^r_{\vec p}}{a^{s\dag}_{\vec q}} = \acom{b^r_{\vec p}}{b^{s\dag}_{\vec q}} = (2\pi)^3\delta^{rs}\df{3}{\vec p-\vec q} \quad\text{and}\quad \acom{a^r_{\vec p}}{b^{s(\dag)}_{\vec q}} = 0 \label{eq:fma} \end{emphalign} % \imp{The proof is similar to the bosonic case above.} \therefore Irreducible Representation = \Emph{Fermionic} Fock space \item \Emph{Problem:} $b^{s\dag}_{\vec p}\ket{0}$ has energy $-\Ep{p}$ \& infinite sum over momenta\\ % \therefore Still no stable vacuum state\\ % \note{(The spectrum of $H$ is still unbounded below due to the sum over momenta.)} \item Fix (?): $b\leftrightarrow b^\dag$ \imp{(we saw above that it changes the sign of the excitation energies)} % \begin{enumerate} \item \Emph{Hamiltonian:} % \begin{emphalign} H&=\sum_s\int\ddp{3}{p}\,\Ep{p}(a_{\vec p}^{s\dag}a_{\vec p}^{s}-b_{\vec p}^{s}b_{\vec p}^{s\dag})\\ &=\sum_s\int\ddp{3}{p}\,\Ep{p}(a_{\vec p}^{s\dag}a_{\vec p}^{s}+b_{\vec p}^{s\dag}b_{\vec p}^{s})-\infty \label{eq:D_H2} \end{emphalign} % \imp{We will drop the infinite constant henceforth.} \aside{(Cross the $-\infty$.)} \item The mode algebra \cref{eq:fma} is \Emph{invariant} under $b\leftrightarrow b^\dag$!\\ % \therefore Unitarity is preserved \Emph{and} Hamiltonian is bounded from below\\ % \imp{\therefore With anticommutation relations, quantization is consistently possible!} \end{enumerate} \item \Emph{Heisenberg picture:} \imp{Now that we have a representation where the Hamiltonian generates a unitary time evolution, we can switch to the Heisenberg picture:} With % \begin{align} e^{iHt}a_{\vec p}^s e^{-iHt}\nte a_{\vec p}^s\,e^{-i\Ep{p}t} \quad\text{and}\quad e^{iHt}b_{\vec p}^s e^{-iHt}\nte b_{\vec p}^s\,e^{-i\Ep{p}t} \end{align} % and $\Psi(x)=e^{iHt}\Psi(\vec x)e^{-iHt}$ we find % \begin{emphalign} \Psi(x) &=\sum_s\int\ddp{3}{p}\frac{1}{\sqrt{2\Ep{p}}}\, \left[a_{\vec p}^su^s(p)e^{-ipx}+b_{\vec p}^{s\dag}v^s(p)e^{ipx}\right] \\ \ol\Psi(x) &=\sum_s\int\ddp{3}{p}\frac{1}{\sqrt{2\Ep{p}}}\, \left[a_{\vec p}^{s\dag}\ol u^s(p)e^{ipx}+b_{\vec p}^{s}\ol v^s(p)e^{-ipx}\right] \end{emphalign} % \imp{These are operator-valued spinor fields, i.e., functions (more precisely: distributions) on Minkowski spacetime that assign to an event $x$ a tuple (``spinor'') of operators that act on the fermionic Fock space where the states of the quantized theory live.} \end{lot} \subsubsection{Continuous symmetries \& Conserved charges} \begin{itemize} \item Time translation \therefore Hamiltonian \imp{(see above)} \item Spatial translations \therefore Momentum operator % \begin{emphalign} \vec P\nte\int\dd{3}{x}\Psi^\dag(-i\nabla)\Psi \nte\sum_s\int\ddp{3}{p}\,\vec p\,(a_{\vec p}^{s\dag}a_{\vec p}^{s}+b_{\vec p}^{s\dag}b_{\vec p}^{s}) \end{emphalign} \item Rotations \therefore Angular momentum operator $\vec J$ \note{% \begin{align} \vec J=\int\dd{3}{x}\Psi^\dag\left[\vec x\times (-i\nabla)+\hlf\vec\Sigma\right]\Psi \quad\text{with}\quad \vec \Sigma = \begin{pmatrix}\vec\sigma & 0 \\ 0 & \vec\sigma\end{pmatrix} \label{eq:J} \end{align} } \item Global phase rotations $e^{i\alpha}\Psi$\\ % \easytoshow Conserved current $j^\mu=\ol\Psi\gmu\Psi$\\ % \therefore Conserved charge: % \begin{emphalign} Q=\int\dd{3}{x}\Psi^\dag\Psi &\nte\sum_s\int\ddp{3}{p}(a_{\vec p}^{s\dag}a_{\vec p}^{s}+b_{-\vec p}^{s}b_{-\vec p}^{s\dag}) \\ &=\sum_s\int\ddp{3}{p}(a_{\vec p}^{s\dag}a_{\vec p}^{s}-b_{\vec p}^{s\dag}b_{\vec p}^{s})+\infty \end{emphalign} % \imp{In QED we will couple the fermions to the EM field; then, $Q$ is the total EM charge of the fermion field.} \note{Recall that in \emph{single-particle quantum mechanics} the global phase rotation symmetry gives rise to a positive density and a current that can be interpreted as probability current; the conserved charge corresponds then to the total probability to find the single particle somewhere. Because of the normal ordering (= dropping the infinite constant) this interpretation does no longer apply as $Q$ can become negative.} \end{itemize} % \note{The operators of conserved charges generate symmetry transformations of the Hamiltonian.} \subsubsection{Excitations = Particles} \begin{emphalign} \begin{array}{rll} a_{\vec p}^{s\dag}\ket{0}: &\text{\Emph{Fermion} with} &\text{energy $\Ep{p}$,}\\ &&\text{momentum $\vec p$,} \\ &&\text{spin $J=\hlf$ (polarization $s$),}\\ &&\text{and charge $Q=+1$} \\[10pt] b_{\vec p}^{s\dag}\ket{0}: &\text{\Emph{Antifermion} with} &\text{energy $\Ep{p}$,}\\ &&\text{momentum $\vec p$,} \\ &&\text{spin $J=\hlf$ (polarization opposite to $s$),}\\ &&\text{and charge $Q=-1$} \end{array} \end{emphalign} % \imp{In QED, the fermions will be \emph{electrons} and the antifermions \emph{positrons}.} \begin{blocknote} \note{% \begin{itemize} \item The two states for $s=1,2$ \emph{suggest} a spin-$\hlf$ representation \item To show this, the action of $\vec J$ (see \cref{eq:J}) on one-particle states must be studied \item One finds for particles at rest: % \begin{align} J_z\,a_{\vec 0}^{s\dag}\ket{0}=\pm\hlf a_{\vec 0}^{s\dag}\ket{0} \quad\text{and}\quad J_z\,b_{\vec 0}^{s\dag}\ket{0}=\mp\hlf b_{\vec 0}^{s\dag}\ket{0} \end{align} % with $\xi^{s=1}=\bispi{1}{0}$ and $\xi^{s=2}=\bispi{0}{1}$. \end{itemize} } \end{blocknote} \subsubsection{Lorentz transformations} \begin{lot} \item \consider Lorentz transformation $\Lambda\in\POLG$ on single particle state $\ket{\vec p,s}_a\equiv \sqrt{2\Ep{p}}\,a_{\vec p}^{s\dag}\ket{0}$:\\ % \imp{(The subscript $a$ ($b$) denotes the state of a(n) (anti-)fermion; we omit it, $\ket{\vec p,s}\equiv\ket{\vec p,s}_a$, when the distinction is not important.)} % \begin{align} \ket{\vec p,s}\mapsto U(\Lambda)\ket{\vec p,s} \end{align} % $U(\Lambda)$: representation of $\POLG$ on Fock space\\ % \imp{For generic rotations/boosts, this mixes the two spin components!} \item \consider \Emph{Special case:} quantization axis parallel to boost and/or rotation axis \therefore Spin polarizations do not mix: % \begin{align} U(\Lambda)\,a_{\vec p}^s\,U^{-1}(\Lambda)=\sqrt{\frac{E_{\Lambda\vec p}}{\Ep{p}}}\,a_{\Lambda \vec p}^s \end{align} % \note{Note that spins mix under generic Lorentz transformations: $a_{\vec p}^1\leftrightarrow a_{\vec q}^2$.} % \note{Note that particles and antiparticles never mix under Lorentz % transformations, $a_{\vec p}^s\nleftrightarrow b_{\vec q}^r$, but % spins generally do, $a_{\vec p}^1\leftrightarrow a_{\vec q}^2$.} \item \imp{Consider this special case, then:} % \begin{align} \braket{\vec p,s}{\vec q,r} =\underbrace{2\Ep{p}(2\pi)^3\df{3}{\vec p-\vec q}\delta^{rs}}_{\text{Lorentz invariant}} =\bra{\vec p,s}U^\dag(\Lambda)U(\Lambda)\ket{\vec q,r} \end{align} % \therefore $U(\Lambda)$ is \Emph{unitary} \item Now we have 3 representations: % \begin{emphalign} \begin{array}{rlll} \Lambda & \text{acts on 4-vectors in $\reals^{1,3}$} & D=4 & \text{not unitary}\\ \Lambda_{\hlf} & \text{acts on bispinors in $\complex^{2}\oplus\complex^{2}$} & D=4 & \text{not unitary}\\ U(\Lambda) & \text{acts on states in fermionic Fock space} & D=\infty & \text{unitary} \end{array} \end{emphalign} \item \imp{Action by conjugation on field operators} \easytoshow % \begin{emphalign} U(\Lambda)\Psi(x)U^{-1}(\Lambda)=\Lambda^{-1}_\hlf\Psi(\Lambda x) \label{eq:UPsiU} \end{emphalign} \end{lot} %%============================================================================== \begin{lecture} \item The spin-statistics theorem \item The Dirac propagator \item Causality \item Discrete symmetries of the Dirac theory \end{lecture} %%============================================================================== %%============================================================================== \begin{problemset} \item The relativistic hydrogen atom \item Parity transformation of Dirac spinors \end{problemset} %%============================================================================== %%============================================================================== \reference{PS:62--71} %%============================================================================== \subsubsection{Spin-statistics theorem} \begin{itemize} \item \Emph{Observation:} % \begin{align} \begin{array}{rlll} \text{Klein-Gordon field $\phi$:} &\text{Spin 0 (scalar)} &\text{\therefore commutator} &\text{\therefore bosonic excitations}\\ \text{Dirac field $\Psi$:} &\text{Spin $\hlf$ (spinor)} &\text{\therefore anticommutator} &\text{\therefore fermionic excitations} \end{array} \end{align} % \imp{This is no coincidence but hints at a more fundamental connection:} \item \Emph{Spin-statistics theorem:} % \begin{emphalign} \left. \begin{array}{r} \text{Lorentz invariance}\\ \text{Causality}\\ \text{Positive energies}\\ \text{Positive norms} \end{array} \right\} \Rightarrow \left\{ \begin{array}{rcl} \text{Integer spin} &\leftrightarrow &\text{Bosons}\\ \text{Half-integer spin} &\leftrightarrow &\text{Fermions} \end{array} \right. \end{emphalign} % \imp{This means, whenever you quantize a relativistic field that transforms under a (projective) half-integer spin representation, the Poisson bracket must be replaced by \emph{anti}commutators. Otherwise unitarity is lost or the vacuum becomes unstable.} \item % \begin{samepage} \Emph{``Proof by picture'':}% % \begin{center} \includegraphics[width=0.9\linewidth]{spin-statistics} \end{center} % \imp{Rigorous proofs are elaborate and quite technical.} \note{\uref\url{http://math.ucr.edu/home/baez/spin_stat.html}} \end{samepage} \end{itemize} \subsubsection{Dirac Propagator} \imp{All that follows is very similar to our discussion of the Klein-Gordon propagator.\\ For details, we refer the student to the corresponding notes.} \begin{lot} \item Propagation amplitudes \aside{(use colors to skip this calculation)}: % \begin{align} \bra{0}\Psi_a(x)\ol\Psi_b(y)\ket{0} &=\int\ddp{3}{p}\frac{1}{2\Ep{p}}e^{-ip(x-y)}\underbrace{\sum_s u^s_a(p) \ol u^s_b(p)}_{(\slashed{p}+m)_{ab}} \\ &=\note{(i\slashed{\partial}_x+m)_{ab}D(x-y)} \\ &\stackrel{x^0>y^0}{=}\int\ddp{4}{p}\frac{i(\slashed{p}+m)_{ab}}{p^2-m^2+i\vep}e^{-ip(x-y)} \end{align} % \begin{align} \bra{0}\ol\Psi_b(y)\Psi_a(x)\ket{0} &=\int\ddp{3}{p}\frac{1}{2\Ep{p}}e^{-ip(y-x)}\underbrace{\sum_s v^s_a(p) \ol v^s_b(p)}_{(\slashed{p}-m)_{ab}} \\ &=\note{-(i\slashed{\partial}_x+m)_{ab}D(y-x)} \\ &\stackrel{x^0y^0$: close contour below \item $x^0y^0\\ -\bra{0}\ol\Psi_b(y)\Psi_a(x)\ket{0} & \text{for}\;x^0t_2$ it is $\T\Psi(t_2)\Psi(t_1)\equiv-\Psi(t_1)\Psi(t_2)$ for fermionic fields! \imp{The Feynman propagator $S_F(x-y)$ of the Dirac field is a $4\times 4$ matrix.} \note{% \item Similarly, one can derive the \emph{Retarded Green's function}: % \begin{align} S_R^{ab}(x-y)\equiv \theta(x^0-y^0)\,\bra{0}\acom{\Psi_a(x)}{\ol\Psi_b(y)}\ket{0} \nte (i\slashed{\partial}_x+m)_{ab}D_R(x-y) \end{align} % Here, $D_R(x-y)$ is the retarded Green's function of the Klein-Gordon field;\\ % $\slashed\partial_x$ denotes derivatives with respect to the variables $x^\mu$ for $\mu=0,1,2,3$ and generates the $\slashed{p}$ in the integral. } \end{lot} \subsubsection{Causality} \begin{lot} \item \Emph{Measurable operators}: % $\hat O(x)=\sum\prod_{i=1}^{\text{even}\,N}(\Psi_i^{(\dag)}(x)\;\vee\; \partial\Psi_i^{(\dag)}(x)\;\vee\;\partial^2\Psi_i^{(\dag)}(x)\;\dots)$ Example: $j^\mu=\ol\Psi\gmu\Psi$ \imp{(check that this is Hermitian!)} (but \Emph{not} $\Psi_a+\Psi_a^\dag$!) \imp{Restricting observables to field polynomials of even degree ensures that space-like separated observables \emph{commute} if space-like separated fields \emph{anticommute} (which is the best we can hope for given our quantization conditions).} \note{The answer to the question \emph{``Why restrict observables to even degree expressions in the fields?''} is therefore: Because these are the only observables that do not violate causality in a theory built from fermionic fields. (There are also more rigorous arguments for this: \uref Superselection.)} \item Causality \imp{for fermionic fields} $\Leftrightarrow$ $\acom{\Psi_a(x)}{\ol\Psi_b(y)}=0$ for $(x-y)^2<0$\\ % \imp{All other anticommutators vanish trivially. Note that here $x=(t,\vec x)$ and $y=(t',\vec y)$, i.e., we consider the anticommutator at \emph{different} times.} We find \imp{(using results from above)} % \begin{align} \acom{\Psi_a(x)}{\ol\Psi_b(y)} &\nte(i\slashed{\partial}_x+m)_{ab}[D(x-y)-D(y-x)]\\ &\mimp{$(x-y)^2<0$}\\ &=(i\slashed{\partial}_x+m)_{ab}[D(x-y)-D(x-y)]=0 \end{align} % \imp{The argument is the same as for the Klein-Gordon field.}\\ % \note{Recall: $D(x-y)=\int\ddp{3}{p}\frac{1}{2\Ep{p}}e^{-ip(x-y)}$} \end{lot} \section{Discrete Symmetries of the Dirac Theory} \subsubsection{Review of the Lorentz group} \imp{Details: \seepset{3}} \begin{itemize} \item \imp{Lorentz group $\LG$ = Lie group with four disconnected components} \item \imp{Continuous Lorentz transformations = \emph{Proper orthochronous Lorentz group} $\POLG$} \item Four components connected by discrete transformations: % \begin{align} \text{Parity}\quad P&:\,(t,\vec x)\mapsto (t,-\vec x)\\ \text{Time reversal}\quad T&:\,(t,\vec x)\mapsto (-t,\vec x) \end{align} % \begin{center} \includegraphics[width=1.0\linewidth]{tikz/structure_lorentz_group} \end{center} \end{itemize} % \begin{sidewaysfigure}[p] % \begin{center} % \includegraphics[width=0.9\linewidth]{poincare-group} % \end{center} % \end{sidewaysfigure} \subsubsection{Parity} \imp{Details: \seepset{4}} \begin{lot} \item Unitary representation on Fock space: % \begin{align} U(P)\,a_{\vec p}^s\,U^{-1}(P)=\underbrace{\eta_a}_{+1}\,a_{-\vec p}^s \quad\text{and}\quad U(P)\,b_{\vec p}^s\,U^{-1}(P)=\underbrace{\eta_b}_{-1}\,b_{-\vec p}^s \end{align} % \imp{Note that we do \emph{not} want spin to change under $P$ because angular momentum $\vec L=\vec r\times\vec p$ also does not pick up a sign under inversion (it is a pseudo-vector).}\\[5pt] % \note{Note that often $U(P)$ is simply written $P$.} \item Equivalent to % \begin{emphalign} U(P)\Psi(t,\vec x)U^{-1}(P)=\underbrace{\gamma^0}_{P_\hlf}\Psi(\underbrace{t,-\vec x}_{Px}) \label{eq:parity_dirac} \end{emphalign} % \note{The $\gamma^0$-matrix exchanges the left- and right-handed Weyl sectors of the bispinor; this makes sense as a parity transformation of space should switch chirality.} \item \note{% Dirac field bilinears (examples): % \begin{alignat}{2} U(P)\ol\Psi\Psi U^{-1}(P)&\nte +\ol\Psi\Psi(t,-\vec x) &&\quad\rightarrow\text{scalar}\\ U(P)\ol\Psi\gamma^5\Psi U^{-1}(P)&\nte -\ol\Psi\gamma^5\Psi(t,-\vec x) &&\quad\rightarrow\text{pseudo-scalar} \end{alignat} % % \item Dirac Lagrangian $\L_D=\ol\f(i\slashed{\partial}-m)\f$ is parity-symmetric: $U(P)\L_D U^{-1}(P)=\L_D$ } \end{lot} \subsubsection{Time Reversal} \begin{lot} \item Time reversal should \dots % \begin{itemize} \item $U(T)\Psi(t,\vec x)U^{-1}(T)= T_\hlf\Psi(-t,\vec x)$, \item $U(T)a^s_{\vec p}U^{-1}(T)=a^?_{-\vec p}$, \item flip spins \imp{(motivated by $\vec L=\vec r\times\vec p\mapsto -\vec L$)}, \item be a symmetry of the Dirac theory: $\com{U(T)}{H}=0$, \item obey $U^{-1}(T)=U^\dag(T)$.\\ % \imp{This is required for any symmetry to preserve overlaps: \uref Wigner's theorem.} \end{itemize} % \note{Note that often $U(T)$ is simply written $T$.} \item \Emph{Problem:} % \begin{alignat}{2} &&\Psi(t,\vec x) &= e^{iHt}\Psi(\vec x) e^{-iHt}\\ &\Rightarrow\quad &U(T)\Psi(t,\vec x)U^{-1}(T) &= e^{iHt}U(T)\Psi(\vec x)U^{-1}(T)e^{-iHt}\\ &\Rightarrow\quad &T_\hlf\Psi(-t,\vec x)\ket{0} &= e^{iHt}T_\hlf\Psi(\vec x)\ket{0}\\ &\Rightarrow\quad &T_\hlf e^{-iHt}\Psi(\vec x)\ket{0} &= e^{iHt}T_\hlf\Psi(\vec x)\ket{0}\\ &\Rightarrow\quad &\underbrace{e^{-2iHt}}_{\text{time-dependent!}}T_\hlf\Psi(\vec x)\ket{0} &= T_\hlf\Psi(\vec x)\ket{0} \end{alignat} % \imp{Here we used that $\com{U(T)}{H}=0$ and $H\ket{0}=0$.} \therefore Not possible \imp{(for invertible $T_\hlf$ and arbitrary times $t$)}! \item \Emph{Solution:} $U(T)$ must be \Emph{antiunitary/antilinear}: % \begin{emphalign} U(T)c=c^*U(T) \quad\text{for}\;c\in\complex \label{eq:antiu} \end{emphalign} % \note{The relation \cref{eq:antiu} makes $U$ \emph{antilinear}; anti\emph{unitarity} means that in addition $\braket{U\psi}{U\phi}=\braket{\psi}{\phi}^\ast$ for all states $\psi$ and $\phi$. Antiunitary operators can be written as $U=V\mathcal{K}$ where $V$ is a unitary operator and $\mathcal{K}$ denotes complex conjugation.} \aside{Highlight the differences with colors in the derivation above: % \begin{alignat}{2} &&\Psi(t,\vec x) &= e^{iHt}\Psi(\vec x) e^{-iHt}\\ &\Rightarrow\quad &U(T)\Psi(t,\vec x)U^{-1}(T) &= e^{-iHt}U(T)\Psi(\vec x)U^{-1}(T)e^{iHt}\\ &\Rightarrow\quad &T_\hlf\Psi(-t,\vec x)\ket{0} &= e^{-iHt}T_\hlf\Psi(\vec x)\ket{0}\\ &\Rightarrow\quad &T_\hlf e^{-iHt}\Psi(\vec x)\ket{0} &= e^{-iHt}T_\hlf\Psi(\vec x)\ket{0}\\ &\Rightarrow\quad &\underbrace{\Id}_{\text{time-independent!}}T_\hlf\Psi(\vec x)\ket{0} &= T_\hlf\Psi(\vec x)\ket{0} \end{alignat} } \item Transformation of \Emph{spin}: % \begin{lot} \item Spinors: \consider Spin basis $\xi^s$ ($s=1,2$) along arbitrary axis $\vec n$: % \begin{align} \xi^1=\bispi{\cos\frac{\theta}{2}}{e^{i\phi}\sin\frac{\theta}{2}} \quad\text{and}\quad \xi^2=\bispi{-e^{-i\phi}\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} \end{align} % \note{That is, $\xi^1=\ket{\uparrow}$ and $\xi^2=\ket{\downarrow}$.}\\ % ``Time-reversed'' (=flipped) spinors: % \begin{align} \ol{\xi^s}\equiv -i\sigma^2(\xi^s)^* \quad\Rightarrow\quad \bicase{\ol{\xi^{1}}}{\ol{\xi^{2}}} =\bicase{\xi^{2}}{-\xi^{1}} \label{eq:xims} \end{align} % \note{% Indeed, if $\vec n\cdot\vec\sigma\,\xi=+\xi$, we have % \begin{align} \vec n\cdot\vec\sigma\,(-i\sigma^2\xi^*) =-i\sigma^2(-\vec n\cdot\vec\sigma)^*\xi^* =i\sigma^2(\xi^*) =-(-i\sigma^2\xi^*) \end{align} % where we used $\vec\sigma\sigma^2=\sigma^2(-\vec\sigma^*)$.\\[10pt] % Note that $T=-i\sigma^2K$ (where $K$ denotes complex conjugation) is the conventional representation of time-reversal symmetry for spinful fermions that you might know from condensed matter physics (e.g., to classify symmetry-protected topological phases). } \item Bispinors: % \begin{eqaligned}[eq:us] u^s(p)\equiv\bispi{\sqrt{p\sigma}\,\xi^s}{\sqrt{p\ol\sigma}\,\xi^s} \quad&\text{and}\quad v^s(p)\equiv\bispi{\sqrt{p\sigma}\,\ol{\xi^{s}}}{-\sqrt{p\ol\sigma}\,\ol{\xi^{s}}} \\ \ol{u^s}(p)\equiv\bispi{\sqrt{p\sigma}\,\ol{\xi^s}}{\sqrt{p\ol\sigma}\,\ol{\xi^s}} \quad&\text{and}\quad \ol{v^s}(p)\equiv\bispi{\sqrt{p\sigma}\,\ol{\ol{\xi^{s}}}}{-\sqrt{p\ol\sigma}\,\ol{\ol{\xi^{s}}}} \end{eqaligned} % \aside{Use colors to skip the second row.} \imp{Note that here $\ol{u^s}$ is \emph{not} the Dirac adjoint $\ol{u}^s$!}\\ % \imp{Recall that the basis $\eta^s$ in the definition of $v^s(p)$ was arbitrary.} \item Define the modes: % \begin{align} \bicase{\ol{\ap{p}^{1}}}{\ol{\ap{p}^{2}}} \equiv \bicase{\ap{p}^{2}}{-\ap{p}^{1}} \quad\text{and}\quad \bicase{\ol{\bp{p}^{1}}}{\ol{\bp{p}^{2}}} \equiv \bicase{\bp{p}^{2}}{-\bp{p}^{1}} \label{eq:ams} \end{align} % \aside{Skip the second part.} \note{Note that this is analog to \cref{eq:xims}!} \item Let $\tilde p\equiv (p^0,-\vec p)$ and show % \begin{eqaligned}[eq:ums] \ol{u^{s}}(\tilde p)&\nte -\gamma^1\gamma^3[u^s(p)]^*\\ \ol{v^{s}}(\tilde p)&\nte -\gamma^1\gamma^3[v^s(p)]^* \end{eqaligned} % Note: $\ol{\ol{\xi^{s}}}=-\xi^s$ used in $\ol{v^s}$ \imp{Use \cref{eq:us} and $\sqrt{\tilde p\sigma}\,\sigma^2=\sigma^2\,\sqrt{p\sigma^*}$ to show this!} \end{lot} \item \Emph{Definition:} % \begin{emphalign} \left. \begin{array}{r} \text{Antilinearity \cref{eq:antiu}}\\[5pt] U(T)\,a_{\vec p}^s U^{-1}(T)\equiv \ol{a^{s}_{-\vec p}}\\[5pt] U(T)\,b_{\vec p}^s U^{-1}(T)\equiv \ol{b^{s}_{-\vec p}} \end{array} \right\} \Rightarrow \begin{aligned}[c] &U(T)\Psi(t,\vec x)U^{-1}(T)\\ \nte &\underbrace{(\gamma^1\gamma^3)}_{T_\hlf}\,\Psi(-t,\vec x) \end{aligned} \end{emphalign} % \imp{Use \cref{eq:ums} and \cref{eq:ams} and $\ol{\ap{p}^2}\,\ol{u^2}(p)=\ap{p}^1 u^1(p)$ etc.\ to show this!} \note{% Note that in Weyl representation % \begin{align} T_{\hlf}=\begin{pmatrix}i\sigma^y & 0 \\ 0 & i\sigma^y\end{pmatrix} \end{align} % i.e., time-reversal acts on spins but does not mix chiralities (as parity did). This makes sense, because (for massless particles) chirality = helicity and helicity is the projection of spin on momentum: $\vec S\cdot\vec p$. Since both spin (angular momentum) $\vec S$ \emph{and} linear momentum $\vec p$ change sign under time-reversal, helicity does not. } \note{% \item Dirac field bilinears (example: $j^\mu=\ol\Psi\gmu\Psi$): % \begin{align} U(T)j^\mu(t,\vec x)U^{-1}(T)&\nte \begin{cases} +j^\mu(-t,\vec x) &\text{for}\;\mu=0\\ -j^\mu(-t,\vec x) &\text{for}\;\mu=1,2,3 \end{cases} \end{align} % \therefore As expected for density ($\mu=0$) and 3-current ($\mu=1,2,3$) } \end{lot} \subsubsection{Charge Conjugation} \begin{lot} \item Discrete, \Emph{non-spacetime} symmetry that exchanges particle and antiparticle: % \begin{emphalign} U(C)\aps{p}{s}U^{-1}(C)=\bps{p}{s} \quad\text{and}\quad U(C)\bps{p}{s}U^{-1}(C)=\aps{p}{s} \end{emphalign} % \imp{Note that there is no representation on Minkowski space as this is an ``internal'' symmetry.} \note{Often $U(C)$ is simply written $C$.} \item \imp{Use \cref{eq:us} to show:} % \begin{align} u^s(p)\nte-i\g^2(v^s(p))^* \quad\text{and}\quad v^s(p)\nte-i\g^2(u^s(p))^* \end{align} \item Then % \begin{align} &U(C)\Psi(x)U^{-1}(C) \\ &\color{notecolor}=\sum_s\int\ddp{3}{p}\frac{1}{\sqrt{2\Ep{p}}} \left[ -i\g^2(v^s(p))^*\bps{p}{s} e^{-ipx} -i\g^2(u^s(p))^*\adps{p}{s} e^{ipx} \right] \\ &\nte-i\g^2(\Psi^\dag)^T=-i(\ol\Psi\g^0\g^2)^T \end{align} \item Therefore: % \begin{emphalign} U(C)\Psi U^{-1}(C)&=-i(\ol\Psi\g^0\g^2)^T\note{(=\underbrace{-i\gamma^2}_{\equiv C_\hlf}\f^*)} \label{eq:charge_conjugation_def}\\ \quad\text{and}\quad U(C)\ol\Psi U^{-1}(C)&\nte-i(\g^0\g^2\Psi)^T \end{emphalign} % \begin{itemize} \item \imp{Note that $C$ essentially exchanges $\Psi\leftrightarrow\ol\Psi$ but is not antiunitary!} \item \note{To show this, recall that $\g^0$ and $\g^2$ are symmetric matrices.} \item \note{It is $C_\hlf^\dag=C_\hlf$, $C_\hlf^2=\Id$ and $C_\hlf\gmu C_\hlf=-{\gamma^\mu}^*$.} \item \note{% Note that the expression in parantheses is only true for the transformation of \emph{classical} (i.e. ``first quantized'') Dirac fields and can be used to show the symmetry of the classical Dirac equation. However, if you take the $^\ast$ to conjugate complex numbers \emph{and} Hilbert space operators, $\Psi^*\equiv(\Psi^\dag)^T$, it is valid for the quantized field as well. } \end{itemize} \note{% \item Dirac field bilinears (examples): % \begin{alignat}{2} U(C)\ol\Psi\Psi U^{-1}(C)&\nte \ol\Psi\Psi &&\quad\text{(Scalar)}\\ U(C)\ol\Psi\gmu\Psi U^{-1}(C)&\nte -\ol\Psi\gmu\Psi &&\quad\text{(Vector)} \end{alignat} } \end{lot} \begin{blocknote} \begin{itemize} \item Any relativistic QFT must be invariant under $\POLG$ \note{($=L^\uparrow_+$)} % \item The free Dirac Lagrangian $\L_D=\ol\Psi(i\gmu\pmu-m)\Psi$ is $\{C,P,T\}$-invariant \item The (classical) Dirac equation $(i\gmu\pmu-m)\Psi=0$ is $\{C,P,T\}$-invariant \item The (quantized) Dirac theory is $\{C,P,T\}$-invariant:\\ % $\com{H}{U(X)}=0$ for $X=P,T,C$ \item Weak interactions \imp{(of the standard model)} violate $C$ and $P$ but preserve $CP$ and $T$ \imp{(\uref Wu experiment)} \item Rare processes \imp{(decay of neutral kaons)} violate $CP$ and $T$ but preserve $CPT$ \item $CPT$ seems to be a perfect symmetry of nature \item \Emph{$CPT$ theorem}: % \begin{align} \left. \begin{array}{r} \text{$\POLG$ invariance}\\ \text{Causality}\\ \text{Locality}\\ \text{Stable vacuum} \end{array} \right\} \Rightarrow \text{$CPT$ symmetry} \end{align} \end{itemize} \end{blocknote}