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\begin{document}
\begin{center}

University of California at San Diego -- 
Department of Physics --
Prof.~John McGreevy

{\Large\bf  Physics 215A QFT Fall 2023}\\
{\Large\bf Assignment 9 \answer{--~~~ Solutions} }
\end{center}


\noindent
\hfill {\bf Due 11:00am { Monday, December 4, 2023}} 

\bigskip
\hrule


\def\aa{{\bf a}}
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\begin{enumerate}
\item {\bf Brain warmers on $\gSO(3)$.}   
\begin{enumerate}
\item Consider the statement that the rotation generators transform as a vector under rotations:
\be \( D_{\(j=1\)}(\theta)\) ^k_j \JJ^j = D_R(\theta)^\dagger \JJ^k D_R(\theta),\ee
where 
$ D_R(\theta)  = e^{ - \ii \theta^i \JJ^i } $ 
and $D_{\(j=1\)}(\theta)  = e^{ - \ii \theta^i J_{(j=1)}^i} $, with $ ( J_{(j=1)}^i )^j_k = - \ii \eps^{ijk}$.  
Show that to leading nontrivial order in $\theta$ (about $\theta =0$) this is equivalent to the $\mathsf{so}(3)$ Lie algebra, 
\be [\JJ^i, \JJ^j] = \ii \eps^{ijk} \JJ^k . \ee


\item Starting from 
the form of the generators in the vector (spin 1) representation,
\be\label{eq:jijk} ({\bf J}^i)^j_k ={-} \ii \eps^{ijk} \ee
(with $\eps^{123}=1$) construct the matrix realizing a rotation by angle $\theta$ about the $z$ axis on a vector.




\item Here let's show that the equation 
\be \label{eq:K-is-a-vector} [J^i, K^j] = \ii \eps^{ijk} K^k 
\ee 
can really be interpreted as the statement that ``$K$ is a vector".  
Let's just think about rotations about one axis $\hat n$.  
Suppose $J^i$ and $K^k$ here are operators acting on the representation $R$.  
Let 
\be K(s)^k \equiv \hat D_R(s)^\dagger K^k \hat D_R(s) \ee
(note that $k$ here is an index, not a power)
where $ \hat D_R(s) \equiv e^{ {\cor -} \ii s \hat n \cdot \vec J } $, so $s$ parametrizes the angle of rotation.  
Show that \eqref{eq:K-is-a-vector} implies that $K(s)$ satisfies the ODE
\be \partial_s K(s)^k = \eps^{kil} n^i K(s)^l \ee
with initial condition $K(0)^k = K^k$.  

Using uniqueness theorems about solutions of linear ODEs that 
\be\label{eq:solution-with-J} K(s)^k = \( D_{(j=1)}(s \hat n)\) ^k {}_j K^j , \ee
where $\( D_{(j=1)}(s \hat n)\) ^k {}_j$ are the matrix elements of the spin one representation
of this rotation.  


\end{enumerate}


\item {\bf Lorentz algebra in $D=3+1$.}

\begin{enumerate}

\item Check the algebra satisfied by rotations and boosts
\be\label{eq:4dlorentz}
[ J^i, J^j] = \ii \eps^{ijk} J^k , ~~~[ J^i, K^j] = \ii \eps^{ijk} K^k,
~~~  [ K^i, K^j] = - \ii \eps^{ijk} J^k \ee
using the explicit matrices in the vector representation given in lecture, namely
\be\label{eq:vector-lorentz-rotation} J^i = \begin{pmatrix} 0 & \cr &  \JJ^i   \end{pmatrix}  \ee
(where the $3\times 3$ matrix ${\bf J^i}$ is $({\bf J}^i)^j_k ={-} \ii \eps^{ijk}$)
%$ \( \JJ^i \)_{jk} = {\cor -}\ii \eps^{ijk} $)
and 
\be\label{eq:vector-lorentz-boost} \( K^i\)^{j}{}_{0} =  \ii \delta^i_j  = \( K^i\)^{0}{}_{j}\ee 
and all other components zero.  

For this purpose, I think typing them into Mathematica and writing $K.J-J.K$ etc.... is perfectly acceptable.


\item Check that 
in terms of the antisymmetric tensor of operators
$$ J^{\mu\nu} = \begin{cases}
\eps^{ijk} J^k, & \mu \nu = ij \\
K^i, & \mu\nu = 0i
\end{cases} ,$$ 
\eqref{eq:4dlorentz} can be rewritten as
\be\label{eq:general-d} [ J^{\mu\nu}, J^{\rho \sigma}] = \ii \( \eta^{\nu\rho} J^{\mu \sigma} 
+ \eta^{\mu\sigma} J^{\nu\rho } - ( \mu \leftrightarrow \nu ) \) ,\ee
which is the form of the $\mathsf{so}(d,1)$ Lie algebra for general $d$.
Make sure you take advantage of symmetries to avoid working too hard.



\item Show that in $D=3+1$, \eqref{eq:4dlorentz} is equivalent to 
$\mathsf{su}(2)_L \times \mathsf{su}(2)_R$: 
$$ [J_+^i, J_-^j] = 0, ~~~~~[J_\pm^i, J_\pm^j ]= \ii \eps^{ijk}J_\pm^k $$
in terms of 
$$ J_\pm^i \equiv \half (J^i \pm \ii K^i). $$


\item{} {\bf The Shayan-Arani operator.} [Bonus problem]
Consider the object $C \equiv \vec K \cdot \vec J$ that Shasha asked about during lecture.   
During lecture I said that it's rotation invariant, but I thought that it might not be invariant under the full Lorentz group.

Show that in fact it {\it is} invariant under the full Lorentz group (i.e. $[ \vec K, C] = 0, [\vec J, C] = 0 $).  So it is in fact a Casimir for the Lorentz group.  
That is, in {\it any} irrep it has to be proportional to the identity operator.  
It will be some function of the quantum numbers $j_L$ and $j_R$.  Find that function by relating it to other Casimirs and/or computing it in some familiar representations.


\end{enumerate}

\item {\bf A representation of the Clifford algebra gives a representation of Lorentz.} [Bonus problem]
Show the following:
Given a collection of $k\times k$ matrices satisfying $ \{ \gamma^\mu, \gamma^\nu\}  = 2 \eta^{\mu\nu} $
(the Clifford algebra, with $\mu, \nu = 0..d$), 
we can make a $k$-dimensional representation of $SO(1,d)$ with generators
$$ J^{\mu\nu} = { \ii \over 4 } [ \gamma^\mu, \gamma^\nu ] . $$
As an intermediate step, it is helpful to show that 
$$[  J^{\mu\nu} , \gamma^\rho] \propto \( \gamma^\mu \eta^{\nu\rho} - \gamma^\nu \eta^{\rho \mu}\) . $$
Convince yourself that this last equation says
that $\gamma^\rho$ transforms as a four-vector, \ie~
$$ { [ \gamma^\rho, J_\text{Dirac}^{\mu\nu} ] = \( J_\text{vector}^{\mu\nu}\)^\rho{}_\sigma \gamma^\sigma }.$$



\item {\bf Bispinors.}

\begin{enumerate}
\item Show that any product of gamma matrices between two spinors 
\be V^{\mu_1 \cdots \mu_n} \equiv  \bar \Psi \gamma^{\mu_1} \cdots \gamma^{\mu_n} \Psi \ee
is a tensor, in the sense that 
\be  V^{\mu_1 \cdots \mu_n}\mapsto \Lambda^{\mu_1}{}_{\nu_1} \cdots
\Lambda^{\mu_n}{}_{\nu_n} V^{\nu_1 \cdots \nu_n}. \ee
(A {\it tensor} is defined to be an object that transforms this way under Lorentz transformations.)



\item 
Let $ \gamma^{\mu\nu} \equiv \half [\gamma^\mu, \gamma^\nu ] $ be just the antisymmetric bit,
and similarly for more indices.
Show that any bispinor $ \Gamma_{ab} \bar \Psi_a \Psi_b $ can be 
decomposed as a sum of these tensors:
$\sum_n  A_{\mu_1 \cdots \mu_n } \bar \Psi \gamma^{\mu_1 \cdots \mu_n} \Psi  $,
where the $A$s are totally antisymmetric.



\end{enumerate}


\item {\bf Gamma matrices in other dimensions.}
\begin{enumerate}
\item Find a collection of $D$ matrices satisfying the Clifford algebra $ \{\gamma^\mu, \gamma^\nu\}  = 2 \eta^{\mu\nu} $ in $D=2$.  Are they the smallest possible?



\item Find a collection of $D$ matrices satisfying the Clifford algebra $ \{\gamma^\mu, \gamma^\nu\}  = 2 \eta^{\mu\nu} $ in $D=3$.  Are they the smallest possible?



\item Find a collection of $D$ matrices satisfying the Clifford algebra $ \{\gamma^\mu, \gamma^\nu\}  = 2 \eta^{\mu\nu} $ in $D=5$.  Are they the smallest possible?


\item{} [bonus problem] Find a collection of $D$ matrices satisfying the Clifford algebra $ \{\gamma^\mu, \gamma^\nu\}  = 2 \eta^{\mu\nu} $ in $D=6$, or any higher dimension.  Are they the smallest possible?


\end{enumerate}


\end{enumerate}

The following problems are about discrete symmetries of scalar field theories. They are a useful warmup to the discussion of discrete symmetries of the Dirac fermion.

\begin{enumerate}[resume]

%\item {\bf Brain warmer.}
\item {\bf Brain warmer: $\IZ_2$ symmetry of real scalar field theory.}

What does the operator 
$$ U \equiv e^{ \ii \pi \sum_k \aa_k^\dagger \aa_k } $$
do to the real scalar field 
$$ \phi(x)  \to \phi'(x) = U \phi(x) U^\dagger $$
whose ladder operators are $ \aa, \aa^\dagger$ ?  

For which Lagrangians is this a symmetry?  


\item {\bf Charge conjugation in complex scalar field theory.} [Bonus problem]

\label{prob:C}

Consider again a free complex Klein-Gordon field $\Phi$.
Define a discrete symmetry operation (charge conjugation) $C$, by
$$ \Phi(x) \mapsto C \Phi(x) C^{-1} = \eta_c \Phi^\dagger(x) $$
where $C$ is a unitary operator, and $\eta_c$ is an arbitrary phase factor.
Assume that the vacuum is invariant under charge conjugation: $ C \ket{0} = \ket{0}$.
\begin{enumerate}
\item Show that the free lagrangian is invariant under $C$, but the particle number current $j^\mu$
changes sign.
\item Show that the annihilation operators satisfy
$$ C \aa_k C^{-1} = \eta_c \bb_k,~~~C \bb_k C^{-1} = \eta_c^\star \aa_k $$
and hence show that $C$ interchanges particle
and antiparticle states, up to a phase.


\end{enumerate}  


\item {\bf Parity symmetry of scalar field theory.} [Bonus problem]

\label{prob:P}
Under the parity transformation
$$ \vec x \mapsto  \vec x' = - \vec x $$
a real Klein-Gordon transforms as 
\be\label{eq:parity-def} \phi(t, \vec x) \mapsto  P \phi(t, x) P^{-1} = \eta_p \phi(t, -\vec x )  \ee
where $P$ is unitary and $\eta_P = \pm 1$ is the {\it intrinsic parity} of the field $\phi$.
Again assume $ P \ket{0} = \ket{0}.$



\begin{enumerate}

\item Show that the parity transformation preserves the free 
Lagrangian (though not the Lagrangian density),
for both values of $\eta_P$.



\item Show that an arbitrary $n$-particle state transforms as 
$$ P \ket{ \vec k_1, \cdots \vec k_n} = \eta_P^n \ket{ - \vec k_1, \cdots, - \vec k_n} . $$

\item Here we give an explicit realization of the parity operator.  
Let
$$ P_1 \equiv e^{ - \ii { \pi \over 2 } \sum_k \aa^\dagger_k \aa^\nd_k } , ~~~
P_2 \equiv e^{ \ii\eta_p { \pi \over 2 } \sum_k \aa^\dagger_k \aa^\nd_{-k} } .$$
Show that 
$$ P_1 \aa_k P_1^{-1} =   \ii \aa_k , ~~~
P_2 \aa_k P_2^{-1} = -\ii \eta_p \aa_{-k} .$$
Hint: Use the following version of the Campbell-Baker-Hausdorff formula
$$ e^{ \ii \alpha A} B e^{ - \ii \alpha A } = \sum_{n=0}^\infty {( \ii \alpha)^n \over n!} B_n $$
where $B_0 \equiv B$ and $ B_n = [ A, B_{n-1} ]$ for $n=1,2....$.  

Show that $P\equiv P_1 P_2 $ is unitary, and satisfies \eqref{eq:parity-def}.


\item {\bf Action on the current of a complex scalar field.}  
Consider now a complex scalar field.  
Using the results from problems \ref{prob:C} and the preceding parts of \ref{prob:P},
find the action of parity on the particle current $j^\mu \mapsto P j^\mu P^{-1} $.
%and $j^\mu \mapsto C j^\mu C^{-1} $.
(You'll have to extend the action of $P$ from the case of a real field to the complex case.)



\end{enumerate}



\end{enumerate}
\end{document}