How to show that the charge conjugation operator reverses the charge(s) of a (fermionic or bosonic) state?
Let us consider a spin-$\frac{1}{2}$ fermionic state of momentum $\textbf{k}$ and spin projection $s$ given by $a_s^\dagger(\textbf{k})|0\rangle$. The normal ordered charge operator associated with the symmetry $\psi\to e^{iq}\psi$ is given by $$Q=q\int d^3\textbf{p}\sum\limits_{s=1,2}\Big[a_s^\dagger(\textbf{p})a_s(\textbf{p})-b_s^\dagger(\textbf{p})b_s(\textbf{p})\Big]$$ where $a^\dagger, b^\dagger$ are respectively the particle and antiparticle creation operators. It is a trivial matter to show that $$Qa_s^\dagger(\textbf{k})|0\rangle=qa_s^\dagger(\textbf{k})|0\rangle~~ \text{and}~~ Qb_s^\dagger(\textbf{k})|0\rangle=-qb_s^\dagger(\textbf{k})|0\rangle.$$
Now, the charge conjugation operator is given by $$C=i\gamma^2\gamma^0.$$ I want to show that $$Q(Ca_s^\dagger(\textbf{k})|0\rangle)=-q(Ca_s^\dagger(\textbf{k})|0\rangle)$$ or something like $$Q(Ca_s^\dagger(\textbf{k})|0\rangle)=b_s^\dagger(\textbf{k})|0\rangle.$$
Am I on the right track? If yes, any hint how to act C on the state $a_s^\dagger(\textbf{k})|0\rangle$?
Answer
Yes, you are on the right track.
The rest of this answer is about the distinction between the charge conjugation operator $C_{op}$ and the charge conjugation matrix $C_{mat}$. If this misses the point of the question, let me know and I'll revise (or delete) the answer.
The field operator $\psi$ is an array of operators — a Dirac spinor whose four components each act as an individual operator on the Hilbert space. A Dirac matrix $\gamma^\mu$ does not act on the Hilbert space; it only mixes the components of the field operator $\psi$. When we write an expression like $\gamma^0\psi|0\rangle$, we are mixing matrix notation (it is a column matrix in which each component is a whole state-vector) and operator notation (because each component of $\gamma^0\psi$ is a different operator acting on $|0\rangle$).
The charge conjugation operator $C_{op}$ exchanges states of the form $$ \int d^3x\ \sum_n f_n(x)\psi_n(x)|0\rangle \tag{1a} $$ with states of the form $$ \int d^3x\ \sum_n g_n(x)\psi^*_n(x)|0\rangle, \tag{1a} $$ where $f$ and $g$ are ordinary functions with the same number of components as $\psi$, and where $\psi^*_n$ denotes the operator adjoint of $\psi_n$. (If the components of $\psi$ are $\psi_n$, then the components of $\psi^\dagger$ are $\psi^*_n$.) We want $C_{op}$ to exchange these because $\psi$ has components that create antiparticles (and components that annihilate particles), and $\psi^\dagger$ has components that create particles (and components that annihilate antiparticles).
The key point is that the charge conjugation matrix $C_{mat}$ is used to define the charge conjugation operator like this: $$ C_{op}\psi C_{op}^{-1} \sim C_{mat}\psi^* \tag{2} $$ where $\psi^*$ denotes the component-wise operator adjoint of $\psi$, without any matrix transpose. Again, this equation mixes matrix notation and operator notation. On the left-hand side, $\psi$ is a column-matrix, and the operator transformation $C_{op}\cdots C_{op}^{-1}$ acts on each individual element of this matrix. On the right-hand side, $\psi^*$ is again a column matrix, the transpose of the row matrix $\psi^\dagger$, and $C_{mat}$ is a square matrix of ordinary numbers; this matrix product mixes the components of $\psi^*$ with each other.
The matrix $C_{mat}$ can be expressed as some combination of Dirac matrices (as shown in the question), but an explicit expression for the operator $C_{op}$ requires more. We could presumably express it as some combination of field operators (or creation and annihilation operators), but usually we just define it through its effect on the field operators, as in equation (2). After writing $\psi$ in terms of creation/annihilation operators, equation (2) tells us how $C_{op}$ affects them, too.
The matrix $C_{mat}$ should be chosen so that the effect of $C_{op}$ on the current $q\overline{\psi}\gamma^\mu\psi$ is equivalent to reversing the sign of the charge $q$, because this is the current that couples to $A_\mu$ in the Lagrangian. According to this requirement, different representations for the $\gamma$-matrices lead to different representations for the matrix $C_{mat}$. One common representation leads to $C_{mat}\propto\gamma^0\gamma^2$, as stated in the question.
When the matrix $C_{mat}$ is chosen to satisfy that requirement, the operator $C_{op}$ will do something like $$ C_{op}a^\dagger C_{op}^{-1} \sim b^\dagger \hskip2cm C_{op}a C_{op}^{-1} \sim b $$ as anticipated by the question. Then, since the vacuum is invariant under $C_{op}$, we have $$ C_{op}a^\dagger C_{op}^{-1}|0\rangle = C_{op}a^\dagger |0\rangle \sim b^\dagger |0\rangle. $$ The fact that $C_{op}$ reverses the charge of the state is guaranteed by the fact that it reverses the sign of the charge density operator: $$ C_{op}QC_{op}^{-1}=-Q \hskip1cm \text{with } Q\sim q\overline{\psi}\gamma^0\psi = q\psi^\dagger\psi\sim q(a^\dagger a - b^\dagger b). $$ I glossed over a lot of details in this answer. The main point is that $C_{op}$ and $C_{mat}$ are two different things — related, but different.
No comments:
Post a Comment