I am working on problem 2.2 Part d of Peskin and Schroeder's An Introduction to Quantum Field Theory. The authors claim that there are three charges based on the three Pauli Matrices and that that these charges are given by:
\begin{align}
Q^a = \frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \bigg[ \alpha_{j\textbf{p}}^\dagger \sigma^a_{jk} \alpha_{k\textbf{p}} – \beta_{j\textbf{p}}^\dagger \sigma^a_{jk} \beta_{k\textbf{p}} \bigg].
\end{align}
The problem statement is
Consider the case of two complex Klein Gordon fields with the same mass. Show that there are four conserved charges and that these charges obey the same commutation relations that angular momenta satisfy. Repeat all this for $n$ identical complex Klein Gordon fields.
However, this is not what I found. Here is what I mean.
The Lagrangian density is given by:
\begin{align*}
L = \partial_\mu \vec{\Phi}^\dagger \partial^\mu \vec{\Phi} – m^2 \vec{\Phi}^\dagger \vec{\Phi}
\end{align*}
We clearly see that the Lagrangian is invariant under $U(2)$ symmetry. The $U(2)$ group has four generators: $1$ and $\frac{1}{2}\sigma^a$, where $\sigma^a$ are the Pauli spin matrices. Now let us perturb the fields as follows:
Hence we have:
\begin{align*}
\Phi_i &\to \big[e^{-i\theta_a \frac{1}{2} \sigma^a}\big]_{ij} \Phi_j \approx \big[1 – i\theta_a \frac{1}{2} \sigma^a \big]_{ij} \Phi_j = \big(\delta_{ij} – i\theta_a \frac{1}{2} \sigma^a_{ij}\big) \Phi_j = \Phi_i – i\theta_a \frac{1}{2} \sigma^a_{ij}\Phi_j\\
\Phi_i &\to \big[e^{-i\theta 1}\big]_{ij} \Phi_j \approx \big[1 – i\theta 1 \big]_{ij} \Phi_j = \big(\delta_{ij} – i\theta \delta_{ij}\big) \Phi_j = \Phi_i – i\theta \Phi_i
\end{align*}
We see that the change in the Lagrangian density is given by:
\begin{align*}
\Delta L &= \partial_\mu \bigg(\frac{\partial L}{\partial (\partial_\mu \Phi_k)} \Delta \Phi_k \bigg) + \partial_\mu \bigg( \frac{\partial L}{\partial (\partial_\mu \Phi_k^*)} \Delta \Phi_k^* \bigg)\\
\end{align*}
Now consider the Noether current for the Pauli matrix generator. We see that the formula for the current is:
\begin{align*}
\theta_a \partial_\mu J^\mu &= \Delta L \Longrightarrow \therefore \partial_\mu\bigg[\theta_a J^\mu – \frac{\partial L}{\partial (\partial_\mu \Phi_k)} \Delta \Phi_k – \frac{\partial L}{\partial (\partial_\mu \Phi_k^*)} \Delta \Phi_k^* \bigg] = 0
\end{align*}
\begin{align*}
(J^\mu)^a &= -g^{\mu\mu} \partial_\mu \Phi_j^* \big(\frac{1}{2}i \sigma^a_{jk} \big) \Phi_k + g^{\mu\mu} \partial_\mu \Phi_j \big(\frac{i}{2} \sigma^{a*}_{jk} \big) \Phi_k^*\\
&= -\partial^\mu \Phi_j^* \big(\frac{1}{2}i \sigma^a_{jk} \big) \Phi_k + \partial^\mu \Phi_j \big(\frac{i}{2} \sigma^{a*}_{jk} \big) \Phi_k^*\\
&= -\frac{i}{2} \bigg[ \partial^\mu \Phi_j^* \sigma^{a}_{jk} \Phi_k – \partial^\mu \Phi_j \sigma^{a*}_{jk} \Phi_k^* \bigg]\\
\end{align*}
Now if the Pauli matrices are $\sigma^1$ or $\sigma^3$, then $\sigma^{a*} = \sigma^a$. In these cases the conserved charges based on the Pauli spin matrices are the following.
\begin{align*}
Q^a &= -\frac{i}{2} \int d^3 x \bigg[ \dot{\Phi}_j^* \sigma^a_{jk} \Phi_k – \Phi_j^* \sigma^a_{jk} \dot{\Phi}_k \bigg] = \frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \bigg[ \alpha_{j\textbf{p}}^\dagger \sigma^a_{jk} \alpha_{k\textbf{p}} – \beta_{j\textbf{p}}^\dagger \sigma^a_{jk} \beta_{k\textbf{p}} \bigg]\\
\end{align*}
However, if we use $\sigma^2$, we note that that $\sigma^{a*} = -\sigma^a$. In this case the Norther charge is:
\begin{align*}
Q^a &= -\frac{i}{2} \int d^3 x \bigg[ \dot{\Phi}_j^* \sigma^a_{jk} \Phi_k + \Phi_j^* \sigma^a_{jk} \dot{\Phi}_k \bigg] = \frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \bigg[ \alpha_{j\textbf{p}}^\dagger \sigma^a_{jk} \alpha_{k\textbf{p}} + \beta_{j\textbf{p}}^\dagger \sigma^a_{jk} \beta_{k\textbf{p}} \bigg]\\
\end{align*}
Question: Is there something I am doing incorrectly? If so what did I do wrong? Although my work seems correct to me, I believe I must be wrong somewhere, since the authors state that all three charges have the same form.
Best Answer
Your steps are very detailed. I believe future readers will benefit a lot from this question.
Your result is correct. To get the same form as Peskin's, notice that the order of $\phi$ and $\phi^*$ is different in $Q^i$ $$Q^i=\int d^3x\frac{i}{2}\left(\phi^*_a(\sigma^i)_{ab}\pi^*_b-\pi_b(\sigma^i)_{ab}\phi_a\right).$$ This hints you to use $$(\sigma^i)^\dagger=\sigma^i\quad\Rightarrow\quad(\sigma^i)^*_{ab}=(\sigma^i)_{ba}.$$ Then your two cases for $1,3$ and $2$ are unified.
Or written out explicitely. $Q^i$ is defined to be $$Q^i:=\pi^*_a\Delta\phi^*_a+\pi_a\Delta\phi_a.$$ Using $$ \Delta\phi_a=-\frac{i}{2}\theta(\sigma^i)_{ab}\phi_b,\quad \Delta\phi^*_a=\frac{i}{2}\theta(\sigma^i)^*_{ab}\phi^*_b=\frac{i}{2}\theta(\sigma^i)_{ba}\phi^*_b.$$ So finally $$Q^i\propto\pi^*_a(\sigma^i)_{ba}\phi^*_b-\pi_a(\sigma^i)_{ab}\phi_b.$$