Return to Question

Under $U(1)$ symmetry of the Dirac Lagrangian I get, up to a constant, a current $$j^\mu=\bar\psi\gamma^\mu\psi.$$ How can I express it using creation and annihilation operators? For example it's easy for $\mu=0$ but for $\mu=i$ I don't know how to proceed.

I can write $$\psi=\int d\phi_k\left(\sum_ia_i(k)u_i(k)e^{-ikx}+\sum_ib^\dagger_i(k)v_i(k)e^{ikx}\right)$$ so $$\bar\psi\gamma^0\psi=\psi^\dagger\psi=\int d\phi_k d\phi_{k'}\left(\sum_{i}a^\dagger_i(k)u^\dagger_i(k)e^{ikx}+\sum_ib_i(k)v^\dagger_i(k)e^{-ikx}\right)\left(\sum_ja_j(k)u_j(k)e^{-ikx}+\sum_jb^\dagger_j(k)v_j(k)e^{ikx}\right)$$

Now I have to multiply and use the identities for example $$u^\dagger_iu_j=2E_k\delta_{ij}$$ $$u_i(k)^\dagger v_j(-k)=0$$ How can I proceed with $\mu=i$? I have to evaluate $$\bar\psi\gamma^i\psi=\psi^\dagger\gamma^0\gamma^i\psi=\int d\phi_k d\phi_{k'}\left(\sum_{i}a^\dagger_i(k)u^\dagger_i(k)e^{ikx}+\sum_ib_i(k)v^\dagger_i(k)e^{-ikx}\right)\gamma^0\gamma^i\left(\sum_ja_j(k)u_j(k)e^{-ikx}+\sum_jb^\dagger_j(k)v_j(k)e^{ikx}\right)$$

Under $U(1)$ symmetry of the Dirac Lagrangian I get, up to a constant, a current $$j^\mu=\bar\psi\gamma^\mu\psi.$$ How can I express it using creation and annihilation operators? For example it's easy for $\mu=0$ but for $\mu=i$ I don't know how to proceed.

I can write $$\psi=\int d\phi_k\left(\sum_ia_i(k)u_i(k)e^{-ikx}+\sum_ib^\dagger_i(k)v_i(k)e^{ikx}\right)$$ so $$\bar\psi\gamma^0\psi=\psi^\dagger\psi=\int d\phi_k d\phi_{k'}\left(\sum_{i}a^\dagger_i(k)u^\dagger_i(k)e^{ikx}+\sum_ib_i(k)v^\dagger_i(k)e^{-ikx}\right)\left(\sum_ja_j(k)u_j(k)e^{-ikx}+\sum_jb^\dagger_j(k)v_j(k)e^{ikx}\right)$$

Now I have to multiply and use the identities for example $$u^\dagger_iu_j=2E_k\delta_{ij}$$ $$u_i(k)^\dagger v_j(-k)=0$$ How can I proceed with $\mu=i$? I have to evaluate $$\bar\psi\gamma^i\psi=\psi^\dagger\gamma^0\gamma^i\psi=\int d\phi_k d\phi_{k'}\left(\sum_{i}a^\dagger_i(k)u^\dagger_i(k)e^{ikx}+\sum_ib_i(k)v^\dagger_i(k)e^{-ikx}\right)\gamma^0\gamma^i\left(\sum_ja_j(k')u_j(k')e^{-ik'x}+\sum_jb^\dagger_j(k')v_j(k')e^{ik'x}\right)$$

Under $U(1)$ symmetry of the Dirac Lagrangian I get, up to a constant, a current $$j^\mu=\bar\psi\gamma^\mu\psi.$$ How can I express it using creation and annihilation operators? For example it's easy for $\mu=0$ but for $\mu=i$ I don't know how to proceed.

Under $U(1)$ symmetry of the Dirac Lagrangian I get, up to a constant, a current $$j^\mu=\bar\psi\gamma^\mu\psi.$$ How can I express it using creation and annihilation operators? For example it's easy for $\mu=0$ but for $\mu=i$ I don't know how to proceed.

I can write $$\psi=\int d\phi_k\left(\sum_ia_i(k)u_i(k)e^{-ikx}+\sum_ib^\dagger_i(k)v_i(k)e^{ikx}\right)$$ so $$\bar\psi\gamma^0\psi=\psi^\dagger\psi=\int d\phi_k d\phi_{k'}\left(\sum_{i}a^\dagger_i(k)u^\dagger_i(k)e^{ikx}+\sum_ib_i(k)v^\dagger_i(k)e^{-ikx}\right)\left(\sum_ja_j(k)u_j(k)e^{-ikx}+\sum_jb^\dagger_j(k)v_j(k)e^{ikx}\right)$$

Now I have to multiply and use the identities for example $$u^\dagger_iu_j=2E_k\delta_{ij}$$ $$u_i(k)^\dagger v_j(-k)=0$$ How can I proceed with $\mu=i$? I have to evaluate $$\bar\psi\gamma^i\psi=\psi^\dagger\gamma^0\gamma^i\psi=\int d\phi_k d\phi_{k'}\left(\sum_{i}a^\dagger_i(k)u^\dagger_i(k)e^{ikx}+\sum_ib_i(k)v^\dagger_i(k)e^{-ikx}\right)\gamma^0\gamma^i\left(\sum_ja_j(k)u_j(k)e^{-ikx}+\sum_jb^\dagger_j(k)v_j(k)e^{ikx}\right)$$

Under U(1) symmetry of the Dirac Lagrangian I get, up to a constant, a current $$j^\mu=\bar\psi\gamma^\mu\psi$$ How can I express it using creation and annihilation operators? For example it's easy for $\mu=0$ but for $\mu=i$ I don't know how to proceed.

Under $U(1)$ symmetry of the Dirac Lagrangian I get, up to a constant, a current $$j^\mu=\bar\psi\gamma^\mu\psi.$$ How can I express it using creation and annihilation operators? For example it's easy for $\mu=0$ but for $\mu=i$ I don't know how to proceed.