This is part 2 of exercise II.1.1 of Zee's QFT in a Nutshell (here's part 1).

This is what I have got:

\begin{align} \bar\psi\gamma^\lambda\psi \mapsto \bar\psi^{\,\prime}\gamma^\lambda\psi^{\,\prime} & = {\psi^{\,\prime}}^\dagger{\gamma^0}^\dagger\gamma^\lambda\psi^{\,\prime} = (S(\Lambda)\psi)^\dagger\gamma^0\gamma^\lambda(S(\Lambda)\psi) \\ & = \psi^\dagger S(\Lambda)^\dagger\gamma^0\gamma^\lambda S(\Lambda)\psi\\ & = \tag{1}\psi^\dagger e^{\frac i4\omega_{\mu\nu}{\sigma^{\mu\nu}}^\dagger}\gamma^0\gamma^\lambda e^{-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}}\psi\\ & = \psi^\dagger\left(1+\frac i4\omega_{\mu\nu}(\sigma^{\mu\nu})^\dagger+O({\omega_{\mu\nu}}^2)\right)\gamma^0\gamma^\lambda\left(1-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}+O({\omega_{\mu\nu}}^2)\right)\psi\\ & = \psi^\dagger\gamma^0\left(\gamma^\lambda+\frac i4\omega_{\mu\nu}[[\gamma^\mu,\gamma^\nu],\gamma^\lambda]+O({\omega_{\mu\nu}}^2)\right)\psi\\ & = \psi^\dagger\gamma^0\left(\gamma^\lambda+\frac 12\omega_{\mu\nu}[\sigma^{\mu\nu},\gamma^\lambda]+O({\omega_{\mu\nu}}^2)\right)\psi \end{align}

Two questions:

• How exactly does the last line prove that $\bar\psi\gamma^\lambda\psi$ transforms as a vector under Lorentz transformations? It certainly looks like a vector transformation to me, because of the commutator between the Lorentz generators and the "vector components" $\gamma^\mu$, but how can I prove this quantitatively?
• Then, a more general question: How can I get rid of the $O({\omega_{\mu\nu}}^2)$? I know that it can be ignored since we consider infinitesimal transformations only, so the transformation behavior is governed by first-order terms only. But what's the mathematical rigorous way to get rid of them? Do I just write a $\simeq$ sign and leave them out at the right-hand side of the $\simeq$? That would not seem right to me, because when dealing with expansions, a $\simeq$ does not imply strict equality, it just implies "equality up to a certain order".

This is part 2 of exercise II.1.1 of Zee's QFT in a Nutshell (here's part 1).

This is what I have got:

\begin{align} \bar\psi\gamma^\lambda\psi \mapsto \bar\psi^{\,\prime}\gamma^\lambda\psi^{\,\prime} & = {\psi^{\,\prime}}^\dagger{\gamma^0}^\dagger\gamma^\lambda\psi^{\,\prime} = (S(\Lambda)\psi)^\dagger\gamma^0\gamma^\lambda(S(\Lambda)\psi) \\ & = \psi^\dagger S(\Lambda)^\dagger\gamma^0\gamma^\lambda S(\Lambda)\psi\\ & = \tag{1}\psi^\dagger e^{\frac i4\omega_{\mu\nu}{\sigma^{\mu\nu}}^\dagger}\gamma^0\gamma^\lambda e^{-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}}\psi\\ & = \psi^\dagger\left(1+\frac i4\omega_{\mu\nu}(\sigma^{\mu\nu})^\dagger+O({\omega_{\mu\nu}}^2)\right)\gamma^0\gamma^\lambda\left(1-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}+O({\omega_{\mu\nu}}^2)\right)\psi\\ & = \psi^\dagger\gamma^0\left(\gamma^\lambda+\frac i4\omega_{\mu\nu}[[\gamma^\mu,\gamma^\nu],\gamma^\lambda]+O({\omega_{\mu\nu}}^2)\right)\psi\\ & = \psi^\dagger\gamma^0\left(\gamma^\lambda+\frac 12\omega_{\mu\nu}[\sigma^{\mu\nu},\gamma^\lambda]+O({\omega_{\mu\nu}}^2)\right)\psi \end{align}

Two questions:

• How exactly does the last line prove that $\bar\psi\gamma^\lambda\psi$ transforms as a vector under Lorentz transformations? It certainly looks like a vector transformation to me, because of the commutator between the Lorentz generators and the "vector components" $\gamma^\mu$, but how can I prove this quantitatively?
• Then, a more general question: How can I get rid of the $O({\omega_{\mu\nu}}^2)$? I know that it can be ignored since we consider infinitesimal transformations only, so the transformation behavior is governed by first-order terms only. But what's the mathematical rigorous way to get rid of them? Do I just write a $\simeq$ sign and leave them out at the right-hand side of the $\simeq$? That would not seem right to me, because when dealing with expansions, a $\simeq$ does not imply strict equality, it just implies "equality up to a certain order".