Trying to rhyme Peskin and Schroeder with Weinberg This is a follow up question of this one.
In the Vol 1, Weinberg derives how a unitary operator $U(\Lambda)$ acts on one-particle states, which is given by equation (2.5.2):
\begin{equation}
U(\Lambda)|p,\sigma\rangle = \sum_{\sigma'} C_{\sigma' \sigma}(\Lambda,p)|\Lambda p, \sigma'\rangle
\end{equation}
Am I correct in thinking that Peskin and Schroeder use this result to derive the Lorentz transformation on a annihilation operator of the quantized Dirac field (given by equation (3.106)):
\begin{equation}
U(\Lambda) a^s_{\mathbf{p}}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}} a_{\Lambda \mathbf{p}}^s
\end{equation}
However, Peskin and Schroeder are only considering Lorentz transformations that do not mix states with different spin indices! In other words, the first equation I wrote becomes:
\begin{equation}
U(\Lambda)|p,\sigma\rangle = C|\Lambda p, \sigma\rangle
\end{equation}
where $C$ is normalization constant which can be determined to be $C=1$. If this is not correct, can you please let me know how they (Peskin and Schroeder) have derived equation (3.109)?
I hope this question is not too confusing.
Edit: for you to see my reasoning. We can now write (for $C=1$):
\begin{equation}
U(\Lambda)|\mathbf{p},s\rangle = \sqrt{2 E_{\mathbf{p}}} U(\Lambda)a^{s\dagger}_\mathbf{p}U^{-1}(\Lambda)|0\rangle=|\Lambda \mathbf{p},s\rangle = \sqrt{2 E_{\Lambda \mathbf{p}}} a^{s\dagger}_{\Lambda \mathbf{p}}|0\rangle 
\end{equation}
which implies:
\begin{equation}
U(\Lambda)a^{s\dagger}_\mathbf{p}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}} a^{s\dagger}_{\Lambda \mathbf{p}}
\end{equation}
Subsequently, using the unitarity of $U(\Lambda)$, we can take the adjoint of the above equation:
\begin{equation}
U(\Lambda)a^{s}_\mathbf{p}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}} a^{s}_{\Lambda \mathbf{p}}
\end{equation}
which is equation (3.109).
 A: I think it is correct. However, just as a pedantic remark, you should prove $U(\Lambda)a^{s\dagger}_\mathbf{p}U^{-1}(\Lambda)$ and $\sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}} a^{s\dagger}_{\Lambda \mathbf{p}}$ act on all the vectors in the same way, not just on the vacuum $|0\rangle$, you need a slight modification of your proof:
$$\sqrt{2 E_{\mathbf{p}}}U(\Lambda)a^{s\dagger}_\mathbf{p}|\mathbf{p_1},s_1;\mathbf{p_2},s_2;\cdots\rangle=U(\Lambda)|\mathbf{p},s;\mathbf{p_1},s_1;\mathbf{p_2},s_2\cdots\rangle\\
=|\Lambda\mathbf{p},s;\Lambda(\mathbf{p_1},s_1);\Lambda(\mathbf{p_2},s_2)\cdots\rangle $$
The second equality is justified since we have assumed $s$ is parallel to the rotation or boost direction, and $s_1,s_2,\cdots$ might not, that is why I labeled the them differently as $\Lambda(\mathbf{p_1},s_1)$. To continue, we have
$$=\sqrt{2 E_{\Lambda \mathbf{p}}} a^{s\dagger}_{\Lambda \mathbf{p}}|\Lambda(\mathbf{p_1},s_1);\Lambda(\mathbf{p_2},s_2)\cdots\rangle\\
=\sqrt{2 E_{\Lambda \mathbf{p}}} a^{s\dagger}_{\Lambda \mathbf{p}}U(\Lambda)|\mathbf{p_1},s_1;\mathbf{p_2},s_2;\cdots\rangle$$
Since the collection $\{|\mathbf{p_1},s_1;\mathbf{p_2},s_2;\cdots\rangle\}$ is a basis of the vector space, 
$$\sqrt{2 E_{\mathbf{p}}}U(\Lambda)a^{s\dagger}_\mathbf{p}=\sqrt{2 E_{\Lambda \mathbf{p}}} a^{s\dagger}_{\Lambda \mathbf{p}}U(\Lambda)$$
holds as an operator identity.
