# 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): $$U(\Lambda)|p,\sigma\rangle = \sum_{\sigma'} C_{\sigma' \sigma}(\Lambda,p)|\Lambda p, \sigma'\rangle$$ 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)): $$U(\Lambda) a^s_{\mathbf{p}}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}} a_{\Lambda \mathbf{p}}^s$$ 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: $$U(\Lambda)|p,\sigma\rangle = C|\Lambda p, \sigma\rangle$$ 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 $C=1$): $$U(\Lambda)|\mathbf{p},s\rangle = \sqrt{2 E_{\mathbf{p}}} U(\Lambda)a^{s\dagger}_\mathbf{p}U^{-1}(\Lambda)|0\rangle=|\Lambda \mathbf{p},s\rangle = \sqrt{2 E_{\Lambda \mathbf{p}}} a^{s\dagger}_{\Lambda \mathbf{p}}|0\rangle$$ which implies: $$U(\Lambda)a^{s\dagger}_\mathbf{p}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}} a^{s\dagger}_{\Lambda \mathbf{p}}$$ Subsequently, using the unitarity of $U(\Lambda)$, we can take the adjoint of the above equation: $$U(\Lambda)a^{s}_\mathbf{p}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}} a^{s}_{\Lambda \mathbf{p}}$$ which is equation (3.109).

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.