# Normalization of the real Klein Gordon Field in Peskin and Schroeder chapter 2

In Peskin & Schroeder's QFT, how do you get from equation 2.35 to 2.37? (In particular, how does the invariant normalization of the Klein-Gordon real field imply that $U(\Lambda)|p> = |\Lambda p>$ ?)

Also, on a more general note, could some explain why for the real Klein-Gordon field we need to make the effort to define invariant normalization? In particular, why do we care if the expression $<q|p>$ is invariant if it is $|<q|p>|^2$ which bears physical meaning?

In answer to the first part of your question, let's work backward. We write \begin{align} U(\Lambda)|\mathbf{p}\rangle &= \sqrt{2E_{\mathbf{p}}}U(\Lambda)a^\dagger_{\mathbf{p}}U^\dagger(\Lambda)U(\Lambda)|0\rangle \\ &= \sqrt{2E_{\mathbf{p}}}[U(\Lambda)a^\dagger_{\mathbf{p}}U^\dagger(\Lambda)]|0\rangle, \end{align} where we used $U(\Lambda)|0\rangle=|0\rangle$ and we have \begin{align} a^\dagger_{\Lambda\mathbf{p}}&=\sqrt{\frac{E_{\mathbf{p}}}{E_{\Lambda\mathbf{p}}}} U(\Lambda)a^\dagger_{\mathbf{p}}U^\dagger(\Lambda), \end{align} which must holds since \begin{align} U(\Lambda) a^\dagger_{\mathbf{p}} \sqrt{E_{\mathbf{p}}}&= a^\dagger_{\Lambda\mathbf{p}}U(\Lambda) \sqrt{E_{\Lambda\mathbf{p}}}. \end{align} Applying this to the vacuum state demonstrates the equation you're asking about. (Note that we've neglected spin throughout.)
• Thanks very much. I have two questions, first, how should I convince myself that $U(\Lambda)A_p^{\dagger} = A_{\Lambda p}^{\dagger} U(\Lambda)$? The only thing I could think of is that "creating a particle with momentum $p$ and then making the transformation $\Lambda$ is like first making the transformation $\Lambda$ and then creating a particle with momentum $\Lambda p$." The second question is, this equation doesn't seem to be compatible with equation 2.38, which involves the energies too. How do you settle that? Finally, about the motivation for normalization, could you elaborate more? – PPR Oct 28 '13 at 19:38
• You're right about Eq.(2.38). I made a mistake -- it should have been $a^\dagger_{\Lambda\mathbf{p}} = \sqrt{ \frac{E_{\mathbf{p}}}{E_{\Lambda\mathbf{p}}} } U(\Lambda)a^\dagger_{\mathbf{p}}U^\dagger(\Lambda)$. And the way you've convinced yourself is essentially the proof. – MarkWayne Oct 29 '13 at 2:38