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

+ 2 like - 0 dislike
2158 views

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?

This post imported from StackExchange Physics at 2014-07-13 04:43 (UCT), posted by SE-user PPR

+ 0 like - 0 dislike

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 $U(\Lambda)|0\rangle$ 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.)

Your second question needs to be clarified. You seem to be conflating Lorentz covariance with the issue of Lorentz invariance. The amplitudes are not invariant. They are covariant.

This post imported from StackExchange Physics at 2014-07-13 04:43 (UCT), posted by SE-user MarkWayne
answered Oct 27, 2013 by (270 points)
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?

This post imported from StackExchange Physics at 2014-07-13 04:43 (UCT), posted by SE-user PPR
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.

This post imported from StackExchange Physics at 2014-07-13 04:43 (UCT), posted by SE-user MarkWayne

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.