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

+ 2 like - 0 dislike

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
asked Oct 27, 2013 in Theoretical Physics by PPR (135 points) [ no revision ]

1 Answer

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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 MarkWayne (270 points) [ no revision ]
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

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