Mistake in eq. (10.7.19) Weinberg Vol I?

Question

In Weinberg Vol I, he writes in equation 10.7.19, $$ \left\langle 0 | \phi (0) | {\mathbf{k}} \right\rangle = ( 2\pi ) ^{ - 3/2} \left( 2 \sqrt{ {\mathbf{k}} ^2 + m ^2 } \right) ^{ - 1/2} N, \tag{10.7.19}$$ where $\phi $ is a unrenormalized scalar field, $ \left| {\mathbf{k}} \right\rangle $ is a one-particle state, and $ N $ is some constant. The $ {\mathbf{k}} $ dependence seems wrong. For one thing, this is often taken as the renormalization condition and set equal to $ 1 $. Furthermore, its easy to show the braket should be $ {\mathbf{k}} $-independent by performing a Lorentz transform: \begin{align*} \left\langle 0 | \phi (0) | {\mathbf{k}} \right\rangle & = \left\langle 0 | U ( \Lambda ) ^\dagger \phi (0) U ( \Lambda ) | {\mathbf{k}} \right\rangle \\ & = \left\langle 0 | \phi (0) | \Lambda {\mathbf{k}} \right\rangle \end{align*} Since I am free to choose $ \Lambda $ as I please, the result can't be dependent on $ {\mathbf{k}} $. Is this a mistake or am I overlooking something?

Answer 1

I think the problem is you are taking the inner product to be Lorentz Invariant. In Weinberg's convention the inner products are not Lorentz Invariante, see Eqn. 2.5.19. For a covariant normalization the $\sqrt{2E}$ factor would be absent and your argument would go through.