I have tried to understand paragraph 10.7 (Kallen-Lehmann Representation) in Weinberg's Quantum theory of fields (vol.1). He calculated matrix element

$$\langle0|\Phi(0)|p\rangle =(2\pi)^{-3/2}\left(2\sqrt{p^{2}+m^{2}}\right)^{-1/2}N.\tag{formula 10.7.19}$$

$N$ is constant. I can't understand how it was obtained. I don't even undestand why it depends on $p$. If we consider the Lorenz invariance of vacuum and operator $\Phi(0)$:

$$U_{\Lambda}|0\rangle=|0\rangle\ \ \text{ and } \ \ U_{\Lambda}\Phi(0)U_{\Lambda}^{-1}=\Phi(0)$$

then we have:




So my first guess was that this matrix element does not depend on $p$ and equals to constant.

Can you help me?


1 Answer 1


First remark: The correct expression is

$$\langle0|\Phi(0)|p\rangle =(2\pi)^{-3/2}\left(2\sqrt{\vec{p}^{2}+m^{2}}\right)^{-1/2}N=(2\pi)^{-3/2}\left(2p^0\right)^{-1/2}N,$$

where $\vec{p}$ represents a spatial vector.

Second remark: Under a homogeneous Lorentz transformation $\Lambda$, the creation operator of a scalar transform as

$$U_0(\Lambda) a(\vec{p}) U_0(\Lambda)^{-1}=\sqrt{(\Lambda p)^0/p^0} a(\vec{p_\Lambda}). $$

Now, the solution:

The idea is to isolate all the terms depending on $p$ from your expression. Using the invariance of the vacuum and the scalar under Lorentz transformations, we can write

$$\langle0|\Phi(0)|p\rangle=\langle0|\Phi(0)a^\dagger(\vec{p})|0\rangle= \langle0|\Phi(0)U_0^{-1}(\Lambda)a^\dagger(\vec{p})U_0(\Lambda)|0\rangle=\sqrt{(\Lambda p)^0/p^0}\langle 0|\Phi(0)a^\dagger(\vec{p_\Lambda})|0\rangle.$$

In this expression we are free to choose a convenient Lorentz transformation $\Lambda$. If I take a boost that brings the particle to its rest frame, then $\vec{p_\Lambda}=0$ and $(\Lambda p)^0=m$. Therefore, if we define $N$ by

$$N=(2\pi)^{3/2} {(2m)}^{1/2}\langle 0|\Phi(0)a^\dagger(0)|0\rangle, $$ then we have the desired result (Note that $N$ does not depend on the four moment $p$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.