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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:

$$\langle0|\Phi(0)|p\rangle=\langle0|U_{\Lambda}^{\dagger}\Phi(0)|p\rangle=\langle0|U_{\Lambda}^{-1}\Phi(0)|p\rangle=\langle0|\Phi(0)U_{\Lambda}^{-1}|p\rangle=\langle0|\Phi(0)|U_{\Lambda}^{-1}p\rangle,$$

so

$$\langle0|\Phi(0)(p)=\langle0|\Phi(0)(U_{\Lambda}^{-1}p).$$

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

Can you help me?

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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 $p$).

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