# Srednicki's QFT: Why $\langle p|\phi(0)|0\rangle$ in the interacting theory is Lorentz invariant?

I am reading Srednicki's QFT and I have met a problem. In its section 5, (5.18) , after deducing the LSZ formula, in order to check whether his supposition "that the creation operators of free field theory would work comparably in the interacting theory" is reasonable, the author considered the number $$\langle p|\phi(x)|0\rangle$$, where $$|p\rangle$$is the one-particle state of momentum $$p$$ and $$\phi$$ is a field operator in the interacting theory. Using the unitary representation of translation we get $$\langle p|\phi(x)|0\rangle=\langle p|e^{-iPx}\phi(0)e^{iPx}|0\rangle=\langle p|\phi(0)|0\rangle e^{-ipx},$$then the author states that since $$\langle p|\phi(0)|0\rangle$$ is a Lorentz invariant number, and the only Lorentz invariant number that we can construct from $$p$$ is $$p^2=-m^2$$,so we can devide $$\phi$$ by a constant number to get the same normalization for one-particle momentum state as free field.

My question is, how can we know that $$\langle p|\phi(0)|0\rangle$$ is a Lorentz invariant number? I guess (not sure) maybe that's because under a Lorentz tranformation $$\Lambda$$, $$\phi(x)$$ becomes $$U(\Lambda)^{-1}\phi (x)U(\Lambda)$$=$$\phi(\Lambda^{-1}x)$$ and $$\Lambda^{-1}0=0$$ so $$\phi(0)$$ does not change,thus $$\langle p|\phi(0)|0\rangle$$ does not change. However, I met a contradiction. Since both $$\langle p|\phi(0)|0\rangle$$ and $$e^{-ipx}$$ are Lorentz invariant, $$\langle p|\phi(x)|0\rangle$$ should be Lorentz invariant ,too. But under the transformation, $$\langle p|\phi(x)|0\rangle$$ becomes $$\langle p|\phi(\Lambda^{-1}x)|0\rangle \neq \langle p|\phi(x)|0\rangle$$, I don't understand where I made a mistake. How do we represent Lorentz transformation in quantum theory's formulation in Heisenberg picture? Should the state vector $$|p\rangle$$ change under transformation?

A function is Lorentz invariant if $$f(p) = f(\Lambda p)$$. Consider the function $$f(p) = \langle p| \phi(0) | 0 \rangle = \langle p| U(\Lambda)^{-1} U(\Lambda) \phi(0) U(\Lambda)^{-1} U(\Lambda) | 0 \rangle$$ where in the last equality, we simply introduced factors of $$1= U(\Lambda)^{-1} U(\Lambda)$$. We now use the fact that $$| 0 \rangle$$ is Lorentz invariant, $$U(\Lambda) | 0 \rangle = | 0 \rangle$$ and $$\langle p|$$ transforms as $$U(\Lambda) | p \rangle = | \Lambda p \rangle \quad \implies \quad \langle p |U(\Lambda)^{-1} = \langle \Lambda p | .$$ Finally $$\phi(x)$$ is a scalar field so $$U(\Lambda) \phi(x) U(\Lambda)^{-1} = \phi(\Lambda x) \quad \implies \quad U(\Lambda) \phi(0) U(\Lambda)^{-1} = \phi(0) .$$ Using all of this, we find $$f(p) = \langle \Lambda p| \phi(0) | 0 \rangle = f(\Lambda p)$$ Consequently, $$f(p)$$ is a Lorentz invariant function.
This is why we use the LIPS (Lorentz Invariant Phase Space) normalization $$\langle {\bf p}|{\bf p}'\rangle = (2\pi)^3 2E_{\bf p}\delta^3({\bf p}-{\bf p}')$$ for the single particle states. Without the $$2E_{\bf p}$$ matrix elements such as $$\langle p|\phi(x)|0\rangle= e^{ipx}$$ would not be Lorentz invariant.