# Quantum field creates particle interpretation

In Peskin & Schroeder's book, the authors say that if $$\phi(x)$$ is the (quantum) Klein-Gordon field operator in Schrödinger's picture and $$|0\rangle$$ is the vacuum state, the application $$\phi(x)|0\rangle = \int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{2E_{p}}e^{-ip\cdot x} |p\rangle$$ is interpreted as creation of a particle at position $$x$$. The argument is that, apart from the factor $$1/2E_{p}$$ in the above formula, the expression is exactly the eigenstate of the position $$|x\rangle$$ for non-relativistic QM.

Can someone help me understand this interpretation?

My points:

(a) Aren't the eigenstates of the position operator $$x$$ delta distributions? The integral on the right hand side of the above expression resembles the wave function of the free particle, not the eigenstates of the position to me.

(b) The authors say that $$\langle x | p\rangle = e^{ip\cdot x}$$ helps the aforementioned interpretation. How come?

In non-relativistic QM, generalized eigenstates of the position operator are delta distributions in the position basis, but are plane waves in the momentum basis, i.e. $$|x\rangle = \int \mathrm dp \ e^{-ipx} |p\rangle$$. See e.g. here, trying to ignore the rather unpleasant typsetting in the second half of the page.
As for $$(b)$$, note that precisely the same relation holds in nonrelativistic QM, see e.g. here.
• It's there becuase the plane wave $|p\rangle$ is normalized so that $\langle p|p'\rangle= (2\pi)^3 2 E_p\delta^3(p-p')$. This is natural as the particle density increases with $E$ because of Lorentz contraction. – mike stone Feb 13 at 12:26