# Interpretation of field operator

Consider a real scalar field operator $\varphi$. It can be written in terms of creation and anihilation operators as $$\varphi(\textbf{x})=\int \tilde{dk}[ a(k)e^{i\textbf{kx}}+a(k)^{\dagger}e^{-i\textbf{kx}}]$$ where $\tilde{dk}$ is a Lorenz-invariant measure. If $\varphi$ is interpreted as creating a particle at $\textbf{x}$ when acting on the vacuum, what is its action on a generic state? It seems to be creating a superposition of a state with one added quantum of energy through the creation operator, and a state with one less quanta of energy through the annihilation operator.

As the formula clearly shows, $\phi(x)$ cannot be interpreted as a pure creation operator of any type. It is a combination of creation and annihilation operators. Creation operators are those called $a(k)^\dagger$ and annihilation operators are called $a(k)$.
So yes, if $\phi(x)$ acts on a generic state with a well-defined number of particles $N$, it produces a linear superposition of states that have $N+1$ and $N-1$ particles, respectively. When it acts on the vacuum, for example, however, the annihilation operator piece drops out and it creates a 1-particle state.
• Thanks, this is very helpful. The origin of my question is that I've often seen in textbooks correlation functions of the type $<0|T\varphi (x_1)\varphi (x_2)|0>$ are often described as a process where a particle is created at $x_2$, travels to $x_1$, and is then annihilated there. – Whelp Sep 24 '12 at 20:14
• Dear @Whelp, that statement is also correct because all the annihilation operators in $\phi(x_2)$ simply annihilate the vacuum ket on the right, and all the creation operators in $\phi(x_1)$ annihilate the vacuum bra on the left, so what is left is only the annihilation part of $\phi(x_1)$ and creation part of $\phi(x_2)$. However, if you had more general states in which the operators are sandwiched, you couldn't drop 1/2 of the terms this easily. – Luboš Motl Oct 3 '12 at 6:10