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.
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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.
It's somewhat hard to understand what you mean by "interpretation". The only right interpretation is the right calculation. It is an operator that gives something if it acts on a state, and all these answers may be calculated. They shouldn't be interpreted, they should be calculated.
answered Sep 24, 2012 at 18:23
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$\begingroup$ 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. $\endgroup$– WhelpCommented Sep 24, 2012 at 20:14
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$\begingroup$ Whelp, that interpretation makes sense if we remember that we can think of each field operator acting on the vacuum bra/ket that they are next to, and that we are then taking the overlap of those two single particle states. $\endgroup$ Commented Sep 24, 2012 at 20:52
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$\begingroup$ @Lubos, well we can interpret things "differently" depending on whether we are thinking e.g. in the language of particle physics versus field integrals. Of course it really means the same thing, so convincing ourselves that an "interpretation" is correct is just a matter of using the dictionary to translate the one we are unsure about into the language we happen to more intuitively understand. $\endgroup$ Commented Sep 24, 2012 at 20:56
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1$\begingroup$ 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. $\endgroup$ Commented Oct 3, 2012 at 6:10