Second quantization of field operators

[grilo]

Homework Statement

(from "Advanced Quantum Mechanics", by Franz Schwabl)
Show, by verifying the relation
\[n(\bold{x})|\phi\rangle = \delta(\bold{x}-\bold{x'})|\phi\rangle\],
that the state
\[|\phi\rangle = \psi^\dagger(\bold{x'})|0\rangle\]
(\[|0\rangle =\]vacuum state) describes a particle with the position \bold{x'}.

Homework Equations

The particle density operator n(\bold{x}) is defined as
n(\bold{x}) = \psi^\dagger(\bold{x})\psi(\bold{x})

The Attempt at a Solution

Acting on the given state with the particle density operator, I got
n(\bold{x})|\phi\rangle = \psi^\dagger(\bold{x})\psi(\bold{x})\psi^\dagger(\bold{x'})|0\rangle = \psi^\dagger(\bold{x})(\delta(\bold{x}-\bold{x'}) \pm \psi^\dagger(\bold{x'})\psi(\bold{x}))|0\rangle
by the (fermion) boson (anti-)commutation rules. Since \psi(\bold{x}) annihilates the vacuum:
n(\bold{x})|\phi\rangle = \delta(\bold{x}-\bold{x'})\psi^\dagger(\bold{x})|0\rangle

which looks like the given equation, but \psi^\dagger(\bold{x})|0\rangle describes a particle at the position \bold{x}.
Integrating the last equation in \bold{x} gives back the original state |\phi\rangle, though.

I'm not sure wheter I misunderstood something or it's just a matter of interpretation.
Can anyone help me?

 

[xepma]

The delta function is only non-zero when x = x', so you are free to set these two parameters equal outside the delta function.

So remember: \delta(x-y) f(y) = \delta(x-y) f(x). We can just pretend x and y are equal outside the delta function. The expression is zero anyway when x and y are not equal.

 

[grilo]

Ah... thanks! I kinda thought that, but was afraid of being lousy. I try to keep my physics as mathematically rigorous as possible, so sometimes I run into those little issues. (:

Thanks again!