# Second quantization of field operators

## 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?

Related Advanced Physics Homework Help News on Phys.org
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.

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!