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## Homework Statement

(from "Advanced Quantum Mechanics", by Franz Schwabl)

Show, by verifying the relation

[tex]\[n(\bold{x})|\phi\rangle = \delta(\bold{x}-\bold{x'})|\phi\rangle\][/tex],

that the state

[tex]\[|\phi\rangle = \psi^\dagger(\bold{x'})|0\rangle\][/tex]

([tex]\[|0\rangle =\][/tex]vacuum state) describes a particle with the position [tex]\bold{x'}[/tex].

## Homework Equations

The particle density operator [tex]n(\bold{x})[/tex] is defined as

[tex]n(\bold{x}) = \psi^\dagger(\bold{x})\psi(\bold{x})[/tex]

## The Attempt at a Solution

Acting on the given state with the particle density operator, I got

[tex]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[/tex]

by the (fermion) boson (anti-)commutation rules. Since [tex]\psi(\bold{x})[/tex] annihilates the vacuum:

[tex]n(\bold{x})|\phi\rangle = \delta(\bold{x}-\bold{x'})\psi^\dagger(\bold{x})|0\rangle[/tex]

which looks like the given equation, but [tex]\psi^\dagger(\bold{x})|0\rangle[/tex] describes a particle at the position [tex]\bold{x}[/tex].

Integrating the last equation in [tex]\bold{x}[/tex] gives back the original state [tex]|\phi\rangle[/tex], though.

I'm not sure wheter I misunderstood something or it's just a matter of interpretation.

Can anyone help me?