Second quantization of field operators

  • Thread starter grilo
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  • #1
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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?
 

Answers and Replies

  • #2
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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: [tex]\delta(x-y) f(y) = \delta(x-y) f(x)[/tex]. 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.
 
  • #3
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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!
 

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