Representation of second quantization

  • Thread starter Petar Mali
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  • #1
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Main Question or Discussion Point

In coordinate representation in QM probality density is:

[tex]\rho(\vec{r})=\psi^*(\vec{r})\psi(\vec{r})[/tex]


in RSQ representation operator of density of particles is

[tex]\hat{n}(\vec{r})=\hat{\psi}^{\dagger}(\vec{r})\hat{\psi}(\vec{r})[/tex]

Is this some relation between this operator and density matrix?

Operator of number of particles is

[tex]\hat{N}=\int d^3\vec{r}\hat{\psi}^{\dagger}(\vec{r})\hat{\psi}(\vec{r})[/tex]

Why I can now use

[tex]\hat{\psi}^{\dagger}(\vec{r})=\sum_k\hat{a}_k^{\dagger}\varphi^*_k(\vec{r})\qquad \hat{\psi}(\vec{r})=\sum_k\hat{a}_k\varphi_k(\vec{r})[/tex] ?

where [tex]\{\varphi_k\}[/tex] is complete ortonormal set.

Thanks
 

Answers and Replies

  • #2
alxm
Science Advisor
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[tex]\hat{n}(\vec{r})=\hat{\psi}^{\dagger}(\vec{r})\hat{\psi}(\vec{r})[/tex]

Is this some relation between this operator and density matrix?
Which operator? What's [tex]\hat{\psi}^{\dagger}[/tex]? Do you mean [tex]a^\dagger|\psi>[/tex]?

Anyway, the matrix elements for creation/annhilation operators are:
[tex]<m|a^\dagger|n> = \sqrt{n+1}\delta_{m,n+1}[/tex]
[tex]<m|a|n> = \sqrt{n}\delta_{m,n-1}[/tex]

They don't form a complete set; the creation/annihilation operators aren't self-adjoint.
 
  • #3
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Which operator? What's [tex]\hat{\psi}^{\dagger}[/tex]? Do you mean [tex]a^\dagger|\psi>[/tex]?

Anyway, the matrix elements for creation/annhilation operators are:
[tex]<m|a^\dagger|n> = \sqrt{n+1}\delta_{m,n+1}[/tex]
[tex]<m|a|n> = \sqrt{n}\delta_{m,n-1}[/tex]

They don't form a complete set; the creation/annihilation operators aren't self-adjoint.

I defined

[tex]
\hat{\psi}^{\dagger}(\vec{r})=\sum_k\hat{a}_k^{\dagger}\varphi^*_k(\vec{r})\qquad \hat{\psi}(\vec{r})=\sum_k\hat{a}_k\varphi_k(\vec{ r})
[/tex]

[tex]\hat{a}_k^{\dagger},\hat{a}_k[/tex] are operators you talking about and [tex]
\{\varphi_k\}
[/tex] form complete set. You did not read my post.
 
  • #4
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I usually think of the density matrix as something like
[tex]
\hat\rho(r,r') = \hat\psi^\dagger(r) \hat\psi(r')
[/tex]
so your density operator would be the diagonal element.

I think your question about the transform relies on the assumption that you can write
[tex]
\delta(r) = \sum_k \phi_k(r)
[/tex]
which certainly works for [tex]\phi_k[/tex] as plane waves but I am not sure about other basis sets.

It's late and I'm tired so I apologize if this doesn't make much sense.
 

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