Anderson Hamiltonian (product of number operators) in 1st quantization?

  • Thread starter AA1983
  • Start date
  • #1
5
0

Main Question or Discussion Point

In the Anderson model, it cost an energy [tex]Un_{\Uparrow}n_{\Downarrow}[/tex] for a quantum dot level to be occupied by two electrons. Here [tex]n_{\Uparrow}[/tex] is the second quantized number operator, counting the number of particles with spin [tex]\Uparrow[/tex]. I need the term [tex]Un_{\Uparrow}n_{\Downarrow}[/tex] in first quantization. Here is what I know:

[tex]Un_{\Uparrow}n_{\Downarrow} =
Ud_{\Uparrow}^{\dagger}d_{\Uparrow}d_{\Downarrow}^{\dagger}d_{\Downarrow}
=
-Ud_{\Uparrow}^{\dagger}d_{\Downarrow}^{\dagger}d_{\Uparrow}d_{\Downarrow}
=
\frac{1}{2}\sum_{\eta_{1}\eta_{2}\eta_{3}\eta_{4}}V_{\eta_{1}\eta_{2}\eta_{3}\eta_{4}}d_{\eta_{1}}^{\dagger}d_{\eta_{2}}^{\dagger}d_{\eta_{3}}d_{\eta_{4}}
[/tex]
where

[tex]
V_{\eta_{1}\eta_{2}\eta_{3}\eta_{4}}=\Big\{
\begin{array}{c}
-2U \qquad \text{for} \qquad \eta_{1}=\eta_{2}=\Uparrow,\: \eta_{2}=\eta_{4}=\Downarrow\\
0 \qquad \text{elsewhere}
\end{array}[/tex].

V is also given by

[tex]
V_{\eta_{1}\eta_{2}\eta_{3}\eta_{4}}=\int dx_{j} dx_{k} \psi_{\eta_{1}}^{\ast}(x_{j})\psi_{\eta_{2}}^{\ast}(x_{k})V(x_{j}-x_{k})
\psi_{\eta_{3}}(x_{j})\psi_{\eta_{4}}(x_{k})[/tex]

Now, what is [tex]V(x_{j}-x_{k})[/tex] ?
 

Answers and Replies

  • #2
5
0
Is it [tex]V=-2U\delta(x_{j}-x_{k})\delta_{\eta_{1}\Uparrow}\delta_{\eta_{2}\Downarrow}\delta_{\eta_{3}\Uparrow}\delta_{\eta_{4}\Downarrow}[/tex] ?
 
  • #3
334
1
I agree in principle, but shouldn't [itex]V_{\eta_{1}\eta_{2}\eta_{3}\eta_{4}}[/itex] be non-zero for other combinations of indices? For example [itex]\eta_{1}=\eta_{4}=\uparrow, \eta_{2}=\eta_{3}=\downarrow[/itex] should probably be allowed, since you're not creating or annihilating two of the same type of spin. Also, since flipping spins means swapping two pairs of fermionic operators in [itex]-Ud_{\Uparrow}^{\dagger}d_{\Downarrow}^{\dagger}d_{ \Uparrow}d_{\Downarrow}[/itex], you won't pick up a minus sign, so it should probably be

[tex]
V=-2U\delta(x_{j}-x_{k})\left(\delta_{\eta_{1}\eta_{3}}\delta_{\eta_{2}\eta_{4}} - \delta_{\eta_{1}\eta4}\delta_{\eta_{2}\eta_{3}}\right)
[/tex]

so [itex]\eta_{1}[/itex] is either the same as [itex]\eta_{3}[/itex] or [itex]\eta_{4}[/itex], and it picks up a minus sign in the latter. Similarly for [itex]\eta_{2}[/itex].
 

Related Threads for: Anderson Hamiltonian (product of number operators) in 1st quantization?

Replies
5
Views
7K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
0
Views
1K
Replies
1
Views
717
  • Last Post
Replies
5
Views
3K
Replies
3
Views
3K
Replies
0
Views
696
  • Last Post
Replies
5
Views
2K
Top