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

  1. Jun 15, 2008 #1
    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} =

    -2U \qquad \text{for} \qquad \eta_{1}=\eta_{2}=\Uparrow,\: \eta_{2}=\eta_{4}=\Downarrow\\
    0 \qquad \text{elsewhere}

    V is also given by

    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})

    Now, what is [tex]V(x_{j}-x_{k})[/tex] ?
  2. jcsd
  3. Jun 15, 2008 #2
    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] ?
  4. Jun 18, 2008 #3
    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

    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)

    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].
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