Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hamiltonian of BCS theory

  1. Oct 8, 2009 #1
    [tex]\hat{H}_{BCS}=\sum_{\vec{p},\sigma}\epsilon(\vec{p})\hat{a}^+_{\vec{p},\sigma}\hat{a}_{\vec{p},\sigma}+\sum_{\vec{p},\vec{p}'}V(\vec{p},\vec{p}')\hat{a}^+_{\vec{p}\uparrow}\hat{a}^+_{-\vec{p}\downarrow}\hat{a}_{-\vec{p}'\downarrow}\hat{a}_{\vec{p}'\uparrow}[/tex]


    What is the meaning of the terms [tex]\hat{a}^+_{\vec{p}\uparrow},\hat{a}^+_{-\vec{p}\downarrow}[/tex]... ?

    If I work mean- field approximation

    [tex]\hat{H}_{BCS}=\hat{H}_0+\hat{H}_2+\delta\hat{H}[/tex]


    What is the procedure to find terms [tex]\hat{H}_0[/tex], [tex]\hat{H}_2[/tex], [tex]\delta\hat{H}[/tex]?
     
  2. jcsd
  3. Oct 8, 2009 #2
    And what is meaning of

    [tex]\sum_{\vec{p},\sigma}\epsilon(\vec{p })\hat{a}^+_{\vec{p},\sigma}\hat{a}_{\vec{p},\sigma}[/tex]

    and second part of the Hamiltonian! Second part is some interraction.

    Can I say that

    [tex]\hat{a}^+_{\vec{p},\sigma}=\hat{a}^+_{\vec{p}}\hat{\xi}_{\sigma}[/tex]?
     
  4. Oct 8, 2009 #3
    The a's are creation operators (the one with the daggers). They create an electron with momentum p and spin up or down.

    The usual procedure in BCS theory is to determine the V(p,p') dependence. You'll notice that the relevant contributions come from the V(p,-p) terms (which is the reason why electrons with opposite momentum are paired up).

    The first term can be seen as the energy carried by an electron. The summation over [tex]a^\dag a[/tex] counts the number of electrons, and the epsilon is the energy packket carried by each electron. You can see view it as sort of a kinetic energy term.

    You cannot make that replacement, since [tex]a^\dag[/tex] is an operator which creates an electron with momentum p and spin sigma.
     
  5. Oct 9, 2009 #4
    Thanks for you're answer. Just to ask in this product of four operators you have [tex]\hat{a}^+_{\vec{p}\uparrow}[/tex] and [tex]\hat{a}^+_{-\vec{p}\downarrow}[/tex] so

    [tex]\hat{a}^+_{\vec{p}\uparrow}\hat{a}^+_{-\vec{p}\downarrow}[/tex]

    Why you don't have product of [tex]\hat{a}^+_{\vec{p}\uparrow}[/tex] and [tex]\hat{a}^+_{\vec{p}\downarrow}[/tex] so

    [tex]\hat{a}^+_{\vec{p}\uparrow}\hat{a}^+_{\vec{p}\downarrow}[/tex]


    And I can't see when I look this Hamiltonian where are Cooper pairs?
     
  6. Oct 9, 2009 #5
    You are summing over all values of p and p', so the sum over [tex]a^\dag_p a^\dag_{-p'}[/tex] is the same as if you write down [tex]a^\dag_p a^\dag_{p'}[/tex]. So the sum also includes the term you mention. It includes all possible pairs of momentum at this stage.

    What's more important is the function V(p,p'). It's this formula that determines what the interaction energy is between two electrons - one carrying momentum p and one carrying momentum p'. What makes BCS theory so special, is that you can show that the function is peaked around V(p,-p). I.e. the binding energy between two electrons carrying opposite momentum is the largest contributions to the system.

    Why this is so: well, you have to look up what the exact shape is of V(p,-p) (it included integrating over the degrees of freedom from the phonon-electron interaction -- very complicated stuff). So no, you cannot see from this Hamiltonian why there are Cooper pairs. You need the exact shape of V(p,p)
     
  7. Oct 9, 2009 #6
    You wanted to say that Hamiltonian I wrote is the same as the Hamiltonian

    [tex]
    \hat{H}_{BCS}=\sum_{\vec{p},\sigma}\epsilon(\vec{p })\hat{a}^+_{\vec{p},\sigma}\hat{a}_{\vec{p},\sigma}+\sum_{\vec{p},\vec{p}'}\hat{a}^+_{\vec{p}\uparrow}\hat{a}^+_{\vec{p}\downarrow}\hat{a}_{\vec{p'}\uparrow}\hat{a}_{\vec{p}'\downarrow}[/tex]

    Can I treat this like

    first sum - the kinetic energy of electrons
    second sum - creation Cooper pair in state with impulse [tex]\vec{p}[/tex] and anihilation Cooper pair in state with impulse [tex]\vec{p}'[/tex]
     
  8. Oct 9, 2009 #7
    Yes, this new Hamiltonian is an approximation to the original one.

    For a treatment on this see Tinkham - Introduction to superconductivity, chapter 3
     
  9. Oct 13, 2009 #8
    Is it easy way to get [tex]\hat{H}_0,\hat{H}_2,\delta\hat{H}[/tex]

    where [tex]\hat{H}_0+\hat{H}_2=\hat{H}_{MF}[/tex] ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Hamiltonian of BCS theory
  1. BCS theory (Replies: 1)

  2. BCS theory (Replies: 7)

  3. BCS theory (Replies: 25)

  4. BCS theory (Replies: 4)

  5. BCS Hamiltonian (Replies: 2)

Loading...