# Hamiltonian of BCS theory

1. Oct 8, 2009

### Petar Mali

$$\hat{H}_{BCS}=\sum_{\vec{p},\sigma}\epsilon(\vec{p})\hat{a}^+_{\vec{p},\sigma}\hat{a}_{\vec{p},\sigma}+\sum_{\vec{p},\vec{p}'}V(\vec{p},\vec{p}')\hat{a}^+_{\vec{p}\uparrow}\hat{a}^+_{-\vec{p}\downarrow}\hat{a}_{-\vec{p}'\downarrow}\hat{a}_{\vec{p}'\uparrow}$$

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

If I work mean- field approximation

$$\hat{H}_{BCS}=\hat{H}_0+\hat{H}_2+\delta\hat{H}$$

What is the procedure to find terms $$\hat{H}_0$$, $$\hat{H}_2$$, $$\delta\hat{H}$$?

2. Oct 8, 2009

### Petar Mali

And what is meaning of

$$\sum_{\vec{p},\sigma}\epsilon(\vec{p })\hat{a}^+_{\vec{p},\sigma}\hat{a}_{\vec{p},\sigma}$$

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

Can I say that

$$\hat{a}^+_{\vec{p},\sigma}=\hat{a}^+_{\vec{p}}\hat{\xi}_{\sigma}$$?

3. Oct 8, 2009

### xepma

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 $$a^\dag a$$ 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 $$a^\dag$$ is an operator which creates an electron with momentum p and spin sigma.

4. Oct 9, 2009

### Petar Mali

Thanks for you're answer. Just to ask in this product of four operators you have $$\hat{a}^+_{\vec{p}\uparrow}$$ and $$\hat{a}^+_{-\vec{p}\downarrow}$$ so

$$\hat{a}^+_{\vec{p}\uparrow}\hat{a}^+_{-\vec{p}\downarrow}$$

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

$$\hat{a}^+_{\vec{p}\uparrow}\hat{a}^+_{\vec{p}\downarrow}$$

And I can't see when I look this Hamiltonian where are Cooper pairs?

5. Oct 9, 2009

### xepma

You are summing over all values of p and p', so the sum over $$a^\dag_p a^\dag_{-p'}$$ is the same as if you write down $$a^\dag_p a^\dag_{p'}$$. 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)

6. Oct 9, 2009

### Petar Mali

You wanted to say that Hamiltonian I wrote is the same as the Hamiltonian

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

Can I treat this like

first sum - the kinetic energy of electrons
second sum - creation Cooper pair in state with impulse $$\vec{p}$$ and anihilation Cooper pair in state with impulse $$\vec{p}'$$

7. Oct 9, 2009

### xepma

Yes, this new Hamiltonian is an approximation to the original one.

For a treatment on this see Tinkham - Introduction to superconductivity, chapter 3

8. Oct 13, 2009

### Petar Mali

Is it easy way to get $$\hat{H}_0,\hat{H}_2,\delta\hat{H}$$

where $$\hat{H}_0+\hat{H}_2=\hat{H}_{MF}$$ ?

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