1
$\begingroup$

Bardeen, Cooper and Schrieffer's (BCS) theory describes spinful Fermions that mutually interact via an attractive contact interaction. The general Hamiltonian reads in second quantization $$H = \sum_{\sigma \in \{\uparrow, \downarrow\}} \int d\mathbf{r} \, \psi_\sigma^\dagger(\mathbf{r}) \left( -\frac{\hbar^2 \nabla^2}{2m} - \mu \right) \psi_\sigma(\mathbf{r}) + V_0 \int d\mathbf{r} \, \psi_\uparrow^\dagger(\mathbf{r}) \psi_\downarrow^\dagger(\mathbf{r}) \psi_\downarrow(\mathbf{r}) \psi_\uparrow(\mathbf{r})$$ with the creation $\psi_\sigma^\dagger(\mathbf{r})$ and annihilation $\psi_\sigma(\mathbf{r})$ field operators, interaction strength $V_0 < 0$ and chemical potential $\mu$. This general ansatz is simplified using the mean-field approximation to obtain the BCS Hamiltonian.

$$H_{BCS} = \sum_{\sigma \in \{\uparrow, \downarrow\}} \int d\mathbf{r} \, \psi_\sigma^\dagger(\mathbf{r}) \left( -\frac{\hbar^2 \nabla^2}{2m} - \tilde{\mu} \right) \psi_\sigma(\mathbf{r}) + \int d\mathbf{r} \, \left( \Delta \psi_\uparrow^\dagger(\mathbf{r}) \psi_\downarrow^\dagger(\mathbf{r}) + \Delta \psi_\downarrow(\mathbf{r}) \psi_\uparrow(\mathbf{r}) \right)$$ with the pairing gap $\Delta = V_0 \langle \psi_\downarrow (\mathbf{r}) \psi_\uparrow (\mathbf{r}) \rangle$ and the redefined chemical potential $\tilde{\mu} = \mu - V_0 \langle \psi_\downarrow^\dagger (\mathbf{r}) \psi_\downarrow (\mathbf{r}) \rangle$. The parameters $\Delta$ and $\tilde{\mu}$ are spatially constant as we consider a homogeneous system.

Usually, the first step from the general ansatz towards the BCS Hamiltonian involves Wick's theorem. As I am not very familiar with Wick's theorem I have the following questions.

  1. As I understand, Wick's theorem is useful to rewrite products of operators as for example $\psi_\uparrow^\dagger (\mathbf{r}) \psi_\downarrow^\dagger (\mathbf{r}) \psi_\downarrow (\mathbf{r}) \psi_\uparrow (\mathbf{r})$ in an exact way. However, how does it help us to apply the mean-field approximation? What are the intermediate steps of the following calculation? To improve readability the position vector $\mathbf{r}$ is omitted.

$$\psi_\uparrow^\dagger \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow \approx % % pairing fields \langle \psi_\downarrow \psi_\uparrow \rangle \, \psi_\uparrow^\dagger \psi_\downarrow^\dagger + \langle \psi_\uparrow^\dagger \psi_\downarrow^\dagger \rangle \psi_\downarrow \psi_\uparrow \\ % % hartree fields + \langle \psi_\uparrow^\dagger \psi_\uparrow \rangle \psi_\downarrow^\dagger \psi_\downarrow + \langle \psi_\downarrow^\dagger \psi_\downarrow \rangle \psi_\uparrow^\dagger \psi_\uparrow \\ % % fock fields - \langle \psi_\uparrow^\dagger \psi_\downarrow \rangle \psi_\downarrow^\dagger \psi_\uparrow - \langle \psi_\downarrow^\dagger \psi_\uparrow \rangle \psi_\uparrow^\dagger \psi_\downarrow$$

  1. Is there an alternate way to apply the mean-field approximation without Wick's theorem? For example by using the idea of the theorem's proof or employing a similar technique as for the Ising model (here). Below you find my unsuccessful attempts.

"Ising model technique"

The idea is to write the product $A B$ of operators with their mean values $\langle A \rangle$ and $\langle B \rangle$ and the deviations $\delta A = A - \langle A \rangle$ and $\delta B = B - \langle B \rangle$.

$$A B = \langle A \rangle \langle B \rangle + \langle A \rangle \delta B + \langle B \rangle \delta A + \delta A \delta B$$

As the mean-field approximation assumes small fluctuations $\langle \delta A \delta B \rangle \ll \langle A \rangle \langle B \rangle$ we can neglect the term quadratic in the deviations. By inserting the definitions of $\delta A$ and $\delta B$ we get

$$A B \approx \langle A \rangle B + \langle B \rangle A - \langle A \rangle \langle B \rangle.$$

When we apply this approximation naively to the interaction term we identify $A = \psi_\uparrow^\dagger \psi_\downarrow^\dagger$ and $B = \psi_\downarrow \psi_\uparrow$ and obtain only the first two terms. The additional constant offset is not observable (here). Another attempt was to rewrite the expression with the fermionic commutation relations as follows before approximating it.

$$\psi_\uparrow^\dagger \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow = % \frac{1}{3} (\psi_\uparrow^\dagger \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow + \psi_\uparrow^\dagger \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow + \psi_\uparrow^\dagger \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow) \\ % = \frac{1}{3} (\psi_\uparrow^\dagger \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow + \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow^\dagger \psi_\uparrow - \psi_\downarrow^\dagger \psi_\uparrow \psi_\uparrow^\dagger \psi_\downarrow + \psi_\downarrow^\dagger \psi_\downarrow)$$ Besides the required terms there is a disturbing summand $\psi_\downarrow^\dagger \psi_\downarrow$ and an overall factor of $1/3$. Another idea is based on the observation that in mean-field $A B \approx B A$.

$$\psi_\uparrow^\dagger \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow = % \frac{1}{6} \cdot 6 \cdot \psi_\uparrow^\dagger \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow \\ % = \frac{1}{6} ( [\psi_\downarrow \psi_\uparrow \psi_\uparrow^\dagger \psi_\downarrow^\dagger + \psi_\uparrow^\dagger \psi_\uparrow - \psi_\downarrow \psi_\downarrow^\dagger] + [\psi_\uparrow^\dagger \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow] \\ % + [\psi_\uparrow^\dagger \psi_\uparrow \psi_\downarrow^\dagger \psi_\downarrow] + [\psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow^\dagger \psi_\uparrow] \\ % + [- \psi_\uparrow^\dagger \psi_\downarrow \psi_\downarrow^\dagger \psi_\uparrow + \psi_\uparrow^\dagger \psi_\uparrow] + [- \psi_\downarrow^\dagger \psi_\uparrow \psi_\uparrow^\dagger \psi_\downarrow + \psi_\downarrow^\dagger \psi_\downarrow]) \\ % % = \frac{1}{6} (\psi_\downarrow \psi_\uparrow \psi_\uparrow^\dagger \psi_\downarrow^\dagger + \psi_\uparrow^\dagger \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow % + \psi_\uparrow^\dagger \psi_\uparrow \psi_\downarrow^\dagger \psi_\downarrow + \psi_\downarrow^\dagger \psi_\downarrow \psi_\uparrow^\dagger \psi_\uparrow \\ % - \psi_\uparrow^\dagger \psi_\downarrow \psi_\downarrow^\dagger \psi_\uparrow - \psi_\downarrow^\dagger \psi_\uparrow \psi_\uparrow^\dagger \psi_\downarrow % + 2 \psi_\uparrow^\dagger \psi_\uparrow - \psi_\downarrow \psi_\downarrow^\dagger + \psi_\downarrow^\dagger \psi_\downarrow) $$

Also here the disturbing terms consisting of two field operators do not vanish.

$\endgroup$
  • $\begingroup$ Hi and welcome to the Physics SE. Can you clarify what you are asking? $\endgroup$ – John Rennie Dec 30 '17 at 6:24
  • $\begingroup$ Hi John, thanks for your comment. Now I enriched my question to make it hopefully clearer. $\endgroup$ – tree frog Dec 30 '17 at 22:43
  • $\begingroup$ I appreciate any hint that clarifies Wick's theorem used to get the BCS mean-field hamiltonian. Please also let me know if something is unclear. Thanks! $\endgroup$ – tree frog Jan 3 '18 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.