# BCS Theory: why does $\langle \hat{\Delta}^{\dagger} \hat{\Delta}\rangle \approx \langle \hat{\Delta}^{\dagger}\rangle \langle \hat{\Delta}\rangle$?

My textbook (Advanced QM by Yuli V. Nazarov) defines the Cooper pair creation operator as $$Z_{k}^{\dagger}=c_{k,\uparrow}^{\dagger}c_{-k,\downarrow}^{\dagger}.$$ The text then goes on to show that the BCS hamiltonian takes the form $$H^{(2)}=\frac{U}{V}\left(\sum_{k}Z_k^{\dagger}\right)\left(\sum_{k}Z_k\right)$$ which is then rewritten in the more tractable form $$H^{(2)}=\frac{V}{U}\hat{\Delta}^{\dagger} \hat{\Delta}$$ where we define $$\hat{\Delta}\equiv \frac{U}{V}\sum_kZ_k$$. when calculating the expectation value of $$H^{(2)}$$ for the ground state, the author makes a few deductions which I find challenging to understand. First, the author makes the approximation $$\langle \hat{\Delta}^{\dagger} \hat{\Delta}\rangle\approx \langle\hat{\Delta}^{\dagger}\rangle \langle\hat{\Delta}\rangle.$$ I am not sure why this approximation is valid though. The authors reasoning is that " because $$\langle \hat{\Delta}^{\dagger} \hat{\Delta}\rangle \propto \sum_{k,k'}\langle Z_k^\dagger Z_{k'}\rangle$$, the dominant contribution to the expectation value comes from the terms where $$k\neq k'$$ , we can safely assume that $$\langle \hat{\Delta}^{\dagger} \hat{\Delta}\rangle\approx \langle\hat{\Delta}^{\dagger}\rangle \langle\hat{\Delta}\rangle$$ ." I do not understand this reasoning though. Any help in explaining it would be most appreciated.

Secondly, once this approximation has been made, the author then goes on to show that $$\langle H^{(2)}\rangle=\frac{V}{U}\langle \hat{\Delta}^{\dagger} \hat{\Delta}\rangle \approx\frac{V}{U} \langle \hat{\Delta}^{\dagger}\rangle \langle\hat{\Delta}\rangle =\frac{V}{U}|\langle \hat{\Delta}\rangle|^2.$$ I also fail to understand the last equality in the above equation. The author seems to have assumed that $$\langle \hat{\Delta}^{\dagger}\rangle=\langle\hat{\Delta} \rangle^{*}$$. I can't understand why this assumption is valid. Once again, any help on this would be most appreciated!

To the second part (since @RogerVadim's answer has already addressed the first), writing the state as $$|\psi\rangle=\sum_ic_i|i\rangle$$ in a basis with $$\langle i|j\rangle=\delta_{ij}$$ (the continuous case is analogous):\begin{align}\left\langle\hat{\Delta}^\dagger\right\rangle-\left\langle\hat{\Delta}\right\rangle^\ast&=\sum_{ij}c_i^\ast c_j\left(\hat{\Delta}^\dagger\right)_{ij}-\left(\sum_{ji}c_j^\ast c_i\hat{\Delta}_{ji}\right)^\ast\\&=\sum_{ij}c_i^\ast c_j\left[\left(\hat{\Delta}^\dagger\right)_{ij}-\left(\hat{\Delta}_{ji}\right)^\ast\right].\end{align}The expression in square brackets is $$0$$ by definition.

• Thanks for this answer! Okay this totally clears up the second issue I brought up. Am I correct in thinking that this reasoning should apply to any operator $\hat{A}$ ? I am still not sure how to solve the first part of my question though. Oct 26 at 9:48
• @SalahTheGoat It's completely general, yes.
– J.G.
Oct 26 at 10:36
• Okay thanks for the great answer! Its helped a lot. I would accept it however I am still getting nowhere with the first part of my question. Oct 26 at 10:38

BCS theory is a Mean-Field theory.

In Mean-Field what you do is finding the best quadratic Hamiltonian that approximate the original non-quadratic Hamiltonian. A quadratic Hamiltonian can always be diagonalized returning a sum of independent single particle excitations. The eigenstates of this Hamiltonian will be Slater determinants of the product of those single particles eigenstates.

When you compute the expectation value of a non-quadratic operator on a Slater determinant you can use Wick theorem that states: $$\langle abcd \rangle = \langle ab \rangle \langle cd \rangle -\langle ac \rangle \langle bd \rangle + \langle ad \rangle \langle bc \rangle$$ Where $$a,b,c,d$$ are fermionic operators. Therefore is you have an Hamiltonian non-quadratic like $$c^\dagger_i c^\dagger_j c_k c_m$$the best quadratic Hamiltonian that will describe this non-quadratic one will be given by contributions that will give the same expectation value like: $$c^\dagger_i c^\dagger_j c_k c_m \rightarrow c^\dagger_i c^\dagger_j \langle c_k c_m \rangle + \langle c^\dagger_i c^\dagger_j \rangle c_k c_m - \langle c^\dagger_i c^\dagger_j \rangle \langle c_k c_m \rangle - \\ c^\dagger_i c_k \langle c^\dagger_j c_m \rangle -\langle c^\dagger_i c_k\rangle c_j^\dagger c_m + \langle c^\dagger_i c_k \rangle \langle c_j^\dagger c_m \rangle + \\ c^\dagger_i c_m \langle c^\dagger_j c_k \rangle + \langle c^\dagger_i c_m \rangle c^\dagger_j c_k - \langle c^\dagger_i c_m\rangle \langle c^\dagger_j c_k \rangle$$

Now when doing this process, often called mean-field decoupling, you can choose to exclude some of those terms that you expect to be zero. In BCS Theory the Hamiltonian is decoupled only with the superconductive channel (terms with $$\langle c^\dagger c^\dagger \rangle$$), the other channels would only give a renormalization of the non-interacting Hamiltonian.

The approximation that the expectation value of the product is equal to the product of the expectation value is just a prescription of the mean-field decoupling, I never heard of an argument like the one you posted but the reason behind the factorisation is just that you are considering the best Ground State that is a Slater determinant

• Thanks for the response! I'm still not sure if I understand how this leads to the approximation in the question title though. I can see how the fact that most terms in the expectation value will have $k\neq k'$ and so we can discard the terms for which $k=k'$. This should result in $\langle \hat{\Delta}^{\dagger} \hat{\Delta}\rangle \approx \langle \sum_{k\neq k'} Z_k^{\dagger} Z_k'\rangle$. But then I'm stuck at this point since I cant reduce this to the product of two separated expectations. Oct 26 at 9:45
• The fact that it factorises as a product of two expectation values comes from the application of Wick's theorem on a ground state (GS) that is a Slater determinant. Oct 27 at 12:24

This is a mean-field approximation in the form suitable for superconducting case. See, e.g., recent discussions here and here. Similar approximation is discussed in many textbooks on QFT methods in condensed matter (e.g., Abrikosov-Gorkov-Dzyaloshinskii or Fetter&Walecka) in the context of the Bose-Einstein condensation - where the creation/annihilation operator corresponding to the boson ground state is replaced by a complex number. The motivation is that

• in case where the number of particles is uncertain this operator has a non-zero expectation value (in fact, it is the order parameter in the transition)
• the macroscopic population of this state justifies neglecting its fluctuations.

Most of these textbook arguments are transferrable to superconductivity mutatis mutandis.

Update
Superconductivity can be viewed as Bose-Einstein condensation of Cooper pairs, if we identify the bosonic operators of creating/annihilating a pair of electrons with opposite spins and zero net momentum as $$a_0=\sum_kc_{-k,\downarrow}c_{k,\uparrow}, a_0^\dagger=\sum_kc_{k,\uparrow}^\dagger c_{-k,\downarrow}^\dagger.$$

Having written the Hamiltonian in terms of these operators we could directly sue the reasoning of the textbooks cited above. Taking F&W as more pedagogical one:

• Thanks for the response! I'm afraid I'm not to sure how I can use the posts you've linked to infer that $\langle \hat{\Delta}^{\dagger} \hat{\Delta}\rangle \approx \langle \hat{\Delta}^{\dagger}\rangle \langle \hat{\Delta}\rangle$ though? Oct 26 at 9:46
• @SalahTheGoat In technical terms, I think you would get more from reading the the sections in the textbooks that I mentioned. It is about justifying the approximation, no mor ethan that. Also, you probably know the Tinkham's book - indespensable reference for working in superconductivity. Oct 26 at 9:51
• Yes I do have Tinkham's book. I have briefly looked at the superconductivity sections of the texts you mentioned and unfortunately they all employ methods involving greens functions which I am not strong with yet. The methodology employed in the text I mentioned in the OP is the most intuitive and simple I have managed to find however the only issue I have with it is the failure to explain why $\langle \hat{\Delta}^{\dagger} \hat{\Delta}\rangle \approx \langle \hat{\Delta}^{\dagger}\rangle \langle \hat{\Delta}\rangle$ Oct 26 at 10:11
• @SalahTheGoat Perhaps, I expressed myself not very clearly in my answer - I meant the sections dealing with Bose-Einstein condensation, where $b_0,b_0^\dagger$ are replaced by $\langle b_0\rangle,\langle b_0^\dagger\rangle$ Oct 26 at 10:13
• Looking at the sections on bose einstein condensation, I still cant seem to see anything that could help me with $\langle \hat{\Delta}^{\dagger} \hat{\Delta}\rangle \approx \langle \hat{\Delta}^{\dagger}\rangle \langle \hat{\Delta}\rangle$ Oct 26 at 10:39