# When can we replace the Cooper pair operator with its expectation value?

In the paper Cooper pair tunnelling into a Quantum Hall fluid, the author models a superconductor in tunnelling contact with a Quantum Hall fluid. The superconductor is modelled

... in terms of a bosonic pair field, $$\hat c$$ , which exhibits long-ranged (vortex-glass) order and has a non-zero condensate, $$\langle \hat c \rangle = \Delta$$.

The author gives a tunnelling Hamiltonian

$$H = t [\hat\psi_{\uparrow}(x=0)^\dagger \hat\psi_{\uparrow}(x=\xi)^\dagger \hat c(x=0) + H.c.]$$

and argues that

Since the superconductor has a non-zero condensate, it is legitimate to simply replace the operator $$\hat c$$ in [the Cooper pair tunnelling Hamiltonian] by the c-number $$\Delta$$.

As I understand it, $$\hat c = \hat\eta_{\uparrow}(x) \hat\eta_{\downarrow}(x^\prime)$$, where $$\hat \eta_\sigma (x)$$ is the electron operator. According to BCS theory,

$$\langle \hat\eta_{\uparrow}(x) \hat\eta_{\downarrow}(x^\prime) \rangle = \Delta,$$

but what is unclear to me is under what circumstances is it OK to just replace the pair operator with its expectation value? What does the author mean by "it is legitimate"? Is it some kind of mean field approximation?

## 2 Answers

The crucial sentence, IMHO, is the one preceding the one you quoted: "Since $$H_{tunn}$$ in Eqn.(3) involves fields near $$x= 0$$, it is useful to integrate out the degrees of freedom for $$x\neq 0$$ in both the edge action above, and in the superconductor." (emphasis by me).

The $$\hat{c}$$ operator is the pair-annihilation operator in the superconductor. It is not the $$\psi_{\sigma}$$ appearing elsewhere in the paper, which are the electrons at the edge of the quantum Hall bar.

The author is interested in low-temperature behavior, where the superconducting gap is large with respect to all other energy scale. And now he wants to integrate out all the degrees of freedom which are far from the edge or high-energy, and remain with an effective low-energy Hamiltonian that describes the edge. This is why the mean-field approximation of replacing $$\hat{c}$$ with its expectation value is justified - the dynamics of $$\hat{c}$$ are of high-energy (involving breaking of Cooper pairs by quasi-particle excitations above the gap), and at low temperatures we can take the order parameter to be uniform at the length-scales of the edge. The mean-field here is the result of the process of integrating-out, with all the correction exponentially small in $$T/\Delta$$.

• Hi, thanks for the answer! I edited my question to distinguish between electrons in the QH fluid and the SC to clarify the second point make, just as a heads up. Mar 27, 2020 at 11:36
• you mean the fields taken at different positions? no. the expectation value which gives $\Delta$ is for a pair creation at a single position. You can get it from the effective SC Hamiltonian $H=\sum_k \epsilon_k \psi^{\dagger}_{\sigma,k} \psi_{\sigma,k} + \Delta \int d^dx \psi_{\uparrow}(x)\psi_{\downarrow}(x)+{\rm{h.c.}}$
– user245141
Mar 27, 2020 at 11:45
• Sorry, there was an error when I was writing the previous comment. Wouldn't a mean field approximation look like $\hat \psi_1 \hat \psi_2 \hat c \simeq \langle \hat \psi_1 \hat \psi_2 \rangle \hat c + \hat \psi_1 \hat \psi_2 \langle \hat c\rangle - \langle \hat \psi_1 \hat \psi_2\rangle \langle \hat c\rangle$? The last term we can ignore, sure, but then the first term is zero because there is no pairing in the QH fluid or what? Mar 27, 2020 at 12:57
• The first term will contribute to the dynamics of $\hat{c}$, which the author integrates out. It is of no interest. He assumes that its effect on the expectation value of $\hat{c}$ itself is negligent or implicitly included in $\Delta$.
– user245141
Mar 27, 2020 at 13:54

There are a few points to note here:

• This is a mean-field approximation, with all the accompanying caveats
• The quantum fluctuations of this quantity are negligible compared to its mean - otherwise we wouldn't be talking about the superconducting state (a macroscopic phase)
• Everybody has been doing so since Bogolyubov wrote the equations ;)