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  Is my simple model for fermi liquid that forms cooper pairs correct?

+ 3 like - 0 dislike
76 views

Assume a 1D Fermi gas/liquid with $N$ electrons and a 2 body interaction: $$V(x_i, x_j) = -|g| \delta(x_i - x_j)$$

The second quantization is given by:
 
$$ H = \sum_{k \sigma}\epsilon_k c_{k\sigma}^\dagger c_{k \sigma} + \int_ {k k' q}  c_{k \sigma}^\dagger c_{k' \sigma' } ^\dagger V(q) c_{k'-q, \sigma} c_{k+q \ \sigma}$$

The Fourier transform of the interaction is: 
$$V(q) =  -|g| \int dx \ e^{iq(x_i -x_j)} \ \delta(x_i-x_j) = -|g|$$ 


Therefore the nth order Feynman diagram interaction in phase space is $(-|g|)^n$.

The energy spectrum will have the correction of $\sum_n (-|g|)^n$, this sum converges for $|g|<1$. 

1. Is it a good "classical" model for cooper pairs? I feel it doesn't because for $|g|\ge 1$ the model explodes.

2. It resembles the analogy between Landau diamagnetism and Pauli paramgnetsim. Like it's said on Pauli paramagnetism's wiki page:

The Pauli susceptibility comes from the spin interaction with the
magnetic field while the Landau susceptibility comes from the spatial
motion of the electrons and it is independent of the spin

3. Is my analysis correct at all? 

4. Is this the Feynman diagram expansion (borrowed from here):

Feynman diagram


 

asked Dec 8, 2017 in Theoretical Physics by lopo (30 points) [ revision history ]

1 Answer

+ 1 like - 0 dislike

I'm not sure if your analysis is correct, mainly because I have trouble understanding where your analysis is. What quantities do you calculate?

A general remark: It looks like you want to consider a small attractive interaction, $-|g|$, and perform a perturbative expansion around the free fermion ground state. However, the trouble is that this is not a good starting point — after all, even for weak interactions, the fermions will join into Cooper pairs. The superconducting ground state that you are interested in is qualitatively different from the non-interacting ground state, you have to use a non-perturbative argument to get there in the first place. In BCS-theory, this non-perturbative argument is the mean-field ansatz, where you introduce an order parameter $\Delta$.

That said, the simple form of the interaction that you consider is useful to understand Cooper pairs. In particular, I found that Ashcroft & Mermin, Chapter 34, Exercise 4 "The Cooper Problem", gives a very nice argument of how Cooper pairs form.

answered Dec 9, 2017 by Greg Graviton (705 points) [ revision history ]

I thought my naive model is interesting, because the existence of a critical $g_c$ for which the gas condense $g_c=1$.  The problem from Ashcroft & Mermin looks similar to what I came up with, in momentum space they also have the interaction $V(k_1,k_2) = -V$ for pairs with $ k_F\le k_i \le k_F + \hbar \omega$, I guess this subtle argument is essential. 

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