Spin Exchange in Fermi Sea - Evaluating S_x \cdot S_{x+1}

[andrewm]

I've found the simplest hopping Hamiltonian for fermions (diagonal in momentum space) has a so-called Fermi sea ground state.

$$H = -2 t \sum_k \cos(k) f^+_k f_k$$

(t is some parameter in units of energy).

How do I evaluate the expected value of the spin exchange operator $$S_x \cdot S_{x+1}$$ in this state? I am having trouble because the ground state obvious in the momentum basis, and it is not written in the position basis.

 

[malawi_glenn]

Fourier transform?

 

[denian]

To evaluate the expected value of the spin exchange operator in the Fermi sea ground state, we need to express the operator in terms of the creation and annihilation operators in the momentum basis. The spin exchange operator can be written as:

S_x · S_{x+1} = (f^+_{k+1} f_k + f^+_k f_{k+1}) · (f^+_{k+1} f_k - f^+_k f_{k+1})

= f^+_{k+1} f_k f^+_{k+1} f_k - f^+_{k+1} f_k f^+_k f_{k+1} + f^+_k f_{k+1} f^+_{k+1} f_k - f^+_k f_{k+1} f^+_k f_{k+1}

= f^+_{k+1} f_k f^+_{k+1} f_k - f^+_k f^+_{k+1} f_k f_{k+1} + f^+_k f_{k+1} f^+_{k+1} f_k - f^+_k f_{k+1} f^+_k f_{k+1}

= f^+_{k+1} f_k (f^+_{k+1} f_k + f^+_k f_{k+1}) - f^+_k f_{k+1} (f^+_{k+1} f_k + f^+_k f_{k+1})

= f^+_{k+1} f_k S_x - f^+_k f_{k+1} S_x

= (f^+_{k+1} f_k - f^+_k f_{k+1}) S_x

= (f^+_{k+1} f_k - f^+_k f_{k+1}) · \frac{1}{2}(c^+_k \sigma_x c_k + c^+_{k+1} \sigma_x c_{k+1})

Where c_k and c^+_k are the annihilation and creation operators for the spin-1/2 fermions in the momentum basis, and \sigma_x is the Pauli spin matrix in the x-direction.

Now, we can use the Ferm