Matrix Product Toolkit | HOWTO / FreeFermion

For free fermions, the 'single mode approximation' produces an exact groundstate.

We test this numerically. First we obtain a wavefunction for m=200 for spinful fermions (fermi-Hubbard model at U=0). I choose half-filling, 2-site unit cell, so quantum numbers 2,0 per unit cell.

Now we use the excitation ansatz

{$$ \ket{k} = \sum_n e^{ikn} c_{\uparrow} \ket{0} $$}

Firstly, the norm is obtained from the expectation value {$\langle k \vert k \rangle$},

#mp-imoments psi "lat:sq(sum_k(sites=2,0.0*2*pi, Cup(0) + exp(i*pi*0.0)*Cup(1)))"\
 --degree 2 -u 1 --cart
#Date: Sat, 23 Jul 2016 21:24:00 +1000
#quantities are calculated per unit cell size of 1 site
#moment #degree #real                   #imag                   
2       1       0.99999297436009        -2.8939912587968e-18    
2       2       0                       0

Not sure why this isn't exactly 1, probably related to the local particle number variance, eg {$(N(0)-1)^2 = 0.4997$}, rather than the exact value of 0.5.

Now for the energy:

#mp-imoments psi "lat:ad(sum_k(sites=2,0.0*2*pi, Cup(0) + exp(i*pi*0.0)*Cup(1))) * H_t * \
(sum_k(sites=2,0.0*2*pi, Cup(0) + exp(i*pi*0.0)*Cup(1)))" --degree 3 -u 1 --cart
#Date: Sat, 23 Jul 2016 21:27:28 +1000
#quantities are calculated per unit cell size of 1 site
#moment #degree #real                   #imag                   
3       1       1.9999444636889         1.1292149123974e-18     
3       2       -1.2731305836167        1.8047273831357e-18     
3       3       0                       0

Dividing by the norm, this gives {$E(N) = -1.273139528236581 + 1.999958514677259$}.

Now the variance:

#mp-imoments psi "lat:ad(sum_k(sites=2,0.0*2*pi, Cup(0) + exp(i*pi*0.0)*Cup(1))) * H_t^2 * \
(sum_k(sites=2,0.0*2*pi, Cup(0) + exp(i*pi*0.0)*Cup(1)))" --degree 3 -u 1 --cart
#Date: Sat, 23 Jul 2016 21:27:31 +1000
#quantities are calculated per unit cell size of 1 site
#moment #degree #real                   #imag                   
3       1       3.9995504874594         6.2411815245896e-17     
3       2       -5.0922528637021        -7.4503672735005e-18    
3       3       1.6208728706093         3.6493891722751e-19

We recall from SMA, that the contribution at 3rd order is the cube of the groundstate energy. At second order, the largest part is {$2E\Delta = -5.092416702000496$}. There is also a term equal to the groundstate variance, which is 0.00016383944952958. There is also a part from {$\bigbraket{0}{H}{k}$}, but this is zero in this case due to symmetry. Summing these up (and remembering to factor in the norm), we get exactly -5.0922528637021, which is good. The 1st order part has two contributions, {$\Delta^2 = 3.999862162021308$}, and the variance due to the excitation, which comes out to be {$\sigma^2_A = -0.000284$}. So we have a reasonably good quality excited state.

Now this excitation cannot be in the tangent space, because we have {$\expect{c^\dagger_i c_j} \neq 0$}. We can test this further by looking at the transfer operator for {$\braket{0}{k}$}. Unfortunately I can't test this immediately as there is a quantum-number bug in the mp-fluctuation program.