HOWTO /
## FreeFermionFor 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. |

Page last modified on July 23, 2016, at 03:07 PM