Table of Contents
Det(i)
wavefunction, read only
takes one argument: i - unsigned integer
for a wavefunction psi psi.Det(i) returns determinant i of psi as a new wavefunction. psi.Det(i) is read only and can not be used to change the value of a determinant.
Example
We can define the function: $$ |\psi\rangle = \left(\frac{1}{\sqrt{4}} a^{\dagger}_0 a^{\dagger}_1 + \frac{1}{\sqrt{4}} a^{\dagger}_0 a^{\dagger}_2 + (1+I)\frac{1}{\sqrt{4}} a^{\dagger}_1 a^{\dagger}_2 \right)|0\rangle. $$ In the example we set the name of the function to “psi” and then print determinant 1, 2 and 3 of the function. Note that operations acting on wavefunctions must not keep the order of determinants fixed. The only safe use of this function is probably in cases where one loops over all determinants in the function.
Input
- Example.Quanty
NF=3 NB=0 psi = NewWavefunction(NF, NB, {{"110",sqrt(1/4)},{"101",sqrt(1/4)},{"011",(1+I)*sqrt(1/4)}}) psi.Name="psi" print(psi.Det(1)) print(psi.Det(2)) print(psi.Det(3))
Result
WaveFunction: psi.Det(1) QComplex = 1 (Real==0 or Complex==1) N = 1 (Number of basis functions used to discribe psi) NFermionic modes = 3 (Number of fermions in the one particle basis) NBosonic modes = 0 (Number of bosons in the one particle basis) # pre-factor +I pre-factor Determinant 1 5.000000000000E-01 0.000000000000E+00 110 WaveFunction: psi.Det(2) QComplex = 1 (Real==0 or Complex==1) N = 1 (Number of basis functions used to discribe psi) NFermionic modes = 3 (Number of fermions in the one particle basis) NBosonic modes = 0 (Number of bosons in the one particle basis) # pre-factor +I pre-factor Determinant 1 5.000000000000E-01 0.000000000000E+00 101 WaveFunction: psi.Det(3) QComplex = 1 (Real==0 or Complex==1) N = 1 (Number of basis functions used to discribe psi) NFermionic modes = 3 (Number of fermions in the one particle basis) NBosonic modes = 0 (Number of bosons in the one particle basis) # pre-factor +I pre-factor Determinant 1 5.000000000000E-01 5.000000000000E-01 011
