https://www.quanty.org/ 2026-04-29T06:50:24+00:00 https://www.quanty.org/ https://www.quanty.org/_media/wiki/dokuwiki.svg text/html 2025-11-20T03:20:46+00:00 Anonymous (anonymous@undisclosed.example.com) https://www.quanty.org/documentation/language_reference/objects/wavefunction/properties/complex?rev=1763608846&do=diff Complex boolean, read onlyFor a wavefunction psi, psi.Complex returns a boolean (true or false) stating if the wavefunction contains complex numbers or not. Example We can define the functions $$ |\psi\rangle = \left(\frac{1}{\sqrt{4}} a^{\dagger}_0 a^{\dagger}_1 + \frac{1}{\sqrt{4}} a^{\dagger}_0 a^{\dagger}_2 + (1+I)\frac{1}{\sqrt{4}} a^{\dagger}_1 a^{\dagger}_2 \right)|0\rangle, $$ and $$ |\psi\rangle = \left(\frac{1}{\sqrt{2}} a^{\dagger}_0 a^{\dagger}_1 + \frac{1}{\sqrt{2}} a^{\dagger}_0 a… text/html 2025-11-20T03:20:47+00:00 Anonymous (anonymous@undisclosed.example.com) https://www.quanty.org/documentation/language_reference/objects/wavefunction/properties/det?rev=1763608847&do=diff Det(i) wavefunction, read onlytakes one argument: i - unsigned integerfor a wavefunction psi psi.Det(i) returns determinant i of psi as a new wavefunction. psi.Det(i) is read only and can not be used to change the value of a determinant. Example We can define the function:$$ |\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. $$properties index text/html 2025-11-20T03:20:47+00:00 Anonymous (anonymous@undisclosed.example.com) https://www.quanty.org/documentation/language_reference/objects/wavefunction/properties/n?rev=1763608847&do=diff N unsigned integer, read onlyreturns the number of determinants saved within the wavefunction. Example We can define the function: $$ |\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, $$ which has three determinants and psi.N thus is equal to three. Input NF=3 NB=0 psi = NewWavefunction(NF, NB, {{"110",sqrt(1/4)},{"101",sqrt(1/4)},{"011",(1+I)*sqrt(1/4)}}) … text/html 2025-11-20T03:20:47+00:00 Anonymous (anonymous@undisclosed.example.com) https://www.quanty.org/documentation/language_reference/objects/wavefunction/properties/name?rev=1763608847&do=diff Name string, read and write.A string representing the name of the wavefunction. The name of the wavefunction is printed during print statements and error messages. Note that the name of the wavefunction is not the same as the variable name that stores the wavefunction.$$ |\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, $$properties index text/html 2025-11-20T03:20:47+00:00 Anonymous (anonymous@undisclosed.example.com) https://www.quanty.org/documentation/language_reference/objects/wavefunction/properties/nb?rev=1763608847&do=diff NB unsigned integer, read and writeCurrent version does not suport bosons, i.e. NB must always be 0.An integer representing the number of Bosons in the basis. For wavefunction psi, index 0 to psi.NF-1 refers to Fermions, index psi.NF to psi.NF+O.NB-1$$ |\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, $$properties index text/html 2025-11-20T03:20:46+00:00 Anonymous (anonymous@undisclosed.example.com) https://www.quanty.org/documentation/language_reference/objects/wavefunction/properties/nbitskey?rev=1763608846&do=diff NBitsKey unsigned integer, read and writeInteger representing the size of the Hash key (size = 2NBitsKey). Wavefunctions are stored in a sparse format. In order to quickly find determinants within the function we use has functions and lookup tables. The size of the lookup table is currently static. This will change in future versions, but for the moment one needs to control this size manually. The larger the key the more memory is needed. If the key is smaller than the total number of determinan… text/html 2025-11-20T03:20:48+00:00 Anonymous (anonymous@undisclosed.example.com) https://www.quanty.org/documentation/language_reference/objects/wavefunction/properties/nf?rev=1763608848&do=diff NF unsigned integer, read and writeAn integer representing the number of Fermionic modes in the basis. For wavefunction psi, index 0 to psi.NF-1 refers to Fermions, index psi.NF to psi.NF+psi.NB-1 refers to Bosons. Changing this number changes the operator. If the new number of Fermions is smaller than the old number all modes referring to Fermions larger than $$ |\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}{\… text/html 2025-11-20T03:20:48+00:00 Anonymous (anonymous@undisclosed.example.com) https://www.quanty.org/documentation/language_reference/objects/wavefunction/properties/start?rev=1763608848&do=diff Properties Wavefunctions have several properties, all can be read, some can be set. Note that setting a property to some value might invoke methods that make large changes to the object. For example changing the number of bits in the hash key will rewrite the full key lookup table.properties index