Quanty - documentation:language_reference:objects:wavefunction:methods
https://www.quanty.org/
2026-04-30T22:05:09+00:00Quanty
https://www.quanty.org/
https://www.quanty.org/_media/wiki/dokuwiki.svgtext/html2025-11-20T03:20:45+00:00Anonymous (anonymous@undisclosed.example.com)Chop
https://www.quanty.org/documentation/language_reference/objects/wavefunction/methods/chop?rev=1763608845&do=diff
Chop
Numerics inside a computer is not exact. Quanty represents numbers by doubles, which can store numbers with about 16 digits accuracy. The fact that you only have 16 digits can lead to number-loss and situations where numbers that should be zero are close to zero but not exactly zero. An example in base 10: If you represent $1/3$$0.3333333333333333$$1-3\times0.3333333333333333 = 0.00000000000000001$$\epsilon$$\epsilon = 2.2 \times 10^{−15}$$$
|\psi\rangle = \left(\frac{1}{\sqrt{4}} a^{\dagge…text/html2025-11-20T03:20:45+00:00Anonymous (anonymous@undisclosed.example.com)Conjugate
https://www.quanty.org/documentation/language_reference/objects/wavefunction/methods/conjugate?rev=1763608845&do=diff
Conjugate
For a wavefunction psi, the method psi.Conjugate() will change the wavefunction psi to its complex conjugate.
Example
We can define the following function:
$$
|\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.
$$
Its conjugate $\psi^*$ can be calculate with the command psi.Conjugate() and is equal to:
$$
|\psi\rangle = \left(\frac{1}{\sqrt{4}} a^{\da…text/html2025-11-20T03:20:46+00:00Anonymous (anonymous@undisclosed.example.com)Normalize
https://www.quanty.org/documentation/language_reference/objects/wavefunction/methods/normalize?rev=1763608846&do=diff
Normalize
for a wavefunction psi the method Normalize() will change the overall prefactor of the wavefunction such that $\langle \psi | \psi \rangle=1$.
Example
We can define the following function:
$$
|\psi\rangle = \left(a^{\dagger}_0 a^{\dagger}_1 + a^{\dagger}_0 a^{\dagger}_2 + (1+I) a^{\dagger}_1 a^{\dagger}_2 \right)|0\rangle.
$$
after normalization it becomes
$$
|\psi\rangle = \left(\frac{1}{\sqrt{4}} a^{\dagger}_0 a^{\dagger}_1 + \frac{1}{\sqrt{4}} a^{\dagger}_0 a^{\dagger}_2 + (1+I)\fr…text/html2025-11-20T03:20:46+00:00Anonymous (anonymous@undisclosed.example.com)Print
https://www.quanty.org/documentation/language_reference/objects/wavefunction/methods/print?rev=1763608846&do=diff
Print
For a wavefunction psi, psi.Print() prints the wavefunction psi. For small wavefunctions print(psi) is preferred over psi.Print(). psi.Print() is more memory efficient and therefore useful for large wavefunctions. psi.Print() returns nill and prints the wavefunction to standard output. Options can be specified if you want to print to file methods indextext/html2025-11-20T03:20:45+00:00Anonymous (anonymous@undisclosed.example.com)Randomize
https://www.quanty.org/documentation/language_reference/objects/wavefunction/methods/randomize?rev=1763608845&do=diff
Randomize
for a wavefunction psi the method Randomize() will change all prefactors to random numbers. Real functions stay real, complex stay complex. Randomize does not add determinants to the basis.
Example
We can define the following function:
$$
|\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.
$$
Randomization will turn $\psi$$$
|\psi\rangle = \left(\alph…text/html2025-11-20T03:20:45+00:00Anonymous (anonymous@undisclosed.example.com)Methods
https://www.quanty.org/documentation/language_reference/objects/wavefunction/methods/start?rev=1763608845&do=diff
Methods
Wavefunctions can be large objects. We here define some methods that can change the object without the need to make a copy of the object that includes the changes.
Available methods
methods index