Quanty - documentation:language_reference:objects:operator:methods
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2026-04-30T21:12:52+00:00Quanty
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https://www.quanty.org/_media/wiki/dokuwiki.svgtext/html2025-11-20T03:20:42+00:00Anonymous (anonymous@undisclosed.example.com)Chop
https://www.quanty.org/documentation/language_reference/objects/operator/methods/chop?rev=1763608842&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}$$$O=3.4+1.2a^{\dagger}_{0}\,a^{\phantom{\dagger}}_{0…text/html2025-11-20T03:20:42+00:00Anonymous (anonymous@undisclosed.example.com)Conjugate
https://www.quanty.org/documentation/language_reference/objects/operator/methods/conjugate?rev=1763608842&do=diff
Conjugate
For an operator O, the method O.Conjugate() will change operator O to its conjugate.
Example
We define:
$$O=3.4+1.2a^{\dagger}_{0}\,a^{\phantom{\dagger}}_{0}+(2.5+I)a^{\dagger}_{1}\,a^{\phantom{\dagger}}_{2},$$
its conjugate, $O^{*}$ is equal to:
$$O^{*}=3.4+1.2a^{\dagger}_{0}\,a^{\phantom{\dagger}}_{0}+(2.5-I)a^{\dagger}_{1}\,a^{\phantom{\dagger}}_{2}.$$
Input
NF=3
NB=0
O = NewOperator(NF,NB,{{3.4},{0,-0,1.2},{1,-2,2.5+I}})
print(O)
O.Conjugate()
print(O)
methods indextext/html2025-11-20T03:20:43+00:00Anonymous (anonymous@undisclosed.example.com)ConjugateTranspose
https://www.quanty.org/documentation/language_reference/objects/operator/methods/conjugatetranspose?rev=1763608843&do=diff
ConjugateTranspose
For an operator O, the method O.ConjugateTranspose() will change operator O to its conjugate transpose.
Example
We define:
$$O=3.4+1.2a^{\dagger}_{0}\,a^{\phantom{\dagger}}_{0}+(2.5+I)a^{\dagger}_{1}\,a^{\phantom{\dagger}}_{2},$$
its conjugate transpose, $O^{\dagger}$ is equal to:
$$O^{\dagger}=3.4+1.2a^{\dagger}_{0}\,a^{\phantom{\dagger}}_{0}+(2.5-I)a^{\dagger}_{2}\,a^{\phantom{\dagger}}_{1}.$$
Input
NF=3
NB=0
O = NewOperator(NF,NB,{{3.4},{0,-0,1.2},{1,-2,2.5+I}})
print(O)
…text/html2025-11-20T03:20:42+00:00Anonymous (anonymous@undisclosed.example.com)Print
https://www.quanty.org/documentation/language_reference/objects/operator/methods/print?rev=1763608842&do=diff
Print
For an operator O, O.Print() prints the operator O. For small operators print(O) is preferred over O.Print(). O.Print() is more memory efficient and therefore useful for large operators. O.Print() returns nill and prints the operator to standard output. Options can be specified if you want to print to file $$O=3.4+1.2a^{\dagger}_{0}\,a^{\phantom{\dagger}}_{0}+(2.5+I)a^{\dagger}_{1}\,a^{\phantom{\dagger}}_{2},$$methods indextext/html2025-11-20T03:20:42+00:00Anonymous (anonymous@undisclosed.example.com)Methods
https://www.quanty.org/documentation/language_reference/objects/operator/methods/start?rev=1763608842&do=diff
Methods
Operators can be large objects. We here define some methods that can change the object without the need to make a copy of the object that includes the changes.
Available methods
methods indextext/html2025-11-20T03:20:42+00:00Anonymous (anonymous@undisclosed.example.com)Transpose
https://www.quanty.org/documentation/language_reference/objects/operator/methods/transpose?rev=1763608842&do=diff
Transpose
For an operator O, the method O.Transpose() will change operator O to its transpose.
Example
We define:
$$O=3.4+1.2a^{\dagger}_{0}\,a^{\phantom{\dagger}}_{0}+(2.5+I)a^{\dagger}_{1}\,a^{\phantom{\dagger}}_{2},$$
its transpose, $O^{T}$ is equal to:
$$O^{T}=3.4+1.2a^{\dagger}_{0}\,a^{\phantom{\dagger}}_{0}+(2.5+I)a^{\dagger}_{2}\,a^{\phantom{\dagger}}_{1}.$$
Input
NF=3
NB=0
O = NewOperator(NF,NB,{{3.4},{0,-0,1.2},{1,-2,2.5+I}})
print(O)
O.Transpose()
print(O)
methods index