Can we define a QN objects using non-conserving quantum number ?

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asked Dec 17, 2017 by JunjieChen (490 points)
edited Dec 17, 2017 by JunjieChen

consider a simple interaction Hamiltonian of spin-1 boson. Each boson has 3 spin states : @@mF=1,0,-1@@. We now use @@a{mF}@@ (@@a{mF}^{\dagger}@@) as the annihilation (creation) operator for different @@mF@@ states.

@@H=a1^{\dagger}a{-1}^{\dagger}a0a0+h.c.@@

Obviously, this Hamiltonian does not conserve @@a{mF}^{\dagger}a{mF}@@ separately, but instead it conserves total @@Sz=a1^{\dagger}a1-a{-1}^{\dagger}a_{-1}@@.

Now I have a problem. When I define a site set file for spin-1 Bose-Hubbard model, I need to define a IQIndex object with (Index, QN) pair. Can I use @@a{mF}^{\dagger}a{mF}@@ as a quantum number for QN objects ? That is, I may define them to be

  Index(nameint("state name",n),1,Site),QN({Np1,1},{Nz0,1},{Nm1,1}));

where Np1=@@a{1}^{\dagger}a1@@, Nz0=@@=a0^{\dagger}a0@@ and Nm1=@@a{-1}^{\dagger}a{-1}@@.

Thank you.

commented Dec 19, 2017 by JunjieChen (490 points)
I think I have known the answer so I try to give it here. The answer is NO if we want to use non-conserving quantum number to define a index-QN pair in IQTensor. IQTensor is designed for sparse tensor with block-structure, which corresponds to specific conserved quantum number of the system. IQTensor makes the calculation, such as contraction of tensor, much faster than general ITensor (dense tensor) because the latter one doesn't consider the symmetry of physical system.

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