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I would like to simplify expressions involving boson commutators using sympy.
The problem is that, using secondquant in sympy, the numerical values of the bosonic commutator [b_0,b^dagger_0]=1 and [b_0,b^dagger_1]=0 is never substituted to the symbolic expression.
In other words, I would like SymPy to know about the commutator identity.

The following code

from sympy import simplify
from sympy.physics.secondquant import Bd, B 
from sympy.physics.quantum import *
comm1=simplify(Commutator(B(0),Bd(0)).doit())
print(comm1)
comm2=simplify(Commutator(B(0),Bd(1)).doit())
print(comm2)

gives

comm1= AnnihilateBoson(0)*CreateBoson(0) - CreateBoson(0)*AnnihilateBoson(0)
comm2= AnnihilateBoson(0)*CreateBoson(1) - CreateBoson(1)*AnnihilateBoson(0)

instead of the expected values:

comm1= 1
comm2= 0

I have tried the code mentioned here How to use sympy.physics.quantum Commutator?

comm2=(Commutator(B(0),Bd(0))._eval_expand_commutator()).doit()
print('comm2=',comm2)

but this gives the same expression as before

comm2= AnnihilateBoson(0)*CreateBoson(0) - CreateBoson(0)*AnnihilateBoson(0)

Moreover I have found this related unanswered question:
Does sympy give me wrong results for the second quantization commutators?

I had a look around line 1682 here:
https://github.com/sympy/sympy/blob/master/sympy/physics/secondquant.py
and according to this the commutator should correctly give a Kronecker delta. However, I still get the symbolic expression reported above.

The problem is caused by

from sympy.physics.quantum import *

This code gives the expected behaviour

from sympy.physics.secondquant import *
from sympy import symbols
a, b = symbols('a,b')
c = Commutator(B(a),Bd(a))
print(c.doit())