# Can one operator commute with four other operators?

I want to know whether I can have a operator $$A$$ which commutes with four other operators $$M_1$$, $$M_2$$, $$M_3$$, and $$M_4$$ (for instance, drawing the operators $$M_j$$ from $$\{H,I,X,Y\}$$).

When can we tell that such an operator $$A$$ exists?

• To be nitpicking, the identity and 0 obviously commute with everything, but that's a trivial case – user2723984 Aug 28 '19 at 13:28
• ha, sorry. except identity cases @user2723984 – Xuemei Gu Aug 28 '19 at 13:29
• Of course. All diagonal operators commute, for instance. – Norbert Schuch Aug 28 '19 at 16:20

If an operator $$A$$ commutes with $$X$$ and $$Y$$ it already must be trivial ($$A = c I, c\in \mathbb{C}$$). One way to prove this is to note that $$A$$ also commutes with $$Z$$ since $$Z = \frac{1}{2}i(YX - XY)$$, and $$A$$ trivially commutes with $$I$$. Since $$A$$ commutes with $$I, X, Y, Z$$, which is Pauli basis of all operators, then $$A$$ commutes with any other operator, hence it must be trivial.
In general, the answer is that you need to find the irreducible representation (irrep) structure of the algebra generated by $$\{M_1,M_2,...\}$$. This will tell you what the commutant of the algebra looks like (commutant = {all operators that commute with all operators of the algebra}) . Then you can pick $$A$$ from the commutant.
In many cases the irrep of the algebra is trivial, like when the generators act irreducibly on the whole space (example: $$\{X,Y\}$$ acting on the space of a qubit) . Then the commutan is just the operators proportional to the identity. Another simple case is when all $$\{M_1,M_2,...\}$$ commute with each other. Then you just need to simultaneously diagonalize them and the commutant is all operators that act arbitrarily on the degenerate sectors.