Maths: Proving ABC=CBA Implies [A,B]=[A,C]=[B,C]=0

[quasar987]

Is it true that $$ABC = CBA$$ implies [A,B]=[A,C]=[B,C]=0 ??

The converse is of course true, and I cannot find a counter-exemple (ex: no 2 of the above commutation relation above are sufficient), but how is this proven??

 

[Dick]

A trivial example is B=0. This can't imply [A,C]=0. Obviously.

 

[quasar987]

Ok, I forgot to add the hypothesis that A,B,C are not null operators.

 

[Dick]

Ok, A=C=[[0,1],[0,0]], B=[[0,0],[1,0]]. ABC=CBA is obvious but [A,B]!=0.

 

[Dick]

Actually no need for explicit matrices. Just take any A and B that don't commute and set C=A.

 

[quasar987]

Sorry for wasting your time Dick but twice you've demonstrated that I am trying to show something stronger than what I really need for this physics problem.

I need to find an iff condition on A, B, C three non equal and non null hermitian operators (they're Sx,Sy and Sz, the spin operators) that makes their product ABC hermitian.

I came up with (ABC)+=C+B+A+=CBA. And now I want to find the iff condition on A,B,C that will make this equal to ABC.

(+ denotes hermitian conjugation)

 

[Dick]

Yeah, 'false' is sort of by definition 'too strong'.

Well, Sx.Sy.Sz is non-hermitian, isn't it? Are A,B,C supposed to be linear combinations of the S's? Ok, so to sum up for the product AB to be hermitian we need that A and B commute. For ABC to be hermitian we need ABC=CBA. I guess I'm fuzzing out on what the actual problem is here... Can you be more specific?

 

[quasar987]

Yes, Sx.Sy.Sz is non-hermitian. Precisely, I need to find for which integers l,m,n is the operator $$S_x^lS_y^mS_z^n$$ an observable. I should have said that in the first place, huh. :p

My hypothesis is that it is hermitian when at most 1 of the exponent is odd.

 

[Dick]

You are clearly right. Since (S_i)^2=1 for i=x,y,z. There are relatively few cases to consider.