Do Functions of Commuting Operators Always Commute?

[McLaren Rulez]

Hi,

If we have two commuting operators A and B, is true that any function of A will commute with any function of B? I have a result which takes [L_{z},r^{2}]=0 and claims that [L_{z},r]=0. How can this be proved? Thank you

 

[grzz]

Hi,

If we have two commuting operators A and B, is true that any function of A will commute with any function of B? I have a result which takes [L_{z},r^{2}]=0 and claims that [L_{z},r]=0. How can this be proved? Thank you

Note that A^{2}B^{2} = AABB = A(AB)B = A(BA)B since A and B commute.
= (AB)AB = (BA)(BA) = B(AB)A = B(BA)A = B^{2}A^{2}

i.e. A^{2} commutes with B^{2}.

In like manner one can prove that A^{n} commutes with B^{m}.

Note also that many functions of A can be expanded as power series in A. Similarly for functions of B.

 

[McLaren Rulez]

Thank you for replying grzz.

I understand how to go from a lower power to a higher power. You showed that if A and B commute, then A^2 and B^2 also commute. But the reverse is what is being done in my question so how can I prove it for decreasing powers?

Thank you.

 

[grzz]

I did not read your qustion carefully!

 

[McLaren Rulez]

Can anyone else help? I have this idea but I am not sure if its correct.

We have Lr^{2}-r^{2}L=0

So, Lr^{2}-rLr+rLr-r^{2}L=0 where I just added and subtracted the middle terms.

Then [L,r]r+r[L,r]=0.

Would it be correct to say that since this is true for arbitrary r, [L,r] must be zero? I am not sure of this argument, especially the last step. Thank you

EDIT: No, that's not right. So any help?

 

[aesir]

It's not true in general.
Counterexample: take the pauli matrices [\sigma_x,\sigma_y^2]=[\sigma_x,1]=0, but [\sigma_x,\sigma_y]=2i\sigma_z \neq0

 

[McLaren Rulez]

Thank you aesir. I get it now.

Looks like there aren't any shortcuts to prove [L,r]=0.