# Commutator property question

## Main Question or Discussion Point

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

## Answers and Replies

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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.

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.

I did not read your qustion carefully!

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?

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

Thank you aesir. I get it now.

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