I Do operators that commute also indicate if their powers commute?

[TheCanadian]

If $A$ and $B$ are two operators that commute (i.e. [$A$,$B$] = 0), does that indicate if $A^m$ and $B^n$ more generally commute where m and n are not necessarily non-negative integers?

 

[fresh_42]

What does it mean, that $[A,B]=0$?

 

[TheCanadian]

What does it mean, that $[A,B]=0$?

AB - BA = 0

 

[fresh_42]

AB - BA = 0

Yep. So $AB = BA$, i.e. you can pull one or many $B$ through how many ever $A$'s there are, step by step.
Edit: I don't know how it is for negative exponents since there must first be defined an inverse. There probably have to be made some assumptions on convergence, too.
Edt2: I think I got it. Let $C = B A^{-1}$, i.e. $B=CA$. Then $AB = ACA = BA = CA^2$.
Thus $ABA^{-1} = CA = B$ or $BA^{-1} = A^{-1}B$.

 

[TheCanadian]

Yep. So $AB = BA$, i.e. you can pull one or many $B$ through how many ever $A$'s there are, step by step.
Edit: I don't know how it is for negative exponents since there must first be defined an inverse. There probably have to be made some assumptions on convergence, too.
Edt2: I think I got it. Let $C = B A^{-1}$, i.e. $B=CA$. Then $AB = ACA = BA = CA^2$.
Thus $ABA^{-1} = CA = B$ or $BA^{-1} = A^{-1}B$.

Thank you for the reply. That helps a lot. One of the other cases I was wondering about was decimals. Does raising two operators to different fractions alter the operators in such a way they no longer commute?

 

[fresh_42]

Thank you for the reply. That helps a lot. One of the other cases I was wondering about was decimals. Does raising two operators to different fractions alter the operators in such a way they no longer commute?

Interesting question. At the moment I'm not sure how to define them in a rigorous manner other than, e.g. $A^3 = B^2$ and then apply the tricks above. What would $A^{e}$ be? Probably a convergent Taylor series or something like that in which case crossing to the limits has to be considered.

 

[George Jones]

What would $A^{e}$ be?

First, what would $A^{e}$ be when $A$ is a positive real number?

 

[fresh_42]

First, what would $A^{e}$ be when $A$ is a positive real number?

The solution of $e \cdot \ln{A} = \ln{x}$ with all terms defined by their Taylor series.

 

[Strilanc]

An easy counter-example via the Pauli matrices not commuting and being their own inverse. Not sure if you'll consider it cheating.

Note that:

$[X, Z] = 2iY$

$[X^2, Z^2] = [I, I] = 0$

$I$ has many square roots, $X$ and $Z$ among them. We can pick different ones for $A$ and $B$ when setting $A=B=I$. So $[A, B] = [I, I] = 0$ but $[A^{\frac{1}{2}}, B^{\frac{1}{2}}] = [X, Z] = 2iY$.

I guess technically $[A^{\frac{1}{2}}, B^{\frac{1}{2}}]$ is not really well defined here. It's a multi-valued function, but some of those values aren't 0.

Your propose rule might hold if we restrict ourselves to the principle powers of a matrix. Require that $A^x$ be interpreted as raising its eigenvalues to $x$. Further require that $(e^{i y})^x$ only allows the $e^{i x y} = \cos(x y) + i \sin(x y)$ solution. Not so sure in that case.

 

[fresh_42]

An easy counter-example via the Pauli matrices not commuting and being their own inverse. Not sure if you'll consider it cheating.

Note that:

$[X, Z] = 2iY$

$[X^2, Z^2] = [I, I] = 0$

$I$ has many square roots, $X$ and $Z$ among them. We can pick different ones for $A$ and $B$ when setting $A=B=I$. So $[A, B] = [I, I] = 0$ but $[A^{\frac{1}{2}}, B^{\frac{1}{2}}] = [X, Z] = 2iY$.

I guess technically $[A^{\frac{1}{2}}, B^{\frac{1}{2}}]$ is not really well defined here. It's a multi-valued function, but some of those values aren't 0.

It might still be the case if we restrict ourselves to the principle powers of a matrix. Require that $A^x$ be interpreted as raising its eigenvalues to $x$. Further require that $(e^{i y})^x$ only allows the $e^{i x y} = \cos(x y) + i \sin(x y)$ solution. Not so sure in that case.

To define $X = A^{\frac{1}{2}}$ is rather edgy, isn't it?

 

[Strilanc]

To define $X = A^{\frac{1}{2}}$ is rather edgy, isn't it?

Well, $X$ is a valid solution to the equation $M^2 = I$. All 2x2 unitary matrices with eigenvalues of $\pm1$ are. So if you want to be exhaustive then you would say that $I_2^{\frac{1}{2}}$ is the set $\{M \in U(2) | M^2 = I\}$ which includes X and Y and Z and H and I and uncountably many other matrices corresponding to 180 degree rotations around an arbitrary axis in 3-d space (and also their negations).

This problem of multiplicity occurs for matrices besides $I$. $X$ has four square roots. $X \otimes X$ has sixteen. $X \otimes I$ has uncountably many.

If you want to restrict yourself to a specific matrix or a subset of matrices out of the set of satisfying results, you need to pick a rule for doing that and specify it.

 

[fresh_42]

If you want to restrict yourself to a specific matrix or a subset of matrices out of the set of satisfying results, you need to pick a rule for doing that and specify it.

I think that is the crucial point. One has to define the algebraic structure first in which the operations "live". Thus one can distinguish between what makes sense and what doesn't. (At least this has been the first time I've read about a Pauli matrix being a square root. But why not. As a $ℂ$-basis of the ring $ℂ^{2 \times 2}$ ...)