# Quantum operators algebra

1. Feb 25, 2014

### prehisto

1. The problem statement, all variables and given/known data
Hi,guys. I have a hard time understanding algebra and tricks of operators.
So i have few examples:
1)[$\hat{p}$2x,xn]
2)[$\hat{l}$z,x],where $\hat{l}$z=x$\hat{p}$y-y$\hat{p}$x
2. Relevant equations

3. The attempt at a solution
1)[$\hat{p}$2x,xn]=
[$\hat{p}$x $\hat{p}$x,xn]=
$\hat{p}$x[$\hat{p}$x,xn]+[$\hat{p}$x,xn]$\hat{p}$x
So xn is not a operator,i dont know what i should do next?

2)[$\hat{l}$z,x],where $\hat{l}$z=x$\hat{p}$y-y$\hat{p}$x
Some tips or ideas here ?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 25, 2014

### Goddar

Hi.
What makes you say that x is not an operator? You've probably seen the relation:
$\hat{x}$|ψ>= x|ψ> in position-eigenstates basis, so xn doesn't contain
anything mysterious, it's just the operator x raised to the n-power (applied n times).
Since you know already the identity: [A,BC] = B[A,C] + [A,B]C, and presumably the values of:
[xi,pj] and [xi,xj] you should manage to get through the problem...

3. Mar 12, 2014

### prehisto

Hello,again . Im looking now at 2. example.
*will not write the operator sign ,its too time consuming .
So [lz,x]=lzx-xlz=
now can i simply multiply lz components by x and then i assume that lz components act on x?
like this:
=xpy(x)-ypx(x)-xlz ?

p.s. the original goal of this example is to find comutator. I have seen that in this kind of examples are functions on which operators act ,for example,[lz,x]f(x). Maybe i need to introduce a function here ?

4. Mar 12, 2014

### Goddar

Ok, if you don't know already the commutation relations between xi and pj then indeed i suggest you derive them by yourself, using a "dummy" function:
[xi,pj]f(x,y,z) = xi{pjf(x,y,z)}–pj{xif(x,y,z)} = ...

But once you know them, this problem doesn't require using a function anymore since, for instance:
[lz,x]=[xpy,x] – [ypx,x] = x[py,x] + [x,x]py – y[px,x] – [y,x]px = ...