How Do You Solve Quantum Operator Commutators?

[prehisto]

Homework Statement

Hi,guys. I have a hard time understanding algebra and tricks of operators.
So i have few examples:
1)[\hat{p}²_x,xⁿ]
2)[\hat{l}_z,x],where \hat{l}_z=x\hat{p}_y-y\hat{p}_x

Homework Equations

The Attempt at a Solution

1)[\hat{p}²_x,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 don't 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 ?

 

[Goddar]

The Attempt at a Solution

1)[\hat{p}²_x,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 don't know what i should do next?

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:
[x_i,p_j] and [x_i,x_j] you should manage to get through the problem...

 

[prehisto]

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:
[x_i,p_j] and [x_i,x_j] you should manage to get through the problem...

Hello,again . I am looking now at 2. example.
*will not write the operator sign ,its too time consuming .
So [l_z,x]=l_zx-xl_z=
now can i simply multiply lz components by x and then i assume that lz components act on x?
like this:
=xp_y(x)-yp_x(x)-xl_z ?

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,[l_z,x]f(x). Maybe i need to introduce a function here ?

 

[Goddar]

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

But once you know them, this problem doesn't require using a function anymore since, for instance:
[l_z,x]=[xp_y,x] – [yp_x,x] = x[p_y,x] + [x,x]p_y – y[p_x,x] – [y,x]p_x = ...