# Homework Help: Can't make [L_z,p^2]=0

1. Dec 5, 2007

### bdforbes

I think [L_z,p^2] is supposed to equal zero, but when I'm getting

= x*(p_y)*(p_x)^2 - y*(p_x)*(p_y)^2

and that doesn't appear to equal zero. Could someone please show a couple steps to help me out?

2. Dec 5, 2007

### cristo

Staff Emeritus
What method have you used? You want to calculate the commutator $[L_z,p^2]=[xp_y-yp_x,p^2][/tex]. You need to expand this and then simplify. 3. Dec 5, 2007 ### mjsd isn't p stands for something like $$p_x = -i\hbar \frac{\partial}{\partial x}$$ ? 4. Dec 5, 2007 ### bdforbes These are the steps I used: [itex] [L_z,p^2]=[L_z,p_x^2+p_y^2+p_z^2]$
$= [L_z,p_x^2]+[L_z,p_y^2]$
$= [xp_y,p_x^2]-[yp_x,p_x^2]+[xp_y,p_y^2]-[yp_x,p_y^2]$
$= p_y x p_x^2-p_x y p_y^2$

5. Dec 5, 2007

### mjsd

note:
$$\left[x_i,p_j\right]= i\hbar \delta_{ij}$$
look what you have done..

6. Dec 5, 2007

### cristo

Staff Emeritus
No, $$p=\sqrt{p_x^2+p_y^2+p_z^2}$$; but the OP seems to have got this part.

7. Dec 5, 2007

### bdforbes

I've used that result to eliminate some terms, but I can't see how it would eliminate the other terms. I could use it to permute the factors of the final terms I have, but that isn't too useful.

8. Dec 5, 2007

### mjsd

are you sure it is p^2 not L^2?

9. Dec 5, 2007

### Avodyne

This is correct, and the 2nd and 3rd terms are zero. But in the 1st term, we can pull out the p_y (because it commutes with everything else), and similarly the p_x in the 4th term; then we have

$$= p_y[x,p_x^2]-p_x[y,p_y^2]$$

Then you need to evaluate the remaining commutators using

$$[A,BC]=[A,B]C + B[A,C]$$

10. Dec 5, 2007

### bdforbes

Thank you, that identity did the trick. That means that

$p_y x p_x^2 = p_x y p_y^2$

but they don't look equal to me! Is there any way other than using the commutators that you could prove the above equality?

11. Dec 5, 2007

### Avodyne

How did you get that? It's not correct. Did you compute the commutator of x and px^2? What did you get?

12. Dec 5, 2007

### bdforbes

$[x,p_x^2]=x p_x^2 - p_x^2 x = x p_x^2$

Oh wait I see what I did wrong! I eliminated the second term because I was hitting x with $p_x^2$, but I forgot that these are operators acting on an unknown function, so it's more like $p_x^2 ( xf )$.
Thanks for the help.