Lz and L^2 are commuting operators

[vballpro]

Homework Statement

Using the definitions of Lz and L^2, show that these two operators commute.

Homework Equations

Lz = -ih_bar * d/d(phi)
L^2 = -(h_bar)^2 {1/sin(theta) * d/d(theta) * [sin(theta) * d/d(theta)] + 1/sin^2(theta) d^2/d(phi)^2}

The Attempt at a Solution

I'm actually not too sure how to begin this. If someone can show me an example of commuting operators that is a bit complex, that would definitely help. My professor only gave us a simple example, which is why I'm confused with this one. Thanks for your help!

 

[gabbagabbahey]

If $[L^2,L_z]=0$ (that is L^2 and L_z commute), what can you say about $L^2(L_z\psi)$ and $L_z(L^2\psi)$ for any wavefunction $\psi$?

 

[CarlB]

$$x\partial_x$$ and $$y\partial_y$$ commute because
$$x\partial_x ( y\partial_y ( f) ) = y\partial_y ( x\partial_x (f))$$ for any function f.

On the other hand, $$x\partial_y$$ does not commute with $$y\partial_x$$

 

[vballpro]

gabba, I would say that LaTeX Code: L^2(L_z\\psi) and LaTeX Code: L_z(L^2\\psi) for any wavefunction LaTeX Code: \\psi would be equivalent.

My problem is more with doing the problem. Do I do the product rule when it comes to -h_bar^2 * 1/sin(theta) * d/d(theta), or is -h_bar^2 pulled out to the front and I just take the derivative of 1/sin(theta)?

 

[gabbagabbahey]

Well, -h_bar^2 is a constant isn't it...don't you always pull the constants out when taking a derivative?

Why don't you post an attempt and I'll see where you might be going wrong...

 

[lanedance]

constants shouldn't matter in differentiation and keep in mind a phi partial derivative means theta is kept constant for teh dervative

might help to remember that the partial derivative cross terms are the same for continuous & differentiable functions ie

for f(u,v) then f_uv = f_vu

(what are the exact conditions for this to apply?)

So maybe make some assumtions about the wavefunction continuity & its derivatives etc...

 

[vballpro]

this is what i have...
Lz(L^2) = ih_bar^3 * {[1/cos(theta) (sin(theta)) + (cos(theta))*1/sin(theta)] + 1/sin^2(theta)}
simplified...ih_bar^3 [sin(theta)/cos(theta) + cos(theta)/sin(theta) + 1/sin^2(theta)]

I did the exact same thing for L^2(Lz), but I realize that making a mistake for each of these will result in the same answer.

 

[gabbagabbahey]

this is what i have...
Lz(L^2) = ih_bar^3 * {[1/cos(theta) (sin(theta)) + (cos(theta))*1/sin(theta)] + 1/sin^2(theta)}
simplified...ih_bar^3 [sin(theta)/cos(theta) + cos(theta)/sin(theta) + 1/sin^2(theta)]

This makes no sense

$L_z$ and $L^2$ are differential operators, so their product should be as well. You should have $$\frac{\partial^2}{\partial \phi \partial \theta}$$ terms in your expression for $L_z(L^2)$.

 

[vballpro]

how do i work with 1/sin(theta)d/d(theta) * (sin(theta)d/d(theta)) part?

 

[vballpro]

do you use the product rule or are they two totally different derivations?

 

[gabbagabbahey]

Use the product rule!

For example,

$$\frac{\partial}{\partial x}\left[f(y)\frac{\partial}{\partial y}\left(g(y)\frac{\partial}{\partial y}\right)\right]=\frac{\partial}{\partial x}\left[f(y)g'(y)\frac{\partial}{\partial y}+f(y)g(y)\frac{\partial^2}{\partial y^2}\right]=f(y)g'(y)\frac{\partial^2}{\partial x \partial y}+f(y)g(y)\frac{\partial^3}{\partial x \partial y^2}$$

 

[vballpro]

2nd time's a charm?
-i*(h_bar)^3 [ cos(theta)/sin(theta) * d^2/d(theta)^2 + d^3/d(theta)^3 + 1/sin^2(theta) * d^3/d^3(theta)

why is it that it's d/dx [ f(y)g'(y) d/dy + f(y)g(y)d^2/dy^2] instead of
d/dx [ f(y)g'(y) d/dy + f'(y)g(y)d^2/dy^2]? (i added in an f ' to the second part. it might be hard to tell with this font.) i thought you said product rule. could you explain?

 

[gabbagabbahey]

2nd time's a charm?
-i*(h_bar)^3 [ cos(theta)/sin(theta) * d^2/d(theta)^2 + d^3/d(theta)^3 + 1/sin^2(theta) * d^3/d^3(theta)

why is it that it's d/dx [ f(y)g'(y) d/dy + f(y)g(y)d^2/dy^2] instead of
d/dx [ f(y)g'(y) d/dy + f'(y)g(y)d^2/dy^2]? (i added in an f ' to the second part. it might be hard to tell with this font.) i thought you said product rule. could you explain?

Because d/dy acts only on the product g(y) d/dy...it doesn't act on f(y)-which comes before the operator.

And 2nd time is definitely not the charm...sorry:...theta and phi are independent variables...but all I see is theta derivatives in your expression...what happened to the phi derivatives?