Commutator problem with momentum operators

[hellomister]

Homework Statement

Find the commutator $$\left[\hat{p_{x}},\hat{p_{y}}\right]$$

Homework Equations

$$\hat{p_{x}}=\frac{\hbar}{i}\frac{\partial}{\partial x}$$

$$\hat{p_{y}}=\frac{\hbar}{i}\frac{\partial}{\partial y}$$

The Attempt at a Solution

$$[\hat{p}_{x}, \hat{p}_{y}]=\hat{p}_{x}\hat{p}_{y}-\hat{p}_{y}\hat{p}_{x}$$

$$=\frac{\hbar}{i}\frac{\partial}{\partial x}\frac{\hbar}{i}\frac{\partial}{\partial y}-\frac{\hbar}{i}\frac{\partial}{\partial y}\frac{\hbar}{i}\frac{\partial}{\partial x}$$

$$=(\frac{\hbar}{i})^2\frac{\partial^2}{\partial x\partial y}-(\frac{\hbar}{i})^2\frac{\partial^2}{\partial y\partial x}$$

Is this the correct commutator? I don't know what else I can do to further complete this. Any help is appreciated.

 

[Dick]

It's fine. But partial derivatives commute in the physics sense. Your two partial derivatives are equal.

 

[hellomister]

Okay thanks for the response, i have a similar problem to that of the first that I put up, but I am very confused.

Problem
Show that $$[\hat{L}_{x},\hat{P}_{y}]=i\hbar\hat{P}_{z}.<br /> (Note: \hat{L}_{x}=y\hat{P}_{z}-z\hat{P}_{y})$$
Televant equations
$$\hat{p_{x}}=\frac{\hbar}{i}\frac{\partial}{\partial x}$$

$$\hat{p_{y}}=\frac{\hbar}{i}\frac{\partial}{\partial y}$$

$$\hat{p_{z}}=\frac{\hbar}{i}\frac{\partial}{\partial z}$$

Attempt
So i started out by just writing out the commutator in the general form:

(1) $$\hat{L}_{x}\hat{P}_{y}-\hat{P}_{y}\hat{L}_{x}$$

Then I slowly substituted some equations
(2) $$(y\hat{P}_{z}-z\hat{P}_{y})\frac{\hbar}{i}\frac{\partial}{\partial y}\psi-\frac{\hbar}{i}\frac{\partial}{\partial y}(y\hat{P}_{z}\psi-z\hat{P}_{y}\psi)$$

(3)$$(y\frac{\hbar}{i}\frac{\partial}{\partial z}-z\frac{\hbar}{i}\frac{\partial}{\partial y})\frac{\hbar}{i}\frac{\partial}{\partial y}\psi-\frac{\hbar}{i}\frac{\partial}{\partial y}(y\frac{\hbar}{i}\frac{\partial}{\partial z}\psi-z\frac{\hbar}{i}\frac{\partial}{\partial y}\psi)$$

I decided to factor out \frac{\hbar}{i}
$$(4)(\frac{\hbar}{i})^2(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})\frac{\partial \psi}{\partial y}-(\frac{\hbar}{i})^2\frac{\partial}{\partial y}(y\frac{\partial \psi}{\partial z}-z\frac{\partial \psi}{\partial y})$$

Now I'm not sure what to do..

 

[Dick]

Now start expanding the derivatives in the second term using the product rule. Look for cancellations with symmetric partial derivatives like dy*dz vs dz*dy.

 

[hellomister]

okay so then i got this

(5) $$(\frac{\hbar}{i})^2(y\frac{\partial}{\partial z}\frac{\partial\psi}{\partial y}-z\frac{\partial}{\partial y}\frac{\partial\psi}{\partial y})-(\frac{\hbar}{i})^2(\frac{\partial\psi}{\partial z}+y\frac{\partial}{\partial y}\frac{\partial\psi}{\partial z}-\frac{\partial z}{\partial y}\frac{\partial\psi}{\partial y}+z\frac{\partial}{\partial y}\frac{\partial\psi}{\partial y})$$

So then would $$y\frac{\partial}{\partial z}\frac{\partial\psi}{\partial y} y\frac{\partial}{\partial y}\frac{\partial\psi}{\partial z}$$ cancel out? I'm not sure what you mean by cancel out haha, i have a really bad math foundation if you could please explain that would help a lot. Also did i do that last line correctly?

$$z\frac{\partial}{\partial y}\frac{\partial\psi}{\partial y}$$ This term would also go away right?

Thanks for all your help!

 

[Altabeh]

okay so then i got this

(5) $$(\frac{\hbar}{i})^2(y\frac{\partial}{\partial z}\frac{\partial\psi}{\partial y}-z\frac{\partial}{\partial y}\frac{\partial\psi}{\partial y})-(\frac{\hbar}{i})^2(\frac{\partial\psi}{\partial z}+y\frac{\partial}{\partial y}\frac{\partial\psi}{\partial z}-\frac{\partial z}{\partial y}\frac{\partial\psi}{\partial y}+z\frac{\partial}{\partial y}\frac{\partial\psi}{\partial y})$$

So then would $$y\frac{\partial}{\partial z}\frac{\partial\psi}{\partial y} y\frac{\partial}{\partial y}\frac{\partial\psi}{\partial z}$$ cancel out? I'm not sure what you mean by cancel out haha, i have a really bad math foundation if you could please explain that would help a lot. Also did i do that last line correctly?

$$z\frac{\partial}{\partial y}\frac{\partial\psi}{\partial y}$$ This term would also go away right?

Thanks for all your help!

1- The last term has a wrong sign.
2- By 'canceling out' one means finding a term with exactly the same form of a given term in the equation but with reversed sign so both would disappear. As in x+y-z-x=0, x and -x cancel out each other and the equation reduces to y-z=0.
2- Remember that derivatives have no order in the sense that $$\frac{\partial}{\partial x}\frac{\partial f(x,y)}{\partial y} = \frac{\partial}{\partial y}\frac{\partial f(x,y)}{\partial x}$$.
4- Now start canceling out terms and I'd recall that the derivative of coordinates wrt each other vanishes.

AB

 

[hellomister]

what about the (partial z/ partial y) (partial psi/ partial y)?
What does that become?

 

[Altabeh]

what about the (partial z/ partial y) (partial psi/ partial y)?
What does that become?

Didn't you notice my fourth remark?!

AB

 

[hellomister]

ahh I see what you mean. Thanks for all the help!