Orbital angular momentum operator

[Laura08]

Hello, sorry I am new to this forum, I hope I found the right category. I have a question about the momentum operator as in Sakurai's "modern quantum mechanics" on p. 196

If I let

$$1-\frac{i}{\hbar} d\phi L_{z} = 1-\frac{i}{\hbar} d\phi (xp_{y}-yp_{x})$$

act on an eigenket $| x,y,z \rangle$

why do I get $| x-yd\phi,y+xd\phi,z \rangle$

and not $| x+yd\phi,y-xd\phi,z \rangle$ ,

with the momentum operators

$$p_{x}=\frac{\hbar}{i}\frac{\partial}{\partial x} , p_{y}=\frac{\hbar}{i}\frac{\partial}{\partial y}$$

Thanks for your help!

 

[malawi_glenn]

can you show us why you think that would yeild:

$| x+yd\phi,y-xd\phi,z \rangle$

?

 

[Laura08]

I just use the operator on each component:

$$[1-\frac{i}{\hbar} d\phi (xp_{y}-yp_{x})] | x,y,z \rangle =$$

$$[1-d\phi (x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x})] | x,y,z \rangle =$$

$$|x-d\phi (x \frac{\partial x}{\partial y}-y \frac{\partial x}{\partial x}),y-d\phi (x \frac{\partial y}{\partial y}-y \frac{\partial y}{\partial x}),z-d\phi (x \frac{\partial z}{\partial y}-y \frac{\partial z}{\partial x}) \rangle =$$

$$|x-d\phi (0-y),y-d\phi (x-0),z-d\phi (0-0) \rangle =$$

$$| x+yd\phi,y-xd\phi,z \rangle$$

 

[malawi_glenn]

Isn't that the correct answer?

 

[Laura08]

Well, I think the calculation is correct, but then I did a backwards rotation, which I didn't intend to do.
The rotation matrix for an infinitesimal rotation about the z-axis is (if I rotate the vector, not the system)

$$R_{z}(d\phi) = \left(\begin{array}{ccc} 1 & -d\phi & 0\\ d\phi & 1 & 0 \\ 0 & 0 & 1 \end{array}\right), R_{z}(d\phi)^{-1} = \left(\begin{array}{ccc} 1 & d\phi & 0\\ -d\phi & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)$$

So $$| x+yd\phi,y-xd\phi,z \rangle$$ = $$R_{z}(d\phi)^{-1}| x,y,z \rangle$$

Yet if you try to determine the quantum mechanical operator for an infinitesimal rotation around the z-axis, starting with

$$\hat{R}| x,y,z \rangle = | x-yd\phi,y+xd\phi,z \rangle$$

(as done e.g. here: http://en.wikipedia.org/wiki/Rotation_operator" , you find

$$\hat{R} = 1-\frac{i}{\hbar} d\phi L_{z}$$

And then inserting this result for $\hat{R}$ leads me back to my problem...