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

[tex] 1-\frac{i}{\hbar} d\phi L_{z} = 1-\frac{i}{\hbar} d\phi (xp_{y}-yp_{x})[/tex]

act on an eigenket [itex]| x,y,z \rangle [/itex]

why do I get [itex]| x-yd\phi,y+xd\phi,z \rangle [/itex]

and not [itex]| x+yd\phi,y-xd\phi,z \rangle [/itex] ,

with the momentum operators

[tex] p_{x}=\frac{\hbar}{i}\frac{\partial}{\partial x} , p_{y}=\frac{\hbar}{i}\frac{\partial}{\partial y}[/tex]

Thanks for your help!

malawi_glenn

can you show us why you think that would yeild:

[itex]| x+yd\phi,y-xd\phi,z \rangle [/itex]

?

Laura08

I just use the operator on each component:

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

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

[tex] |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 =[/tex]

[tex] |x-d\phi (0-y),y-d\phi (x-0),z-d\phi (0-0) \rangle =[/tex]

[tex] | x+yd\phi,y-xd\phi,z \rangle [/tex]

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)

[tex]
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)[/tex]

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

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

[tex]\hat{R}| x,y,z \rangle = | x-yd\phi,y+xd\phi,z \rangle [/tex]

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

[tex]\hat{R} = 1-\frac{i}{\hbar} d\phi L_{z}[/tex]

And then inserting this result for [itex]\hat{R}[/itex] leads me back to my problem...