Why Does the Momentum Operator Yield Different Results in Rotations?

[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)

<br /> R_{z}(d\phi) = \left(\begin{array}{ccc}<br /> 1 &amp; -d\phi &amp; 0\\<br /> d\phi &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \end{array}\right), R_{z}(d\phi)^{-1} = \left(\begin{array}{ccc}<br /> 1 &amp; d\phi &amp; 0\\<br /> -d\phi &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 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...