Matrix represntation of angular momentum operator (QM)

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joker_900
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Homework Statement


The matrix R(q) for rotating an ordinary vector by q around the z-axis is given by@

cosq -sinq 0
sinq cosq 0
0 0 1

From R calculate the matrix J(z).

Homework Equations


-

The Attempt at a Solution



All I know is that U(q) = exp[-iJ(z)q] is the unitary operator which rotates a system, and I believe that

R(q) x = U(dagger)(q) x U(q)

Where x is the position vector.

I have no idea where to go from here, other than expanding U with a taylor series but this didn't seem to go anywhere.

Thanks
 
on Phys.org
You will probably find page 315 and 316 of Shankar's Principles of Quantum Mechanic helpful.
 
AEM said:
You will probably find page 315 and 316 of Shankar's Principles of Quantum Mechanic helpful.

Thanks, but I still don't see how to do the question at all. Are the pages right - do they change with editions?
 
AEM said:
You will probably find page 315 and 316 of Shankar's Principles of Quantum Mechanic helpful.

My edition of Shankar is from 1981. By chance I was reviewing this topic the day before your original post. I'll spend some time thinking about it and see if I can help you out.
 
joker_900 said:

Homework Statement


The matrix R(q) for rotating an ordinary vector by q around the z-axis is given by@

cosq -sinq 0
sinq cosq 0
0 0 1

From R calculate the matrix J(z).

Homework Equations


-


The Attempt at a Solution



All I know is that U(q) = exp[-iJ(z)q] is the unitary operator which rotates a system, and I believe that

R(q) x = U(dagger)(q) x U(q)

Where x is the position vector.

I have no idea where to go from here, other than expanding U with a taylor series but this didn't seem to go anywhere.

Thanks

I think that the correct relation is simply R(q) = exp[-i Jz q].

Just work to first order in q. Then [itex]e^{-iJ_z q} \approx 1 - i J_z q[/itex] (where 1 here is the unit matrix) . Replace cos(q) by 1 and sin(q) by q in the matrix R(q) and you will get the expression for Jz easily.
 
nrqed said:
I think that the correct relation is simply R(q) = exp[-i Jz q].

Just work to first order in q. Then [itex]e^{-iJ_z q} \approx 1 - i J_z q[/itex] (where 1 here is the unit matrix) . Replace cos(q) by 1 and sin(q) by q in the matrix R(q) and you will get the expression for Jz easily.

I just logged on to make similar comments. Take the angle q to be very small, or infinitesimal, leads to the replacements given above.