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Matrix represntation of angular momentum operator (QM)

  • Thread starter joker_900
  • Start date
64
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1. 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).

2. Homework Equations
-


3. 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
 

AEM

Gold Member
360
0
You will probably find page 315 and 316 of Shankar's Principles of Quantum Mechanic helpful.
 
64
0
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

Gold Member
360
0
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.
 

nrqed

Science Advisor
Homework Helper
Gold Member
3,540
181
1. 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).

2. Homework Equations
-


3. 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.
 

AEM

Gold Member
360
0
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.
 

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