# Transformation of $| JM\rangle$ under the group of rotations

I am following the Quantum Mechanics I, Galindo A., Pascual P. and in page 207 explaining the matrix representations of the Rotation Operators in the angular momentum it appears the next (obvious) equation:

Being $\alpha,\beta,\gamma$ the Euler angles, $J$ the value for the total angular momentum (orbital+spin) and $M$ corresponding to the eigen value of $J_z$.

I don't understand the physical meaning of the equations. I mean, on the left part it is applying the rotation operator to the state vector $|JM\rangle$. As $J$ is invariant under rotations the right part of the equation should be obvious. So, what is the right part of the equation telling me? Maybe I'm getting confused with the Dirac notation. Any tip?

• The r.h.s. is telling you that $R$ doesn't change the value of $J$, but only mixes the $M$ components. – AccidentalFourierTransform May 4 '16 at 18:02
• So basicly the equation is telling me that a rotation applied to a state $|JM\rangle$ it is equal to some specific rotations above all $|JM'\rangle$? – Marc C May 4 '16 at 18:10
• ...is equal to some linear combination of all $|JM'\rangle$ – AccidentalFourierTransform May 4 '16 at 18:15
• Oh, now it makes sense: when a rotation it's applied to a state vector it can be expressed as a linear combination of $|JM'\rangle$. I was confused by the notation. Thank you. – Marc C May 4 '16 at 18:22

The left hand: rotate the state $|JM\rangle$ by applying a rotation $R$ on it. Right hand side:
1. insert completeness condition $\sum_{M'} |JM'\rangle\langle JM'|$
2. $D$ is the matrix representation of rotation matrix $R$ in basis ${|JM\rangle}$. The rotated state is expanded in terms of basis ${|JM\rangle}$ with coefficient $D$ in terms of rotation matrix.