How Does Angular Momentum Operate in Exponential Form?

[StephvsEinst]

Hey!

How does the operator of angular momentum operates in exponential form?

$$ e^{-i\theta J}\vert l, m \rangle = ?? $$

where $$J\vert \Psi \rangle = J\vert l, m \rangle$$
and
$$J^2\vert \Psi\rangle = \hbar^2 l(l+1)\vert \Psi\rangle $$

Also, how do you operate $$J_-$$
and $$J_+$$
in exponential form?

Thanks,
Jorge.

 

[HomogenousCow]

They're defined in terms of the taylor series for the exponential function.

 

[vanhees71]

First of all, your notation doesn't make sense. What do you want to calculate? If you like to find the representation of a general rotation, that's given in terms of the Euler angles $(\alpha,\beta,\gamma)$ by the unitary operator
$$\hat{D}(\alpha,\beta,\gamma)=\exp(-\mathrm{i} \alpha \hat{J}_z) \exp(-\mathrm{i} \beta \hat{J}_y) \exp(-\mathrm{i} \gamma \hat{J}_z).$$
You can calculate its matrix elements in terms of the usual eigenstates $|j,m \rangle$, leading to the Wigner D-matrices:
http://en.wikipedia.org/wiki/Wigner_D-matrix

 

[StephvsEinst]

First of all, your notation doesn't make sense. What do you want to calculate? If you like to find the representation of a general rotation, that's given in terms of the Euler angles $(\alpha,\beta,\gamma)$ by the unitary operator
$$\hat{D}(\alpha,\beta,\gamma)=\exp(-\mathrm{i} \alpha \hat{J}_z) \exp(-\mathrm{i} \beta \hat{J}_y) \exp(-\mathrm{i} \gamma \hat{J}_z).$$
You can calculate its matrix elements in terms of the usual eigenstates $|j,m \rangle$, leading to the Wigner D-matrices:
http://en.wikipedia.org/wiki/Wigner_D-matrix

I was talking about a operator that rotates the physical system and leaves the axis fixed (active viewpoint). This rotation is invariant to let us study the properties of the particle without worrying about the angular momentum of the particle (because it is conserved). I wanted to know how the rotation operator worked on the state $$\vert l, m \rangle$$ of the particle, which is given by the Wigner D-matrix. So your answer was what I was looking for. Thanks!