Deriving how angular momentum is the generator of rotations

Question

I'm trying to demonstrate the following identity:

$$R_z(\theta) = e^{\frac{-i}{\hbar} L_z\theta}$$

I know the more general relation has $J_z$ and not $L_z$, but I'm only interested in showing it for the latter. If I write down Schrödinger's equation:

$$i\hbar\frac{\partial \Psi}{\partial t} = H\Psi$$

it is clear I can pass the term $i\hbar$ to the right hand side of the equation, and it is also quite clear the solution can be expressed by exponentiating the right hand side:

$$\Psi = e^{\frac{-i}{\hbar} H t}\Psi_0$$

where $U=e^{\frac{-i}{\hbar} H t}$ is the time evolution unitary operator. Let us now recall that $L_z=-i\hbar\frac{\partial}{\partial \theta}$ , so we can write the equation: $$\frac{\partial \Psi}{\partial \theta} = \frac{\pmb{+}i}{\hbar}L_z\Psi$$

Essentially, this is just another Schrödinger equation, where instead of time evolution we now have a rotation. Proceeding as before, the solution can be written as the matrix resulting from exponentiating the right hand side acting on some initial state $\Psi_0$:

$$\Psi = e^{\frac{\pmb{+}i}{\hbar} L_z\theta}\Psi_0$$

therefore, the rotation operator I get has the opposite sign compared to the one I should be getting. Why is this? I' thinking maybe by writing the problem this way I'm performing a rotation around the $-z$ axis instead of the $z$ axis, which would explain the $+$ sign I'm getting instead of the $-$ sign I should be getting. However, since I'm taking the derivative with respect to positive $\theta$ (that is to say, how my wavefunction evolves as I rotate it anticlockwise, which corresponds to the $+z$ direction), I would expect to get the corresponding negative sign in the exponent.

Answer 1

It's the difference between an active and passive rotation - i.e., are you rotating the wave function or your coordinate system?

In solving your Schrödinger-like equation, you are asking how $\psi$ changes when $\theta$ is varied - this corresponds to the passive point of view, where you rotate your coordinate system.

In contrast, $R(\theta) = e^{-i J_z \theta}$ is how the rotation operator looks from the active point of view, where you rotate the wave function, but keep your coordinates fixed.