Time evolution of a quantum state is defined by the Hamiltonian $\hat{H}$ through its role in the Schrödinger equation $i\,\hbar\,\mathrm{d}_t\,\psi =\hat{H} \,\psi$ in the Schrödinger picture, or the observable evolution equation $\mathrm{d}_t\,\hat{A} = \frac{i}{\hbar}[\hat{H},\,\hat{A}]$ in the Heisenberg picture. This is the definition of time evolution in quantum mechanics.

So, unless the Hamiltonian's action on your state $\psi$ is related to the action of $\hat{L}_z$ by $\hat{H}\,\psi = \omega\,\hat{L}_z\,\psi$, the uniform rotation $\exp\left(-\frac{i\,\omega\,t}{\hbar}\,\hat{L}_z\,\right)\,\psi$ is not a valid time evolution and $\exp\left(-\frac{i\,\theta}{\hbar}\,\hat{L}_z\,\right)\,\psi$ simply defines a co-ordinate rotation.

One instance where things do indeed work as you are thinking is when we think of Maxwell's equations as the one-photon Schrödinger equation, as I discuss in my answer here and here. Here the one-photon quantum state is uniquely defined by a vector field (the pair of positive frequnecy parts of the Riemann-Silbertein vectors $\vec{F}^\pm = \sqrt{\epsilon}\vec{E}\pm i\sqrt{\mu}\vec{H}$) and the one-photon Schrödinger equation is then:

$$i\,\hbar\,\partial_t\,\vec{F}^\pm = \pm\,\hbar\,c\nabla\times\vec{F}^\pm$$

(Note that here $\vec{E}$ and $\vec{H}$ do not stand for electric and magnetic fields, simply vector fields that define the photon's state just as the Dirac equation defines the first quantised electron's state as a spinor field). In momentum co-ordinates (wavevector space), this equation for a plane wave with wavevector $(k_x,k_y,k_z)$becomes:

$$i\,\hbar\,\partial_t\,\vec{F}^\pm= \pm\,\,c\,\left(k_x\,\hat{L}_z+k_y\,\hat{L}_y+k_z\,\hat{L}_z\right)\,\vec{F}^\pm$$

where here the angular momentum operators are of course:

$$\hat{L}_x=i\,\hbar\,\left(\begin{array}{ccc}0&0&0\\0&0&-1\\0&1&0\end{array}\right)\;\;\hat{L}_y=i\,\hbar\,\left(\begin{array}{ccc}0&0&1\\0&0&0\\-1&0&0\end{array}\right)\;\;\hat{L}_z=i\,\hbar\,\left(\begin{array}{ccc}0&-1&0\\1&0&0\\0&0&0\end{array}\right)\;\;$$

So that if the wavevector is directed along the $z$ axis, you do indeed have $\vec{F}^\pm = \exp\left(-\frac{i\,\omega\,t}{\hbar}\,\hat{L}_z\,\right)\,\vec{F}^\pm$ where $\omega = k\,c$, defining left and right hand circular polarisation states.

Time evolution of a quantum state is defined by the Hamiltonian $\hat{H}$ through its role in the Schrödinger equation $i\,\hbar\,\mathrm{d}_t\,\psi =\hat{H} \,\psi$ in the Schrödinger picture, or the observable evolution equation $\mathrm{d}_t\,\hat{A} = \frac{i}{\hbar}[\hat{H},\,\hat{A}]$ in the Heisenberg picture. This is the definition of time evolution in quantum mechanics.

So, unless the Hamiltonian's action on your state $\psi$ is related to the action of $\hat{L}_z$ by $\hat{H}\,\psi = \omega\,\hat{L}_z\,\psi$, the uniform rotation $\exp\left(-\frac{i\,\omega\,t}{\hbar}\,\hat{L}_z\,\right)\,\psi$ is not a valid time evolution and $\exp\left(-\frac{i\,\theta}{\hbar}\,\hat{L}_z\,\right)\,\psi$ simply defines a co-ordinate rotation.

One instance where things do indeed work as you are thinking is when we think of Maxwell's equations as the one-photon Schrödinger equation, as I discuss in my answer here and here. Here the one-photon quantum state is uniquely defined by a vector field (the pair of positive frequnecy parts of the Riemann-Silbertein vectors $\vec{F}^\pm = \sqrt{\epsilon}\vec{E}\pm i\sqrt{\mu}\vec{H}$) and the one-photon Schrödinger equation is then:

$$i\,\hbar\,\partial_t\,\vec{F}^\pm = \pm\,\hbar\,c\nabla\times\vec{F}^\pm$$

(Note that here $\vec{E}$ and $\vec{H}$ do not stand for electric and magnetic fields, simply vector fields that define the photon's state just as the Dirac equation defines the first quantised electron's state as a spinor field). In momentum co-ordinates (wavevector space), this equation for a plane wave with wavevector $(k_x,k_y,k_z)$becomes:

$$i\,\hbar\,\partial_t\,\vec{F}^\pm= \pm\,\,c\,\left(k_x\,\hat{L}_z+k_y\,\hat{L}_y+k_z\,\hat{L}_z\right)\,\vec{F}^\pm$$

where here the angular momentum operators are of course:

$$\hat{L}_x=i\,\hbar\,\left(\begin{array}{ccc}0&0&0\\0&0&-1\\0&1&0\end{array}\right)\;\;\hat{L}_y=i\,\hbar\,\left(\begin{array}{ccc}0&0&1\\0&0&0\\-1&0&0\end{array}\right)\;\;\hat{L}_z=i\,\hbar\,\left(\begin{array}{ccc}0&-1&0\\1&0&0\\0&0&0\end{array}\right)\;\;$$

So that if the wavevector is directed along the $z$ axis, you do indeed have $\vec{F}^\pm = \exp\left(-\frac{i\,\omega\,t}{\hbar}\,\hat{L}_z\,\right)\,\vec{F}^\pm$ where $\omega = k\,c$, defining left and right hand circular polarisation states.