How does a rotation operator change the angular momentum of a quantum system?

Question

I think I'm having some problems understanding the role of rotation operator $$\mathscr {R} = e^{-i/\hbar \hat{L_z} \theta}.$$ Suppose we have a quantum system in the state $\vert \psi_0 \rangle$ which is an eigenstate of $\hat{L_z}$ with eigenvalue $l_z$, $$\hat{L_z}\vert \psi_0 \rangle = l_z \vert \psi_0 \rangle,$$ where $\hat{L_z}$ is the projection of angular momentum along $z$ axis. If we "rotate" the whole system around $z$ axis by an angle $\theta$, the "rotated" state should be: $\vert \psi_1 \rangle$ = $\mathscr{R} \vert \psi_0 \rangle$. Then we have $$\hat{L_z} \vert \psi_1 \rangle = \hat{L_z}\mathscr{R} \vert \psi_0 \rangle = \mathscr{R}\hat{L_z}\vert \psi_0 \rangle=\mathscr{R}l_z\vert \psi_0 \rangle = l_z\mathscr{R}\vert \psi_0 \rangle=l_z \vert \psi_1 \rangle,$$ which means the rotated system is still in an eigenstate of $\hat{L_z}$ with eigenvalue $l_z$.

This part I can understand. But if we rotate the system by an angle $\theta = \omega t$, which means the new system is "rotating" around $z$ axis with angular velocity $\omega$. But if we repeat the above procedure and simply substitute $\theta$ with $\omega t$, we still get the result that the rotated system is still in an eigenstate of $\hat{L_z}$ with eigenvalue $l_z$. This means the rotating system has the same amount of angular momentum along $z$ axis as the original one. This part I cannot understand. Because intuitively we should expect the rotated system to have more angular momentum than the original one because it is "rotating" with angular velocity $\omega$.

Answer 1

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.