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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 $\omega$.

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  • $\begingroup$ The problem is probably that you are not really implementing a time-evolution this way. In other words, for what concerns the formalism, you are still considering stationary, time-independent states, only calling "$\omega t$" the rotation angle. But I hope someone will be able to better address the question $\endgroup$ – glS Dec 29 '14 at 12:24
  • $\begingroup$ The variable $t$ is more like a parametrisation of the rotation, i.e. the specification of a one-parameter subgroup of the proper rotation group. For quantised systems, one cannot increase the momentum by an arbitrary real number $\hbar\omega$, so it would be hard to imagine that the action of $\mathscr R$ is actually affecting the total angular momentum. $\endgroup$ – Phoenix87 Dec 29 '14 at 12:53
  • $\begingroup$ I understand that you apply to your system a global rotation with angular velocity $\omega$. Then I would treat the problem by using two frames of reference, one that rotates together with your system, and the lab frame. So, you have here two angular momenta, one with respect to the axes of the system, and another one due to the rotation of the system as a whole, and they can be added. But we have to do with a quantum system and the additional angular momentum may be quantized. Is $\omega$ arbitrary? $\endgroup$ – Sofia Dec 29 '14 at 16:03
  • $\begingroup$ @Sofia, yes $\omega$ is arbitrary. In fact I thought "rotating" the quantum system is equivalent to "rotating" the reference frame in the opposite direction. Since the rotation of reference frame is arbitrary, I assume that $\omega$ is arbitrary. The the question becomes "are the angular momentum of the system the same seen in the rotating frame and in the stationary frame ?". I expect they are not, but the above deduced lead me to the result that they are the same, which confuses me. Thanks for your answer -:)) $\endgroup$ – B.S.Gao Dec 29 '14 at 16:41
  • $\begingroup$ @B.S.Gao: it's not the same to rotate the system of coordinates, and to rotate an atom. The atom is a quantum system, and its movements are quantized. Savanna seems to say that the case is even worse, that you simply can't do it. I am not sure whether I understand properly that explanation, but indeed, how you do this rotation? How do you rotate an atom? $\endgroup$ – Sofia Dec 29 '14 at 16:53
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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.

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    $\begingroup$ @glance: I am not sure, s.t. I withdrew my answer temporarily. I am not sure that the additional angular momentum can be classical, because the object is quantum. I think that the system can be rotated, but I doubt that $\omega$ can be arbitrary, and about the additional angular momentum, I believe that it should be also quantized. $\endgroup$ – Sofia Dec 29 '14 at 16:14
  • $\begingroup$ @Sofia You are correct: AM is always quantized, because rotations live in a compact domain ($e^{i\,\theta}$ lies on a circle, not a noncompact line) so AM is quantized for the same reason that classical vibrations on a circular object or in a compact resonator are quantized. LionelBrits gives a great explanation of these ideas in his answer: physics.stackexchange.com/a/154626/26076 $\endgroup$ – WetSavannaAnimal Dec 29 '14 at 21:10

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