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In Classical Mechanics it is "easy", if some system has some angular momentum at the beginning then conservation of angular momentum means that no matter what happens, the angular momentum vector at the end will be the same.

In Quantum Mechanics it is different because we cannot even know the angular momentum of a system. We can only know its magnitude $L^2$ and one of its components, say $L_z$. So we cannot talk about the conservation of angular momentum as implying that the initial vector must be the same as the final vector. We don't know such vector.

So how do we apply the conservation of angular momentum?

  • The mean value of the vector $\langle \vec{L} \rangle$, of which we know exactly the three components, is the magnitude that is conserved? What about other quantities related to angular momentum, such as its dispersion $\sigma_L$?
  • Is the angular momentum ket of a system the thing that is conserved? I mean, if a system is in some eigenstate of angular momentum $|j,m\rangle$, is this property of the system the one that does not change due to conservation?
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In QM, angular momentum is conserved as an operator.

Deal with it.


This is best seen in the Heisenberg picture, where the equation of motion for $\hat{\mathbf L}$ in a spherically-symmetric hamiltonian $\hat H$, $$ i\hbar\frac{\mathrm d}{\mathrm dt}\hat{\mathbf L} = [\hat{\mathbf L},\hat H] = 0, $$ so $\hat{\mathbf L}$ is conserved as an operator as $t$ goes from $t=0$ onwards. In particular, this means that:

  • If the system is in an eigenstate of a function of $\hat{\mathbf L}$ (including the magnitude $\hat{L}^2$ or any components) then it stays in that eigenspace for all time.
  • If the system is in a superposition of such eigenstates, then the weights and relative phases of that superposition are maintained.
  • This then implies that all moments of all distributions of observables that are a function of $\hat{\mathbf L}$ are constant in time.

These conclusions also hold equally well in the Schrödinger picture.

(On the other hand, if you know that the system is in some eigenstate $|\psi⟩$ of $\hat L^2$ and $\hat L_z$ and it evolves under a spherically symmetric hamiltonian $\hat H$ in the Schrödinger picture, you can conclude that it will remain as an eigenstate of $\hat L^2$ and $\hat L_z$ with its same eigenvalues, but you cannot conclude that the state does not change, as there is typically radial motion which can obviously change.)

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There is some confusion here. Quantization of an observable does not change the need of conservation .For example energy is always conserved in the quantized energy levels of the atom, the input and output , including potential energy, have to be exactly matched. Just the possible values are constrained.

The same holds true for momentum and angular momentum:

In the process of solving the Schrodinger equation for the hydrogen atom, it is found that the orbital angular momentum is quantized according to the

$L^2=l(l+1){h^2/(2π)}^2$

For bound states the angular momentum is quantized. Experimental measurements in particle physics have forced the concept of spin in order to keep conservation of angular momentum valid .

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  • $\begingroup$ So if the atom is in $|l,m\rangle$ state before some process and angular momentum is conserved, how do you apply the angular momentum conservation? The ket must be the same after the process? Or just the mean value? Or what is the thing that must be the same before and after? $\endgroup$ – user171780 Aug 7 '18 at 13:31
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    $\begingroup$ $\mathbf{L}(t)= \mathbf{L}(0)$. Whatever the r.h.side does to your ket, the l.h.side will do too. The eigenvalues of $L_z$ and $\mathbf{L}^2 $ will stay the same. Transition matrix elements of such operators will be the same. It is not true that you may not discuss and describe operators whose components you may not measure simultaneously. $\endgroup$ – Cosmas Zachos Aug 7 '18 at 13:42

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