# Question about angular momentum and its absorption by quantum particles

I was thinking a little about how the absorption of angular momentum occurs from the point of view of QM. For example, suppose we have an atom A and an electron $$e^-$$.

The electron $$e^-$$ is ejected from a source radially in direction of the center of the atom. Suppose that the atom has net angular momentum $$= 0$$ and it absorbs the electron, my question is what will happen now.

I mean, the electron has angular momentum $$\hbar /2$$, but since it was emitted from a source in a randomly way, it is equally probable that the spin of the electron (if we measure it while it is on the path to the atom) be collapsed in any direction.

So, my interpretation is that the atom, after absorbing the electron, will maintain its angular momentum = 0, since in an average way the electron has angular momentum = 0.

Now, suppose we create a source that emits electron in such way that the direction of the spin is the z axis. The electron now is in the state $$|\psi \rangle = |+ \rangle/\sqrt{2} + |- \rangle/\sqrt{2}$$

Supposing that the atom is enclosed by a box with an open hole in its surface that allows the electron to pass, but that does not allow us to see inside it. Is it right to say that the atom now is in the state $$|\psi \rangle = |+ \rangle/\sqrt{2} + |- \rangle/\sqrt{2}$$? In other words, it is equally probably the atom angular momentum $$\hbar/2$$ in the + or - direction?

I would like to know if my statement in bold is right, and if it is right to interpret the italic way.

• What are $\lvert+\rangle$ and $\lvert-\rangle$? Are they eigenstates of $S_z$, $S_x$, or something else? Aug 18, 2021 at 6:52
• This may or may not help: from a classical standpoint there is more than the spin. And i know that includes bohr model too. When it leave it takes its spin and takes its revolution. (Thats also true on a very simplistic level of a plant leaving an orbit tangentially; it loses its angular momentum that way. And the sun/planet system do too.) Both are rotation and revolution are angular momentum. Same thing when is captured. Aug 20, 2021 at 6:29

I suppose you are measuring the angular momentum of the composite atom (e- with the old atom as a whole) in the box. Then this is analogous to the canonical example of 2 coupled spin-1/2 particles, where you have a singlet state and a triplet states. In this case the total Hamiltonian is given by $$H \propto \vec{S_1} \cdot \vec{S_2} = S_1^x \otimes S_2^x + S_1^y \otimes S_2^y + S_1^z \otimes S_2^z$$ where $$S_1,S_2$$ are defined by 3 Pauli matrices in their own Hilbert space. After diagonlizing it you'll get four states: singlet and triplet.
Now in your case the electron $$S_1$$ is spin-1/2, but the atom is spin-0 whose Hilbert space structure of angular momentum $$S_2$$ is trivial, there is nothing for $$S_1$$ to couple to. So in the respect of angular momentum, the composite system of the electron and the old atom in the box will be the same as the single electron itself - the whole system is in a superposition of the two orthogonal states.