# Angular momentum and QM

• LCSphysicist
In summary, the conversation discusses the absorption of angular momentum from the point of view of quantum mechanics. The question is raised about what happens when an atom with net angular momentum of 0 absorbs an electron with an angular momentum of $\hbar/2$. It is suggested that the atom will maintain its angular momentum of 0, but if the electron is emitted in a specific direction, the atom's angular momentum will change to $\hbar/2$. The question of whether this change in angular momentum can be observed is also raised.

#### LCSphysicist

Homework Statement
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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.

Consider a sodium ion (Na+) that has zero spin and orbital angular momentum, ##S=0## and ##L=0##. Its total angular momentum is ##J=L+S=0.## Say it captures an electron. The atom now has ##S=\frac{1}{2}## and ##L=0.## Its total angular momentum is not zero, it is ##J=L+S=\frac{1}{2}.## If you have a collection of such atoms and you pass them through a Stern-Gerlach machine, ideally half of them will come out "spin up" relative to the machine and the rest "spin down".