What happens to an atom's angular momentum when it absorbs an electron?

AI Thread Summary
When an atom with zero angular momentum absorbs an electron, the resulting angular momentum depends on the electron's spin state. If the electron is emitted randomly, its average contribution to the atom's angular momentum is zero, allowing the atom to maintain a total angular momentum of zero. However, if the electron's spin is aligned along a specific axis, such as the z-axis, the atom's total angular momentum becomes non-zero, specifically 1/2. In a scenario where multiple such atoms are analyzed, a Stern-Gerlach experiment would show an equal distribution of spin states, with half of the atoms exhibiting spin up and half spin down. Thus, the absorption of the electron alters the atom's angular momentum based on the electron's spin orientation.
LCSphysicist
Messages
644
Reaction score
162
Homework Statement
.
Relevant Equations
.
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.
 
Physics news on Phys.org
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".
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top