According to Pauli’s exclusion principle, an $s$ orbital contains at most two electrons with the opposite spin (up and down). Why can't an $s$ orbital contain a third electron whose state is the linear combination of spin up and down?
$\begingroup$
$\endgroup$
0
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
2
Electrons are magnets, they have magnetic fields. Those fields have only two possible orientations, and a single orbital can only be occupied by two electrons if those orientations are mutually opposed.
OK, but why? Why are electrons magnets? Why are there only two possible orientations for the fields? Why can the two electrons in the orbital only exist with opposite directions for the magnetic moments?
Well, there are answers to those questions, but they don't make sense a lot of sense. The magnetic moment of electrons is due to a property called "spin"; the spin value of an electron is 1/2, so it can adopt one of two spin quantum states, +1/2 and -1/2, which correspond to the "up" and "down" orientations of the magnetic moment. Since it has a half-integer spin, an electron is a member of a class of sub-atomic particles called "fermions" which obey rules called the Pauli Exclusion Principle and Fermi-Dirac statistics - one key result of these rules is that no two identical fermions can simultaneously occupy the same quantum state: you cannot have two electrons in the same orbital if they have the same spin orientation, so if one is +1/2, the other must be -1/2, and no more can be added because a third would have to adopt the same quantum state as the one of the first two.
OK, but why? Why does an electron have a spin value of 1/2? Why can't two fermions adopt the same quantum state (when integer spin particles, called bosons, can do so)? Why does the property of "spin" even exist? You need quantum mechanics to even begin to answer those questions -- again, that's more math than anybody is prepared to show you for at least five years -- and even then, the "why" of the Pauli Exclusion Principle is an empirical observation (it has no theoretical basis, it's just an after-the-fact thing that happens) without relativity, but there is a firm theoretical rationale and a reason "why" if one revisits the postulates of quantum mechanics and takes into general relativity into account. But relativistic quantum field theory is more math than any mortal should be forced to endure.
OK, but why? Why is quantum mechanics how the universe behaves? Why is general relativity the way the universe behaves?
You can play the "why" game for a very long time. None of your high school teachers will be able to give you more than one or two repetitions. University professors will be able to go a lot farther. Eventually, you get to questions to which the only possible answer is "nobody knows.... yet". And those are the questions that scientists -- including those university professors -- spend their day trying to answer.
Which is terrific fun. So keep asking "why", even to the answers your teachers give you. And if they can't answer, go ask someone else, or read about it, or go to university. And if they don't know, be a scientist, and figure it out yourself, and tell the rest of us.
-
1
-
1$\begingroup$ I really loved this answer, in particular the last paragraph. Very nice motivation :) ! $\endgroup$ Feb 7, 2017 at 13:03
$\begingroup$
$\endgroup$
1
Because we can have only one electron per quantum state. Spin up and spin down are two different states. A linear combination of the two is not a new independent state. It is obviously formed from the spin up and spin down states.
-
3$\begingroup$ -1. First, it does not answer the question, it just paraphrases it in a different way. Secondly, it is just plain wrong. A linear combination of two states is a new state, this new state is obviously linearly dependent with the two states, but it is a different state. $\endgroup$– WildcatJan 19, 2015 at 9:59
$\begingroup$
$\endgroup$
The complete wave function of a two electron system is written using slater determinants. So, for helium it would look like this:
$$\begin{pmatrix}1s(1)\alpha(1) & 1s(1)\beta(1)\\\ 1s(2)\alpha(2) & 1s(2)\beta(2)\end{pmatrix}$$
For lithium, in what you are saying, it would look like this:
$$\begin{pmatrix}1s(1)\alpha(1) & 1s(1)\beta(1) & 1s(1)(\alpha(1)+\beta(1))\\\ 1s(2)\alpha(2) & 1s(2)\beta(2) & 1s(2)(\alpha(2)+\beta(2)) \\\ 1s(3)\alpha(3) & 1s(3)\beta(3) & 1s(3)(\alpha(3)+\beta(3))\end{pmatrix}$$
This obviously is zero, as can be seen from basic principles of matrices and determinants, which is not possible, as the wavefunction obviously cannot be zero. (Note: 1s(i) refers to the spatial part of the wave function and $ \alpha,\beta $ represents the spin part.)
