# Pauli exclusion principle and position-state restriction

## Main Question or Discussion Point

Can one deduce from Pauli's exclusion principle (through the Slater Determinant) that two electrons with different spins in the same energy level, can't have the same position?

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Mentor
Why do you think this would be true at all?

Bill_K
Page 3 of the following link says so?
http://www.physics.metu.edu.tr/~sturgut/slater.pdf
It says, "No two electrons in a system can be in the same one-particle state... Note that in the statement “one-particle state” refers to both space and spin parts."

The OP says the spins are different. With an antisymmetric spin part, the space part of the two-electron wavefunction must be symmetric.

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Let me explain my problem clearly. In the expression $\psi _\alpha (x)$ for wave function of an electron, $\alpha$ is the state and $x$ includes both position and spin. I don't know whether $\alpha$ includes the spin or not and if it includes spin, is this spin the spin existed in the $x$?

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cgk
If you have a state which is a Slater determinant, you can derive with the Pauli principle that no two electrons with the *same* spin are in the same position. That is, if you have a N-particle wave function Θ, which is a Slater determinant, then the value Θ(x1, x2, x3,..., xN) = 0 whenever any two different coordinates i, j coincide (i.e., xi = xj). Here xi = (ri, si) is the combined space spin coordinate of the electron.

In the case of a Slater determinant, however, electronic coordinates with opposite spin are actually uncorrelated. That is, two electrons with opposite spin can happily sit on top of each other. Nothing in either theory or practice (this actually happens in real life) stops them from doing this.

 the orbitals entering the determinant are, strictly speaking, functions which take one of those combined coordinates and return a complex number.

Let me explain my problem clearly. In the expression $\psi _\alpha (x)$ for wave function of an electron, $\alpha$ is the state and $x$ includes both position and spin. I don't know whether $\alpha$ includes the spin or not and if it includes spin, is this spin the spin existed in the $x$?

cgk
It works like this: In reality, there is no wave function for an electron. There is only a single wave function for the N-electron system. In the Hartree-Fock approximation, this is a single Slater determinant Θ (and in the general case, it can be written as a linear combination of Slater determinants).

This Slater determinant is a function of N variables xi = (ri,si), which include both space and spin coordinates. The determinant itself is simply an anti-symmetrized product of N one-particle wave functions φ_k (the so called spin-orbitals):
$$\Phi(x_1,x_2,\ldots,x_N) = \frac{1}{N!}\sum_{p\in S_N} \mathrm{sgn}(p)\varphi_1(x_{p(1)}) \varphi_2(x_{p(2)}) \ldots \varphi_N(x_{p(N)})$$
where $S_N$ is the permutation group of order N, and sgn(p) is the permutation p's sign.
What can we see from that? For this to work, the orbitals φ_k must be mathematical functions $(\mathbb{R}^3\times\mathbb{Z_2})\mapsto \mathbb{C}$, which take one coordinate x and map it to one complex number. Formally, they can do that in any way they like: they just are functions which take three continuous and one discrete sub-variables and map it to a single scalar.

But how is this done in reality? For example, there are many "closed-shell" molecules. In that case, the molecular determinant can be written in terms of pairs of spin-orbitals φ_{Ak} and φ_{Bk}, one for alpha- and one for beta spin, with the same spatial part φ_{k}. That is, these spin-orbitals φ_{Sk} are chosen in such a way that they consist of one part φ_{k} which only depends on the spatial coordinate (i.e., is a normal 3d function), and one part S, which depends only on the spin coordinate.:
$$\varphi_{kS}(\vec x) = \varphi_{kS}(\vec r, s) = \varphi_k(\vec r) S(s)$$
S here is a 'discrete function' of one variable s, which can only have to values: s=A and s=B (or up and down or how you would like to call them). In practice one just uses the basis functions of this space; that means one spin function "A(s)" which returns 1 for s=A and 0 for s=B, and one "B(s)" which returns 1 for s=B and 0 for s=A.

So, to sum it up: One spatial "molecular orbital" (or state if you prefer that name) leads to two possible spin orbitals, which have the same spatial part and different spin parts. The orbitals are not mapped one to one to electronic coordinates; on the contrary, any coordinate x=(r,s) can go into *any* orbital (see anti-symmetrizer above). However, for some combinations of orbtial/coordinate, the contribution to the wave function vanishes (e.g., if you put a coordinate x=(r,A) into a spin orbital φ_{Bk}, then φ_{Bk}(x) = 0). Note again that the orbital φ_k and electronic coordinate x_k are not connected: any electron coordinate goes into any orbital.

This might be a bit confusing at first, but if you look through it step by step, then you just understood *the* basis of practical many-body methods.

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