What does the Pauli Exclusion Principle mean in terms of the wave equation? In general chemistry, they teach you that the Pauli Exclusion Principle stipulates that two electrons cannot have the same four quantum numbers. However, I know that the real definition has something to do with the wave equations of two particles that are "near" each other. Can somebody provide me with a more detailed definition of the Pauli Exclusion Principle?
 A: This has to do with multi-particles wavefunctions.
Suppose we have two particles with states $\psi_{1}(\vec{r}_{1})$ and $\psi_{2}(\vec{r}_{2})$. Then the probability distributions are of course $P_{1}(\vec{r}_{1})=|\psi_{1}(\vec{r}_{1})|^2$ and $P_{2}(\vec{r}_{2})=|\psi_{2}(\vec{r}_{2})|^2$. It is tempting to think that the wavefunction for both particles is just
$$\phi(\vec{r}_{1},\vec{r}_{2})=\psi_{1}(\vec{r}_{1})\psi_{2}(\vec{r}_{2})$$
since then you get that the probability for finding each particles in a certain position is the product of the individual probabilities. But there is a fundamental fact: you can't distinguish between those particles, and as a result the probability neither has to. Thus we must require that
$$|\phi(\vec{r}_{1},\vec{r}_{2})|^2=|\phi(\vec{r}_{2},\vec{r}_{1})|^2$$
which means that
$$\phi(\vec{r}_{1},\vec{r}_{2})=e^{i\theta}\phi(\vec{r}_{2},\vec{r}_{1})$$
A priori, $\theta$ can be a function of the spatial coordinates. But, it turns out that there is a connection between this phase and the spin of the particles in question. For spin half particles, or fermions, this phase is $-1$. For integer spin particles, or bosons, this phase is $1$. This is called the spin-statistics theorem. In other words, fermions' wavefunction is anti-symmetric under particle exchange while bosons' wavefunction is symmetric.
Lets now consider the case of two electrons. Electrons are fermions, and therefore their mutual wavefunction must be anti-symmetric
$$\phi(\vec{r}_{1},\vec{r}_{2})=\frac{1}{\sqrt{2}}(\psi_{1}(\vec{r}_{1})\psi_{2}(\vec{r}_{2})-\psi_{1}(\vec{r}_{2})\psi_{2}(\vec{r}_{1}))$$
In particular, if the quantum numbers of both electrons are the same, then they are in the same state $\psi_{1}=\psi_{2}$ which leads to
$$\phi(\vec{r}_{1},\vec{r}_{2})=0$$
so it can't happen.
Notice: I omitted the spin coordinate from the above derivation for easier writing, but keep in mind that it should be there also. You can alternatively think that $\vec{r}=(x,y,z,s)$ where $s$ is the spin.
A: Let us denote the wave function of a system of two electrons (for simplicity) as : $\Psi(\vec{r}_{1},\sigma_{1};\vec{r}_{2},\sigma_{2})$ where $\vec{r}$ stands for position coordinate and $\sigma$ stand for spin coordinate. Then as electrons are fermions, the wave function is antisymmetric (spin-statistics theorem). i.e., $\Psi(\vec{r}_{2},\sigma_{2};\vec{r}_{1},\sigma_{1})=-\Psi(\vec{r}_{1},\sigma_{1};\vec{r}_{2},\sigma_{2})$. Now if we assumes the wave function of the form (mean-field or Hartree-Fock or non-interacting, a simplest choice satisfying anti-symmetric nature, which is hidden in your question) $\Psi(\vec{r}_{1},\sigma_{1};\vec{r}_{2},\sigma_{2})=\frac{1}{\sqrt{2!}}\det\begin{pmatrix}\psi(\{q_{1}\},\vec{r}_{1}^{}) & \psi(\{q_{1}\},\vec{r}_{2}^{}) \\ \psi(\{q_{2}\},\vec{r}_{1}^{}) & \psi(\{q_{2}\},\vec{r}_{2}^{}) \end{pmatrix}$. It follows that $\{q_{1}\}=\{q_{2}\}$ $\Rightarrow$ $\Psi(\vec{r}_{1},\sigma_{1};\vec{r}_{2},\sigma_{2})=0$ (which is not a reasonable wave function to describe two electrons). Here $\{q\}$ stand for quantum numbers specifying the single orbitals of electrons.
