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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?

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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.

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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.

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