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Consider the totally antisymmetric entangled spin-triplet wavefunction $$\psi=\left(\psi_\alpha({\vec r}_1)\psi_\beta(\vec{r}_2)-\psi_\alpha({\vec r}_2)\psi_\beta({\vec r}_1)\right)\left(|\uparrow_1\downarrow_2\rangle+|\downarrow_1\uparrow_2\rangle\right).$$ In this state, two electrons have different spin quantum numbers $m_s$; one is spin-up and the other is spin-down. See this answer.

The other quantum numbers (collectively specified by the indices $\alpha,\beta$) are also different, in general. In the special case, when $\alpha=\beta$, the wavefunction vanishes. But suppose $\alpha\neq \beta$ in which case, if we set, ${\vec r}_1={\vec r}_2$, the wavefunction vanishes though all quantum numbers, in general, are different. I want to understand this.

First of all, is it meaningful to set ${\vec r}_1={\vec r}_2$ and worry about the vanishing of the wavefunction? If this limit makes sense, then does the vanishing of wavefunction in this limit have a physical significance?

Isn't this vanishing of wavefunction a consequence antisymmetry of the wavefunction rather than of exclusion principle (because this vanishing can happen will all quantum numbers being different)?

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Yes, it is meaningful to think about this and actually really important. Let's first look what we are actually talking about when looking at $\psi(\vec{r_1},\vec{r_2})$. This function defines the probability to find one electron in $\vec{r_1}$ and the other one at $\vec{r_2}$ at the same time. So if we set $\vec{r_1}=\vec{r_2}$ in the wavefunction, this is the probability to find both electrons at that point. And for an antisymmetric spatial wavefunction, this probability is zero, we will never find both electrons at the same point (this is what's special about this antisymmetric wavefunction). If we pick any other $\vec{r_1}\neq\vec{r_2}$ near the nucleus, there will mostly be a probability $\neq 0$ for the two electrons to be in these respective positions.

The physical significance of this is that electrons in a state with antisymmetrical spatial(!) wavefunction tend to be (on average) further away from each other, as due to the construction of the antisymmetric wavefunction, roughly speaking, the probability vanishes the closer they get. This has far-reaching implications: (I will dwelve a bit into atomic physics here as i guess this is also where you treat this problem right now). As the electrons are on average further away from each other, there will be less repulsion between them. This also effects their energies: Less repulsion between them means a stronger binding to the nucleus, and a lower ground state energy. As electrons always want to be in the lowest energy state, they prefer states with wavefunctions that are spatially "more antisymmetric" and with the connection to the spin e.g. triplet>singlet states, but also quintet>triplets. And this determines actually in which order the atomic shells (higher multiplicity states first) are filled and leads to Hund's first rule!

To answer your final question, yes the vanishing of the spatial wavefunction is a direct result of the antisymmetry. But that we have a antisymmetric wavefunction for the symmetric spin triplet state in the first place is caused by the Pauli exclusion principle, which tells us that the product of the two HAS to be antisymmetric. This is actually also the full statement of the Pauli principle, and that we can't have two electrons in the same state is just an indirect implication of it (as if both spin and other quantum numbers $\alpha, \beta$ are equal, an antisymmetrically constructed wavefunction for either spin or spatial part would be zero).

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