[Physics] Helium ground state spatial component of the wave function symmetric

Question

I do not understand why the spatial component of the wave function in the ground state for the helium is necessarily symmetric.
In that case the spin component is antisymmetric according to pauli exclusion principle.

Answer 1

Note: I'm not sure exactly what background you have so I have included lots of information, some of which may be superfluous but which hopefully helps other readers put the whole picture together.

Helium has two electrons, so (as you mentioned) one must consider the Pauli Exclusion Principle (aka the PEP) [0], [1]. The $total$ eigenfunctions of the system are the products of spatial eigenfunctions and spin wavefunctions [2].

Frist, let's consider the spin wavefunctions. We pick some axis to define the components of the spins, as usual let's choose the $z$ axis. We want to construct the two-particle spin wavefunction $\chi(1,2)$ from the single particle wavefunctions $\chi(1), \chi(2)$. The (first order) Hamiltonian for the Helium atom doesn't include any spin-dependent interactions, so the simplest way to construct the two-particle wavefunctions is just to stick the two spins together. Each spin can be 'up' or 'down' (along the $z$ axis), so our first guess would be: $$ \chi_1(1,2) = |up>_1|up>_2 \\ \chi_2(1,2) = |up>_1|down>_2 \\ \chi_3(1,2) = |down>_1|up>_2 \\ \chi_4(1,2) = |down>_1|down>_2.$$

The problem is, these wavefunctions aren't symmetrized. We $need$ the spin wavefunctions $and$ the spatial wavefunctions to be either symmetric or antisymmetric so that we have a well-defined symmetry when we multiply them together to get the eigenfunctions for the whole system.

So, let's symmetrize these spin wavefunctions [3]. We can see that $\chi_1$ and $\chi_4$ are already symmetric under exchange of the labels, so all we need to do is form symmetric/antisymmetric combinations of the $\chi_2$ and $\chi_3$. If you stare at it a little bit you can see that you can make two combinations, one which is symmetric and one which is antisymmetric: $$ \chi_{symm}(1,2) = \frac{1}{\sqrt(2)}(|up>_1|down>_2+|down>_1|up>_2) \\ \chi_{antisymm}(1,2) = \frac{1}{\sqrt(2)}(|up>_1|down>_2-|down>_1|up>_2).$$

Note: now we have 3 symmetric spin wavefunctions (a $triplet$) and a single antisymmetric wavefunction (a $singlet$).

Ok now let's consider the spatial eigenfunctions. These come directly from solving the Hamiltonian for the two-electron-one-nucleus system. I won't go through it here, but if you ignore everything except Coulomb interactions you end up with spatial wavefunctions described by two quantum numbers: $n,l$, where $n$ describes the electronic energy level and $l$ the orbital angular momentum. Intuitively, then, the ground state spatial wavefunction has the smallest $n$ and the smallest $l$: $n,l=1,0$. States with $l=0$ angular momentum are often called 's-states' and the wavefunction is spherically symmetric about the origin (nucleus) [hence, 's']. So an $n,l=1,0$ spatial wavefunction is $symmetric$.

Your question remains: why is $l=0$ (which corresponds to the lowest 'momentum' energy) symmetric?

A first approximation to the solution of the full two-electron Hamiltonian is the 'independent particle model'. As the name implies, you assume the particles are basically independent and solve the Schrodinger equation separately for each electron. We know how to do this, it's the solution to the Hydrogen atom. Considering only Coulomb interactions to start, this solution has eigenfucntions which are products of radial functions and spherical harmonics. The lowest-energy spherical harmonics have $l=0$, and spherical harmonic solutions with $l=0$ are spherical (thus, symmetric): https://en.wikipedia.org/wiki/Spherical_harmonics. This is a the 'mathematical' reason. Perhaps more intuitively, the $l$ quantum number corresponds to orbital angular momentum, and the lowest $l$ being spherical is kind of like saying that the electron stays 'closest', on average, to the nucleus, therefore having the lowest energy.

Since the ground state spatial wavefunction is symmetric, we have to pick the antisymmetric spin wavefunction when we construct the total wavefunction, as you mentioned in your question.

[0] I also like to call the Heisenberg Uncertainty Principle the HUP.

[1] The PEP requires that the $total$ wavefunction of a system of N fermions must be antisymmetric. In this case, that means that the wavefunction must change sign if $all$ coordinates (spatial and spin) are interchanged.

[2] This entire answer is paraphrased from 'Physics of Atoms and Molecules', Bransden and Joachain

[3] Once the spin wavefunctions are symmetrized, they also become eigenfucntions of the $S^2$ operator and mutually orthogonal, which is important/useful.