So just thinking about Spin and not the helium atom for the moment,
$$S=S_1+S_2, \quad S^2=S_1^2+2S_1\cdot S_2+S_2^2, \quad S_z=S_{1z}+S_{2z}$$
We can also use that $S_+=S_x+iS_y$ and $S_-=S_x-iS_y$ that
$$S_1\cdot S_2=S_{1z}S_{2z}+\frac{1}{2}\left(S_{1+}S_{2-}+S_{1-}S_{2+}\right)$$
Anyway what will be the old spin basis will be common eigenvectors of the commuting observables $$\{S_1^2, S_2^2,S_{1z},S_{2z}\}$$
These could be notated e.g. $\left|\uparrow\uparrow\right\rangle=\left|\uparrow\right\rangle\otimes\left|\uparrow\right\rangle=\chi^+(1)\chi^+(2)$. In the new spin basis we take the vectors to be common eigenvectors of the commuting observables
$$\{S_1^2, S_2^2,S^2,S_{z}\}$$
We just change basis and as you have it in the new basis the eigenstates are
\begin{align}
\left|1,1\right\rangle&=\left|\uparrow\uparrow\right\rangle\\
\left|1,0\right\rangle&=\frac{1}{\sqrt{2}}\left(\left|\uparrow\downarrow\right\rangle+\left|\downarrow\uparrow\right\rangle\right)\\
\left|1,-1\right\rangle&=\left|\downarrow\downarrow\right\rangle\\
\left|0,0\right\rangle&=\frac{1}{\sqrt{2}}\left(\left|\uparrow\downarrow\right\rangle-\left|\downarrow\uparrow\right\rangle\right)\\
\end{align}
You can check using the definitions of $S_z$ and $S^2$ that theses are eigenvectors, e.g.
$$S_z\left|1,1\right\rangle= (S_{1z}+S_{2z})\left|\uparrow\uparrow\right\rangle=S_{1z}\left|\uparrow\uparrow\right\rangle+S_{2z}\left|\uparrow\uparrow\right\rangle$$
$$S_1\left|\uparrow\uparrow\right\rangle=\left(S_1\left|\uparrow\right\rangle\right)\otimes \left|\uparrow\right\rangle=\frac{\hbar}{2}\left|\uparrow\uparrow\right\rangle$$
$$S_2\left|\uparrow\uparrow\right\rangle=\left|\uparrow\right\rangle\otimes \left(S_2\left|\uparrow\right\rangle\right)=\frac{\hbar}{2}\left|\uparrow\uparrow\right\rangle$$
So,
$$S_z\left|1,1\right\rangle= \hbar\left|\uparrow\uparrow\right\rangle=\hbar \left|1,1\right\rangle$$
Another example is $S^2$,
\begin{align}
S^2 \left|1,1\right\rangle&=\left(S_1^2+S_2^2 +2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+}\right)\left|\uparrow,\uparrow\right\rangle\\
&=\left(\frac{3}{4}\hbar^2+\frac{3}{4}\hbar^2 +2\frac{\hbar}{2}\frac{\hbar}{2}+0+0\right)\left|\uparrow,\uparrow\right\rangle\\
&=\left(\frac{8}{4}\hbar^2\right)\left|\uparrow,\uparrow\right\rangle\\
&=2\hbar^2\left|1,1\right\rangle\\
&=(s)(s+1)\hbar^2\left|1,1\right\rangle\quad \text{where } s=1.
\end{align}
You get the second line similar to the last example and using how the single spin operators act on a single spin, in short:
\begin{align}
S^2\left|\uparrow\right\rangle&=\frac{3}{4}\hbar^2\left|\uparrow\right\rangle\\
S_z\left|\uparrow\right\rangle&=\frac{\hbar}{2}\left|\uparrow\right\rangle\\
S_+\left|\uparrow\right\rangle&=0\\
S_-\left|\downarrow\right\rangle&=0
\end{align}
Anyway, back to the helium atom, depending on the levels of approximation (order in perturbation theory etc) you might try write say the ground state as
$$\psi=\phi_{\text{orbital}}\cdot\chi_{\text{spin}}$$
or with kets,
$$\left|\psi\right\rangle=\left|\phi_{\text{orbital}}\right\rangle\otimes\left|\chi_{\text{spin}}\right\rangle$$
Written as such the spin operator on acts on the $\chi$ part of the wave function, so if this is a spin singlet or triplet then it is and eigenvalue of $S^2$ and $S_z$, e.g. if $\left|\chi\right\rangle=\left|s,m\right\rangle$ then
\begin{align}
S^2\left|\psi\right\rangle&=S^2\left(\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\right)\\
&=\left|\phi_{\text{orbital}}\right\rangle\otimes\left(S^2\left|s,m\right\rangle\right)\\
&=s(s+1)\hbar^2\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\\
\end{align}
and
\begin{align}
S_z\left|\psi\right\rangle&=S_z\left(\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\right)\\
&=\left|\phi_{\text{orbital}}\right\rangle\otimes\left(S_z\left|s,m\right\rangle\right)\\
&=m\hbar\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\\
\end{align}
I don't know all of the details of the exact calculations of Helium ground and excited states, or the energies however.
What is the difference between parallel and antiparallel spins for a pair of nucleons?
It simply means that nucleons can be parallel in spin or opposite nature of spin- example of two nucleons
( up , up ) or ( down , down ) or ( up , down ) or (down, up)
if they are parallel total spin S = s(1) +s(2) = 1 a triplet state
if they are anti -parallel
total spin S = s(1) +s(2) = 0 a singlet state
the spin-spin or spin-orbit interactions will be different for two states.
which may affect the nuclear force.
the scattering cross sections for two states will give different results.
As nucleons are added in a nucleus in a specific state the nucleons get paired and the final spin state of the nucleus is decided by the odd nucleon (its spin). In case of even number nucleus the total spin is zero.
A characteristic of the collection of protons and neutrons (which are fermions) is that a nucleus of odd mass number A will have a half-integer spin and a nucleus of even A will have integer spin. The suggestion that the angular momenta of nucleons tend to form pairs is supported by the fact that all nuclei with even Z and even N have nuclear spin I=0.
Ref.-Reference-http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/nspin.html
Spin-spin coupling is the coupling of the intrinsic angular momentum (spin) of different particles.
Examples are-
Such coupling between pairs of nuclear spins is an important feature of (NMR) spectroscopy as it can provide detailed information about the structure and conformation of molecules.
Spin-spin coupling between nuclear spin and electronic spin is responsible for hyperfine structure in atomic spectral transitions.
Am I understanding this correctly? - I'm thinking I might be missing something in regards to the Pauli Exclusion Principle. How can two nucleons with the same spin state pair?
Due to the similarity in mass and nuclear properties between the proton and neutron, they are sometimes considered as two symmetric types of the same object, a nucleon.
The symmetry relating the proton and neutron is known as isospin and denoted I (or sometimes T).
Isospin is an SU(2) symmetry, like ordinary spin, so is completely analogous to it.
The proton and neutron form doublet, with a downstate (↓) being a neutron, and an up (↑) state is a proton.
A pair of nucleons can either be in an antisymmetric state of isospin called singlet or in a symmetric state called triplet.
An example-
A nucleus with one proton and one neutron, i.e. a deuterium nucleus.
and thus consists of three types of nuclei, which are supposed to be symmetric: a deuterium nucleus (actually a highly excited state of it), a nucleus with two protons, and a nucleus with two neutrons. The latter two nuclei are not stable or nearly stable, and therefore so is this type of deuterium (meaning that it is indeed a highly excited state of deuterium).
The deuteron wavefunction must be antisymmetric if the isospin representation is used (since a proton and a neutron are not identical particles, the wavefunction need not be antisymmetric in general). Apart from their isospin, the two nucleons also have spin and spatial distributions of their wavefunction. The latter is symmetric if the deuteron is symmetric under parity (i.e. have an "even" or "positive" parity), and antisymmetric if the deuteron is antisymmetric under parity (i.e. have an "odd" or "negative" parity). The parity is fully determined by the total orbital angular momentum of the two nucleons: if it is even then the parity is even (positive), and if it is odd then the parity is odd (negative).
The deuteron, being an isospin singlet, is antisymmetric under nucleons exchange due to isospin, and therefore must be symmetric under the double exchange of their spin and location.
Ref. < https://en.wikipedia.org/wiki/Deuterium#Spin_and_energy
Therefore one may try constructing the wave function of deuteron holding the Pauli principle and the ground state of deuteron will come to a triplet state of nucleons with Binding of about 2.2 MeV.
Best Answer
The spins are not antiparallel in this state as @my2cts answer correctly states (as demonstrated by $\vec S_1 \cdot \vec S_2$ having a positive eigenvalue).
I was confused by this at first too, especially the argument, that the measurements of $S_{1z}$ and $S_{2z}$ will point to opposite directions due to entanglement seemed convincing to me.
But it just looks that way because the $S_z$ basis is the wrong basis to look at. They have to point in opposite directions in this basis, because the total $m_z = 0$. So you are just asking a spin state in the $x$-$y$-plane for its $z$-component. This way you can't find out whether the spins are parallel or antiparallel in that plane.
To make the parallel orientation of the spins manifest in the expression for the state, you'll have to rewrite the state in a basis in the plane.
We rewrite our state terms of the eigenstates $\lvert \pm \rangle$ of $S_x$: $$ S_x \lvert \pm \rangle = \pm \frac \hbar 2 \lvert \pm \rangle$$ In terms of those states we have $$ \lvert \uparrow / \downarrow \rangle = \frac{1}{\sqrt{2}} \big( \lvert + \rangle \pm \lvert - \rangle \big)$$ in terms of these states the two product states become $$ \lvert \uparrow \downarrow \rangle = \frac 1 2 \big( \lvert + \rangle + \lvert - \rangle \big) \otimes \big( \lvert + \rangle - \lvert - \rangle \big) = \frac 1 2 \big( \lvert++\rangle - \lvert+-\rangle + \lvert -+ \rangle - \lvert -- \rangle \big) $$ $$ \lvert \downarrow \uparrow \rangle = \frac 1 2 \big( \lvert++\rangle - \lvert-+\rangle + \lvert +- \rangle - \lvert -- \rangle \big) $$ So the triplet state with $m_z = 0$ is given by the following in this basis: $$ \frac{\lvert \uparrow \downarrow \rangle + \lvert \downarrow \uparrow \rangle}{\sqrt{2}} = \frac{\lvert ++ \rangle - \lvert -- \rangle}{\sqrt{2}}$$
And in this form we can see manifestly that the spins are in parallel. (Similarly, this could be done for the eigenstates of $S_y$.)