Combination of 3 spin $\frac{1}{2}$ particles to yield a state of net spin $\frac{1}{2}$

Question

I did some prior research to this question on stackexchange before posting my question. Due to my limited knowledge in this field, I am not sure if my question is unique since there has been ample questions related to this problem. The most widely cited question I have seen was Adding 3 electron spins.

My question revolves around the construction of the $|\frac{3}{2},m⟩$ and $|\frac{1}{2},m⟩$ states.

To construct the $|\frac{3}{2},m⟩$ state, I am aware that we can start from constructing $|\frac{3}{2},\frac{3}{2}⟩$, which is fairly simple, as this necessitates that the 3 half spin particles must all spin up and that their spins don't cancel. To achieve this, all particles must spin up, thereby the state can be represented by $|↑↑↑⟩$. To achieve the remaining $|\frac{3}{2},m⟩$, we apply the ladder operator. My first question is that how the magnitude of spin is being calculated from the state $|↑↑↑⟩$. For instance, I know that from $|↑↑↑⟩$, the spin in the z direction is found by counting the number of arrows pointing upwards multiplied by half, which will give us a value of $\frac{3}{2}$.

Then now here comes the next challenging problem for me. I have entirely no clue of how to construct the $|\frac{1}{2},m⟩$ state since I don't know how is the net spin computed and thus have no idea how to rearrange the arrows to give me a state that will yield a net spin of $\frac{1}{2}$. According to https://physics.stackexchange.com/a/29552/327428, I need to couple a pair of $\frac{1}{2}$ spin particles then take the product of an up state, which I have do not understand why coupling gives a net spin of 0, and why a product of an upstate will increase its net spin by $\frac{1}{2}$. Citing one of the states of net spin $\frac{3}{2}$, where

$$ |3/2, 1/2⟩ = \frac{|↑↑↓⟩+|↑↓↑⟩+|↓↑↑⟩}{\sqrt{3}}$$

does this imply that for the second term, I associate a $|↑⟩$ to $|↑↓⟩$ term I get $|↑↓↑⟩$, so the spin for this term is $\frac{1}{2}$?

Lastly, I have a question regarding the degeneration of the half spin states. For this case, there exists 2 linearly independent solution to the half spin states. My question is what is the physical interpretation of this.

Sidenote: Mathematically, I understand that these half spin states exist as eigenvectors of $S^2$ with corresponding eigenvalues. Once one of the doublet state is being found, the second one can be derived by using orthogonalisation with other eigenstates, since we need one more eigenvector to span the eigenspace, and we can exploit the fact that eigenvectors corresponding to different eigenvalues will be orthogonal.

It is a bit mouthful since the question I ask is based on something I don't understand very well so I am welcome to any feedback for clarification.

Answer 1

I guess the question I ask can be explained in a simpler manner eh?

The narrow question can be answered by elementary linear algebra, of course, but the point of such questions is that you understand angular momentum composition, which such narrow questions and answers may help you chase away prejudices about... In any case, your teacher might have shown you this:

My first question is how the magnitude of spin is calculated from the state $|↑↑↑⟩$. For instance, I know that from $|↑↑↑⟩$, the spin in the z direction is found by counting the number of arrows pointing upwards multiplied by half, which will give us a value of $\frac{3}{2}$.

You found that state has $S_z=3/2$, but, crucially, moreover, you found that is the maximum spin, since it cannot be raised by the raising ladder operator. Consequently, it must be the highest $S_z$ state of a given s, which must be, ipso facto, s=3/2. You might also apply the $S^2=_+_−+_(_−\mathbb{1})$ on that state, and compare the eigenvalue to the s(s+1) standard expression, but you committed to avoidance of the real McCoy.

Then, here comes the next challenging problem for me. I have entirely no clue of how to construct the $|\frac{1}{2},m⟩$ state, since I don't know how is the net spin computed and thus have no idea how to rearrange the arrows to give me a state that will yield a net spin of $\frac{1}{2}$. [...] $$ |3/2, 1/2⟩ = \frac{|↑↑↓⟩+|↑↓↑⟩+|↓↑↑⟩}{\sqrt{3}}$$

You don't need to add the three spins piecemeal. You have the above 3-vector, completely symmetric in its entries, and there are two normalized 3-vectors orthogonal to it and mutually orthogonal. Find them, assisted by the answers you quote! (Consider rephasings of $~({|↑↑↓⟩+|↑↓↑⟩-2|↓↑↑⟩})/{\sqrt{6}}$ and $~({|↑↑↓⟩-|↑↓↑⟩ })/{\sqrt{2}}$. You may check they are annihilated by a raising operator, and also by a succession of two lowering operators, so you are done: you have two disjoint spin doublets at hand. You may thus eschew application of your dreaded $S^2$ that most anyone else would use...)

Lastly, I have a question regarding the degeneracy of the half-spin states. For this case, there exist two linearly independent solutions to the half-spin states. My question is what is the physical interpretation of this.

Not clear what you expect in the way of "physical interpretation"... surely not a "shortcut to bad math", as Phil Morse emphasized... You have two independent s=1/2 doublets, usable in whatever physics you put into them. One needs more information and answerable questions to provide something meaningful.