An essential exercise involving spin composition 
Suppose we have two particles with spin $1/2$. They have $S^{tot}=1$ and $S^{tot}_y=0$.
How can we write the state of the system in terms of the eigenstates of $S_{1z},S_{2z}$?

My attempt:
I would like to divide the problem in two: firstly we write the state of the system in terms of the states of $S_{1y},S_{2y}$ and then we re-write the state in terms of $S_{1z},S_{2z}$:
We know that it's always true that:
$$S^{tot}_y=S_{1y}+S_{2y}$$
this is a fundamental rule of the spin's algebra, right? So:
$$0=S_{1y}+S_{2y}$$
but of course this last two can only be $+1/2$ or $-1/2$, so the possible states are:
$$|+1/2,-1/2\rangle  \ \ \ or \ \ \ |-1/2,+1/2\rangle$$
I then state that the more general solution is the linear combination of this two possibilities, and so (let me call $+1/2$ simply $+$ and $-1/2$ simply $-$):
$$|S^{tot}=1,S^{tot}_y=0\rangle =a|+,-\rangle+b|-,+\rangle$$
For me this should be the solution. But: in my lecture notes it's stated that the true solution is:
$$|S^{tot}=1,S^{tot}_y=0\rangle =\frac{1}{\sqrt{2}}\left[|+,-\rangle+|-,+\rangle\right] \tag{1}$$
I don't get why this must be true, remember that the particles are not identical. This is my first problem.
But suppose that we do not have this problem, let's assume (1) to be correct. Then how can I translate (1) from $y$ to $z$? What is the link? How does the link work? This is my second problem.
I would like to understand what I am getting right and what I am getting wrong.
 A: Let us first fix clearer notation. We call the spin operators on the $i$-th particle $S_{ix}$ (and so on) and their eigenvalues $s_{ix}$ (and so on). $S_i^2 = S_{ix}^2 + S_{iy}^2 + S_{iz}^2$ is the Casimir operator of the rotation algebra $\mathfrak{su}(2)$ has a single eigenvalue $s_i(s_i + 1)$ on irreducible representations of the rotation algebra. For the "total" operators and eigenvalues on the combined two-particle space we just drop the $i$ subscript. In this notation your question asks you to find the two-particle state with $s = 1$ and $s_y = 0$ in terms of eigenstates of $S_{z}$.

*

*You have correctly determined that a general eigenstate of $S_y$ has the form $a\lvert +,-\rangle + b\lvert -,+\rangle$ in terms of eigenstates of the $S_{iy}$. But you also need it to be an eigenstate of $S^2$ with the eigenvalue $1(1+1) = 2$.


*Since the eigenstates of each of the $S_{ix},S_{iy},S_{iz}$ form a basis for the one-particle space, the eigenstates of one of these operators can be expressed as a linear combination of the eigenstates of another of these operators. The notation $\lvert +,-\rangle$ is just shorthand for $\lvert +\rangle \otimes \lvert -\rangle$, so by linearity of the tensor product the expressions from the one-particle space straightforwardly extend to the two-particle space - just plug them in and multiply out.
A: (Brief introduction about spin for a $1/2$ particle)
If for a particle with spin $s=1/2$ the set of commuting operators $\left\{S_y,\boldsymbol{S}^2\right\}$ is chosen as basis, then one may label the states according to the $\pm 1/2$ eigenstates of $S_y$ (I will use the following notation instead of your simplified one because it will be useful when addressing your second point of the question)
$$S_y|S_y;\pm\rangle=\pm\frac{1}{2}|S_y;\pm\rangle\tag{1}\label{sy}$$
with total angular momentum
$$\boldsymbol{S}^2|S_y;\pm\rangle=\frac{3}{4}|S_y;\pm\rangle\tag{2}\label{s2}$$
One may introduce raising and lowering operators as $S_\pm=S_x\pm \mathrm{i}S_z$ whose actions yield
$$S_+|S_y;+\rangle=0,\quad S_+|S_y;-\rangle=|S_y;+\rangle\tag{3}\label{s+}$$
along with
$$S_-|S_y;-\rangle=0,\quad S_-|S_y;+\rangle=|S_y;-\rangle\tag{4}\label{s-}$$
(Moving to a system of two particles of spin $1/2$)
The total spin along $Oy$, whose eigenvalue you seek to be null, that it $s_y^\mathrm{tot}=0$ (notice I denoted the eigenvalue with small letter in order to not cause any confusion) would be given by
$$S_y^\mathrm{tot}=S_y^1\otimes\mathsf{I}_2+\mathsf{I}_1\otimes S_y^2\tag{5}\label{sytot}$$
This may seem just spin with extra steps but I always find it useful when I know on which space the operators act (the identity was denoted by $\mathsf{I}$).
There aren't many possibilities. Direct computations show that states of the type $|S_y;+\rangle|S_y;+\rangle$ would yield $s_y^{tot}=1$, states $|S_y;-\rangle|S_y;-\rangle$ give $s_y^{tot}=-1$. Nevertheless, the states $|S_y;+\rangle|S_y;-\rangle$ and $|S_y;-\rangle|S_y;+\rangle$ result in, via direct computation using \eqref{sytot}, the eigenvalue $s_y^{tot}=0$. Therefore, a state with $s_y^{tot}=0$ must be a linear combination of the type
$$|s_y^{tot}=0\rangle=a|S_y;+\rangle|S_y;-\rangle+b|S_y;-\rangle|S_y;+\rangle\tag{6}\label{state}$$
where the coefficients must of course satisfy $a^2+b^2=1$ as a normalization condition. This is a result which you also obtained but I just wanted to go through it with my notation.
On the other hand, the total spin of a composite system may be expanded as
$$\boldsymbol{S}^2=(\boldsymbol{S}^1)^2\otimes\mathsf{I}_2+\mathsf{I}_1\otimes(\boldsymbol{S}^2)^2+2\boldsymbol{S}^1\otimes\boldsymbol{S}^2\tag{7}\label{s2tot}$$
It is useful to express the term $\boldsymbol{S}^1\boldsymbol{S}^2$ (I'm dropping the direct product term since too much rigurosity is slowing my typesetting) as
$$\boldsymbol{S}^1\boldsymbol{S}^2=\underbrace{S_x^1S_x^2+S_z^1S_z^2}_{\displaystyle \dfrac{1}{2}(S_+^1S_-^2+S_-^1S_+^2)}+S_y^1S_y^2\tag{8}\label{s1s2}$$
Now one may proceed to act with the total spin from \eqref{s2tot}, along with the term from \eqref{s1s2}, on the proposed state \eqref{state}. The computation is not difficult, it just uses repeatedly the action of the operators from \eqref{sy}, \eqref{s2}, \eqref{s+} and \eqref{s-}. There might be a shorter more elegant way to derive it than with raising and lowering operators. Nevertheless, the final result reads
$$\boldsymbol{S}^2|s_y^{tot}=0\rangle=(a+b)|S_y;+\rangle|S_y;-\rangle+(a+b)|S_y;-\rangle|S_y;+\rangle $$
One is interested in states having $s^\mathrm{tot}=1$ which imply
$$\boldsymbol{S}^2|s_y^{tot}=0\rangle=\underbrace{s^\mathrm{tot}(s^\mathrm{tot}+1)}_{\displaystyle 2}|s_y^{tot}=0\rangle\stackrel{\eqref{state}}=2a|S_y;+\rangle|S_y;-\rangle+2b|S_y;-\rangle|S_y;+\rangle$$
Therefore $a=b$ and along with $a^2+b^2=1$ and assuming $a,b\in\mathbb{R}$, the state characterized by $s_y^{tot}=0$ and $s^{tot}=1$ is
$$|s_y^{tot}=0, s^{tot}=1\rangle=\frac{1}{\sqrt{2}}(|S_y;+\rangle|S_y;-\rangle+|S_y;-\rangle|S_y;+\rangle)\tag{9}\label{finalstate}$$
Now that we have the state, the following question was how to express it in the basis (for each particle) labeled by the eigenvalues of $\left\{S_z,\boldsymbol{S}^2\right\}$ (at least this is how I understood it, correct me if I am mistaken).
This would somehow assume that you now do a measurement of spin along another axis, which should erase the information measured before. It is then fair to assume that it is equally probable to obtain projections of spin along $Oz$ of $\pm 1/2$, irrespective of what you obtained before along $Oy$, thus
$$|\langle S_y;+|S_z,+\rangle|^2=|\langle S_y;-|S_z,+\rangle|^2=\frac{1}{2}$$
Consequently, one may expand
$$|S_z,+\rangle=\underbrace{\langle S_y;+|S_z;+\rangle}_{\displaystyle\dfrac{1}{\sqrt{2}}}|S_y;+\rangle+\underbrace{\langle S_y;-|S_z;+\rangle}_{\displaystyle\dfrac{1}{\sqrt{2}}}|S_y;-\rangle$$
and similarly (but with a change in sign which will be explained immediately)
$$|S_z,-\rangle=\underbrace{\langle S_y;+|S_z;-\rangle}_{\displaystyle\dfrac{1}{\sqrt{2}}}|S_y;+\rangle+\underbrace{\langle S_y;-|S_z;-\rangle}_{\displaystyle-\dfrac{1}{\sqrt{2}}}|S_y;-\rangle$$
where the change in sign is chosen in order to assure
$$|\langle S_z;+|S_z,-\rangle|=0$$
Therfore, you have your $\left\{|S_y;\pm\rangle\right\}$ basis expressed in terms of $\left\{|S_z;\pm\rangle\right\}$ as
$$|S_y;+\rangle=\frac{1}{2}(|S_z;+\rangle)+|S_z;-\rangle),\quad|S_y;-\rangle=\frac{1}{2}(|S_z;+\rangle)-|S_z;-\rangle)$$
which allow you to immediately express your state from \eqref{finalstate} in this basis now labeled by the eigenstates of $S_z^1,S_z^2$.
