I have a question about this equation: $$\frac{1}{2}\otimes\frac{1}{2}=1\oplus0.$$
I'm a bit confused by the right-hand side. Should '1' and '0' be interpreted as the total spin? If so, if there're two particles both have spin $-\frac{1}{2}$, why there's no -1 on the right-hand side? Could '1' and '0' be treated like two orthogonal subspaces of the total Hilbert space? Also, why there's a 3 by 3 block corresponds to 1 and 1 by 1 block corresponds to 0? Thanks!!
This equation is an equation for representations of the Lie algebra of rotations. It is a fact proven in every quantum mechanics book that the irreducible representations of the angular momentum algebra $[J_i,J_j]=\epsilon_{ijk}J_k$ are spin $j$ systems where $j\in \mathbb{N}/2=\{0,1/2,1,3/2,2,\dots\}$. We thus call this representation by the integer $j$. Now, we can take tensor products and direct sum of representations. This is what the formula $1/2\otimes 1/2=0\oplus 1$ refers to. Physically, it says that a composite system consisting of two spin $1/2$ particles has states of total spin $0$ and states of total spin $1$. In particular, in this system the states corresponding to total spin $0$ and those of total spin $1$ are orthogonal. This is part of the definition of direct sum of unitary representations.
Now, note that a spin $j=1/2$ particle can have two states, one in which the spin is up and one in which the spin is down. The representation $j=1/2$ is spanned by these 2 states $m=1/2,-1/2$.
Finally, when people say that there is a 3 by 3 block corresponding to $1$ they are referring to the fact that the representation $1$ is three dimensional. It has states with $m=-1,0,1$. On the other hand, the representation $0$ is 1 dimensional. Its only state has $m=0$.
So the related question is not so much help to you because you have a much more fundamental misunderstanding. Once we clear that up, any further questions should be answered by the above.
There is no such $-1$ or $-1/2$ system.
For this we have to be clear on the difference between two different angular momentum quantum numbers, $\ell$ and $m$. They come from two different operators, $$\begin{align} \langle L^2\rangle &= \hbar^2 \ell (\ell + 1),\\ \langle L_z \rangle &= \hbar m, \end{align} $$for some arbitrary axis $z$ which we choose to co-diagonalize with $L^2$. Now there is an obvious relationship where if $|m| = \ell + 1$ then you would have $\langle L^2\rangle < \langle L_z^2\rangle$ which would be unphysical, so there is a clear restriction that $|m|\le 1$ must be maintained. Within that you have integer steps of $m$, so an $\ell = 1$ system allows $m = -1, 0, 1$ inside of that total-angular-momentum.
To the spin story we add that half-integers are allowed and if $\ell$ is a half-integer then $m$ is also a half-integer, so for an $\ell = \frac12$ system then $m = -\frac12, +\frac12.$
But the key thing is that you are describing the world in terms of $m$ but the equation that you are looking at describes the world in terms of $\ell.$ There are no spin-negative-one-half systems, there are just spin-one-half systems which happen to be most-pointed-along the negative $z$-axis for whatever $z$ we happened to use to diagonalize.
It is worth also re-emphasizing these words “most-pointed-along.” When we say that the spin is $-1/2$ we are saying that the total amount of the spin is a definite $\sqrt{3/4}~\hbar$ but that only $\hbar/2$ of it is pointed in the $-z$-direction. This is why if you follow up with an $x$-measurement, say, you see $+x$ with 50% probability and $-x$ with 50% probability, the vector is actually in some sense distributed along a “ring” $(L_x, L_y, L_z) = (\sqrt{1/2} \cos\theta, \sqrt{1/2} \sin\theta, -1/2)$ for $0 \le \theta < 2\pi.$
