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In quantum mechanics, we know that the spin 1/2 matrices are:

$$S_x = \frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad S_y = \frac{\hbar}{2} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad S_z = \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$

While I am pretty sure I understand how we got these, it is still fuzzy for me. Thus, as an application of this (and as part of homework), I am trying to understand how to get the matrices for higher spin levels.

Thus, with the spin 1/2 matrices, how do we obtain the spin 1 or greater matrices?

Qmechanic
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TheAkashain
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    Hint: there will be 2s+1 values of m or 3 eigenstates (1,0,0) (0,1,0) and (0,0,1) since s=1. Find the action of $S_z S_+ S_-$ on these states and form the matrices as you would have done in the spin 1/2 case. We cannot answer homework questions. – joseph h Oct 31 '20 at 10:22
  • Take the symmetric part of the tensor product of your matrices with themselves. – Prof. Legolasov Oct 31 '20 at 10:45
  • @Prof.Legolasov No, these are generators. You need their coproduct, instead. – Cosmas Zachos Oct 31 '20 at 15:16
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    @NiharKarve sure you can! Take the standard coproduct, and reduce the 4x4 answer. – Cosmas Zachos Oct 31 '20 at 15:19
  • @CosmasZachos Ah of course - by 'reduce' do you mean eliminating the trivial rep part? – Nihar Karve Oct 31 '20 at 15:21
  • @NiharKarve Of course; with a simple change of basis, you isolate and eliminate the singlet. See here. – Cosmas Zachos Oct 31 '20 at 15:23
  • @CosmasZachos I meant the matrix tensor product, I’m not sure what you mean by coproduct but I think we are talking about the same thing – Prof. Legolasov Oct 31 '20 at 15:42
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    @Prof Legolasov. No, we are not talking about the same thing. The coproduct of generators is not a tensor product of them. You probably have their exponentials in mind... – Cosmas Zachos Oct 31 '20 at 15:45

3 Answers3

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You don't really derive higher-spin matrices from lower-spin ones. Rather, they are derived from the algebra of spin operators and from how they act oon the chosen spin basis.

The basis chosen is usually the basis of eigenvectors of operator $S_z$, and are often denoted $|j,m\rangle$, $2j\in\mathbb{N}$, $m\in\{-j,-j+1\dots,j\}$. Operator $S_z$ act on them as follows:

$$ S_z|j,m\rangle = \hbar m |j,m\rangle $$ We also have operators $S_+=S_x+iS_y$ and $S_-=S_x-iS_y$. Using the algebra of spin operators it can be proven that $$ S_+|j,m\rangle = \hbar \sqrt{(j-m)(j+m+1)}|j,m+1\rangle $$ $$ S_-|j,m\rangle = \hbar \sqrt{(j+m)(j-m+1)}|j,m-1\rangle $$ (try to do it yourself, or ask if you need help). These relations are enough to write down the matrices of $S_z$, $S_+$ and $S_-$ in the basis of vectors $|j,m\rangle$, and from $S_+$ and $S_-$ you can obtain $S_x$ and $S_y$.

nemo
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Adam Latosiński
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    That's one way to derive the arbitrary-spin irreps, yes, but it is also possible to derive them from tensoring the fundamental irrep with itself. For example, $V_{1/2} \otimes V_{1/2} = V_0 \oplus V_1$, so tensoring the matrices OP has in their question with themselves and then factoring out one copy of the trivial representation will leave you with spin-1 matrices that are $3 \times 3$. It is also possible to do this for arbitrary spin: spin-$j$ is the symmetric part of the tensor product of $2 j$ copies of spin-$1/2$. – Prof. Legolasov Oct 31 '20 at 10:43
  • Thank you so much! This gave me the information I needed to research the rest. For those wondering in the future how to do this: $|j,m> = |j,m+1>$ is equivalent to $ \delta_{m,m+1} $ on some matrix indexed by the first and second m, meaning only indices like (1,0) and (0,-1) work. I marked your answer as the correct one for giving me what I needed to figure it out! – TheAkashain Oct 31 '20 at 15:21
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You don't build the spin $1$ matrices from the spin $\frac 12$ matrices. You rather repeat the whole procedure, which you learned with $2\times 2$ matrices for spin $\frac 12$. But now you do it with $3\times 3$ matrices for spin $1$. As you already know from spin $\frac 12$ the 3 matrices are not unique. So there is much freedom in choosing a possible set of 3 matrices.

Begin with $S_z$. You know it has 3 eigenvalues: $+\hbar$, $0$, $-\hbar$. So you can choose $S_z$ with these eigenvalues along the diagonal and $0$ everywhere else. $$S_z=\hbar\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} \tag{1}$$ To find possible $S_x$ and $S_y$ matrices, you need to make sure they are Hermitian and satisfy the commutation relations. $$\begin{align} [S_x,S_y]&=i\hbar S_z \\ [S_y,S_z]&=i\hbar S_x \\ [S_z,S_x]&=i\hbar S_y \end{align} \tag{2}$$ (By the way: these relations are required for every set of angular momentum operators, not only for spin $\frac 12$ or spin $1$.)

You may find $S_x$ with $S_y$ by using high sophisticated math as given in the other answers.

But actually it is not too difficult to find them just by trial and error. Getting some inspirations from the $S_x$ and $S_y$ from spin $\frac 12$, you find that $$S_x=\frac{\hbar}{\sqrt{2}} \begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix} \tag{3}$$ $$S_y=\frac{\hbar}{\sqrt{2}} \begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix} \tag{4}$$ together with $S_z$ from (1) satisfy all the commutation relations (2).

hft
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Thomas Fritsch
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Hints :

REFERENCE : Total spin of two spin- 1/2 particles. $=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$

We could derive the spin-1 matrices adding two spin-1/2. But this addition is not so easy as you may expect since you must study first about product spaces and product transformations in detail (see SECOND ANSWER and THIRD ANSWER in above REFERENCE respectively).

For the present in the main answer of above REFERENCE you could watch from equation (01) to equation (10) how to derive the $S^{\left(j\boldsymbol{=}1\right)}_z$ matrix \begin{equation} S^{\left(j\boldsymbol{=}1\right)}_z\boldsymbol{=}\sqrt{\tfrac{1}{2}} \begin{bmatrix} \:1\: & \:0\: & \:0\: \\ \:0\: & \:0\: & \:0\: \\ \:0\: & \:0\: & \!\!\!-1 \end{bmatrix} \tag{01}\label{01} \end{equation} combining (adding) two $S^{\left(j\boldsymbol{=}1/2\right)}_z$ matrices \begin{equation} S^{\left(j\boldsymbol{=}1/2\right)}_z\boldsymbol{=}\tfrac{1}{2} \begin{bmatrix} \:1\: & \:0\: \\ \:0\: & \!\!\!-1 \end{bmatrix} \tag{02}\label{02} \end{equation}

As an other example you could watch how to derive the $S^{\left(j\boldsymbol{=}3/2\right)}_k$ matrices combining (adding) the $S^{\left(j\boldsymbol{=}1/2\right)}_k$ matrices with the $S^{\left(j\boldsymbol{=}1\right)}_k$ ones, see equations (Ex-25),(Ex-25.1),(Ex-25.2) and (Ex-25.3) in FIFTH ANSWER of above REFERENCE.

Frobenius
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