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How do you represent $S_x$ and $S_y$ and $S_z$ as a 3D matrix? Can someone explain how $$\left[ J_x,J_y \right] = i\hbar\epsilon_{ijk}J_k,$$ comes out in 3D also? How does it relate to $S_x$ $S_y$ $S_z$? And how can I write $S_x$ and $S_y$ and $S_z$ in Dirac notation in 3D.

Here is the 3D matrix representation for $S_x$ and $S_y$ and $S_z$.

$$S_x=\frac{\hbar}{\sqrt{2}} \begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \\ S_y=\frac{\hbar}{\sqrt{2}} \begin{bmatrix}0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{bmatrix} \\ S_z=\hbar \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} \\ H=\hbar^2 \begin{bmatrix}A & 0 & B \\ 0 & 0 & 0 \\ B & 0 & A \end{bmatrix} $$

I have asked my teacher through the mail, he said I should use the relation $S_\pm = S_x \pm iS_y$, I think it relates to how to build the $H$ representation for a $$ H = A S^{2}_z + B(S_x^2 - S_y^2) $$

confused wether $B(S_x^2 - S_y^2)$ gives $B S_z$?

I don't know how I can apply this. I know how I do represent the $S_x$ $S_y$ $S_z$ in 2D. But I just don't know why.

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  • $\begingroup$ I am not sure what your teacher meant with that suggestion. If you want I can tell you how to get 3x3 matrices that satisfy the algebra you wrote, but I don't know if this is what you need or not. $\endgroup$
    – DelCrosB
    Oct 19, 2016 at 21:51
  • $\begingroup$ Eq. 5.15 on page 27 and eq. 5.22 on page 28. $\endgroup$ Oct 19, 2016 at 21:51
  • $\begingroup$ I see, I think. So in this case if we say it is a spin 1 system this means that L+S = J = 1, then we can apply the operator in 5.15 such to "build" up this matrix, using the Clebsh equation. Right ? $\endgroup$ Oct 19, 2016 at 22:32
  • $\begingroup$ The states $|j,m\rangle$ form the basis of a $2l+1$ dim., irreducible representation of $SU(2)$ and eq. (5.27) can be used to construct the generators $J_i$ of $SU(2)$ in that dim., which will implement the Lie algebra $[J_i,J_j]=i \epsilon_{ijk}J_k$. One only needs eq. (5.27) and the basis vectors (for a given $J$ and $M=-J,-J+1,\dots,+J$). For $J=1$ this is shown in eq. (5.30) and (5.31). To see that the $SU(2)$ has the given Lie algebra with the structure constant $\epsilon_{ijk}$ one can calculate the commutators using the fundamental representation of $SU(2)$: $J_i=\frac 1 2 \sigma_i$. $\endgroup$
    – N0va
    Oct 19, 2016 at 23:01
  • $\begingroup$ You may verify easily that the angular momentum Lie algebra has the generic representations given for all dimensions. $\endgroup$ Oct 20, 2016 at 0:01

1 Answer 1

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It sounds like what you're asking is: how do you construct a representation of SU(2) in terms of 3x3 matrices on a real 3-dimensional vector space? (This representation is also known as the "spin-1" representation, as it's used to describe the spin of spin-1 particles.)

The H you mention, which appears to be some kind of Hamiltonian, is irrelevant to the above question. I assume it is part of a longer homework question which isn't described fully here, so I'll ignore it.

As your teacher mentions, a simple way to construct $S_x$, $S_y$, and $S_z$ is to start with the raising and lowering operators $S_+$ and $S_-$.

If you work in the $S_z$ basis, then you know what the action of $S_z$ is on each of the 3 $S_z$ eigenstates:

$S_z \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix} = \hbar \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$

$S_z \begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix} = 0$

$S_z \begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix} = -\hbar \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$

So the 3x3 matrix form of $S_z$ in this basis must be:

$S_z = \hbar\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix}$

And you also know the actions of the raising and lowering operators on these $S_z$ eigenstates (up to an undetermined constant):

$S_+$ $\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix} = c\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}$

$S_+$ $\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix} = c\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$

$S_+$ $\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix} = 0$

$S_-$ $\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix} = 0$

$S_-$ $\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix} = c\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}$

$S_-$ $\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix} = c\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}$

If you take those actions and write them in matrix form, you get:

$S_+$ = c$\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$

$S_-$ = c$\begin{bmatrix}0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}$

Then, you can write down $S_x$ and $S_y$ just by taking the right linear combinations of $S_+$ and $S_-$:

$S_x = \frac{1}{2}(S_+ + S_-)$

$S_y = \frac{1}{2i}(S_+ - S_-)$

The only final step required is to determine the constant c. This can be determined by finding the eigenvalues of the $S_x$ and $S_y$ matrices. You want them to be $-\hbar$, $0$, and $\hbar$. You can accomplish this by setting $c = \hbar\sqrt{2}$.

As for the commutation relations $[J_i,J_j] = i\hbar\epsilon_{ijk} J_k$, it just means that:

$[S_x,S_y] = i\hbar S_z$

$[S_y,S_z] = i\hbar S_x$

$[S_z,S_x] = i\hbar S_y$

You can verify these directly using matrix multiplication, for example by showing that $S_x S_y - S_y S_x = i\hbar S_z$

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    $\begingroup$ This is great but in general $c$ will not be the same constant for $S_+$ and $S_-$, spin 1/2 and spin 1 are special cases. I think the method should still work, though; you just gotta use two different constants. And if you can remember that it's $\sqrt{j(j+1)-m(m\pm 1)}$, even better. $\endgroup$
    – Javier
    Oct 20, 2016 at 2:13

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