I'm trying to see if there is a way to geometrically derive a general form for the quantum mechanics spin operators. I'm trying to deduce their commutation relations without using any knowledge of angular momentum, and solely using geometry (No knowledge of physics required for this problem, also all physical constants are taken out of the equations).
In QM we define the three self-adjoint $2\times 2$ matrices $S_x, S_y, S_z$. We also know that each operator has two eigenvalues: $1$ and $-1$. As well, the operators are to satisfy the commutation relations below: ($a,b,c \in \{x,y,z\}$)
$$ [S_a, S_b] = \epsilon_{abc} 2i S_c $$
Any triplet of operators satisfying the conditions above, can constitute a valid choice for spin operators in QM. A common choice for these operators are the Pauli matrices, which by default satisfy the above commutation relation and also have $\pm 1$ eigenvalues.
I find the commutation relation defined above to be abstruse, so I'm trying to motivate it using a geometrical argument.
I start by characterizing each operator by only its (normalized) eigenvectors, which are clearly orthogonal as each $S_i$ is self-adjoint. From physical observations, we know that the component of each eigenvector of $S_i$ along the eigenvectors of $S_j$ where $i\neq j$, has a magnitude of $1/\sqrt{2}$. In other words, eigenvectors of $S_i$ have equal components (with a possible phase difference) along the eigenvectors of $S_j$.
So hypothetically, to find a choice for the three spin operators, all we have to do (without knowing the commutation relation) is to find 3 pairs of normalized vectors $(\vec{a}_{i1}, \vec{a}_{i2})$ in $\mathbb{C}^2$, where the $i$th pair constitute the eigenvectors of $S_i$. The three pairs must satisfy the following property: $$ | (\vec{a}_{im}, \vec{a}_{jn}) | = \dfrac{1}{\sqrt{2}}\ \ \ \forall i \neq j, \ \ m,n \in \{1, 2\} $$
If we can find any 3 pairs of vectors satisfying the above, whilst aditionally each pair being orthogonal, they should be a valid choice for the eigenvectors of our spin operators. After that, considering we know the eigenvalues are $\pm 1$, we can extract the matrices themselves by the relation $S_n = \sum_i \lambda_i v_i v_i^{\dagger}$ where $\lambda_i$ and $v_i$ are the $i$th eigenvalue and eigenvector respectively.
The questions I'm struggling with are the following:
- What would be the general form of the solution to the above problem? What choices of $\vec{a}_{i,j}$ are possible?
- Can we prove that for any valid choice of the three said pairs, the commutation relation stands for the operators we can extract from the three pairs?
For the choices below, the Pauli matrices are retrieved.
$$ \vec{a}_{z1} = \begin{bmatrix}1 \\ 0\end{bmatrix} $$ $$ \vec{a}_{z2} = \begin{bmatrix}0 \\ 1\end{bmatrix} $$ $$ \vec{a}_{x1} = \dfrac{1}{\sqrt{2}}\begin{bmatrix}1 \\ 1\end{bmatrix} $$ $$ \vec{a}_{x2} = \dfrac{1}{\sqrt{2}}\begin{bmatrix}1 \\ -1\end{bmatrix} $$ $$ \vec{a}_{y1} = \dfrac{1}{\sqrt{2}}\begin{bmatrix}1 \\ i\end{bmatrix} $$ $$ \vec{a}_{y2} = \dfrac{1}{\sqrt{2}}\begin{bmatrix}1 \\ -i\end{bmatrix} $$