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The only crucial point is the degeneracy of eigenspaces. Consider the finite dimensional Hilbert space $\cal H$ (the extension to the infinite dimensional case is more difficult also because a part of continuous spectrum may appear) and a pair of commuting Hermitian operators $A$ and $B$ on that space such that the following requirement is satisfied. R.: ...


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Consider self-adjoint operators $A$ and $B$, on a Hilbert space $\mathscr{H}$. Roughly speaking (forgetting about domains of definition), it is possible only if your Hilbert space can be written as $\mathscr{H}=\mathscr{H}_1\otimes \mathscr{H}_2$; where $\mathscr{H}_1$ and $\mathscr{H}_2$ are Hilbert spaces such that: $A=A_1\otimes 1$ and $B=1\otimes B_2$, ...


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A very explicit argument: $$\phi(\xi)=(\xi-c_1)\dots (\xi-c_{r-1})(\xi-c_r)(\xi-c_{r+1})\dots(\xi-c_n)$$ where $r\leq n$ and $c_k> c_l$ whenever $k>l$. We can order the $c_i$ like this because the $c_i$ are real. Now, $$X_r(\xi)=(\xi-c_1)\dots (\xi-c_{r-1})(\xi-c_{r+1})\dots(\xi-c_n) $$ Clearly, the set of zeros of $X_r(\xi)$ is $\{c_i\ \bigl|\ ...


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If all the c's are different, then since $X_r(\xi)$ is the quotient when $\phi(\xi)$ is divided by $(\xi-c_r)$, $(\xi-c_r)$ cannot be a factor of $X_r(\xi)$ else two of the c's would equal $c_r$. $X_r(c_r)$ is, by the remainder theorem (I imagine this has some other name), the remainder when $X_r(\xi)$ is divided by $(\xi-c_r)$. Since $(\xi-c_r)$ is not a ...


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This is because there are just two possible values to the spin in any direction, $-\frac{\hbar}{2}$ and $\frac{\hbar}{2}$, they just differ in a sign, so when you square it you get a single value $\frac{\hbar^2}{4}$. Think about this, the only possible value when you measure the square of $S_z$ is $\frac{\hbar^2}{4}$ for any state, so $$ ...


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OP asks: Is there any physical meaning to this? Yes, the Pauli matrix $\sigma_j$ represents (up to a proportionality factor) the spin in the $j$th direction of a spin $\frac{1}{2}$ system. Such system has only two spin states: $\uparrow$ and $\downarrow$, with opposite eigenvalues. The square $\sigma_j^2$ can no longer see the sign, so it only has one ...



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