# Tag Info

0

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.: ...

2

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 $B=1\otimes B_2$, ...

2

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 $\{c_i\ \bigl|\ ... 2 If all the c's are different, then since$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)$is not a ... 2 This is because there are just two possible values to the spin in any direction,$-\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$$... 9 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 ...

Top 50 recent answers are included