Quantum Mechanical Observables

Complete Sets of Commuting Observables

$A | φ_{n} ⟩ = a_{n} | φ_{n} ⟩$

For non-degenerate eigenstates, $a_{n}$ relates to one $| φ_{n} ⟩$ . Then $A$ is a Complete Set of Commuting Observables (CSCO).

For degenerate eigenstates, some $a_{n}$ relates to some set of states, ${| φ_{n}^{i} ⟩} \in E_{n}$

$L_{z}^{2} ≐ (\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix})$ , $| L_{z}^{2} = 1, 1 ⟩, | L_{z}^{2} = 1, 2 ⟩$ . Then ${L_{z}^{2}}$ is not a CSCO.

If $A$ is not a CSCO, look for $B$ such that $[A, B] = 0$ .

Example

Given $A \Rightarrow a_{1} = 0, a_{2, 3} = 1$ . $| a_{1} ⟩ = (100)^{T}, | a_{2} ⟩ = (010)^{T}, | a_{3} ⟩ = (001)^{T}$ .

Let $[A, B] = 0$ with $b_{1} = 0, b_{2} = 1, b_{3} = - 1$ , $| b_{1} ⟩ = (100)^{T}, | b_{2} ⟩ = (010)^{T}, | b_{3} ⟩ = (001)^{T}$ .

Then, $B A | ψ ⟩ = A B | ψ ⟩ = \sum_{i} c_{i} a_{i} b_{i} | b_{i} ⟩$ . We can also then write, $| a_{2} ⟩ = | a = 1, b = 1 ⟩$ and $| a_{3} ⟩ = | a = 1, b = - 1 ⟩$ .

Alternatively, if we had $b_{1} = 1, b_{2} = 1, b_{3} = 0$ , $| b_{1} ⟩ = (100)^{T}, | b_{2} ⟩ = (010)^{T}, | b_{3} ⟩ = (001)^{T}$ . Then, $B A | ψ ⟩ = A B | ψ ⟩ = \sum_{i} c_{i} a_{i} b_{i} | b_{i} ⟩$ . We can also then write, $| a_{1} ⟩ = | a = 0, b = 1 ⟩, | a_{2} ⟩ = | a = 1, b = 1 ⟩$ and $| a_{3} ⟩ = | a = 1, b = 0 ⟩$ .