# Quantum Mechanical Observables

## Complete Sets of Commuting Observables

$$A|\varphi_n\rangle = a_n|\varphi_n\rangle$$

For non-degenerate eigenstates, $$a_n$$ relates to one $$|\varphi_n\rangle$$. Then $$A$$ is a Complete Set of Commuting Observables (CSCO).

For degenerate eigenstates, some $$a_n$$ relates to some set of states, $$\{|\varphi_n^i\rangle\} \in\mathcal{E}_n$$

$$L_z^2 \doteq \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}$$, $$|L_z^2=1,1\rangle,|L_z^2=1,2\rangle$$. Then $$\{L_z^2\}$$ is not a CSCO.

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

### Example

Given $$A\Rightarrow a_1=0,a_{2,3}=1$$. $$|a_1\rangle = (1 0 0)^T,|a_2\rangle = (0 1 0)^T,|a_3\rangle=(0 0 1)^T$$.

Let $$[A,B]=0$$ with $$b_1=0,b_2=1,b_3=-1$$, $$|b_1\rangle = (1 0 0)^T,|b_2\rangle=(0 1 0)^T,|b_3\rangle=(0 0 1)^T$$.

Then, $$BA|\psi\rangle = AB|\psi\rangle = \sum_i c_ia_ib_i|b_i\rangle$$. We can also then write, $$|a_2\rangle = |a=1,b=1\rangle$$ and $$|a_3\rangle=|a=1,b=-1\rangle$$.

Alternatively, if we had $$b_1=1,b_2=1,b_3=0$$, $$|b_1\rangle = (1 0 0)^T,|b_2\rangle=(0 1 0)^T,|b_3\rangle=(0 0 1)^T$$. Then, $$BA|\psi\rangle = AB|\psi\rangle = \sum_i c_ia_ib_i|b_i\rangle$$. We can also then write, $$|a_1\rangle = |a=0,b=1\rangle,|a_2\rangle = |a=1,b=1\rangle$$ and $$|a_3\rangle=|a=1,b=0\rangle$$.

Created: 2023-06-25 Sun 02:30

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