Compatible observables and simultaneous eigenkets

# Compatible observables

$\hat A$ and $\hat B$ are compatible when they commute:

$$ [\hat A, \hat B] = 0 $$

and incompatible when $[\hat A, \hat B] \neq 0$.

Assume that the ket space is spanned by the eigenkets of $\hat A$.

$$ \implies |\alpha\rangle = \sum_{a'}\langle a'|\alpha\rangle|a'\rangle $$

We can also say that the same ket space is spanned by the eigenkets of $\hat B$.

How is $|a'\rangle$ related to $|b'\rangle$ when $[\hat A, \hat B] = 0$?

# Degeneracy

If two or more linearly independent eigenkets of $\hat A$ have the same eigenvalue, those eigenvalues are degenerate.

$\implies |a'\rangle \to$ not sufficient (to label a state). $\langle a''|a'\rangle = 0$ iff they are non-degenerate.

If the dim. of the ket space > no. of distinct eigenvalues of $\hat A$, we’re cooked! How can $\sum_{a'}c_{a'}|a'\rangle$ be true?

Ignore them (for now).

# Theorem

Theorem: Suppose $[\hat A, \hat B] = 0$ and the eigenvalues of $\hat A$ are non-degenerate. Then the matrix elements $\langle a''|\hat B|a'\rangle$ are all diagonal. And the matrix elements of $\hat A$ are already diagonal if the basis kets are $\lbrace|a'\rangle\rbrace$.

$$ \begin{aligned} \langle a''|[\hat A, \hat B]|a'\rangle &= \langle a''|\hat A\hat B - \hat B\hat A|a'\rangle \\ &= a''\langle a''|\hat B|a'\rangle - \langle a''|\hat B|a'\rangle a' \\ &= (a'' - a')\langle a''|\hat B|a'\rangle \end{aligned} $$

(assuming $a''$ is real, which it is since $\hat A$ is Hermitian). The left side is $0$ since the commutator is $0$:

$$ (a'' - a')\langle a''|\hat B|a'\rangle = 0 $$

$\therefore a' = a''$ or $\langle a''|\hat B|a'\rangle = 0$

• $a' = a''$ for two different kets would mean degeneracy $\implies$ not true.
• so $\langle a''|\hat B|a'\rangle = 0$ must be true.

Note that the way we write the matrix is $\langle a''|\hat B|a'\rangle$, with $a''$ the row and $a'$ the column.

$\therefore$ for every row $\neq$ column, element $= 0$, and the diagonal elements are $\delta_{a'a''}\langle a'|\hat B|a'\rangle$.

$\therefore$ We can represent the matrix $\hat B$ with the same set of base kets as $\hat A$: $\lbrace|a'\rangle\rbrace$

$$ \implies \hat B = \sum_{a''}|a''\rangle\langle a''|\hat B|a''\rangle\langle a''| $$

If $\hat B$ acts on $|a'\rangle$:

$$ \hat B|a'\rangle = |a'\rangle\langle a'|\hat B|a'\rangle $$

(survives only for that eigenket). But this is the eigenvalue equation of $\hat B$, with eigenvalue $\langle a'|\hat B|a'\rangle$

$\implies |a'\rangle$ is an eigenket of both $\hat B$ and $\hat A$.

$\therefore$ we can use $|a',b'\rangle$ to denote this

Compatible operators have simultaneous eigenkets.

$$ \begin{aligned} \hat A|a',b'\rangle &= a'|a',b'\rangle \\ \hat B|a',b'\rangle &= b'|a',b'\rangle \end{aligned} $$

(different eigenvalues). Or, we can use a collective index $|K'\rangle = |a',b'\rangle$.

# Maximal set of commuting observables

Suppose we have a maximal set of commuting observables

$$ [\hat A, \hat B] = [\hat B, \hat C] = \dots = 0 $$

The eigenvalues of the individual $\hat A, \hat B$ etc. may have degeneracies, but if we specify a combination $(a', b', c', \dots)$ then the simultaneous eigenket of $\hat A, \hat B, \hat C, \dots$ is unique.

$$ |K'\rangle = |a', b', c', \dots\rangle $$

They’re orthonormal:

$$ \langle K''|K'\rangle = \delta_{K'K''} = \delta_{a'a''}\delta_{b'b''}\dots $$

and closure:

$$ \sum_{K'}|K'\rangle\langle K'| = 1 = \sum_{a'}\sum_{b'}\dots|a',b',\dots\rangle\langle a',b',\dots| $$

# Measuring compatible observables

What about measurements of $\hat A$ and $\hat B$ when $[\hat A, \hat B] = 0$?

$$ |\alpha\rangle \xrightarrow{\hat A} |a',b'\rangle \xrightarrow{\hat B} |a',b'\rangle \xrightarrow{\hat A} |a',b'\rangle $$

when the eigenvalues are non-degenerate. Measuring one doesn’t mess up the other.

When there is degeneracy, the system (an arbitrary state ket) after the first $\hat A$ is

$$ \sum_i c_{a'}^{(i)}|a', b^{(i)}\rangle $$

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