Introduction: When Two Observables Coexist
Can a particle have definite position AND definite momentum? The answer depends on whether the operators commute. Commuting observables can be simultaneously measured; non-commuting ones cannot.
The Compatibility Theorem
Theorem: Two observables \(\hat{A}\) and \(\hat{B}\) have a complete set of simultaneous eigenstates if and only if \([\hat{A}, \hat{B}] = 0\).
If they commute:
- There exists a basis \(\{|a, b\rangle\}\) where \(\hat{A}|a,b\rangle = a|a,b\rangle\) and \(\hat{B}|a,b\rangle = b|a,b\rangle\)
- Measuring \(A\) then \(B\) gives the same statistics as measuring \(B\) then \(A\)
- Both quantities can be known precisely simultaneously
Proof Sketch
If \(\hat{A}|a\rangle = a|a\rangle\) and \([\hat{A}, \hat{B}] = 0\):
\[\hat{A}(\hat{B}|a\rangle) = \hat{B}\hat{A}|a\rangle = \hat{B}(a|a\rangle) = a(\hat{B}|a\rangle)\]So \(\hat{B}|a\rangle\) is also an eigenvector of \(\hat{A}\) with eigenvalue \(a\). If \(a\) is non-degenerate, \(\hat{B}|a\rangle \propto |a\rangle\), making \(|a\rangle\) an eigenvector of \(\hat{B}\) too.
Worked Examples
Example 1: Compatible Observables
\(\hat{L}^2\) and \(\hat{L}_z\) commute: \([\hat{L}^2, \hat{L}_z] = 0\)
Common eigenstates: \(|l, m\rangle\) with:
- \(\hat{L}^2|l,m\rangle = \hbar^2 l(l+1)|l,m\rangle\)
- \(\hat{L}_z|l,m\rangle = \hbar m|l,m\rangle\)
Example 2: Incompatible Observables
\(\hat{L}_x\) and \(\hat{L}_z\) don't commute: \([\hat{L}_x, \hat{L}_z] = -i\hbar\hat{L}_y \neq 0\)
No simultaneous eigenstates exist (except trivial \(l = 0\)).
Example 3: Complete Set of Commuting Observables (CSCO)
For the hydrogen atom, a CSCO is \(\{\hat{H}, \hat{L}^2, \hat{L}_z, \hat{S}_z\}\). These all commute, giving unique quantum numbers \((n, l, m_l, m_s)\) for each state.
The Quantum Connection
Finding a Complete Set of Commuting Observables is how we label quantum states uniquely. For a hydrogen atom, \(n, l, m, s\) come from operators that all commute with each other. Position and momentum don't commute, so there's no state with both definite position and momentum—this is the uncertainty principle restated algebraically.