A student proposes using the operator (p̂ + ix̂) to represent a measurable physical quantity, where p̂ and x̂ are both Hermitian. Why is this operator unsuitable as an observable?
AThe combination of position and momentum operators violates the uncertainty principle by construction
B(p̂ + ix̂) is not Hermitian — its conjugate transpose is (p̂ − ix̂) ≠ (p̂ + ix̂) — so its eigenvalues may be complex and cannot represent real measurement outcomes
CHermitian operators cannot be added or combined in quantum mechanics
DThe operator has no physical interpretation, so it fails on interpretive rather than mathematical grounds
The Hermitian condition  = † is required for real eigenvalues. (p̂ + ix̂)† = p̂† + (ix̂)† = p̂ − ix̂ ≠ p̂ + ix̂. Since (p̂ + ix̂) ≠ (p̂ + ix̂)†, it is not Hermitian. Its eigenvalues can be complex numbers, which cannot represent physical measurement results. Note that p̂ alone and x̂ alone are individually Hermitian and valid observables — but forming the combination with a factor of i breaks the Hermitian property.
Question 2 Multiple Choice
Why must the eigenstates of a Hermitian observable form an orthonormal basis for the Hilbert space?
ABecause quantum measurements must be repeatable, and repeatability requires eigenstates to be stable solutions
BBecause orthogonality ensures the Born-rule probabilities |⟨aₙ|ψ⟩|² sum to 1 for any normalized state |ψ⟩, preserving the probabilistic interpretation
CBecause the Schrödinger equation requires that energy eigenstates be mutually perpendicular
DBecause non-orthogonal eigenstates would violate the Heisenberg uncertainty principle
The probabilistic interpretation of quantum mechanics requires that measurement probabilities sum to 1. If the eigenstates of an observable form an orthonormal basis {|aₙ⟩}, then any state |ψ⟩ = Σ cₙ|aₙ⟩ with Σ|cₙ|² = 1, and the probabilities |⟨aₙ|ψ⟩|² = |cₙ|² sum to 1 automatically. If the eigenstates were not orthogonal, overlapping expansion coefficients would double-count probability, the sum could exceed 1, and the Born rule would be inconsistent. Hermitian operators guarantee orthogonality of eigenstates with distinct eigenvalues, which is exactly why they — and only they — can represent observables.
Question 3 True / False
For a Hermitian operator Â, the calculation aₙ = ⟨aₙ|Â|aₙ⟩ = ⟨†aₙ|aₙ⟩ = aₙ* forces each eigenvalue aₙ to be real — this follows directly from the condition  = †.
TTrue
FFalse
Answer: True
This is the standard one-line proof. Starting with Â|aₙ⟩ = aₙ|aₙ⟩ and taking the inner product with ⟨aₙ|: aₙ = ⟨aₙ|Â|aₙ⟩. Using  = †: ⟨aₙ|Â|aₙ⟩ = ⟨†aₙ|aₙ⟩ = ⟨aₙ|aₙ⟩* · aₙ* = aₙ*. So aₙ = aₙ*, which means aₙ is real. This proof works for any Hermitian operator on any Hilbert space — it is a consequence purely of  = †, not of any particular physical system or operator.
Question 4 True / False
The raising operator ↠is a valid quantum mechanical observable for the harmonic oscillator, because it has well-defined, predictable action on nearly every energy eigenstate.
TTrue
FFalse
Answer: False
↠is not Hermitian: (â†)† = â ≠ â†. Therefore it is not an observable — you cannot measure it directly. Although ↠has well-defined action on energy eigenstates (â†|n⟩ = √(n+1)|n+1⟩), its eigenvalues are complex numbers (it belongs to the family of operators with coherent state eigenstates, but these form an overcomplete non-orthogonal set). Observable operators are exactly the Hermitian ones; non-Hermitian operators like â, â†, and their combinations appear in calculations and in defining Hamiltonians, but do not themselves represent measurable quantities.
Question 5 Short Answer
Why can't any linear operator on a Hilbert space represent a physical observable? What does the Hermitian property guarantee, and why are both guarantees necessary for the Born rule to work?
Think about your answer, then reveal below.
Model answer: Two guarantees are needed: (1) real eigenvalues, so that measurement outcomes are real numbers, and (2) orthogonal eigenstates, so that the Born-rule probabilities sum to 1. The Hermitian condition  = † delivers both. Real eigenvalues ensure the measured value is physically meaningful. Orthogonality ensures the probability of each outcome — |⟨aₙ|ψ⟩|² — sums to 1 across all possible outcomes when the eigenstates form a complete orthonormal basis. Without either condition, the probability interpretation breaks down: imaginary eigenvalues are uninterpretable as measurement results, and non-orthogonal eigenstates cause probabilities to exceed 1.
This is why the Hermitian condition is not arbitrary mathematical convention — it is the precise condition that makes the Born rule self-consistent. A non-Hermitian operator might have complex eigenvalues (failing condition 1) or non-orthogonal eigenstates (failing condition 2) or both. Either failure makes the statistical interpretation incoherent. The mathematical elegance of Hermitian operators is that a single algebraic condition ( = †) simultaneously guarantees both physical requirements.