A quantum system is in state |ψ⟩ = c₁|a₁⟩ + c₂|a₂⟩, where |a₁⟩ and |a₂⟩ are eigenstates of observable A with eigenvalues a₁ and a₂. A single measurement of A is performed. What result is obtained?
AThe value a₁ + a₂, since the state is a superposition of both eigenstates
BEither a₁ or a₂, with probabilities |c₁|² and |c₂|² respectively
CThe expectation value c₁a₁ + c₂a₂, which is the quantum average
DEither a₁ or a₂ with equal probability, regardless of the coefficients
Measurement always yields one of the operator's eigenvalues — never a superposition value or weighted average. The probability of obtaining aₙ is |cₙ|² = |⟨aₙ|ψ⟩|². The expectation value ⟨ψ|Â|ψ⟩ = Σ aₙ|cₙ|² is the average over many measurements, not the result of any single one. After measurement, the state collapses to the corresponding eigenstate.
Question 2 Multiple Choice
Why must operators representing physical observables be Hermitian?
ABecause Hermitian operators always commute with each other, ensuring simultaneous measurability
BBecause Hermitian operators have real eigenvalues and their eigenstates form a complete orthonormal basis
CBecause non-Hermitian operators cannot be applied to ket vectors in Hilbert space
DBecause Hermitian operators produce complex eigenvalues, which represent phase information
Two physical requirements force Hermiticity. First, measurement outcomes must be real numbers — a complex energy or position is physically meaningless. Hermitian operators guarantee real eigenvalues. Second, any quantum state must be expressible as a combination of measurement outcomes — the spectral theorem guarantees that Hermitian operators' eigenstates form a complete orthonormal basis for the Hilbert space. Option 0 is wrong: Hermitian operators do not necessarily commute; commuting Hermitian operators can be simultaneously measured, but Hermiticity alone does not require commutativity.
Question 3 True / False
If two observables A and B have commuting operators ([Â, B̂] = 0), it is very difficult to simultaneously know the exact values of both.
TTrue
FFalse
Answer: False
The opposite is true: when operators commute, they share a common eigenbasis, meaning a state can simultaneously be an eigenstate of both. This allows both observables to have definite values at once. It is non-commuting operators ([Â, B̂] ≠ 0) — like position and momentum ([X̂, P̂] = iℏ) — that preclude simultaneous definite values, giving rise to uncertainty relations.
Question 4 True / False
The expectation value ⟨ψ|Â|ψ⟩ gives the probability-weighted average of all possible measurement outcomes for a system in state |ψ⟩.
TTrue
FFalse
Answer: True
If |ψ⟩ = Σ cₙ|aₙ⟩, then ⟨ψ|Â|ψ⟩ = Σ aₙ|cₙ|², which is exactly a probability-weighted average of the eigenvalues. This is the quantum analog of the classical expected value. The expectation value is not what you observe in a single measurement — that gives one eigenvalue — but rather what you observe as the average over many identically prepared measurements.
Question 5 Short Answer
Why does the non-commutativity of position and momentum ([X̂, P̂] = iℏ ≠ 0) imply that a particle cannot have simultaneously definite position and momentum? What would commutativity have implied instead?
Think about your answer, then reveal below.
Model answer: Non-commuting operators have no shared eigenbasis: no state can be simultaneously an eigenstate of both X̂ and P̂. Any state with definite position is a superposition of momentum eigenstates (and vice versa), so measuring momentum on a state with definite position yields a spread of outcomes — the uncertainty principle. If [X̂, P̂] = 0, the operators would share a common eigenbasis, meaning states could simultaneously have definite position and momentum, and classical determinism would be recovered.
The commutator [X̂, P̂] = iℏ is not just a computational fact — it encodes the physical incompatibility of position and momentum. The Heisenberg uncertainty relation ΔxΔp ≥ ℏ/2 follows directly. This is why quantum mechanics is fundamentally different from classical mechanics: the operators for canonically conjugate variables do not commute, making the classical phase-space picture of definite (x, p) points impossible.