A particle is prepared in an equal superposition of two energy eigenstates |E₁⟩ and |E₂⟩ with E₁ = 1 eV and E₂ = 3 eV. The expectation value ⟨H⟩ = 2 eV. What does this tell you about the result of a single energy measurement?
AThe measurement will return exactly 2 eV, since that is the expectation value
BThe measurement will return either 1 eV or 3 eV with equal probability, averaging to 2 eV over many measurements
CThe measurement is indeterminate and could return any value between 1 eV and 3 eV
DThe expectation value is only meaningful for eigenstates, so ⟨H⟩ = 2 eV is not physically interpretable here
The expectation value is a long-run average, not a prediction for a single measurement. A single energy measurement can only return an eigenvalue of H — here, either 1 eV or 3 eV. Since the superposition is equal-weight, each occurs with probability 1/2, and the average over many measurements is (1 + 3)/2 = 2 eV. The expectation value 2 eV is never actually observed in a single measurement; it is a statistical property of the state.
Question 2 Multiple Choice
For an eigenstate |φₙ⟩ of observable A with eigenvalue aₙ, what is the variance ⟨(ΔA)²⟩ = ⟨A²⟩ − ⟨A⟩²?
Aaₙ², since squaring the eigenvalue gives the mean-square
Baₙ, since the expectation value equals the eigenvalue
CZero, because every measurement of A returns aₙ with certainty
DIt depends on the specific observable and cannot be determined without knowing the full spectrum
For an eigenstate, every measurement returns exactly aₙ — there is no spread. ⟨A⟩ = aₙ and ⟨A²⟩ = aₙ², so the variance is aₙ² − aₙ² = 0. Zero variance means zero uncertainty: ΔA = 0. This is the quantitative expression of the fact that eigenstates are states of definite value for their observable. It also connects to the Heisenberg uncertainty principle: an eigenstate of position would have zero position uncertainty but maximal momentum uncertainty.
Question 3 True / False
The uncertainty ΔA in observable A reflects the imprecision of the measuring apparatus — a better instrument would reduce ΔA toward zero.
TTrue
FFalse
Answer: False
Quantum mechanical uncertainty is a property of the *state*, not of the measuring apparatus. ΔA = √(⟨A²⟩ − ⟨A⟩²) is computed from the wavefunction and tells you the intrinsic spread in outcomes for that state, even with a perfect measurement device. An eigenstate has ΔA = 0 regardless of the apparatus; a superposition has ΔA > 0 regardless of how good the instrument is. The Heisenberg uncertainty principle ΔxΔp ≥ ℏ/2 is a statement about states, not about experimental limitations.
Question 4 True / False
Ehrenfest's theorem shows that the expectation values of position and momentum obey Newton's second law, which is why macroscopic objects follow classical trajectories even though they obey quantum mechanics.
TTrue
FFalse
Answer: True
Ehrenfest's theorem states d⟨x⟩/dt = ⟨p⟩/m and d⟨p⟩/dt = −⟨∂V/∂x⟩. When a wavepacket is narrow enough that the potential varies slowly across it, ⟨∂V/∂x⟩ ≈ (∂V/∂x)|_{⟨x⟩}, and the quantum equations reduce to the classical F = ma. Macroscopic objects have extremely narrow wavefunctions relative to the scales over which forces vary, so their expectation values track the classical trajectory with negligible spread. This is the quantum-classical correspondence: quantum mechanics reduces to classical mechanics in the appropriate limit.
Question 5 Short Answer
What is the physical meaning of the expectation value ⟨A⟩, and why does it not tell you what result any single measurement will return?
Think about your answer, then reveal below.
Model answer: ⟨A⟩ is the long-run average of measurement outcomes: if you prepare many identical copies of the state |ψ⟩ and measure observable A on each, ⟨A⟩ is the mean of the results. It does not predict a single outcome because quantum measurement is inherently probabilistic — each measurement yields one of A's eigenvalues, with probabilities given by the Born rule. Unless |ψ⟩ is already an eigenstate of A, different measurements of the same state return different eigenvalues. The expectation value summarizes the distribution of those outcomes, just as a probability distribution's mean doesn't tell you the result of a single draw.
This distinction — statistical average versus single-event prediction — is fundamental to understanding quantum mechanics. The expectation value connects the quantum formalism to experimental practice: it's what an experimentalist measures when they repeat the same preparation and average many results.