A quantum state is prepared as |ψ⟩ = (1/√2)|a₁⟩ + (1/√2)|a₂⟩, a superposition of two eigenstates of observable  with eigenvalues a₁ and a₂. What does a single measurement of  yield?
AThe average value (a₁ + a₂)/2, since the state is an equal superposition
BEither a₁ or a₂, each with probability 1/2
CAn undefined result, because the state is not an eigenstate of Â
DBoth a₁ and a₂ simultaneously, since both components are present
A measurement of an observable always yields one of the operator's eigenvalues — never an intermediate average. The probabilities are |c₁|² and |c₂|², which here are both 1/2. Option A is the most common misconception: the expectation value (a₁ + a₂)/2 is the *average over many measurements*, not any single outcome. The operator defines possible outcomes and probabilities; it does not determine which eigenvalue is realized on a given measurement.
Question 2 Multiple Choice
Why must physical observables in quantum mechanics correspond to Hermitian operators rather than arbitrary linear operators?
AHermitian operators are computationally simpler to apply to state vectors
BHermitian operators guarantee that all eigenvalues are real numbers, and measurement outcomes must be real
CNon-Hermitian operators cannot be expressed in Dirac notation
DHermitian operators always commute with each other, ensuring consistent measurements
Physical measurement outcomes must be real numbers — you cannot measure an imaginary position or momentum. Hermitian operators († = Â) are guaranteed to have real eigenvalues, making them the only valid candidates for physical observables. Option D is false: Hermitian operators generally do *not* commute (in fact, non-commutativity encodes the uncertainty principle). The Hermiticity requirement follows directly from demanding real measurement outcomes, not computational convenience.
Question 3 True / False
Applying the momentum operator p̂ to any wavefunction returns a real number representing the particle's momentum.
TTrue
FFalse
Answer: False
Applying p̂ = −iℏ(d/dx) to an arbitrary wavefunction returns *another wavefunction*, not a number. Only when the wavefunction is an eigenstate of p̂ — specifically, a plane wave e^(ikx) — does the result equal a number (the eigenvalue ℏk) times the original state. For a general superposition, the operator maps one state vector to another, encoding the probability distribution over momentum outcomes rather than a single definite value.
Question 4 True / False
A particle in an eigenstate of the position operator x̂ with eigenvalue x₀ will yield x₀ with certainty upon position measurement.
TTrue
FFalse
Answer: True
This is the defining property of eigenstates: Â|ψ⟩ = a|ψ⟩ means measuring observable  on state |ψ⟩ yields eigenvalue a with probability 1. A position eigenstate (a Dirac delta function in position space) is maximally localized — measuring position returns x₀ every time. The trade-off, embodied in the uncertainty principle, is that such a state is completely delocalized in momentum space.
Question 5 Short Answer
Explain why the momentum operator takes the form p̂ = −iℏ(d/dx) rather than simply being multiplication by a position-like variable. What physical reasoning connects differentiation to momentum?
Think about your answer, then reveal below.
Model answer: The form follows from the de Broglie relation p = ℏk and the structure of plane waves. A state of definite momentum p is a plane wave e^(ikx) with k = p/ℏ. Differentiating: d/dx e^(ikx) = ik · e^(ikx), so −iℏ(d/dx) e^(ikx) = ℏk · e^(ikx) = p · e^(ikx). The plane wave is an eigenstate of −iℏ(d/dx) with eigenvalue p. The differential operator is necessary because momentum in quantum mechanics is linked to spatial frequency (the rate of phase oscillation), not position. Multiplication by x encodes position information; differentiation with respect to x encodes how rapidly the wavefunction oscillates, which encodes momentum.
The momentum operator being differential is not an arbitrary convention — it is forced by the requirement that plane waves (states of definite momentum) be eigenstates of the momentum operator, which follows from de Broglie. This reflects the deep connection between momentum and spatial translation symmetry: in quantum mechanics, the momentum operator is the generator of spatial translations, and generators of continuous symmetries always appear as differential operators.