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Operators, eigenvalues and quantum measurement

Science · Quantum & Modern Physics · ages 20-22

Name ______________________ Date ____________

For a nonzero state psi, what makes it an eigenstate of an operator A?
- A psi = a psi for one constant a
- The input state is zero everywhere
- A psi = x psi with a varying factor x
In the position representation, what does the position operator do to a wavefunction?
- Takes a time derivative
- Squares the wavefunction
- Multiplies by x
A normalized equal superposition of orthonormal energy eigenstates at 2 eV and 10 eV always gives its 6 eV expectation value in one ideal energy measurement.
Circle one: True False
A normalized state superposes two orthonormal energy eigenstates with distinct energies E1 and E2. The amplitude for E2 has magnitude 1/2. What is the probability of measuring E2?
- 1/2
- 1
- 1/4
For a nonzero state psi, applying H returns 3 eV times psi. What do you conclude?
- Not an eigenstate; 3 eV is only an average
- Energy eigenstate with eigenvalue 3 eV
- Energy eigenstate with eigenvalue 1.5 eV
What is the one-dimensional canonical momentum operator in the position representation?
- -iℏ ∂/∂x
- x times the wavefunction
- iℏ ∂/∂t
An ideal measurement of a discrete energy spectrum gives E1 from an unknown preparation. Why does this alone not establish that the state was an E1 eigenstate before measurement?
- Every energy measurement returns the mean
- An energy eigenstate cannot return E1
- A superposition can also return E1
Why can a normalizable state not have both position and momentum spreads equal to zero?
- A detector always destroys all momentum data
- Their spreads obey Δx Δp >= ℏ/2
- Position and momentum are the same observable

Answer key

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Operators, eigenvalues and quantum measurement W1-mt_E9C_9DlwD2-s1

A psi = a psi for one constant a · An eigenstate satisfies A psi = a psi with one constant eigenvalue a. The state must be nonzero, although a may be zero.
Multiplies by x · The position operator sends psi(x) to x psi(x). It does not differentiate or square the wavefunction.
False · The possible outcomes are 2 eV and 10 eV, each with probability 1/2. Their mean is 6 eV, but 6 eV is not an outcome for this state.
1/4 · For a discrete outcome, square the magnitude of its amplitude: (1/2)² = 1/4.
Energy eigenstate with eigenvalue 3 eV · H psi = 3 eV psi passes the eigenstate test. The constant multiplying the same nonzero state is its energy eigenvalue, 3 eV.
-iℏ ∂/∂x · Canonical momentum differentiates with respect to x and multiplies the result by -iℏ.
A superposition can also return E1 · An E1 eigenstate gives E1 with certainty, but a superposition with a nonzero E1 component can also give E1. This result does not identify the earlier state uniquely.
Their spreads obey Δx Δp >= ℏ/2 · Both spreads being zero would give a zero product, contradicting Δx Δp >= ℏ/2. This is a state constraint, not a detector limitation.