Question and first prediction
Can two devices give the same detector counts and leave opposite final states?
A POVM specifies positive effects E₊ and E₋ and therefore outcome probabilities. A full quantum instrument also describes the state transformation associated with each recorded outcome. Different instruments may have exactly the same effects while implementing very different physical backactions. This laboratory compares a minimally disturbing Lüders choice with a device that records the same effect but resets the qubit to a fixed pure state after every outcome.
Recommended background: Density matrices, Kraus operators, selective measurement updates and trace-preserving quantum channels.
Results that must agree with the equations
Computed observables
Fixed experimental statistics
POVM outcome probabilities
The two instruments share the same pair of positive effects, so their raw detector statistics coincide.
E±=(I±ησz)/2 and p±=(1±η cosθ)/2. The entire prediction is independent of the chosen state-update instrument.
State transformations
Different outputs for identical measured effects
Compare the output x- and z-components for your chosen outcome or discarded record.
Lüders Kraus updates preserve the source state at η=0. A measure-and-prepare device always outputs |x+〉, so detector counts alone cannot determine its backaction.
Mathematical explanation
Effects encode probabilities; instruments encode state changes
Use effects E±=(I±ησz)/2 with 0≤η≤1. For the prepared pure qubit cos(θ/2)|0〉+sin(θ/2)|1〉, the observed + probability is p+=(1+ηcosθ)/2. The effect operators are positive and sum to identity.
A Lüders instrument chooses M±=√E± and conditional state ρ±=M±ρM±/p±. For + and p+>0, the conditional Bloch coordinates are rx′=√(1−η²)sinθ/(1+ηcosθ), rz′=(cosθ+η)/(1+ηcosθ). The negative branch follows by η→−η.
Now define an alternative outcome map I±(ρ)=Tr(E±ρ)|x+〉〈x+|. This is a completely positive measure-and-prepare instrument: Kraus operators may be chosen K±,j=√e±,j |x+〉〈j|, with E±=ΣⱼK±,j†K±,j. The outcome probabilities match the Lüders choice at every setting, yet the conditional and nonselective output state is always |x+〉〈x+|.
Test an exactly solvable limit
At η=0 each effect equals I/2: both instruments report fair outcome probabilities. The Lüders instrument applies M±=I/√2 and leaves the system unchanged, while the measure-and-prepare instrument resets it to |x+〉. At η=1 the POVM is a projective z measurement; the two physical devices still differ in the states produced after recording an outcome.
Specify the model's validity
An ideal qubit, effects diagonal in the z basis and preparation restricted to the real x–z great circle. Kraus decompositions describe mathematically valid channels, while microscopic implementations and thermodynamic reset costs are outside the model.
Follow a self-directed learning path
Three quantitative investigations
- Hold effects fixed. Set η=.2 and toggle instrument mode. Confirm that P(+) remains identical while the output Bloch vector changes.
- Inspect no-information backaction. Set η=0 with input |z+〉 and compare both nonselective outputs. Explain why identical detector statistics impose no uniqueness on conditional states.
- Study the sharp limit. Raise η toward one. Determine when the selected + branch becomes impossible, and explain the conditional normalization rule in that case.
Check your understanding: Which mathematical object specifies the conditional state following a POVM outcome?
Extension: verify explicitly that the two alternative Kraus families both satisfy ΣⱼKⱼ†Kⱼ=I once the outcome branches are included.
Connect back to quantum theory
What you can conclude
POVM effects govern information in the measurement outcomes, while an instrument governs the physical state transformation. This distinction is central to weak measurements, quantum error correction, feedback control and thermodynamic consequences of measurement.
Lecture notes: Composite-system measurement, effects and instruments · Projective-measurement foundations.