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LESSON 7 OF 12

Lesson 07: Quantum states, completion flags and memory

Phase-sensitive states and readable outcomes require different descriptions. A quantum instrument specifies both the probability of a record and the change caused by recording it.

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The central connection

A complete instrument includes all outcomes. Completion can be unbiased with respect to an unknown payload while still depending on source preparation and apparatus resources.

What you will learn

  • Calculate probabilities and conditional states in a two-outcome measurement.
  • Compare coherent evolution before and after registration.
  • Derive retry probabilities with explicit recovery and independence assumptions.

Before you begin: Lessons 01–06. This lesson introduces two-component quantum states and supplies the measurement rule it uses.

Supply the quantum framework explicitly

A normalized two-level pure state is ψ=a|0⟩+b|1⟩, where a and b are complex amplitudes and |a|²+|b|²=1. In the measurement rule used here, a detector of the displayed basis reports zero with probability |a|² and one with probability |b|². We are supplying this quantum rule to study a completion mechanism; the preceding phase construction alone did not derive it.

A density matrix ρ describes pure preparations and probabilistic mixtures in one language. For a pure state it is |ψ⟩⟨ψ|. Its trace equals one. Off-diagonal entries carry coherence relative to the selected basis, which can affect later phase-sensitive tests even when the immediate basis probabilities coincide.

Worked example: write every outcome

Prepare ψ=(√3/2)|0⟩+(1/2)|1⟩. The projectors P0=|0⟩⟨0| and P1=|1⟩⟨1| define unnormalized outcome maps Ik(ρ)=PkρPk. Their traces are 3/4 and 1/4, which sum to one. Conditional on the first result, the normalized state is |0⟩⟨0|; conditional on the second, it is |1⟩⟨1|.

If the apparatus performs the measurement but its result is ignored, the remaining state is diag(3/4,1/4). The input's off-diagonal entries √3/4 have vanished. Keeping only successful zero records would instead report a pure zero state together with success probability 3/4. Omitting that probability would hide the discarded trials and misdescribe the operation.

A normalized two-component input is measured by projectors onto zero and one. The two recorded outcomes have probabilities p and one minus p, and their Kraus effects sum to the identity. Selecting outcome zero succeeds with probability p and produces state zero when p is positive.
Completion instruments and records. Exact normalized projective-instrument example. Calling one outcome a completion success is an additional operational choice, and postselection must retain its success probability.
Open the full-size diagram →

Completion requires the entire instrument

More generally, outcome operators Mk obey ΣMk†Mk=I. This identity ensures that the sum of outcome probabilities is one for every input. The unnormalized map, its trace and the normalized conditional state are three distinct objects. Conditional normalization is meaningful only when the selected outcome has nonzero probability.

In DTT terms, a work register can evolve internally before a completion flag exposes a result. The flag must be produced by the specified dynamics and correlated with an actual completion condition. Calling a projector completion does not explain how a physical apparatus produces or retains the flag. The state of the work register, accepted and rejected outcomes, and later memory readout all belong to the model.

Registration changes what later work can use

Let |+⟩=(|0⟩+|1⟩)/√2 and let H be the Hadamard operation, which sends |+⟩ to |0⟩. Write D for an unread measurement in the zero–one basis. Compare D followed by H with H followed by D, using the same initial state and final basis detector.

In the first protocol, D removes coherence and gives I/2. A unitary H leaves I/2 unchanged, so the final outputs are balanced. In the second protocol, H first produces |0⟩, which D leaves unchanged; the final result is certainly zero. The difference is standard quantum registration physics. A distinctive source proposal must predict how its actual registration operation couples to inner work, rather than treating this familiar contrast as evidence by itself.

A completion that preserves an unknown payload

Choose Macc=√p V and Mfail=√(1−p)W, with fixed unitaries V,W and 0<p≤1. The squared effects sum to I, and acceptance has probability p for every normalized payload. The accepted conditional state is VρV†. On failure, applying W† restores the original state, including correlations with a passive reference.

With fresh independent apparatus and the same p on each restored attempt, first success at attempt n has probability (1−p)n−1p. The mean attempt count is 1/p. These statements require the failure recovery and fresh preparation just specified. An arbitrary rejected branch can leak information irreversibly, in which case restoring the payload and applying this geometric retry law are not justified.

A result must remain available

After successful writing, test whether the record remains readable at declared later delays. A perfect initial flag and a rapidly overwritten memory do not provide the same operational resource as a protected record. Completion accuracy, payload disturbance and retention lifetime need separate measurements, even when a common source supplies all three mechanisms.

The course now connects its two meanings of sufficient information. Lesson 03 asked whether a record predicts future tests. This lesson asks how an instrument prepares that record and changes the state. The completion guide expands the instrument and recovery examples. A common source explanation becomes stronger when one fixed dynamics constrains both its coherent phase and its readable outcomes.

Try the exercises

Work through each question, then open the answer to compare the steps.

Exercise 1

For recoverable independent attempts with p=1/4, calculate the probability of first success on attempt three, the probability of success within three attempts, and the mean attempt count.

Show the worked answer

Two failures followed by success have probability (3/4)²(1/4)=9/64. Failure on all three attempts has probability (3/4)³=27/64, so success within three has probability 1−27/64=37/64. The mean count over all possible attempts is 1/p=4. This mean is not a guarantee of success by attempt four.

Exercise 2

Start with |+⟩ and apply an unread zero–one measurement. What remains, and why can a final Hadamard not recover a certain zero output?

Show the worked answer

The input density matrix is one half times the matrix whose four entries are one. Unread measurement sets its off-diagonal entries to zero, giving I/2. Conjugation by H gives H(I/2)H†=I/2 because HH†=I. The final zero probability is therefore 1/2, whereas applying H directly to the coherent |+⟩ would give zero with probability one. The missing coherence cannot be restored by that unitary alone.

Scope of this guideExact two-level calculations within supplied quantum mechanics. The probability rule, unitary controls, recoverable failures and fresh trials are assumptions, not consequences of completion terminology.

Continue with the articles

  • A quantum source for generation and completion

    A positive phase–dilation Hamiltonian ties one phase encoder to a powered coordinate and controlled work. It gives an explicit quantum realization of generation and completion while retaining its constitutive interfaces as inputs.

  • Completion without biasing an unknown input

    A factorized accepted channel gives the same success probability for every admitted unknown input, including inputs entangled with a reference. The failure channel and the amplification of errors by conditioning require separate checks.

  • Undoing rejected partial work

    A failed completion attempt can be coherently undone by reversing its actual source-controlled work prefix. The recovery restores an unknown payload under explicit inverse-access, memory, and control assumptions.

  • Reliable completion by repeated attempts

    Fresh recoverable attempts produce a geometric success law and an explicit reliability bound for a finite archive. Improved completion probability consumes prepared sources, flags, control, and runtime.

  • A record that carries its own completion certificate

    An autonomous history Hamiltonian can generate both a data record and a flag tied to its completed write. The exact construction retains incomplete outcomes, coherent intermediate stages, and finite observation windows.

  • Keeping an observed label stable

    A transported record can preserve an already observed label under a declared class of later readings. Exact projector identities and channel-error bounds distinguish genuine retention from unchanged one-time statistics.