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Every finite instrument induces a unique POVM
Statement
Let be a finite instrument from a nonzero finite-dimensional complex inner-product space to a finite-dimensional complex inner-product space . There is a unique finite POVM on such that Consequently, for every density operator on , the instrument probability equals the POVM weight .
Facts & Assumptions
Given: The finite instrument and its finite nonempty outcome set .
Finite instruments and conditional states: every is completely positive and for every .
Kraus and Choi characterizations of complete positivity: every completely positive finite-dimensional map between full operator spaces has a finite Kraus representation.
Finite Kraus representation: a finite Kraus family represents when for every input operator .
Instrument branch Kraus families satisfy total normalization: for finite branch Kraus families of an instrument, .
The adjoint is characterised by and Adjoints satisfy , , , and : , , and .
Non-negative and positive operators: non-negativity means self-adjointness and a non-negative quadratic form.
Finite positive effect family and Finite POVMs and their outcome law: a finite POVM is a nonempty finite family of non-negative operators summing to identity, and its outcome weights are .
Trace is a linear functional on and For and , : trace is linear and for compatible rectangular maps , .
The trace pairing on a finite endomorphism space is nondegenerate: if for every , then .
Proof
For each , [F1] makes completely positive, so [F2] supplies a finite family representing it as in [F3]. If the supplied family is empty, append one zero operator; this preserves the formula and makes each index list nonempty. Since is finite, the lists can be selected one at a time by finite induction; no arbitrary-index choice is used.
Define . Concatenating the finite lists in step 1.1 gives a Kraus representation of . These are exactly the finite branch Kraus families of the instrument in [F1], so [F4] gives .
For each outcome define . By [F5], each summand is self-adjoint and . Thus each is non-negative by [F6]. Step 2.1 gives , so is a finite POVM by [F7].
For every , the Kraus formula [F3], trace linearity, and rectangular cyclicity [F8] give . This proves the displayed outcome-law identity for every density operator as well.
If another family has the same identity for every , then for all . By [F8], this equals ; [F9] implies . This holds for each , so the induced POVM is unique.
Depends on
- Finite instruments and conditional states
- Kraus and Choi characterizations of complete positivity
- Finite Kraus representation
- Instrument branch Kraus families satisfy total normalization
- Finite positive effect family
- Finite POVMs and their outcome law
- Non-negative and positive operators
- The adjoint $T^*:W\to V$ is characterised by $\langle Tv,w\rangle_W=\langle v,T^*w\rangle_V$
- Adjoints satisfy $(S+T)^*=S^*+T^*$, $(\lambda T)^*=\overline\lambda T^*$, $(ST)^*=T^*S^*$, and $T^{**}=T$
- The basis-independent trace of an endomorphism of a finite-dimensional vector space
- Trace is a linear functional on $M_n(F)$
- For $A\in M_{m\times n}(F)$ and $B\in M_{n\times m}(F)$, $\operatorname{tr}(AB)=\operatorname{tr}(BA)$
- The trace pairing on a finite endomorphism space is nondegenerate
Used by
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Sources
- John Watrous, The Theory of Quantum Information (2018 author-hosted draft), §§2.2.2 and 2.3.2 (standard reference, not scraped)
- John Preskill, Quantum Computation, Chapter 3: Foundations II: Measurement and Evolution (standard reference, not scraped)