Parseval for the character transform #
The combining stage of the Pauli appendix moves between two descriptions of the same data: an
F_q-valued measurement {P_a} and its binary observables
O^r = sum_a (-1)^{tr(ar)} P_a, one for each r in F_q. The two are Fourier transforms of each
other, and what the estimates need is that the transform is an isometry: a bound on the
observables, averaged over r, is a bound on the measurement elements, summed over a, and
conversely.
This file is that dictionary, for vectors and for families of matrices applied to a state:
fourierVecis the transform of a family of vectors indexed byV = F_q^n, against the characters of the trace form, andsum_norm_fourierVec_sqis Parseval --- an identity, not an estimate;trObsis the observable of anF_q-indexed family at a parameterr,trObs_fourierinverts it, andsum_stateSqNorm_eq_avg_trObsis Parseval in the form the appendix uses: the sum over outcomes of the squared state-norms of the elements is the average overrof the squared state-norms of the observables.
The one-field character orthogonality it all rests on is MIPRE.Weyl.sum_sgn_trMul.
The transform of a family of vectors #
The character transform of a family of vectors indexed by V = F_q^n.
Equations
- MIPRE.fourierVec T e = (↑(Fintype.card (n → F)))⁻¹ • ∑ a : n → F, MIPRE.sgn (MIPRE.Weyl.trDot a e) • T a
Instances For
The unnormalized transform: the same sum without the 1/|V|.
Equations
- MIPRE.fourierVecRaw T e = ∑ a : n → F, MIPRE.sgn (MIPRE.Weyl.trDot a e) • T a
Instances For
The character sum, on the sesquilinear form. Only the diagonal survives.
Parseval for the character transform. The transform is an isometry up to the normalization: the sum of the squared norms of the transform is the average of the squared norms of the family. No hypothesis on the family at all.
The one-field transform, and its inverse #
The appendix's other direction: an F_q-valued family P has binary observables
O^r = sum_a (-1)^{tr(ar)} P_a, and P is recovered from O by the same transform, averaged.
The binary observable of an F_q-indexed family of operators at the probe r.
Equations
- MIPRE.trObs P r = ∑ a : F, MIPRE.sgn ((Algebra.trace (ZMod 2) F) (a * r)) • P a
Instances For
The transform back: the a-th element recovered from the observables.
Equations
- MIPRE.trFourier A a = (↑(Fintype.card F))⁻¹ • ∑ r : F, MIPRE.sgn ((Algebra.trace (ZMod 2) F) (a * r)) • A r
Instances For
Coarse-grainings and the transform #
A measurement is usually presented as a POVM whose outcomes are then read by a function; the transform has to see through that reading. Both lemmas below are the same regrouping of a sum along the fibres of the reading.
The one-field transform sees through a coarse-graining.
The two-field transform of a product #
The combined measurement is indexed by a pair (a, b), and the family it transforms is a
product of one operator depending on r and one depending on s. The transform factorizes,
which is why the combined measurement is the product of the two marginals whenever the two
commute.
The pair (r, s) as a vector of F_q^2.
Instances For
A sum over pairs is a sum over F_q^2.
Parseval over one copy of the field, and over two #
The combining stage coarse-grains an F_q x F_q-valued measurement by the linear form
(a, b) |-> alpha a + beta b, and the fibres have q elements -- so a per-fibre triangle
inequality would cost a factor q. Parseval is what avoids that, and it is needed in three
shapes: over F_q (to turn the sum over fibres into an average over one probe), over
F_q x F_q (to turn that average back into a sum over outcome pairs), and the cancellation of
the zero probe.
The unnormalized character transform of a family of vectors indexed by F_q itself.
Equations
- MIPRE.trVecRaw T r = ∑ c : F, MIPRE.sgn ((Algebra.trace (ZMod 2) F) (c * r)) • T c
Instances For
Parseval over two copies of the field, in the product-indexed shape the combining stage uses.
Coarse-graining by a linear form #
Parseval for a coarse-graining by a linear form. A family of vectors indexed by
F_q x F_q that sums to zero has, on average over (alpha, beta) uniform, fibre sums under
(a, b) |-> alpha a + beta b whose total squared norm is 1 - 1/q times the family's own. In
particular it is at most the family's: the fibres have q elements, and a per-fibre triangle
inequality would cost exactly that factor, which this avoids.
The zero-sum hypothesis is what kills the r = 0 character; without it the statement is false
(take U supported on one point).