1.10 Relative phase is absent from one histogram and present in another
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For a normalized state \(|v\rangle=\sum_nv(n)|n\rangle\), the standard-basis probabilities are \(|v(n)|^2\). In the basis analyzed by \(F^\dagger\), the probabilities are \(|(F^\dagger v)(k)|^2\).
These describe two measurement procedures on separately prepared copies. Measuring the first histogram on a particular copy does not preserve that copy as a coherent input for the second procedure.
Theorem QA13: Equal initial probabilities and orthogonal states
The states \(\psi\) and \(\chi\) have identical probabilities in the original basis and are orthogonal:
Proof
Squaring a real number or its absolute value gives the same result, proving the first claim. For orthogonality,
The positive entries contribute \(16+4+4=24\), and the negative entries contribute \(-4-16-4=-24\). Their sum is zero. \(\square\)
This is a strong information boundary. The states are distinct even after the physically irrelevant overall phase is disregarded. Orthogonal normalized states cannot be scalar multiples of one another. Yet their original-basis probability histograms are identical.
Theorem QA14: Complete arithmetic phase separation
After applying \(F^\dagger\),
More explicitly,
whereas
Proof
QA12 gives the first amplitude table. For the second, set \(a(n)=|c(n)|\). Its vector is
It has period six. Pairing \(n\) and \(n+6\) in its twelve-point transform produces a factor \(1+(-1)^k\), so every odd transform coefficient vanishes. For \(k=2\ell\), its transform is twice the six-point transform of \((4,0,2,0,2,0)\). The doubled transform is sixteen when \(\ell\) is divisible by three and four otherwise, as one sees from
When \(\ell\equiv0\pmod3\), the bracket is eight. Otherwise the two cubic phases sum to minus one and the bracket is two. Thus
State normalization and the unitary inverse transform divide this table by \(\sqrt{48}\sqrt{12}=24\), producing the second amplitude table. Its support is even, while every label in \(U_{12}\) is odd. The selected probability is therefore zero. The signed state has four probabilities of \(1/4\), with total one. \(\square\)
The full probability comparison is:
| Label | Initial probability, both states | After \(F^\dagger\), signed state | After \(F^\dagger\), unsigned state |
|---|---|---|---|
| 0 | \(1/3\) | \(0\) | \(4/9\) |
| 1 | \(0\) | \(1/4\) | \(0\) |
| 2 | \(1/12\) | \(0\) | \(1/36\) |
| 3 | \(0\) | \(0\) | \(0\) |
| 4 | \(1/12\) | \(0\) | \(1/36\) |
| 5 | \(0\) | \(1/4\) | \(0\) |
| 6 | \(1/3\) | \(0\) | \(4/9\) |
| 7 | \(0\) | \(1/4\) | \(0\) |
| 8 | \(1/12\) | \(0\) | \(1/36\) |
| 9 | \(0\) | \(0\) | \(0\) |
| 10 | \(1/12\) | \(0\) | \(1/36\) |
| 11 | \(0\) | \(1/4\) | \(0\) |
Every column sums to one. The Fourier probability supports are disjoint. Negative amplitudes in the original arithmetic signal therefore encode observable relative phase, even though their signs do not appear in the first probability column.
Theorem QA15: No inverse from the initial histogram alone
There is no function from original-basis probability histograms to state vectors that reconstructs both \(\psi\) and \(\chi\). There is also no reconstruction of both states up to overall phase from that histogram alone.
Proof
The two histograms are equal by QA13. A function assigns the same output to equal inputs, so it cannot output the two distinct vectors. For the phase-insensitive formulation, QA13 shows that the states are orthogonal and normalized. If \(\chi=\lambda\psi\) for any complex scalar of modulus one, then \(\langle\psi,\chi\rangle=\lambda\ne0\), a contradiction. \(\square\)
The first formulation is formalized in the later publication module. The strengthening to equivalence up to global phase is a prose corollary proved here. It is not counted as an additional inherited Lean declaration.