Register endianness and a two-qubit partial trace

Given: The big-endian encoding, the ordered product basis, the Bell vector $\mathrm{\mid }{\mathrm{\Phi }}^{+}⟩$, its outer-product operator $R$, and the coordinate definition of partial trace.

${\mathrm{enc}}_n(x)={\sum }_{k<n}{x}_k{2}^{n-1-k}$, with the first bit most significant (Big-endian encoding of ordered bit strings).

The encoding is a bijection to the slots $\{0,\dots ,2^n-1\}$ (Big-endian bit-string indices are bijective).

The one-qubit computational basis is orthonormal and the two-factor product basis uses the stated factor order (The computational basis is orthonormal, The product basis is orthonormal and tensor dimensions multiply).

A ket is its vector and the outer product acts by $\mathrm{\mid }x⟩⟨y\mathrm{\mid }z=⟨z,y{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}x$ (Dirac Notation Conversion).

In orthonormal coordinates, the matrix of $\mathrm{\mid }x⟩⟨y\mathrm{\mid }$ is $[x][y{]}^{†}$ (Dirac Notation Coordinate Identity).

In the ordered product basis, $({\mathrm{Tr}}_{B}R{)}_{a{a}^{\mathrm{\prime }}}={\sum }_{b<2}{R}_{(a,b),(a^{\mathrm{\prime }},b)}$ (Partial trace over a finite tensor factor).

This coordinate contraction is the unique trace-characterized partial trace and is independent of the orthonormal basis of the second factor (The partial trace is characterized by the trace pairing and is basis-independent).

The trace is the sum of diagonal matrix entries (The basis-independent trace of an endomorphism of a finite-dimensional vector space, The trace $\mathrm{tr}(A)$ as the sum of the diagonal entries).