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Registers, Tensor Products, and Partial Trace — Examples

1 · Prerequisites

2 · Summary

This example checks the first-bit-most-significant convention on a three-bit word, identifies the matrix of the two-qubit Bell projector in the ordered product basis, and computes its partial trace over the second factor.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Register endianness and a two-qubit partial trace

Example

Take two qubit factors with ordered bases $(e_{0}, e_{1})$ , and order the product basis lexicographically as $(e_{0} \otimes e_{0}, e_{0} \otimes e_{1}, e_{1} \otimes e_{0}, e_{1} \otimes e_{1})$ . Let $∣ Φ^{+} ⟩ = (∣ 00 ⟩ + ∣ 11 ⟩) / \sqrt{2}$ and $R = ∣ Φ^{+} ⟩ ⟨ Φ^{+} ∣$ . Its matrix in this basis is $R = \frac{1}{2} (\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{array}) .$ The big-endian index of $101$ is $5$ , and tracing out the second factor gives $diag (1 / 2, 1 / 2)$ , whose trace is $1$ .

Facts & Assumptions

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

[F1]

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

[F2]

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

[F3]

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).

[F4]

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

[F5]

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

[F6]

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

[F7]

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).

[F8]

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

Verification

1.1givenF1F2algebra

By [F1], ${enc}_{3} (1, 0, 1) = 1 \cdot 2^{2} + 0 \cdot 2^{1} + 1 \cdot 2^{0} = 5$ . By [F2], $5$ is one of the eight three-bit register slots.

1.2givenF3algebra

In the ordered product basis, $∣ Φ^{+} ⟩$ has coordinate column $v = (1, 0, 0, 1)^{T} / \sqrt{2}$ . Orthonormality [F3] gives $⟨ Φ^{+}, Φ^{+} ⟩ = 1$ .

2.1givenF4F5step 1.2algebra

By [F4]–[F5], the coordinate matrix of $R$ is $v v^{†}$ . Its only nonzero entries are $1 / 2$ at $(00, 00), (00, 11), (11, 00), (11, 11)$ , so it is the displayed matrix.

3.1givenF6step 2.1algebra

By [F6] and this matrix, $({Tr}_{B} R)_{00} = R_{00, 00} + R_{01, 01} = 1 / 2$ and $({Tr}_{B} R)_{01} = R_{00, 10} + R_{01, 11} = 0$ .

4.1givenF6step 2.1step 3.1algebra

The other entries are $({Tr}_{B} R)_{10} = R_{10, 00} + R_{11, 01} = 0$ and $({Tr}_{B} R)_{11} = R_{10, 10} + R_{11, 11} = 1 / 2$ . Hence ${Tr}_{B} R = diag (1 / 2, 1 / 2)$ .

5.1step 4.1F7F8algebra∎

By [F7], this contraction is the intrinsic partial trace, and [F8] gives the output trace $1 / 2 + 1 / 2 = 1$ .