Paper VII: Quantum circuits as packet paths

I.4 Quantum circuits as packet paths

On this page · Permalink · Source line 273

Let

\[ U=U_mU_{m-1}\cdots U_1 \]

be a circuit with \(m\ge1\), and let \(\{P_\alpha\}\) be a complete packet system. For the empty circuit \(m=0\), set \(U=I\). Its boundary block is \(P_\beta P_\alpha=\delta_{\beta\alpha}P_\alpha\).

Permalink · Source line 275

Theorem I.4.1: Exact circuit path expansion

For source packet \(\alpha_0\) and target packet \(\alpha_m\),

\[ \boxed{ P_{\alpha_m}UP_{\alpha_0} = \sum_{\alpha_1,\ldots,\alpha_{m-1}} P_{\alpha_m}U_mP_{\alpha_{m-1}} \cdots P_{\alpha_1}U_1P_{\alpha_0}. } \]

Summing over the source and target packets reconstructs the complete circuit operator.

Read proof · Read statement and proof side by side · Permalink · Source line 281 · Claim record · Lean coverage · Code · Worked examples

Proof of Theorem I.4.1

Insert the identity resolution \(I=\sum_\alpha P_\alpha\) between every adjacent pair of gates. Distributing the finite sums gives the path expansion. Summing the left and right boundary projectors reconstructs \(U\).

Back to statement · Read statement and proof side by side · Permalink · Source line 296 · Claim record · Lean coverage · Code · Worked examples

Theorem I.4.2: Packetwise unitarity conservation

If \(U\) is unitary and \(U_{\beta\alpha}=P_\beta UP_\alpha\), then

\[ \boxed{ \sum_\beta U_{\beta\alpha}^*U_{\beta\gamma} = \delta_{\alpha\gamma}P_\alpha. } \]

Similarly,

\[ \sum_\alpha U_{\beta\alpha}U_{\delta\alpha}^* = \delta_{\beta\delta}P_\beta. \]

Read proof · Read statement and proof side by side · Permalink · Source line 300 · Claim record · Lean coverage · Code · Worked examples

Proof of Theorem I.4.2

Using completeness and unitarity,

\[ \sum_\beta U_{\beta\alpha}^*U_{\beta\gamma} = P_\alpha U^*\left(\sum_\beta P_\beta\right)UP_\gamma = P_\alpha P_\gamma. \]

For the second identity, sum over the source packets and use unitarity in the reverse order.

Back to statement · Read statement and proof side by side · Permalink · Source line 317 · Claim record · Lean coverage · Code · Worked examples

Example I.4.3: Bell seed circuit

In this example only, \(H\) denotes the Hadamard gate, not a Hamiltonian. Starting from \({\left|00\right\rangle}\),

\[ (H\otimes I){\left|00\right\rangle} = \frac{{\left|00\right\rangle}+{\left|10\right\rangle}}{\sqrt2}, \]

and then

\[ \operatorname{CNOT}(H\otimes I){\left|00\right\rangle} = \frac{{\left|00\right\rangle}+{\left|11\right\rangle}}{\sqrt2}. \]

The parity packets show that the final state lies entirely in the even sector, while the intermediate circuit path passes through both even and odd addresses.

Permalink · Source line 329

Bell-state coordinate probabilities are one half at 00 and 11. — The Bell state generated from the seed \({\left|00\right\rangle}\).

Packet paths through the seed, Hadamard, and controlled-NOT stages. — Circuit reconstruction as a sum over packet paths.

Permalink · Source line 345

Numerical certificate I.4.4: Bell and path reconstruction

The original numerical implementation reported residuals of \(0\) for both the Bell state and its complete reconstruction from parity paths.

Permalink · Source line 361 · Claim record · Lean coverage · Code · Worked examples

Source and conversion record

Source SHA-256: f4fb9d650bb8f6a7a9b6ccdd04882b4c165566f81c962f4765983b0ce77714b9

Unsupported prose commands stop the build. The mathematical expression sequence is checked against the source, with manuscript macros expanded for display.

Conversion records

Read and explore

Proofs and implementation

About this work

Other resources