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Quantum Teleportation Protocol

Advanced quantum

Alice’s state ∣ψ⟩|\psi\rangle is parameterised on the Bloch angles. Stepped playback advances the register through entanglement, Bell measurement, and the classically controlled Xm2/Zm1X^{m_2}/Z^{m_1} corrections; hovering any gate reveals its action in Dirac notation.

12 components · 11 connections · stepped three-qubit protocol

the aha

You never move the qubit. You move its description — and the original is destroyed in the act.

why it works
The no-cloning theorem forbids copying an unknown ∣ψ⟩=α∣0⟩+β∣1⟩|\psi\rangle = \alpha|0\rangle + \beta|1\rangle, so teleportation does something stranger. Alice entangles her mystery qubit with half of a shared Bell pair and measures both in the Bell basis. Her measurement destroys the original and yields two classical bits m1,m2m_1, m_2 — pure randomness that says nothing about α,β\alpha, \beta. She phones those two bits to Bob, who applies Xm2Zm1X^{m_2}Z^{m_1} to his half, and the exact state re-materialises on his side. Nothing carrying α\alpha and β\beta ever crossed the gap; the entanglement was pre-laid, and two classical bits finished the job. Fidelity F=∣⟨ψ∣ψout⟩∣2=1F = |\langle\psi|\psi_{out}\rangle|^2 = 1 — bit-for-bit, provably the same state.
governing model
∣ψ⟩=α∣0⟩+β∣1⟩  ⋅  α=cos⁡θ2, β=eiϕsin⁡θ2|\psi\rangle = \alpha|0\rangle + \beta|1\rangle \;\cdot\; \alpha = \cos\tfrac{\theta}{2},\ \beta = e^{i\phi}\sin\tfrac{\theta}{2}
structural inventory
  • 3×3\times qubit trace (ψ,A,B)(\psi,A,B)
  • HH + CNOT preparation
  • Bell-basis measurement
  • X/ZX/Z correction gates
boundaries
  • Bloch θ∈[0,π], ϕ∈[0,2π]\theta\in[0,\pi],\ \phi\in[0,2\pi]
  • 66-step protocol
  • fidelity F=∣⟨ψ∣ψout⟩∣2F=|\langle\psi|\psi_{out}\rangle|^2
fault relay classical correction link cut

  1. Dial an arbitrary ∣ψ⟩|\psi\rangle on the Bloch sphere.

    Pick any θ,ϕ\theta, \phi. This is Alice’s unknown state — and neither she nor Bob is allowed to read it.

  2. Step through the protocol and read the fidelity.

    After the Bell measurement and Bob’s Xm2Zm1X^{m_2}Z^{m_1} correction, the output fidelity locks to F=1F = 1. The state arrived, exactly, on two classical bits of help.

  3. Cut the classical correction link.

    Withhold m1,m2m_1, m_2 and Bob’s qubit is left in a random rotation of the original — fidelity collapses. Entanglement alone is not enough; the classical channel is load-bearing.
causal signal trace paused

Prepare and ancillas The unknown state and two ancillas enter on separate rails.

step 1 / 6 Prepare ψ and ancillas

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