QuantumCore

Lesson 25 · 75 min · core

Quantum Teleportation

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Alice holds an unknown qubit ∣ψ⟩=α∣0⟩+β∣1⟩\lvert\psi\rangle=\alpha\lvert0\rangle+\beta\lvert1\rangle and shares the Bell pair ∣Φ+⟩\lvert\Phi^+\rangle with Bob. With one CNOT\mathrm{CNOT}, one HH, two measurements and two classical bits she transfers ∣ψ⟩\lvert\psi\rangle to Bob's qubit, while her own copy is destroyed. This lesson derives the four measurement branches term by term, checks the corrections Zm0Xm1Z^{m_0}X^{m_1} in each branch, proves the principle of deferred measurement with projector algebra, derives the measurement-free final state ∣+⟩∣+⟩∣ψ⟩\lvert{+}\rangle\lvert{+}\rangle\lvert\psi\rangle, computes Bob's reduced state 12I\tfrac12I before the classical bits arrive (no signalling), and counts the resources: one ebit and two classical bits per teleported qubit.

Prerequisites: Entanglement & the Bell States

Objectives and key results

  • Expand ∣ψ⟩0∣Φ+⟩12\lvert\psi\rangle_0\lvert\Phi^+\rangle_{12} and derive, term by term, the state after CNOT0→1\mathrm{CNOT}_{0\to1} and H0H_0 as a sum over Alice's four outcomes ∣m0m1⟩\lvert m_0m_1\rangle.

  • Show that the correction Zm0Xm1Z^{m_0}X^{m_1} maps Bob's branch state Xm1Zm0∣ψ⟩X^{m_1}Z^{m_0}\lvert\psi\rangle back to ∣ψ⟩\lvert\psi\rangle for each of the four outcomes.

  • Prove the principle of deferred measurement: ∑m∣m⟩⟨m∣⊗Um\sum_m\lvert m\rangle\langle m\rvert\otimes U^m commutes with the projectors ∣k⟩⟨k∣⊗I\lvert k\rangle\langle k\rvert\otimes I, so classically controlled gates after a measurement equal quantum-controlled gates before it.

  • Derive the final state ∣+⟩∣+⟩∣ψ⟩\lvert{+}\rangle\lvert{+}\rangle\lvert\psi\rangle of the measurement-free protocol with CNOT1→2\mathrm{CNOT}_{1\to2} and CZ0→2\mathrm{CZ}_{0\to2}.

  • Compute Bob's reduced state ρ2=14(σ+XσX+ZσZ+YσY)=12I\rho_2=\tfrac14(\sigma+X\sigma X+Z\sigma Z+Y\sigma Y)=\tfrac12I before the bits arrive and explain why teleportation cannot transmit information faster than light.

  • State the resource identity 1 ebit+2 cbits≥1 qubit1\ \text{ebit}+2\ \text{cbits}\ge1\ \text{qubit} and explain why the two classical bits are uniformly random and carry no information about α,β\alpha,\beta.

formula · Teleportation identity
(H⊗I⊗I)(CNOT0→1⊗I) ∣ψ⟩0∣Φ+⟩12=12∑m0,m1∈{0,1}∣m0m1⟩⊗Xm1Zm0∣ψ⟩(H\otimes I\otimes I)(\mathrm{CNOT}_{0\to1}\otimes I)\,\lvert\psi\rangle_0\lvert\Phi^+\rangle_{12}=\tfrac12\sum_{m_0,m_1\in\{0,1\}}\lvert m_0m_1\rangle\otimes X^{m_1}Z^{m_0}\lvert\psi\rangle
identity · Corrections
Zm0Xm1⋅Xm1Zm0∣ψ⟩=∣ψ⟩for all m0,m1∈{0,1}Z^{m_0}X^{m_1}\cdot X^{m_1}Z^{m_0}\lvert\psi\rangle=\lvert\psi\rangle\quad\text{for all }m_0,m_1\in\{0,1\}
theorem · Principle of deferred measurement
[∑m∣m⟩⟨m∣⊗Um, ∣k⟩⟨k∣⊗I]=0⟹measure-then-Um ≡ controlled-U-then-measure\Big[\textstyle\sum_{m}\lvert m\rangle\langle m\rvert\otimes U^{m},\ \lvert k\rangle\langle k\rvert\otimes I\Big]=0\quad\Longrightarrow\quad\text{measure-then-}U^{m}\ \equiv\ \text{controlled-}U\text{-then-measure}
formula · Deferred protocol output
CZ0→2 CNOT1→2 H0 CNOT0→1 ∣ψ⟩0∣Φ+⟩12=∣+⟩0∣+⟩1∣ψ⟩2\mathrm{CZ}_{0\to2}\,\mathrm{CNOT}_{1\to2}\,H_0\,\mathrm{CNOT}_{0\to1}\,\lvert\psi\rangle_0\lvert\Phi^+\rangle_{12}=\lvert{+}\rangle_0\lvert{+}\rangle_1\lvert\psi\rangle_2
identity · No signalling
ρ2=14(σ+XσX+ZσZ+YσY)=12Ifor every σ=∣ψ⟩⟨ψ∣\rho_2=\tfrac14\big(\sigma+X\sigma X+Z\sigma Z+Y\sigma Y\big)=\tfrac12I\quad\text{for every }\sigma=\lvert\psi\rangle\langle\psi\rvert
theorem · Resource count
1 ebit+2 cbits ≥ 1 qubit1\ \text{ebit}+2\ \text{cbits}\ \ge\ 1\ \text{qubit}

The teleportation identity#

Three qubits take part. Qubit 00 belongs to Alice and carries the unknown state ∣ψ⟩=α∣0⟩+β∣1⟩\lvert\psi\rangle=\alpha\lvert0\rangle+\beta\lvert1\rangle with ∣α∣2+∣β∣2=1\lvert\alpha\rvert^2+\lvert\beta\rvert^2=1. Qubits 11 (Alice) and 22 (Bob) hold the pre-shared Bell pair ∣Φ+⟩12=12(∣00⟩+∣11⟩)\lvert\Phi^+\rangle_{12}=\tfrac1{\sqrt2}(\lvert00\rangle+\lvert11\rangle), prepared earlier by CNOT1→2H1∣00⟩\mathrm{CNOT}_{1\to2}H_1\lvert00\rangle. Qubit 00 is the most significant bit, so ∣abc⟩\lvert abc\rangle means qubit 00 in ∣a⟩\lvert a\rangle, qubit 11 in ∣b⟩\lvert b\rangle, qubit 22 in ∣c⟩\lvert c\rangle. Alice knows neither α\alpha nor β\beta; measuring her qubit directly would only return one bit and destroy the state.

Definition

Teleportation protocol

(i) Alice applies CNOT0→1\mathrm{CNOT}_{0\to1} and then H0H_0. (ii) She measures qubits 00 and 11 in the computational basis, obtaining bits m0,m1m_0,m_1. (iii) She sends (m0,m1)(m_0,m_1) to Bob over a classical channel. (iv) Bob applies Xm1X^{m_1} and then Zm0Z^{m_0} to qubit 22, i.e. the operator Zm0Xm1Z^{m_0}X^{m_1}. Steps (i)–(ii) together are a measurement in the Bell basis of qubits 0,10,1, because (H⊗I)CNOT(H\otimes I)\mathrm{CNOT} is the inverse of the Bell-state circuit CNOT(H⊗I)\mathrm{CNOT}(H\otimes I).

Derivation

Expanding the input and applying CNOT₀₁ and H₀

  1. Step 1:
    ∣ψ⟩0∣Φ+⟩12=(α∣0⟩+β∣1⟩)⊗12(∣00⟩+∣11⟩)\lvert\psi\rangle_0\lvert\Phi^+\rangle_{12}=(\alpha\lvert0\rangle+\beta\lvert1\rangle)\otimes\tfrac1{\sqrt2}(\lvert00\rangle+\lvert11\rangle)
    Input state: Alice's qubit 00 next to the shared pair on qubits 1,21,2.
  2. Step 2:
    =12(α∣000⟩+α∣011⟩+β∣100⟩+β∣111⟩)=\tfrac1{\sqrt2}\big(\alpha\lvert000\rangle+\alpha\lvert011\rangle+\beta\lvert100\rangle+\beta\lvert111\rangle\big)
    Distribute the tensor product over both sums; ∣a⟩⊗∣bc⟩=∣abc⟩\lvert a\rangle\otimes\lvert bc\rangle=\lvert abc\rangle.
  3. Step 3:
    →CNOT0→112(α∣000⟩+α∣011⟩+β∣110⟩+β∣101⟩)\xrightarrow{\mathrm{CNOT}_{0\to1}}\tfrac1{\sqrt2}\big(\alpha\lvert000\rangle+\alpha\lvert011\rangle+\beta\lvert110\rangle+\beta\lvert101\rangle\big)
    CNOT0→1∣a,b,c⟩=∣a,b⊕a,c⟩\mathrm{CNOT}_{0\to1}\lvert a,b,c\rangle=\lvert a,b\oplus a,c\rangle: the two α\alpha terms have a=0a=0 and are unchanged; in the β\beta terms qubit 11 flips, ∣100⟩↦∣110⟩\lvert100\rangle\mapsto\lvert110\rangle and ∣111⟩↦∣101⟩\lvert111\rangle\mapsto\lvert101\rangle.
  4. Step 4:
    →H012[α(∣0⟩+∣1⟩)∣00⟩+α(∣0⟩+∣1⟩)∣11⟩+β(∣0⟩−∣1⟩)∣10⟩+β(∣0⟩−∣1⟩)∣01⟩]\xrightarrow{H_0}\tfrac12\Big[\alpha(\lvert0\rangle+\lvert1\rangle)\lvert00\rangle+\alpha(\lvert0\rangle+\lvert1\rangle)\lvert11\rangle+\beta(\lvert0\rangle-\lvert1\rangle)\lvert10\rangle+\beta(\lvert0\rangle-\lvert1\rangle)\lvert01\rangle\Big]
    H∣0⟩=12(∣0⟩+∣1⟩)H\lvert0\rangle=\tfrac1{\sqrt2}(\lvert0\rangle+\lvert1\rangle) and H∣1⟩=12(∣0⟩−∣1⟩)H\lvert1\rangle=\tfrac1{\sqrt2}(\lvert0\rangle-\lvert1\rangle) on qubit 00; the two factors 12\tfrac1{\sqrt2} combine into 12\tfrac12.
  5. Step 5:
    =12[α∣000⟩+α∣100⟩+α∣011⟩+α∣111⟩+β∣010⟩−β∣110⟩+β∣001⟩−β∣101⟩]=\tfrac12\Big[\alpha\lvert000\rangle+\alpha\lvert100\rangle+\alpha\lvert011\rangle+\alpha\lvert111\rangle+\beta\lvert010\rangle-\beta\lvert110\rangle+\beta\lvert001\rangle-\beta\lvert101\rangle\Big]
    Multiply out each bracket: eight basis kets, each with coefficient ±12α\pm\tfrac12\alpha or ±12β\pm\tfrac12\beta.
  6. Step 6:
    ∣00⟩01: α∣000⟩+β∣001⟩=∣00⟩(α∣0⟩+β∣1⟩)\lvert00\rangle_{01}:\ \alpha\lvert000\rangle+\beta\lvert001\rangle=\lvert00\rangle(\alpha\lvert0\rangle+\beta\lvert1\rangle)
    Collect the kets whose first two bits are 0000: α∣000⟩\alpha\lvert000\rangle and β∣001⟩\beta\lvert001\rangle.
  7. Step 7:
    ∣01⟩01: α∣011⟩+β∣010⟩=∣01⟩(α∣1⟩+β∣0⟩)\lvert01\rangle_{01}:\ \alpha\lvert011\rangle+\beta\lvert010\rangle=\lvert01\rangle(\alpha\lvert1\rangle+\beta\lvert0\rangle)
    First two bits 0101: α∣011⟩\alpha\lvert011\rangle and β∣010⟩\beta\lvert010\rangle; Bob's amplitudes appear swapped.
  8. Step 8:
    ∣10⟩01: α∣100⟩−β∣101⟩=∣10⟩(α∣0⟩−β∣1⟩)\lvert10\rangle_{01}:\ \alpha\lvert100\rangle-\beta\lvert101\rangle=\lvert10\rangle(\alpha\lvert0\rangle-\beta\lvert1\rangle)
    First two bits 1010: α∣100⟩\alpha\lvert100\rangle and −β∣101⟩-\beta\lvert101\rangle; the minus sign comes from H∣1⟩H\lvert1\rangle.
  9. Step 9:
    ∣11⟩01: α∣111⟩−β∣110⟩=∣11⟩(α∣1⟩−β∣0⟩)\lvert11\rangle_{01}:\ \alpha\lvert111\rangle-\beta\lvert110\rangle=\lvert11\rangle(\alpha\lvert1\rangle-\beta\lvert0\rangle)
    First two bits 1111: α∣111⟩\alpha\lvert111\rangle and −β∣110⟩-\beta\lvert110\rangle; both the swap and the sign appear.

Result

12[∣00⟩(α∣0⟩+β∣1⟩)+∣01⟩(α∣1⟩+β∣0⟩)+∣10⟩(α∣0⟩−β∣1⟩)+∣11⟩(α∣1⟩−β∣0⟩)]\boxed{\tfrac12\Big[\lvert00\rangle(\alpha\lvert0\rangle+\beta\lvert1\rangle)+\lvert01\rangle(\alpha\lvert1\rangle+\beta\lvert0\rangle)+\lvert10\rangle(\alpha\lvert0\rangle-\beta\lvert1\rangle)+\lvert11\rangle(\alpha\lvert1\rangle-\beta\lvert0\rangle)\Big]}
Derivation

Bob's four branch states as Paulis applied to |ψ⟩

  1. Step 1:
    X∣ψ⟩=αX∣0⟩+βX∣1⟩=α∣1⟩+β∣0⟩X\lvert\psi\rangle=\alpha X\lvert0\rangle+\beta X\lvert1\rangle=\alpha\lvert1\rangle+\beta\lvert0\rangle
    XX swaps ∣0⟩\lvert0\rangle and ∣1⟩\lvert1\rangle: this is the branch 0101.
  2. Step 2:
    Z∣ψ⟩=αZ∣0⟩+βZ∣1⟩=α∣0⟩−β∣1⟩Z\lvert\psi\rangle=\alpha Z\lvert0\rangle+\beta Z\lvert1\rangle=\alpha\lvert0\rangle-\beta\lvert1\rangle
    Z∣0⟩=∣0⟩Z\lvert0\rangle=\lvert0\rangle, Z∣1⟩=−∣1⟩Z\lvert1\rangle=-\lvert1\rangle: this is the branch 1010.
  3. Step 3:
    XZ∣ψ⟩=X(α∣0⟩−β∣1⟩)=α∣1⟩−β∣0⟩XZ\lvert\psi\rangle=X(\alpha\lvert0\rangle-\beta\lvert1\rangle)=\alpha\lvert1\rangle-\beta\lvert0\rangle
    Apply ZZ first and then XX: this is the branch 1111.
  4. Step 4:
    ⇒ ∣Ψ1⟩=12∑m0,m1∣m0m1⟩⊗Xm1Zm0∣ψ⟩\Rightarrow\ \lvert\Psi_1\rangle=\tfrac12\sum_{m_0,m_1}\lvert m_0m_1\rangle\otimes X^{m_1}Z^{m_0}\lvert\psi\rangle
    With X0=Z0=IX^0=Z^0=I the branch 0000 is ∣ψ⟩\lvert\psi\rangle itself, so all four lines have the form Xm1Zm0∣ψ⟩X^{m_1}Z^{m_0}\lvert\psi\rangle.

Quick check

Alice measures qubits 00 and 11 of ∣Ψ1⟩\lvert\Psi_1\rangle. What is the probability of the outcome m0m1=10m_0m_1=10 (it does not depend on α,β\alpha,\beta)?

tolerance ±0.000001

Bob's corrections in all four branches#

After Alice measures (m0,m1)(m_0,m_1), Bob's qubit is in Xm1Zm0∣ψ⟩X^{m_1}Z^{m_0}\lvert\psi\rangle. Since X2=Z2=IX^2=Z^2=I, the inverse of Xm1Zm0X^{m_1}Z^{m_0} is Zm0Xm1Z^{m_0}X^{m_1}: Bob applies XX if m1=1m_1=1, then ZZ if m0=1m_0=1. We check this branch by branch with explicit amplitudes.

Proposition

The corrections restore |ψ⟩

For every outcome (m0,m1)∈{0,1}2(m_0,m_1)\in\{0,1\}^2,   Zm0Xm1(Xm1Zm0∣ψ⟩)=∣ψ⟩\;Z^{m_0}X^{m_1}\big(X^{m_1}Z^{m_0}\lvert\psi\rangle\big)=\lvert\psi\rangle.

Proof (4 steps)
  1. Step 1:
    (m0,m1)=(0,0):I (α∣0⟩+β∣1⟩)=α∣0⟩+β∣1⟩(m_0,m_1)=(0,0):\quad I\,(\alpha\lvert0\rangle+\beta\lvert1\rangle)=\alpha\lvert0\rangle+\beta\lvert1\rangle
    No correction is needed; Bob already holds ∣ψ⟩\lvert\psi\rangle.
  2. Step 2:
    (m0,m1)=(0,1):X(α∣1⟩+β∣0⟩)=α∣0⟩+β∣1⟩(m_0,m_1)=(0,1):\quad X(\alpha\lvert1\rangle+\beta\lvert0\rangle)=\alpha\lvert0\rangle+\beta\lvert1\rangle
    XX swaps the basis kets back: X∣1⟩=∣0⟩X\lvert1\rangle=\lvert0\rangle, X∣0⟩=∣1⟩X\lvert0\rangle=\lvert1\rangle.
  3. Step 3:
    (m0,m1)=(1,0):Z(α∣0⟩−β∣1⟩)=α∣0⟩+β∣1⟩(m_0,m_1)=(1,0):\quad Z(\alpha\lvert0\rangle-\beta\lvert1\rangle)=\alpha\lvert0\rangle+\beta\lvert1\rangle
    ZZ flips the sign of the ∣1⟩\lvert1\rangle component a second time: −β⋅(−1)=β-\beta\cdot(-1)=\beta.
  4. Step 4:
    (m0,m1)=(1,1):ZX(α∣1⟩−β∣0⟩)=Z(α∣0⟩−β∣1⟩)=α∣0⟩+β∣1⟩(m_0,m_1)=(1,1):\quad ZX(\alpha\lvert1\rangle-\beta\lvert0\rangle)=Z(\alpha\lvert0\rangle-\beta\lvert1\rangle)=\alpha\lvert0\rangle+\beta\lvert1\rangle
    First XX (because m1=1m_1=1) swaps the kets, then ZZ (because m0=1m_0=1) removes the sign.

∎

Quick check

Alice obtains m0m1=11m_0m_1=11. Which operation must Bob apply to qubit 22?

Select one answer

The principle of deferred measurement#

A classically controlled gate "apply UU to the target if the measured bit is 11" can be replaced by the quantum controlled gate CU=∣0⟩⟨0∣⊗I+∣1⟩⟨1∣⊗UC_U=\lvert0\rangle\langle0\rvert\otimes I+\lvert1\rangle\langle1\rvert\otimes U acting before the measurement. In teleportation this turns "XX on qubit 22 if m1=1m_1=1" into CNOT1→2\mathrm{CNOT}_{1\to2} and "ZZ on qubit 22 if m0=1m_0=1" into CZ0→2\mathrm{CZ}_{0\to2}, which gives a circuit with no mid-circuit measurement at all.

Theorem

Principle of deferred measurement

Let CU=∑m∈{0,1}∣m⟩⟨m∣⊗UmC_U=\sum_{m\in\{0,1\}}\lvert m\rangle\langle m\rvert\otimes U^m with UU unitary on the target register, and let Pk=∣k⟩⟨k∣⊗IP_k=\lvert k\rangle\langle k\rvert\otimes I be the measurement projectors of the control. For every state ρ\rho of control and target, (a) CUPk=PkCUC_UP_k=P_kC_U; (b) "measure the control, then apply UkU^k on outcome kk" and "apply CUC_U, then measure the control" give the same outcome probabilities and the same post-measurement states; (c) the reduced state of the target is the same whether or not the control is measured at the end.

Proof (6 steps)
  1. Step 1:
    CUPk=∑m(∣m⟩⟨m∣k⟩⟨k∣)⊗Um=∣k⟩⟨k∣⊗UkC_UP_k=\sum_m\big(\lvert m\rangle\langle m\vert k\rangle\langle k\rvert\big)\otimes U^m=\lvert k\rangle\langle k\rvert\otimes U^k
    Multiply the tensor factors separately and use ⟨m∣k⟩=δmk\langle m\vert k\rangle=\delta_{mk}.
  2. Step 2:
    PkCU=∑m(∣k⟩⟨k∣m⟩⟨m∣)⊗Um=∣k⟩⟨k∣⊗Uk=CUPkP_kC_U=\sum_m\big(\lvert k\rangle\langle k\vert m\rangle\langle m\rvert\big)\otimes U^m=\lvert k\rangle\langle k\rvert\otimes U^k=C_UP_k
    Same computation with the projector on the left: (a) holds.
  3. Step 3:
    (I⊗Uk)Pk=∣k⟩⟨k∣⊗Uk=CUPk=PkCU(I\otimes U^k)P_k=\lvert k\rangle\langle k\rvert\otimes U^k=C_UP_k=P_kC_U
    The classically controlled operation applied after outcome kk is I⊗UkI\otimes U^k, which acts on the measured branch exactly as CUC_U does.
  4. Step 4:
    pkclassical=Tr(Pkρ)=Tr(PkCUρCU†Pk)=pkquantump_k^{\mathrm{classical}}=\mathrm{Tr}(P_k\rho)=\mathrm{Tr}(P_kC_U\rho C_U^\dagger P_k)=p_k^{\mathrm{quantum}}
    Tr(PkCUρCU†Pk)=Tr(CUPkρPkCU†)=Tr(PkρPk)=Tr(Pkρ)\mathrm{Tr}(P_kC_U\rho C_U^\dagger P_k)=\mathrm{Tr}(C_UP_k\rho P_kC_U^\dagger)=\mathrm{Tr}(P_k\rho P_k)=\mathrm{Tr}(P_k\rho) by (a), cyclicity of the trace, CU†CU=IC_U^\dagger C_U=I and Pk2=PkP_k^2=P_k.
  5. Step 5:
    (I⊗Uk)PkρPk(I⊗Uk)†pk=PkCUρCU†Pkpk\frac{(I\otimes U^k)P_k\rho P_k(I\otimes U^k)^\dagger}{p_k}=\frac{P_kC_U\rho C_U^\dagger P_k}{p_k}
    The unnormalised post-measurement states agree because (I⊗Uk)Pk=PkCU(I\otimes U^k)P_k=P_kC_U from the third line; dividing by the equal probabilities gives (b).
  6. Step 6:
    Trc(∑kPkCUρCU†Pk)=∑k⟨k∣CUρCU†∣k⟩=Trc(CUρCU†)\mathrm{Tr}_{\mathrm{c}}\Big(\sum_kP_kC_U\rho C_U^\dagger P_k\Big)=\sum_k\langle k\rvert C_U\rho C_U^\dagger\lvert k\rangle=\mathrm{Tr}_{\mathrm{c}}\big(C_U\rho C_U^\dagger\big)
    Averaging over outcomes gives the non-selective state ∑kPk(⋅)Pk\sum_kP_k(\cdot)P_k; the partial trace over the control c\mathrm{c} only uses the diagonal blocks ⟨k∣⋅∣k⟩\langle k\rvert\cdot\lvert k\rangle, which PkP_k leaves untouched. Hence (c): the final measurement of the control can even be dropped.

∎

Derivation

The two corrections as controlled gates

  1. Step 1:
    [X2 if m1=1] ⟼ CNOT1→2=∣0⟩⟨0∣1⊗I2+∣1⟩⟨1∣1⊗X2[X_2\ \text{if}\ m_1=1]\ \longmapsto\ \mathrm{CNOT}_{1\to2}=\lvert0\rangle\langle0\rvert_1\otimes I_2+\lvert1\rangle\langle1\rvert_1\otimes X_2
    Apply the theorem with control qubit 11 and U=XU=X on qubit 22 (identity on qubit 00).
  2. Step 2:
    [Z2 if m0=1] ⟼ CZ0→2=∣0⟩⟨0∣0⊗I2+∣1⟩⟨1∣0⊗Z2[Z_2\ \text{if}\ m_0=1]\ \longmapsto\ \mathrm{CZ}_{0\to2}=\lvert0\rangle\langle0\rvert_0\otimes I_2+\lvert1\rangle\langle1\rvert_0\otimes Z_2
    Apply the theorem with control qubit 00 and U=ZU=Z on qubit 22. CZ\mathrm{CZ} is symmetric in its two qubits, since it is diag(1,1,1,−1)\mathrm{diag}(1,1,1,-1) on them.
  3. Step 3:
    measure 0,1 last (or never)\text{measure }0,1\ \text{last (or never)}
    Both controls are measured only at the end; by part (c) Bob's reduced state does not depend on whether those final measurements happen.

Measurement-free teleportation: the final state#

Proposition

Output of the deferred protocol

CZ0→2 CNOT1→2 H0 CNOT0→1 ∣ψ⟩0∣Φ+⟩12=∣+⟩0∣+⟩1∣ψ⟩2\mathrm{CZ}_{0\to2}\,\mathrm{CNOT}_{1\to2}\,H_0\,\mathrm{CNOT}_{0\to1}\,\lvert\psi\rangle_0\lvert\Phi^+\rangle_{12}=\lvert{+}\rangle_0\lvert{+}\rangle_1\lvert\psi\rangle_2, with ∣+⟩=12(∣0⟩+∣1⟩)\lvert{+}\rangle=\tfrac1{\sqrt2}(\lvert0\rangle+\lvert1\rangle).

Proof (6 steps)
  1. Step 1:
    ∣Ψ1⟩=12[∣00⟩∣ψ⟩+∣01⟩X∣ψ⟩+∣10⟩Z∣ψ⟩+∣11⟩XZ∣ψ⟩]\lvert\Psi_1\rangle=\tfrac12\Big[\lvert00\rangle\lvert\psi\rangle+\lvert01\rangle X\lvert\psi\rangle+\lvert10\rangle Z\lvert\psi\rangle+\lvert11\rangle XZ\lvert\psi\rangle\Big]
    The teleportation identity of the first section.
  2. Step 2:
    →CNOT1→212[∣00⟩∣ψ⟩+∣01⟩X2∣ψ⟩+∣10⟩Z∣ψ⟩+∣11⟩X2Z∣ψ⟩]\xrightarrow{\mathrm{CNOT}_{1\to2}}\tfrac12\Big[\lvert00\rangle\lvert\psi\rangle+\lvert01\rangle X^2\lvert\psi\rangle+\lvert10\rangle Z\lvert\psi\rangle+\lvert11\rangle X^2Z\lvert\psi\rangle\Big]
    XX acts on qubit 22 exactly in the branches with qubit 11 in ∣1⟩\lvert1\rangle, i.e. 0101 and 1111.
  3. Step 3:
    =12[∣00⟩∣ψ⟩+∣01⟩∣ψ⟩+∣10⟩Z∣ψ⟩+∣11⟩Z∣ψ⟩]=\tfrac12\Big[\lvert00\rangle\lvert\psi\rangle+\lvert01\rangle\lvert\psi\rangle+\lvert10\rangle Z\lvert\psi\rangle+\lvert11\rangle Z\lvert\psi\rangle\Big]
    X2=IX^2=I.
  4. Step 4:
    →CZ0→212[∣00⟩∣ψ⟩+∣01⟩∣ψ⟩+∣10⟩Z2∣ψ⟩+∣11⟩Z2∣ψ⟩]\xrightarrow{\mathrm{CZ}_{0\to2}}\tfrac12\Big[\lvert00\rangle\lvert\psi\rangle+\lvert01\rangle\lvert\psi\rangle+\lvert10\rangle Z^2\lvert\psi\rangle+\lvert11\rangle Z^2\lvert\psi\rangle\Big]
    ZZ acts on qubit 22 exactly in the branches with qubit 00 in ∣1⟩\lvert1\rangle, i.e. 1010 and 1111.
  5. Step 5:
    =12(∣00⟩+∣01⟩+∣10⟩+∣11⟩)⊗∣ψ⟩=\tfrac12\big(\lvert00\rangle+\lvert01\rangle+\lvert10\rangle+\lvert11\rangle\big)\otimes\lvert\psi\rangle
    Z2=IZ^2=I; every branch now carries the same Bob state, so it factors out.
  6. Step 6:
    =12(∣0⟩+∣1⟩)⊗12(∣0⟩+∣1⟩)⊗∣ψ⟩=∣+⟩∣+⟩∣ψ⟩=\tfrac1{\sqrt2}(\lvert0\rangle+\lvert1\rangle)\otimes\tfrac1{\sqrt2}(\lvert0\rangle+\lvert1\rangle)\otimes\lvert\psi\rangle=\lvert{+}\rangle\lvert{+}\rangle\lvert\psi\rangle
    12(∣00⟩+∣01⟩+∣10⟩+∣11⟩)\tfrac12(\lvert00\rangle+\lvert01\rangle+\lvert10\rangle+\lvert11\rangle) is the product ∣+⟩⊗∣+⟩\lvert{+}\rangle\otimes\lvert{+}\rangle, as expanding the right-hand side shows.

∎

Circuit · 3 qubits

Quantum circuit on 3 qubitsRY(1.10) on q0; RZ(0.60) on q0; H on q1; X on q2 controlled by q1; X on q1 controlled by q0; H on q0; X on q2 controlled by q1; Z on q2 controlled by q0q0q1q2RY(1.10)RZ(0.60)HH

Deferred teleportation of ∣ψ⟩=RZ(0.6)RY(1.1)∣0⟩\lvert\psi\rangle=R_Z(0.6)R_Y(1.1)\lvert0\rangle: preparation of ∣ψ⟩\lvert\psi\rangle and ∣Φ+⟩12\lvert\Phi^+\rangle_{12}, Alice's CNOT0→1\mathrm{CNOT}_{0\to1} and H0H_0, Bob's quantum-controlled corrections CNOT1→2\mathrm{CNOT}_{1\to2} and CZ0→2\mathrm{CZ}_{0\to2}. The output is ∣+⟩∣+⟩⊗e−0.3i∣ψ⟩\lvert{+}\rangle\lvert{+}\rangle\otimes e^{-0.3i}\lvert\psi\rangle: the eight amplitudes are 12e−0.3icos⁡0.55\tfrac12e^{-0.3i}\cos0.55 on even indices and 12e0.3isin⁡0.55\tfrac12e^{0.3i}\sin0.55 on odd indices.

∣ψ⟩=(0.4072−0.126i)∣000⟩+(0.2497+0.0772i)∣001⟩+(0.4072−0.126i)∣010⟩+(0.2497+0.0772i)∣011⟩+(0.4072−0.126i)∣100⟩+(0.2497+0.0772i)∣101⟩+(0.4072−0.126i)∣110⟩+(0.2497+0.0772i)∣111⟩\lvert\psi\rangle = \left(0.4072 - 0.126i\right)\lvert 000\rangle + \left(0.2497 + 0.0772i\right)\lvert 001\rangle + \left(0.4072 - 0.126i\right)\lvert 010\rangle + \left(0.2497 + 0.0772i\right)\lvert 011\rangle + \left(0.4072 - 0.126i\right)\lvert 100\rangle + \left(0.2497 + 0.0772i\right)\lvert 101\rangle + \left(0.4072 - 0.126i\right)\lvert 110\rangle + \left(0.2497 + 0.0772i\right)\lvert 111\rangle

Bob's state before the bits arrive: no signalling#

Until the classical message arrives, Bob does not know (m0,m1)(m_0,m_1). His description of qubit 22 is the reduced state of ∣Ψ1⟩\lvert\Psi_1\rangle, equivalently the average of the four branch states with probabilities 14\tfrac14. Write σ=∣ψ⟩⟨ψ∣=(∣α∣2αβˉαˉβ∣β∣2)\sigma=\lvert\psi\rangle\langle\psi\rvert=\begin{pmatrix}\lvert\alpha\rvert^2&\alpha\bar\beta\\ \bar\alpha\beta&\lvert\beta\rvert^2\end{pmatrix}.

Proposition

Bob's reduced state is I/2

ρ2=Tr01∣Ψ1⟩⟨Ψ1∣=14∑m0,m1Xm1Zm0 σ Zm0Xm1=12I\rho_2=\mathrm{Tr}_{01}\lvert\Psi_1\rangle\langle\Psi_1\rvert=\tfrac14\sum_{m_0,m_1}X^{m_1}Z^{m_0}\,\sigma\,Z^{m_0}X^{m_1}=\tfrac12I for every input ∣ψ⟩\lvert\psi\rangle.

Proof (5 steps)
  1. Step 1:
    ρ2=∑m0,m1⟨m0m1∣Ψ1⟩⟨Ψ1∣m0m1⟩=14∑m0,m1Xm1Zm0∣ψ⟩⟨ψ∣Zm0Xm1\rho_2=\sum_{m_0,m_1}\langle m_0m_1\vert\Psi_1\rangle\langle\Psi_1\vert m_0m_1\rangle=\tfrac14\sum_{m_0,m_1}X^{m_1}Z^{m_0}\lvert\psi\rangle\langle\psi\rvert Z^{m_0}X^{m_1}
    Partial trace over qubits 0,10,1 in the computational basis; ⟨m0m1∣Ψ1⟩=12Xm1Zm0∣ψ⟩\langle m_0m_1\vert\Psi_1\rangle=\tfrac12X^{m_1}Z^{m_0}\lvert\psi\rangle and X,ZX,Z are Hermitian.
  2. Step 2:
    XσX=(∣β∣2αˉβαβˉ∣α∣2),ZσZ=(∣α∣2−αβˉ−αˉβ∣β∣2)X\sigma X=\begin{pmatrix}\lvert\beta\rvert^2&\bar\alpha\beta\\ \alpha\bar\beta&\lvert\alpha\rvert^2\end{pmatrix},\qquad Z\sigma Z=\begin{pmatrix}\lvert\alpha\rvert^2&-\alpha\bar\beta\\ -\bar\alpha\beta&\lvert\beta\rvert^2\end{pmatrix}
    XσXX\sigma X swaps both rows and columns; ZσZZ\sigma Z multiplies the off-diagonal entries by (+1)(−1)=−1(+1)(-1)=-1.
  3. Step 3:
    (XZ)σ(XZ)†=X(ZσZ)X=(∣β∣2−αˉβ−αβˉ∣α∣2)=YσY(XZ)\sigma(XZ)^\dagger=X(Z\sigma Z)X=\begin{pmatrix}\lvert\beta\rvert^2&-\bar\alpha\beta\\ -\alpha\bar\beta&\lvert\alpha\rvert^2\end{pmatrix}=Y\sigma Y
    Swap rows and columns of ZσZZ\sigma Z; since XZ=−iYXZ=-iY, conjugation by XZXZ equals conjugation by YY (the phases (−i)(+i)(-i)(+i) cancel).
  4. Step 4:
    σ+XσX+ZσZ+YσY=(2(∣α∣2+∣β∣2)αβˉ+αˉβ−αβˉ−αˉβαˉβ+αβˉ−αˉβ−αβˉ2(∣α∣2+∣β∣2))\sigma+X\sigma X+Z\sigma Z+Y\sigma Y=\begin{pmatrix}2(\lvert\alpha\rvert^2+\lvert\beta\rvert^2)&\alpha\bar\beta+\bar\alpha\beta-\alpha\bar\beta-\bar\alpha\beta\\ \bar\alpha\beta+\alpha\bar\beta-\bar\alpha\beta-\alpha\bar\beta&2(\lvert\alpha\rvert^2+\lvert\beta\rvert^2)\end{pmatrix}
    Add the four matrices entry by entry: each diagonal entry collects ∣α∣2\lvert\alpha\rvert^2 twice and ∣β∣2\lvert\beta\rvert^2 twice; the off-diagonal entries cancel in pairs.
  5. Step 5:
    =(2002)⇒ρ2=14⋅2I=12I=\begin{pmatrix}2&0\\0&2\end{pmatrix}\quad\Rightarrow\quad\rho_2=\tfrac14\cdot2I=\tfrac12I
    Normalisation ∣α∣2+∣β∣2=1\lvert\alpha\rvert^2+\lvert\beta\rvert^2=1.

∎

The same conclusion follows without any algebra about σ\sigma: before Alice acts, Bob's qubit is half of ∣Φ+⟩\lvert\Phi^+\rangle, whose marginal is 12I\tfrac12I (lesson on Bell states). Alice's operations act only on qubits 0,10,1, and for any unitary VV on Alice's side TrA[(V⊗I)ρ(V†⊗I)]=TrAρ\mathrm{Tr}_A\big[(V\otimes I)\rho(V^\dagger\otimes I)\big]=\mathrm{Tr}_A\rho by cyclicity of the partial trace over AA; her non-selective measurement ρ↦∑kPkρPk\rho\mapsto\sum_kP_k\rho P_k also leaves TrA\mathrm{Tr}_A unchanged, as in part (c) of the deferred-measurement theorem. Nothing Alice does locally changes ρ2\rho_2, so Bob cannot learn anything until the two bits arrive, and those travel no faster than light.

Quick check

What is the trace distance T(12I,σ)=12∥12I−σ∥1T(\tfrac12I,\sigma)=\tfrac12\lVert\tfrac12I-\sigma\rVert_1 between Bob's state before the bits arrive and a pure target σ\sigma?

tolerance ±0.000001

Resources and what teleportation does not do#

Proposition

One ebit and two classical bits per qubit

Teleportation transfers one qubit using one shared ebit ∣Φ+⟩\lvert\Phi^+\rangle, which is consumed, and two classical bits. The two bits are uniformly distributed and independent of ∣ψ⟩\lvert\psi\rangle.

Proof (3 steps)
  1. Step 1:
    after the protocol: qubits 0,1 in ∣m0m1⟩, qubit 2 in ∣ψ⟩\text{after the protocol: qubits }0,1\text{ in }\lvert m_0m_1\rangle,\ \text{qubit }2\text{ in }\lvert\psi\rangle
    Alice's measurement collapses qubits 0,10,1 to a product basis state, so the entanglement between qubits 11 and 22 is gone: the ebit is used up.
  2. Step 2:
    P(m0,m1)=∥12Xm1Zm0∣ψ⟩∥2=14P(m_0,m_1)=\big\lVert\tfrac12X^{m_1}Z^{m_0}\lvert\psi\rangle\big\rVert^2=\tfrac14
    Xm1Zm0X^{m_1}Z^{m_0} is unitary, so every branch has norm 12\tfrac12: the distribution is uniform, carries 22 bits of entropy, and has no dependence on α,β\alpha,\beta.
  3. Step 3:
    H(m0,m1)=−∑m0,m114log⁡214=2 bitsH(m_0,m_1)=-\sum_{m_0,m_1}\tfrac14\log_2\tfrac14=2\ \text{bits}
    Both bits are needed. Knowing only m0m_0, Bob can undo Zm0Z^{m_0} but must average over m1m_1, since Zm0Xm1Zm0=±Xm1Z^{m_0}X^{m_1}Z^{m_0}=\pm X^{m_1}: he holds 12(σ+XσX)\tfrac12(\sigma+X\sigma X), whose Bloch vector is (rx,0,0)(r_x,0,0) instead of (rx,ry,rz)(r_x,r_y,r_z), so it differs from σ\sigma whenever ⟨Y⟩ψ\langle Y\rangle_\psi or ⟨Z⟩ψ\langle Z\rangle_\psi is non-zero. Knowing only m1m_1, he undoes Xm1X^{m_1} and holds 12(σ+ZσZ)\tfrac12(\sigma+Z\sigma Z) with Bloch vector (0,0,rz)(0,0,r_z), wrong whenever ⟨X⟩ψ\langle X\rangle_\psi or ⟨Y⟩ψ\langle Y\rangle_\psi is non-zero.

∎

TypeScriptrunnable
import { Statevector, reducedStateFromPure, traceDistance, matrix } from '@/lib/quantum-simulator';

// |psi> = cos(0.55)|0> + e^{0.6i} sin(0.55)|1>, prepared as R_Z(0.6) R_Y(1.1)|0> (global phase e^{-0.3i})
const c = Math.cos(0.55), s = Math.sin(0.55);
const psi = [{ re: c, im: 0 }, { re: s * Math.cos(0.6), im: s * Math.sin(0.6) }];
const sigma = matrix.fromRows([
  [[c * c, 0], [c * s * Math.cos(0.6), -c * s * Math.sin(0.6)]],
  [[c * s * Math.cos(0.6), c * s * Math.sin(0.6)], [s * s, 0]],
]);
const halfI = matrix.scale(matrix.identity(2), 0.5);

// Qubit 0 = |psi>, qubits 1, 2 = |Phi+>; then Alice applies CNOT(0 -> 1) and H(0)
const sv = Statevector.fromBitstring('000').applyGate('RY', [0], [1.1]).applyGate('RZ', [0], [0.6])
  .applyGate('H', [1]).applyGate('X', [2], [], [1])
  .applyGate('X', [1], [], [0]).applyGate('H', [0]);
console.log('before the bits arrive: T(rho_2, I/2) =', traceDistance(reducedStateFromPure(sv, [2]), halfI).toFixed(6));

// Branch by branch: Bob's state is 2 <m0 m1|Psi>; apply X^{m1}, then Z^{m0}, and compare with |psi>
for (const m0 of [0, 1]) {
  for (const m1 of [0, 1]) {
    let a = sv.amplitude(4 * m0 + 2 * m1), b = sv.amplitude(4 * m0 + 2 * m1 + 1);
    const p = (a.re * a.re + a.im * a.im) + (b.re * b.re + b.im * b.im);
    if (m1 === 1) [a, b] = [b, a];
    if (m0 === 1) b = { re: -b.re, im: -b.im };
    const ov = { re: psi[0].re * a.re + psi[0].im * a.im + psi[1].re * b.re + psi[1].im * b.im,
      im: psi[0].re * a.im - psi[0].im * a.re + psi[1].re * b.im - psi[1].im * b.re };
    console.log('m0 m1 = ' + m0 + m1 + ': p = ' + p.toFixed(4) + '  fidelity after Z^m0 X^m1 = ' + ((ov.re * ov.re + ov.im * ov.im) / p).toFixed(6));
  }
}

// Deferred corrections CNOT(1 -> 2) and CZ(0 -> 2): output |+>|+>|psi>
sv.applyGate('X', [2], [], [1]).applyGate('Z', [2], [], [0]);
console.log('after deferred corrections: T(rho_2, sigma) =', traceDistance(reducedStateFromPure(sv, [2]), sigma).toFixed(6));
const plusPlus = matrix.fromRows([0, 1, 2, 3].map(() => [0, 1, 2, 3].map(() => [0.25, 0])));
console.log('Alice register: T(rho_01, |++><++|) =', traceDistance(reducedStateFromPure(sv, [0, 1]), plusPlus).toFixed(6));

Expected output

before the bits arrive: T(rho_2, I/2) = 0.000000
m0 m1 = 00: p = 0.2500  fidelity after Z^m0 X^m1 = 1.000000
m0 m1 = 01: p = 0.2500  fidelity after Z^m0 X^m1 = 1.000000
m0 m1 = 10: p = 0.2500  fidelity after Z^m0 X^m1 = 1.000000
m0 m1 = 11: p = 0.2500  fidelity after Z^m0 X^m1 = 1.000000
after deferred corrections: T(rho_2, sigma) = 0.000000
Alice register: T(rho_01, |++><++|) = 0.000000

References

  1. C. H. Bennett, G. Brassard, C. Crépeau, R. Jozsa, A. Peres and W. K. Wootters, Teleporting an unknown quantum state via dual classical and Einstein-Podolsky-Rosen channels, Physical Review Letters 70, 1895–1899 (1993).doi:10.1103/PhysRevLett.70.1895
  2. M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), §1.3.7 (quantum teleportation) and §4.4 (measurement, principle of deferred measurement).doi:10.1017/CBO9780511976667

Hands-on labs

  • Teleport a qubit

    2 steps · auto-graded

    Teleport the state ∣ψ⟩=RZ(0.6)RY(1.1)∣0⟩≃cos⁡0.55 ∣0⟩+e0.6isin⁡0.55 ∣1⟩\lvert\psi\rangle=R_Z(0.6)R_Y(1.1)\lvert0\rangle\simeq\cos0.55\,\lvert0\rangle+e^{0.6i}\sin0.55\,\lvert1\rangle from Alice's qubit 00 to Bob's qubit 22 through the pre-shared pair ∣Φ+⟩12\lvert\Phi^+\rangle_{12}. First implement the protocol with the corrections moved before measurement (principle of deferred measurement), then with real mid-circuit measurements and classically conditioned corrections. The grader compares Bob's reduced density matrix with σ=∣ψ⟩⟨ψ∣\sigma=\lvert\psi\rangle\langle\psi\rvert in trace distance.

    Open lab