QuantumCore

Lesson 24 · 65 min · core

The No-Cloning Theorem

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No physical device copies an unknown quantum state: if a unitary UU maps ∣ψ⟩∣s⟩↦∣ψ⟩∣ψ⟩\lvert\psi\rangle\lvert s\rangle\mapsto\lvert\psi\rangle\lvert\psi\rangle and ∣ϕ⟩∣s⟩↦∣ϕ⟩∣ϕ⟩\lvert\phi\rangle\lvert s\rangle\mapsto\lvert\phi\rangle\lvert\phi\rangle, then ⟨ψ∣ϕ⟩=⟨ψ∣ϕ⟩2\langle\psi\vert\phi\rangle=\langle\psi\vert\phi\rangle^2, so the two states are orthogonal or equal. We prove this from unitarity and, independently, from linearity, compute the copy fidelity F=c6+s6+2c3s3cos⁡ϕF=c^6+s^6+2c^3s^3\cos\phi of the CNOT "copier" for every Bloch state, extend the theorem to cloners that use an ancilla or environment, and show that a cloner would let a shared Bell pair transmit a bit faster than light. The same inner-product argument is the reason an eavesdropper cannot read BB84 key bits without disturbing them.

Prerequisites: Entanglement & the Bell States

Objectives and key results

  • Prove the no-cloning theorem from the preservation of inner products: a unitary cloner of ∣ψ⟩\lvert\psi\rangle and ∣ϕ⟩\lvert\phi\rangle forces ⟨ψ∣ϕ⟩∈{0,1}\langle\psi\vert\phi\rangle\in\{0,1\}.

  • Prove it again from linearity alone: a device that copies ∣0⟩\lvert0\rangle and ∣1⟩\lvert1\rangle maps ∣+⟩∣0⟩\lvert+\rangle\lvert0\rangle to ∣Φ+⟩≠∣+⟩∣+⟩\lvert\Phi^+\rangle\neq\lvert+\rangle\lvert+\rangle.

  • Derive the copy fidelity of the CNOT copier, F(θ,ϕ)=c6+s6+2c3s3cos⁡ϕF(\theta,\phi)=c^6+s^6+2c^3s^3\cos\phi with c=cos⁡θ2c=\cos\tfrac\theta2, s=sin⁡θ2s=\sin\tfrac\theta2, and evaluate it at the poles, at ∣+⟩\lvert+\rangle, ∣+i⟩\lvert+i\rangle, ∣−⟩\lvert-\rangle and on average over the sphere.

  • Extend the theorem to cloners with an ancilla or environment, U∣ψ⟩∣s⟩∣e⟩=∣ψ⟩∣ψ⟩∣eψ⟩U\lvert\psi\rangle\lvert s\rangle\lvert e\rangle=\lvert\psi\rangle\lvert\psi\rangle\lvert e_\psi\rangle, and hence to every physical operation.

  • Explain, with an explicit protocol, why a cloner would allow superluminal signalling through a shared Bell pair, and why no-cloning protects quantum key distribution.

theorem · No-cloning theorem (Wootters–Zurek, Dieks)
U∣ψ⟩∣s⟩=∣ψ⟩∣ψ⟩, U∣ϕ⟩∣s⟩=∣ϕ⟩∣ϕ⟩ ⟹ ⟨ψ∣ϕ⟩=⟨ψ∣ϕ⟩2 ⟹ ⟨ψ∣ϕ⟩∈{0,1}U\lvert\psi\rangle\lvert s\rangle=\lvert\psi\rangle\lvert\psi\rangle,\ U\lvert\phi\rangle\lvert s\rangle=\lvert\phi\rangle\lvert\phi\rangle\ \Longrightarrow\ \langle\psi\vert\phi\rangle=\langle\psi\vert\phi\rangle^2\ \Longrightarrow\ \langle\psi\vert\phi\rangle\in\{0,1\}
identity · Linearity obstruction
U∣+⟩∣0⟩=12(∣00⟩+∣11⟩)=∣Φ+⟩≠∣+⟩∣+⟩=12(∣00⟩+∣01⟩+∣10⟩+∣11⟩)U\lvert+\rangle\lvert0\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)=\lvert\Phi^+\rangle\neq\lvert+\rangle\lvert+\rangle=\tfrac12\big(\lvert00\rangle+\lvert01\rangle+\lvert10\rangle+\lvert11\rangle\big)
formula · CNOT copier fidelity
F(θ,ϕ)=∣⟨ψψ∣ CNOT ∣ψ0⟩∣2=c6+s6+2c3s3cos⁡ϕ=1−34sin⁡2θ+14sin⁡3θcos⁡ϕF(\theta,\phi)=\big\lvert\langle\psi\psi\vert\,\mathrm{CNOT}\,\vert\psi0\rangle\big\rvert^2=c^6+s^6+2c^3s^3\cos\phi=1-\tfrac34\sin^2\theta+\tfrac14\sin^3\theta\cos\phi
theorem · No-cloning with an environment
U∣ψ⟩∣s⟩∣e⟩=∣ψ⟩∣ψ⟩∣eψ⟩ ⟹ ⟨ψ∣ϕ⟩=⟨ψ∣ϕ⟩2⟨eψ∣eϕ⟩ ⟹ ∣⟨ψ∣ϕ⟩∣∈{0,1}U\lvert\psi\rangle\lvert s\rangle\lvert e\rangle=\lvert\psi\rangle\lvert\psi\rangle\lvert e_\psi\rangle\ \Longrightarrow\ \langle\psi\vert\phi\rangle=\langle\psi\vert\phi\rangle^2\langle e_\psi\vert e_\phi\rangle\ \Longrightarrow\ \lvert\langle\psi\vert\phi\rangle\rvert\in\{0,1\}
formula · Sphere averages of the CNOT copier
F‾=14π∫F dΩ=12,F‾single=14π∫(c4+s4) dΩ=23\overline{F}=\frac1{4\pi}\int F\,d\Omega=\frac12,\qquad\overline{F}_{\text{single}}=\frac1{4\pi}\int\big(c^4+s^4\big)\,d\Omega=\frac23

The cloning problem#

Classical information can be copied at will: the reversible bit operation (a,b)↦(a,a⊕b)(a,b)\mapsto(a,a\oplus b) with b=0b=0 writes a copy of aa into a blank register. Its quantum version is the CNOT gate, CNOT∣a⟩∣b⟩=∣a⟩∣a⊕b⟩\mathrm{CNOT}\lvert a\rangle\lvert b\rangle=\lvert a\rangle\lvert a\oplus b\rangle, which indeed maps ∣0⟩∣0⟩↦∣0⟩∣0⟩\lvert0\rangle\lvert0\rangle\mapsto\lvert0\rangle\lvert0\rangle and ∣1⟩∣0⟩↦∣1⟩∣1⟩\lvert1\rangle\lvert0\rangle\mapsto\lvert1\rangle\lvert1\rangle. The question of this lesson is whether some fixed device copies every qubit state, including superpositions it knows nothing about. The answer, found independently by Wootters and Zurek and by Dieks in 1982, is no.

Definition

Quantum cloning machine

Let S\mathcal S be a set of normalised single-qubit states. A cloning machine for S\mathcal S is a unitary UU on C2⊗C2\mathbb C^2\otimes\mathbb C^2 together with a fixed normalised blank state ∣s⟩\lvert s\rangle such that

U(∣ψ⟩⊗∣s⟩)=∣ψ⟩⊗∣ψ⟩for every ∣ψ⟩∈S.U\big(\lvert\psi\rangle\otimes\lvert s\rangle\big)=\lvert\psi\rangle\otimes\lvert\psi\rangle\qquad\text{for every }\lvert\psi\rangle\in\mathcal S .

The machine is universal if S\mathcal S contains every pure qubit state. Both UU and ∣s⟩\lvert s\rangle are fixed in advance: they may not depend on the unknown input ∣ψ⟩\lvert\psi\rangle, since a device that is told ∣ψ⟩\lvert\psi\rangle can simply prepare a second copy from scratch.

CNOT=∣0⟩⟨0∣⊗I+∣1⟩⟨1∣⊗X=(1000010000010010)\mathrm{CNOT}=\lvert0\rangle\langle0\rvert\otimes I+\lvert1\rangle\langle1\rvert\otimes X=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}
CNOT (control qubit 0, target qubit 1)4×4

Unitary verified: U†U=IU^\dagger U = I (max deviation 0.0e+0)

Proof from unitarity#

A unitary preserves inner products, ⟨Uu∣Uv⟩=⟨u∣U†U∣v⟩=⟨u∣v⟩\langle Uu\vert Uv\rangle=\langle u\vert U^\dagger U\vert v\rangle=\langle u\vert v\rangle. Cloning, on the other hand, squares inner products, because ⟨ψψ∣ϕϕ⟩=⟨ψ∣ϕ⟩2\langle\psi\psi\vert\phi\phi\rangle=\langle\psi\vert\phi\rangle^2. The only complex numbers equal to their own square are 00 and 11.

Theorem

No-cloning theorem (inner-product form)

Let UU be a unitary on H⊗H\mathcal H\otimes\mathcal H, let ∣s⟩∈H\lvert s\rangle\in\mathcal H be normalised, and let ∣ψ⟩,∣ϕ⟩∈H\lvert\psi\rangle,\lvert\phi\rangle\in\mathcal H be normalised states with U∣ψ⟩∣s⟩=∣ψ⟩∣ψ⟩U\lvert\psi\rangle\lvert s\rangle=\lvert\psi\rangle\lvert\psi\rangle and U∣ϕ⟩∣s⟩=∣ϕ⟩∣ϕ⟩U\lvert\phi\rangle\lvert s\rangle=\lvert\phi\rangle\lvert\phi\rangle. Then ⟨ψ∣ϕ⟩∈{0,1}\langle\psi\vert\phi\rangle\in\{0,1\}: the two states are either orthogonal or identical. In particular no universal cloning machine exists.

Proof (7 steps)
  1. Step 1:
    (⟨ψ∣⊗⟨s∣)(∣ϕ⟩⊗∣s⟩)=⟨ψ∣ϕ⟩ ⟨s∣s⟩=⟨ψ∣ϕ⟩\big(\langle\psi\rvert\otimes\langle s\rvert\big)\big(\lvert\phi\rangle\otimes\lvert s\rangle\big)=\langle\psi\vert\phi\rangle\,\langle s\vert s\rangle=\langle\psi\vert\phi\rangle
    The inner product of product vectors factorises, ⟨a⊗b∣c⊗d⟩=⟨a∣c⟩⟨b∣d⟩\langle a\otimes b\vert c\otimes d\rangle=\langle a\vert c\rangle\langle b\vert d\rangle, and the blank is normalised, ⟨s∣s⟩=1\langle s\vert s\rangle=1.
  2. Step 2:
    (⟨ψ∣⊗⟨s∣)(∣ϕ⟩⊗∣s⟩)=(⟨ψ∣⊗⟨s∣) U†U (∣ϕ⟩⊗∣s⟩)\big(\langle\psi\rvert\otimes\langle s\rvert\big)\big(\lvert\phi\rangle\otimes\lvert s\rangle\big)=\big(\langle\psi\rvert\otimes\langle s\rvert\big)\,U^\dagger U\,\big(\lvert\phi\rangle\otimes\lvert s\rangle\big)
    Insert U†U=IU^\dagger U=I, which holds because UU is unitary.
  3. Step 3:
    =(U(∣ψ⟩⊗∣s⟩))†(U(∣ϕ⟩⊗∣s⟩))=\Big(U\big(\lvert\psi\rangle\otimes\lvert s\rangle\big)\Big)^\dagger\Big(U\big(\lvert\phi\rangle\otimes\lvert s\rangle\big)\Big)
    The bra of the vector U∣v⟩U\lvert v\rangle is ⟨v∣U†\langle v\rvert U^\dagger, since (AB)†=B†A†(AB)^\dagger=B^\dagger A^\dagger.
  4. Step 4:
    =(⟨ψ∣⊗⟨ψ∣)(∣ϕ⟩⊗∣ϕ⟩)=⟨ψ∣ϕ⟩ ⟨ψ∣ϕ⟩=⟨ψ∣ϕ⟩2=\big(\langle\psi\rvert\otimes\langle\psi\rvert\big)\big(\lvert\phi\rangle\otimes\lvert\phi\rangle\big)=\langle\psi\vert\phi\rangle\,\langle\psi\vert\phi\rangle=\langle\psi\vert\phi\rangle^2
    Substitute the two cloning equations, then factorise the inner product of product vectors as in the first step.
  5. Step 5:
    x:=⟨ψ∣ϕ⟩ ⟹ x=x2 ⟹ x(1−x)=0 ⟹ x∈{0,1}x:=\langle\psi\vert\phi\rangle\ \Longrightarrow\ x=x^2\ \Longrightarrow\ x(1-x)=0\ \Longrightarrow\ x\in\{0,1\}
    The first and last lines are equal. C\mathbb C is a field, so a product vanishes only if one factor does: x=0x=0 or x=1x=1.
  6. Step 6:
    x=1 ⟹ ∥∣ψ⟩−∣ϕ⟩∥2=⟨ψ∣ψ⟩+⟨ϕ∣ϕ⟩−2 Re ⟨ψ∣ϕ⟩=1+1−2=0x=1\ \Longrightarrow\ \big\lVert\lvert\psi\rangle-\lvert\phi\rangle\big\rVert^2=\langle\psi\vert\psi\rangle+\langle\phi\vert\phi\rangle-2\,\mathrm{Re}\,\langle\psi\vert\phi\rangle=1+1-2=0
    Expand the squared norm; a vector of norm 00 is the zero vector, so x=1x=1 means ∣ϕ⟩=∣ψ⟩\lvert\phi\rangle=\lvert\psi\rangle.
  7. Step 7:
    ⟨0∣+⟩=⟨0∣12(∣0⟩+∣1⟩)=12∉{0,1}\langle0\vert+\rangle=\langle0\rvert\tfrac1{\sqrt2}\big(\lvert0\rangle+\lvert1\rangle\big)=\tfrac1{\sqrt2}\notin\{0,1\}
    A universal cloner would clone both ∣0⟩\lvert0\rangle and ∣+⟩\lvert+\rangle; their overlap 12\tfrac1{\sqrt2} is neither 00 nor 11, contradicting the conclusion. Hence no universal cloner exists.

∎

The converse also holds: every orthonormal pair can be cloned. For {∣0⟩,∣1⟩}\{\lvert0\rangle,\lvert1\rangle\} the CNOT does it. For {∣+⟩,∣−⟩}\{\lvert+\rangle,\lvert-\rangle\} take U=(H⊗H) CNOT (H⊗I)U=(H\otimes H)\,\mathrm{CNOT}\,(H\otimes I) with blank ∣0⟩\lvert0\rangle:

∣±⟩∣0⟩→  H⊗I  ∣a⟩∣0⟩→  CNOT  ∣a⟩∣a⟩→  H⊗H  ∣±⟩∣±⟩,\lvert\pm\rangle\lvert0\rangle\xrightarrow{\;H\otimes I\;}\lvert a\rangle\lvert0\rangle\xrightarrow{\;\mathrm{CNOT}\;}\lvert a\rangle\lvert a\rangle\xrightarrow{\;H\otimes H\;}\lvert\pm\rangle\lvert\pm\rangle ,

with a=0a=0 for ∣+⟩\lvert+\rangle and a=1a=1 for ∣−⟩\lvert-\rangle, because H∣+⟩=∣0⟩H\lvert+\rangle=\lvert0\rangle, H∣−⟩=∣1⟩H\lvert-\rangle=\lvert1\rangle and H∣0⟩=∣+⟩H\lvert0\rangle=\lvert+\rangle, H∣1⟩=∣−⟩H\lvert1\rangle=\lvert-\rangle. What cannot be copied is an unknown state from a set that contains two non-orthogonal members. Copying classical data (orthogonal codewords) is never in conflict with quantum mechanics.

Quick check

Which pair of states can be cloned by one fixed unitary with one fixed blank state?

Select one answer

Proof from linearity#

The second proof does not even use unitarity. Cloning is a quadratic map of the amplitudes,

α∣0⟩+β∣1⟩ ⟼ α2∣00⟩+αβ∣01⟩+αβ∣10⟩+β2∣11⟩,\alpha\lvert0\rangle+\beta\lvert1\rangle\ \longmapsto\ \alpha^2\lvert00\rangle+\alpha\beta\lvert01\rangle+\alpha\beta\lvert10\rangle+\beta^2\lvert11\rangle ,

whereas quantum evolution is linear. A linear map is fixed by what it does to a basis, so once it copies ∣0⟩\lvert0\rangle and ∣1⟩\lvert1\rangle its action on every superposition is already decided, and the result is not a copy.

Proposition

A linear copier of the basis states does not copy superpositions

Let UU be any linear map on C2⊗C2\mathbb C^2\otimes\mathbb C^2 with U∣0⟩∣0⟩=∣00⟩U\lvert0\rangle\lvert0\rangle=\lvert00\rangle and U∣1⟩∣0⟩=∣11⟩U\lvert1\rangle\lvert0\rangle=\lvert11\rangle. Then U∣+⟩∣0⟩=∣Φ+⟩=12(∣00⟩+∣11⟩)U\lvert+\rangle\lvert0\rangle=\lvert\Phi^+\rangle=\tfrac1{\sqrt2}(\lvert00\rangle+\lvert11\rangle), which is not equal to ∣+⟩∣+⟩\lvert+\rangle\lvert+\rangle, not even up to a global phase, and is not a product state at all.

Proof (6 steps)
  1. Step 1:
    ∣+⟩∣0⟩=12(∣0⟩+∣1⟩)⊗∣0⟩=12(∣00⟩+∣10⟩)\lvert+\rangle\lvert0\rangle=\tfrac1{\sqrt2}\big(\lvert0\rangle+\lvert1\rangle\big)\otimes\lvert0\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert10\rangle\big)
    The tensor product is bilinear, so it distributes over the sum in the first factor.
  2. Step 2:
    U∣+⟩∣0⟩=12(U∣00⟩+U∣10⟩)=12(∣00⟩+∣11⟩)=∣Φ+⟩U\lvert+\rangle\lvert0\rangle=\tfrac1{\sqrt2}\big(U\lvert00\rangle+U\lvert10\rangle\big)=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)=\lvert\Phi^+\rangle
    Linearity of UU, then the two cloning equations for the basis states.
  3. Step 3:
    ∣+⟩∣+⟩=12(∣0⟩+∣1⟩)⊗(∣0⟩+∣1⟩)=12(∣00⟩+∣01⟩+∣10⟩+∣11⟩)\lvert+\rangle\lvert+\rangle=\tfrac12\big(\lvert0\rangle+\lvert1\rangle\big)\otimes\big(\lvert0\rangle+\lvert1\rangle\big)=\tfrac12\big(\lvert00\rangle+\lvert01\rangle+\lvert10\rangle+\lvert11\rangle\big)
    Expand the product of the two sums term by term; the prefactor is 12⋅12=12\tfrac1{\sqrt2}\cdot\tfrac1{\sqrt2}=\tfrac12.
  4. Step 4:
    ∣Φ+⟩↔(12,0,0,12)T,∣++⟩↔(12,12,12,12)T\lvert\Phi^+\rangle\leftrightarrow\big(\tfrac1{\sqrt2},0,0,\tfrac1{\sqrt2}\big)^T,\qquad\lvert{+}{+}\rangle\leftrightarrow\big(\tfrac12,\tfrac12,\tfrac12,\tfrac12\big)^T
    Coefficient vectors in the order ∣00⟩,∣01⟩,∣10⟩,∣11⟩\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle. The ∣01⟩\lvert01\rangle entries are 00 and 12\tfrac12, so ∣Φ+⟩≠eiγ∣++⟩\lvert\Phi^+\rangle\neq e^{i\gamma}\lvert{+}{+}\rangle for every phase γ\gamma, because eiγ⋅12≠0e^{i\gamma}\cdot\tfrac12\neq0.
  5. Step 5:
    ⟨++∣Φ+⟩=12⋅12+12⋅0+12⋅0+12⋅12=12,∣⟨++∣Φ+⟩∣2=12\langle{+}{+}\vert\Phi^+\rangle=\tfrac12\cdot\tfrac1{\sqrt2}+\tfrac12\cdot0+\tfrac12\cdot0+\tfrac12\cdot\tfrac1{\sqrt2}=\tfrac1{\sqrt2},\qquad\big\lvert\langle{+}{+}\vert\Phi^+\rangle\big\rvert^2=\tfrac12
    Sum of the products of corresponding (real) coefficients; the overlap of the output with the desired two-copy state has squared modulus 12\tfrac12.
  6. Step 6:
    ψ00ψ11−ψ01ψ10=12⋅12−0⋅0=12≠0\psi_{00}\psi_{11}-\psi_{01}\psi_{10}=\tfrac1{\sqrt2}\cdot\tfrac1{\sqrt2}-0\cdot0=\tfrac12\neq0
    A product state (a0,a1)⊗(b0,b1)(a_0,a_1)\otimes(b_0,b_1) has amplitudes ψjk=ajbk\psi_{jk}=a_jb_k, so ψ00ψ11−ψ01ψ10=a0b0a1b1−a0b1a1b0=0\psi_{00}\psi_{11}-\psi_{01}\psi_{10}=a_0b_0a_1b_1-a_0b_1a_1b_0=0. For ∣Φ+⟩\lvert\Phi^+\rangle this determinant is 12\tfrac12, so the output is entangled: the "copy" is correlated with the original instead of being an independent replica.

∎

Circuit · 2 qubits

Quantum circuit on 2 qubitsH on q0; X on q1 controlled by q0q0q1H

The CNOT copier on ∣+⟩∣0⟩\lvert+\rangle\lvert0\rangle (HH prepares ∣+⟩\lvert+\rangle on qubit 0). The statevector panel shows (12,0,0,12)\big(\tfrac1{\sqrt2},0,0,\tfrac1{\sqrt2}\big), the Bell state ∣Φ+⟩\lvert\Phi^+\rangle: the amplitudes of ∣01⟩\lvert01\rangle and ∣10⟩\lvert10\rangle that a true copy ∣++⟩\lvert{+}{+}\rangle would need are exactly zero.

∣ψ⟩=12∣00⟩+12∣11⟩\lvert\psi\rangle = \tfrac{1}{\sqrt{2}}\lvert 00\rangle + \tfrac{1}{\sqrt{2}}\lvert 11\rangle

Quick check

A linear device copies ∣0⟩\lvert0\rangle and ∣1⟩\lvert1\rangle into a blank ∣0⟩\lvert0\rangle. What does it output on ∣−⟩∣0⟩\lvert-\rangle\lvert0\rangle?

Select one answer

How well does the CNOT copy?#

The CNOT is a perfect copier of ∣0⟩\lvert0\rangle and ∣1⟩\lvert1\rangle and, by the two proofs above, an imperfect copier of everything else. We quantify the imperfection by the copy fidelity

F(ψ)=∣⟨ψψ∣ CNOT ∣ψ0⟩∣2,F(\psi)=\big\lvert\langle\psi\psi\vert\,\mathrm{CNOT}\,\vert\psi0\rangle\big\rvert^2 ,

the Born-rule probability that the output passes the test "is the two-qubit state ∣ψψ⟩\lvert\psi\psi\rangle?". We first treat real amplitudes and then the general Bloch state.

Derivation

Real amplitudes: F=(α3+β3)2F=(\alpha^3+\beta^3)^2

  1. Step 1:
    ∣ψ⟩=α∣0⟩+β∣1⟩,α,β∈R,α2+β2=1\lvert\psi\rangle=\alpha\lvert0\rangle+\beta\lvert1\rangle,\qquad\alpha,\beta\in\mathbb R,\quad\alpha^2+\beta^2=1
    A real input state.
  2. Step 2:
    CNOT∣ψ0⟩=α CNOT∣00⟩+β CNOT∣10⟩=α∣00⟩+β∣11⟩\mathrm{CNOT}\lvert\psi0\rangle=\alpha\,\mathrm{CNOT}\lvert00\rangle+\beta\,\mathrm{CNOT}\lvert10\rangle=\alpha\lvert00\rangle+\beta\lvert11\rangle
    Linearity, with CNOT∣00⟩=∣00⟩\mathrm{CNOT}\lvert00\rangle=\lvert00\rangle and CNOT∣10⟩=∣11⟩\mathrm{CNOT}\lvert10\rangle=\lvert11\rangle (the target flips when the control is 11).
  3. Step 3:
    ∣ψψ⟩=α2∣00⟩+αβ∣01⟩+βα∣10⟩+β2∣11⟩\lvert\psi\psi\rangle=\alpha^2\lvert00\rangle+\alpha\beta\lvert01\rangle+\beta\alpha\lvert10\rangle+\beta^2\lvert11\rangle
    Expand (α∣0⟩+β∣1⟩)⊗(α∣0⟩+β∣1⟩)(\alpha\lvert0\rangle+\beta\lvert1\rangle)\otimes(\alpha\lvert0\rangle+\beta\lvert1\rangle) by bilinearity.
  4. Step 4:
    ⟨ψψ∣ CNOT ∣ψ0⟩=α2⋅α+αβ⋅0+βα⋅0+β2⋅β=α3+β3\langle\psi\psi\vert\,\mathrm{CNOT}\,\vert\psi0\rangle=\alpha^2\cdot\alpha+\alpha\beta\cdot0+\beta\alpha\cdot0+\beta^2\cdot\beta=\alpha^3+\beta^3
    The basis {∣jk⟩}\{\lvert jk\rangle\} is orthonormal, so the inner product is the sum of products of matching coefficients; complex conjugation does nothing to real numbers.
  5. Step 5:
    F=(α3+β3)2F=\big(\alpha^3+\beta^3\big)^2
    Square the real overlap.
  6. Step 6:
    ∣+⟩: α=β=12 ⇒ α3+β3=2⋅2−3/2=12 ⇒ F=12\lvert+\rangle:\ \alpha=\beta=\tfrac1{\sqrt2}\ \Rightarrow\ \alpha^3+\beta^3=2\cdot2^{-3/2}=\tfrac1{\sqrt2}\ \Rightarrow\ F=\tfrac12
    (12)3=2−3/2(\tfrac1{\sqrt2})^3=2^{-3/2} and 2⋅2−3/2=2−1/22\cdot2^{-3/2}=2^{-1/2}; this reproduces ∣⟨++∣Φ+⟩∣2=12\lvert\langle{+}{+}\vert\Phi^+\rangle\rvert^2=\tfrac12 from the linearity proof.
  7. Step 7:
    ∣−⟩: α=12, β=−12 ⇒ α3+β3=2−3/2−2−3/2=0 ⇒ F=0\lvert-\rangle:\ \alpha=\tfrac1{\sqrt2},\ \beta=-\tfrac1{\sqrt2}\ \Rightarrow\ \alpha^3+\beta^3=2^{-3/2}-2^{-3/2}=0\ \Rightarrow\ F=0
    The output ∣Φ−⟩\lvert\Phi^-\rangle is orthogonal to ∣−−⟩\lvert{-}{-}\rangle.

Result

F=(α3+β3)2,F(∣+⟩)=12,F(∣−⟩)=0\boxed{F=\big(\alpha^3+\beta^3\big)^2,\qquad F(\lvert+\rangle)=\tfrac12,\qquad F(\lvert-\rangle)=0}
Derivation

General Bloch state: F=c6+s6+2c3s3cos⁡ϕF=c^6+s^6+2c^3s^3\cos\phi

  1. Step 1:
    ∣ψ⟩=α∣0⟩+β∣1⟩,α=c=cos⁡θ2,β=s eiϕ=sin⁡θ2 eiϕ,θ∈[0,π]\lvert\psi\rangle=\alpha\lvert0\rangle+\beta\lvert1\rangle,\qquad\alpha=c=\cos\tfrac\theta2,\quad\beta=s\,e^{i\phi}=\sin\tfrac\theta2\,e^{i\phi},\qquad\theta\in[0,\pi]
    Bloch parametrisation with the global phase chosen so that α≥0\alpha\ge0; then c,s≥0c,s\ge0 and c2+s2=1c^2+s^2=1.
  2. Step 2:
    CNOT∣ψ0⟩=α∣00⟩+β∣11⟩\mathrm{CNOT}\lvert\psi0\rangle=\alpha\lvert00\rangle+\beta\lvert11\rangle
    The same linearity step as for real amplitudes; nothing in it used that α,β\alpha,\beta are real.
  3. Step 3:
    ∣ψψ⟩=α2∣00⟩+αβ∣01⟩+βα∣10⟩+β2∣11⟩ ⟹ ⟨ψψ∣=αˉ2⟨00∣+αˉβˉ⟨01∣+βˉαˉ⟨10∣+βˉ2⟨11∣\lvert\psi\psi\rangle=\alpha^2\lvert00\rangle+\alpha\beta\lvert01\rangle+\beta\alpha\lvert10\rangle+\beta^2\lvert11\rangle\ \Longrightarrow\ \langle\psi\psi\rvert=\bar\alpha^2\langle00\rvert+\bar\alpha\bar\beta\langle01\rvert+\bar\beta\bar\alpha\langle10\rvert+\bar\beta^2\langle11\rvert
    The bra is the conjugate transpose of the ket: every coefficient is complex-conjugated.
  4. Step 4:
    ⟨ψψ∣ CNOT ∣ψ0⟩=αˉ2α+βˉ2β=∣α∣2αˉ+∣β∣2βˉ\langle\psi\psi\vert\,\mathrm{CNOT}\,\vert\psi0\rangle=\bar\alpha^2\alpha+\bar\beta^2\beta=\lvert\alpha\rvert^2\bar\alpha+\lvert\beta\rvert^2\bar\beta
    Only ⟨00∣00⟩=⟨11∣11⟩=1\langle00\vert00\rangle=\langle11\vert11\rangle=1 contribute; then αˉα=∣α∣2\bar\alpha\alpha=\lvert\alpha\rvert^2 and βˉβ=∣β∣2\bar\beta\beta=\lvert\beta\rvert^2.
  5. Step 5:
    =c2⋅c+s2⋅s e−iϕ=c3+s3e−iϕ=c^2\cdot c+s^2\cdot s\,e^{-i\phi}=c^3+s^3e^{-i\phi}
    ∣α∣2=c2\lvert\alpha\rvert^2=c^2, αˉ=c\bar\alpha=c, ∣β∣2=s2\lvert\beta\rvert^2=s^2 and βˉ=s e−iϕ\bar\beta=s\,e^{-i\phi}.
  6. Step 6:
    F=∣c3+s3e−iϕ∣2=(c3+s3e−iϕ)(c3+s3eiϕ)F=\big\lvert c^3+s^3e^{-i\phi}\big\rvert^2=\big(c^3+s^3e^{-i\phi}\big)\big(c^3+s^3e^{i\phi}\big)
    ∣z∣2=zzˉ\lvert z\rvert^2=z\bar z, and the conjugate of c3+s3e−iϕc^3+s^3e^{-i\phi} is c3+s3eiϕc^3+s^3e^{i\phi} because c,sc,s are real.
  7. Step 7:
    =c6+c3s3(eiϕ+e−iϕ)+s6e−iϕeiϕ=c6+s6+2c3s3cos⁡ϕ=c^6+c^3s^3\big(e^{i\phi}+e^{-i\phi}\big)+s^6e^{-i\phi}e^{i\phi}=c^6+s^6+2c^3s^3\cos\phi
    Multiply out the four terms; e−iϕeiϕ=1e^{-i\phi}e^{i\phi}=1 and, by Euler's formula, eiϕ+e−iϕ=2cos⁡ϕe^{i\phi}+e^{-i\phi}=2\cos\phi.
  8. Step 8:
    c6+s6=(c2+s2)3−3c2s2(c2+s2)=1−3c2s2c^6+s^6=\big(c^2+s^2\big)^3-3c^2s^2\big(c^2+s^2\big)=1-3c^2s^2
    The binomial expansion (c2+s2)3=c6+3c4s2+3c2s4+s6=c6+s6+3c2s2(c2+s2)(c^2+s^2)^3=c^6+3c^4s^2+3c^2s^4+s^6=c^6+s^6+3c^2s^2(c^2+s^2), together with c2+s2=1c^2+s^2=1.
  9. Step 9:
    cs=cos⁡θ2sin⁡θ2=12sin⁡θ ⇒ c2s2=14sin⁡2θ,c3s3=18sin⁡3θcs=\cos\tfrac\theta2\sin\tfrac\theta2=\tfrac12\sin\theta\ \Rightarrow\ c^2s^2=\tfrac14\sin^2\theta,\quad c^3s^3=\tfrac18\sin^3\theta
    Double-angle formula sin⁡θ=2sin⁡θ2cos⁡θ2\sin\theta=2\sin\tfrac\theta2\cos\tfrac\theta2; raise it to the second and third powers.
  10. Step 10:
    F(θ,ϕ)=1−34sin⁡2θ+14sin⁡3θcos⁡ϕF(\theta,\phi)=1-\tfrac34\sin^2\theta+\tfrac14\sin^3\theta\cos\phi
    Substitute: 1−3⋅14sin⁡2θ+2⋅18sin⁡3θcos⁡ϕ1-3\cdot\tfrac14\sin^2\theta+2\cdot\tfrac18\sin^3\theta\cos\phi. For ϕ=0\phi=0 this is (c3+s3)2(c^3+s^3)^2, the real case with α=c\alpha=c, β=s\beta=s; for ϕ=π\phi=\pi it is (c3−s3)2(c^3-s^3)^2, the real case with β=−s\beta=-s.

Result

F(θ,ϕ)=c6+s6+2c3s3cos⁡ϕ=1−34sin⁡2θ+14sin⁡3θcos⁡ϕ\boxed{F(\theta,\phi)=c^6+s^6+2c^3s^3\cos\phi=1-\tfrac34\sin^2\theta+\tfrac14\sin^3\theta\cos\phi}
Derivation

Special values: F=1F=1 only at the poles

  1. Step 1:
    θ=0: c=1, s=0 ⇒ F=1;θ=π: c=0, s=1 ⇒ F=1\theta=0:\ c=1,\ s=0\ \Rightarrow\ F=1;\qquad\theta=\pi:\ c=0,\ s=1\ \Rightarrow\ F=1
    At the poles ∣0⟩\lvert0\rangle and ∣1⟩\lvert1\rangle the cross term vanishes and c6+s6=1c^6+s^6=1: basis states are copied perfectly, whatever ϕ\phi is.
  2. Step 2:
    θ=π2: sin⁡θ=1 ⇒ F=1−34+14cos⁡ϕ=14(1+cos⁡ϕ)\theta=\tfrac\pi2:\ \sin\theta=1\ \Rightarrow\ F=1-\tfrac34+\tfrac14\cos\phi=\tfrac14\big(1+\cos\phi\big)
    On the equator c=s=12c=s=\tfrac1{\sqrt2}, so c6=s6=c3s3=18c^6=s^6=c^3s^3=\tfrac18.
  3. Step 3:
    F(∣+⟩)=14(1+1)=12,F(∣+i⟩)=14(1+0)=14,F(∣−⟩)=14(1−1)=0,F(∣−i⟩)=14F(\lvert+\rangle)=\tfrac14(1+1)=\tfrac12,\quad F(\lvert+i\rangle)=\tfrac14(1+0)=\tfrac14,\quad F(\lvert-\rangle)=\tfrac14(1-1)=0,\quad F(\lvert-i\rangle)=\tfrac14
    ϕ=0,π2,π,3π2\phi=0,\tfrac\pi2,\pi,\tfrac{3\pi}2 give cos⁡ϕ=1,0,−1,0\cos\phi=1,0,-1,0.
  4. Step 4:
    CNOT∣+i⟩∣0⟩=12(∣00⟩+i∣11⟩),⟨+i,+i∣=12(⟨00∣−i⟨01∣−i⟨10∣−⟨11∣)\mathrm{CNOT}\lvert{+i}\rangle\lvert0\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle+i\lvert11\rangle\big),\quad\langle{+i},{+i}\vert=\tfrac12\big(\langle00\rvert-i\langle01\rvert-i\langle10\rvert-\langle11\rvert\big)
    Direct check for ∣+i⟩=12(∣0⟩+i∣1⟩)\lvert+i\rangle=\tfrac1{\sqrt2}(\lvert0\rangle+i\lvert1\rangle): the two-copy ket has coefficients 12(1,i,i,i2)=12(1,i,i,−1)\tfrac12(1,i,i,i^2)=\tfrac12(1,i,i,-1), and the bra conjugates them.
  5. Step 5:
    ⟨+i,+i∣ CNOT ∣+i,0⟩=12⋅12(1⋅1+(−1)⋅i)=1−i22,∣1−i22∣2=28=14\langle{+i},{+i}\vert\,\mathrm{CNOT}\,\vert{+i},0\rangle=\tfrac12\cdot\tfrac1{\sqrt2}\big(1\cdot1+(-1)\cdot i\big)=\tfrac{1-i}{2\sqrt2},\qquad\Big\lvert\tfrac{1-i}{2\sqrt2}\Big\rvert^2=\tfrac{2}{8}=\tfrac14
    Only the ∣00⟩\lvert00\rangle and ∣11⟩\lvert11\rangle coefficients overlap; ∣1−i∣2=2\lvert1-i\rvert^2=2. This agrees with the formula.
  6. Step 6:
    F≤1−34sin⁡2θ+14sin⁡3θ=1−14sin⁡2θ (3−sin⁡θ)F\le1-\tfrac34\sin^2\theta+\tfrac14\sin^3\theta=1-\tfrac14\sin^2\theta\,\big(3-\sin\theta\big)
    Use cos⁡ϕ≤1\cos\phi\le1 and sin⁡3θ≥0\sin^3\theta\ge0 for θ∈[0,π]\theta\in[0,\pi].
  7. Step 7:
    3−sin⁡θ≥2>0 ⇒ F=1  ⟺  sin⁡θ=0  ⟺  θ∈{0,π}3-\sin\theta\ge2>0\ \Rightarrow\ F=1\iff\sin\theta=0\iff\theta\in\{0,\pi\}
    The subtracted term 14sin⁡2θ(3−sin⁡θ)\tfrac14\sin^2\theta(3-\sin\theta) is non-negative and vanishes only when sin⁡θ=0\sin\theta=0: the CNOT copies perfectly exactly the two basis states, as the no-cloning theorem allows.

Result

F(∣0⟩)=F(∣1⟩)=1,F(∣+⟩)=12,F(∣±i⟩)=14,F(∣−⟩)=0\boxed{F(\lvert0\rangle)=F(\lvert1\rangle)=1,\quad F(\lvert+\rangle)=\tfrac12,\quad F(\lvert\pm i\rangle)=\tfrac14,\quad F(\lvert-\rangle)=0}
Derivation

Averages over the Bloch sphere

  1. Step 1:
    f‾=14π∫02π ⁣ ⁣∫0πf(θ,ϕ) sin⁡θ dθ dϕ,1‾=14π⋅2π⋅[−cos⁡θ]0π=2π⋅24π=1\overline{f}=\frac1{4\pi}\int_0^{2\pi}\!\!\int_0^{\pi}f(\theta,\phi)\,\sin\theta\,d\theta\,d\phi,\qquad\overline{1}=\frac1{4\pi}\cdot2\pi\cdot\big[-\cos\theta\big]_0^\pi=\frac{2\pi\cdot2}{4\pi}=1
    Uniform average over pure states: the area element of the unit sphere is dΩ=sin⁡θ dθ dϕd\Omega=\sin\theta\,d\theta\,d\phi and the total area is 4π4\pi.
  2. Step 2:
    ∫02πcos⁡ϕ dϕ=[sin⁡ϕ]02π=0 ⇒ sin⁡3θcos⁡ϕ‾=0\int_0^{2\pi}\cos\phi\,d\phi=\big[\sin\phi\big]_0^{2\pi}=0\ \Rightarrow\ \overline{\sin^3\theta\cos\phi}=0
    The integrand factorises into a θ\theta part and a ϕ\phi part, and the ϕ\phi integral vanishes.
  3. Step 3:
    ∫0πsin⁡3θ dθ=∫0π(1−cos⁡2θ)sin⁡θ dθ=∫−11(1−u2) du=2−23=43\int_0^\pi\sin^3\theta\,d\theta=\int_0^\pi\big(1-\cos^2\theta\big)\sin\theta\,d\theta=\int_{-1}^{1}\big(1-u^2\big)\,du=2-\tfrac23=\tfrac43
    Write sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta and substitute u=cos⁡θu=\cos\theta, du=−sin⁡θ dθdu=-\sin\theta\,d\theta; the limits θ=0,π\theta=0,\pi become u=1,−1u=1,-1 and the sign of dudu flips them back.
  4. Step 4:
    sin⁡2θ‾=14π⋅2π⋅43=23\overline{\sin^2\theta}=\frac1{4\pi}\cdot2\pi\cdot\frac43=\frac23
    sin⁡2θ⋅sin⁡θ=sin⁡3θ\sin^2\theta\cdot\sin\theta=\sin^3\theta, the ϕ\phi integral gives 2π2\pi.
  5. Step 5:
    F‾=1−34⋅23+14⋅0=12\overline{F}=1-\tfrac34\cdot\tfrac23+\tfrac14\cdot0=\tfrac12
    Average the closed form F=1−34sin⁡2θ+14sin⁡3θcos⁡ϕF=1-\tfrac34\sin^2\theta+\tfrac14\sin^3\theta\cos\phi term by term (the average is linear).
  6. Step 6:
    ρcopy=Tr0(∣χ⟩⟨χ∣), ∣χ⟩=α∣00⟩+β∣11⟩ ⇒ ρcopy=∣α∣2∣0⟩⟨0∣+∣β∣2∣1⟩⟨1∣\rho_{\text{copy}}=\mathrm{Tr}_0\big(\lvert\chi\rangle\langle\chi\rvert\big),\ \lvert\chi\rangle=\alpha\lvert00\rangle+\beta\lvert11\rangle\ \Rightarrow\ \rho_{\text{copy}}=\lvert\alpha\rvert^2\lvert0\rangle\langle0\rvert+\lvert\beta\rvert^2\lvert1\rangle\langle1\rvert
    Partial trace over the original: ∑a(⟨a∣⊗I)∣χ⟩⟨χ∣(∣a⟩⊗I)\sum_{a}(\langle a\rvert\otimes I)\lvert\chi\rangle\langle\chi\rvert(\lvert a\rangle\otimes I); a=0a=0 leaves α∣0⟩\alpha\lvert0\rangle and a=1a=1 leaves β∣1⟩\beta\lvert1\rangle, so no cross terms ∣0⟩⟨1∣\lvert0\rangle\langle1\rvert survive. By symmetry the original qubit ends in the same mixed state.
  7. Step 7:
    Fsingle=⟨ψ∣ρcopy∣ψ⟩=∣α∣2∣α∣2+∣β∣2∣β∣2=c4+s4=1−2c2s2=1−12sin⁡2θF_{\text{single}}=\langle\psi\vert\rho_{\text{copy}}\vert\psi\rangle=\lvert\alpha\rvert^2\lvert\alpha\rvert^2+\lvert\beta\rvert^2\lvert\beta\rvert^2=c^4+s^4=1-2c^2s^2=1-\tfrac12\sin^2\theta
    ⟨ψ∣0⟩=αˉ\langle\psi\vert0\rangle=\bar\alpha and ⟨ψ∣1⟩=βˉ\langle\psi\vert1\rangle=\bar\beta; then c4+s4=(c2+s2)2−2c2s2c^4+s^4=(c^2+s^2)^2-2c^2s^2 and c2s2=14sin⁡2θc^2s^2=\tfrac14\sin^2\theta. The single-copy fidelity measures how close each output qubit is, on its own, to ∣ψ⟩\lvert\psi\rangle.
  8. Step 8:
    F‾single=1−12⋅23=23\overline{F}_{\text{single}}=1-\tfrac12\cdot\tfrac23=\tfrac23
    Use sin⁡2θ‾=23\overline{\sin^2\theta}=\tfrac23 again.

Result

F‾=12,F‾single=23\boxed{\overline{F}=\tfrac12,\qquad\overline{F}_{\text{single}}=\tfrac23}

Quick check

What is the copy fidelity F=∣⟨ψψ∣ CNOT ∣ψ0⟩∣2F=\lvert\langle\psi\psi\vert\,\mathrm{CNOT}\,\vert\psi0\rangle\rvert^2 of the CNOT copier for ∣ψ⟩=∣+i⟩=12(∣0⟩+i∣1⟩)\lvert\psi\rangle=\lvert+i\rangle=\tfrac1{\sqrt2}(\lvert0\rangle+i\lvert1\rangle)? Give a decimal.

tolerance ±0.001

Quick check

On the equator, F=14(1+cos⁡ϕ)F=\tfrac14(1+\cos\phi). What is FF at ϕ=π3\phi=\tfrac\pi3? Give a decimal.

tolerance ±0.001

Cloning with an ancilla or environment#

A cleverer machine might use extra workspace: an ancilla or environment EE that starts in a fixed state ∣e⟩\lvert e\rangle and is allowed to end in a state ∣eψ⟩\lvert e_\psi\rangle that depends on the input. This also covers machines that measure, compute, and prepare: every physical operation can be realised as a unitary on the system together with a large enough environment. The inner-product argument survives, with one extra factor.

Theorem

No-cloning with an ancilla or environment

Let UU be a unitary on H⊗H⊗HE\mathcal H\otimes\mathcal H\otimes\mathcal H_E, let ∣s⟩\lvert s\rangle and ∣e⟩\lvert e\rangle be normalised, and suppose U∣ψ⟩∣s⟩∣e⟩=∣ψ⟩∣ψ⟩∣eψ⟩U\lvert\psi\rangle\lvert s\rangle\lvert e\rangle=\lvert\psi\rangle\lvert\psi\rangle\lvert e_\psi\rangle and U∣ϕ⟩∣s⟩∣e⟩=∣ϕ⟩∣ϕ⟩∣eϕ⟩U\lvert\phi\rangle\lvert s\rangle\lvert e\rangle=\lvert\phi\rangle\lvert\phi\rangle\lvert e_\phi\rangle for normalised states ∣ψ⟩,∣ϕ⟩\lvert\psi\rangle,\lvert\phi\rangle and normalised environment states ∣eψ⟩,∣eϕ⟩\lvert e_\psi\rangle,\lvert e_\phi\rangle. Then ∣⟨ψ∣ϕ⟩∣∈{0,1}\lvert\langle\psi\vert\phi\rangle\rvert\in\{0,1\}: the inputs are orthogonal or equal up to a global phase.

Proof (8 steps)
  1. Step 1:
    ⟨ψ∣ϕ⟩=⟨ψ∣ϕ⟩ ⟨s∣s⟩ ⟨e∣e⟩=⟨ψ,s,e∣ϕ,s,e⟩\langle\psi\vert\phi\rangle=\langle\psi\vert\phi\rangle\,\langle s\vert s\rangle\,\langle e\vert e\rangle=\big\langle\psi,s,e\big\vert\phi,s,e\big\rangle
    Factorisation of the inner product of triple products, with ⟨s∣s⟩=⟨e∣e⟩=1\langle s\vert s\rangle=\langle e\vert e\rangle=1.
  2. Step 2:
    =⟨ψ,s,e∣ U†U ∣ϕ,s,e⟩=(U∣ψ,s,e⟩)†(U∣ϕ,s,e⟩)=\big\langle\psi,s,e\big\vert\,U^\dagger U\,\big\vert\phi,s,e\big\rangle=\big(U\lvert\psi,s,e\rangle\big)^\dagger\big(U\lvert\phi,s,e\rangle\big)
    Insert U†U=IU^\dagger U=I and use (U∣v⟩)†=⟨v∣U†(U\lvert v\rangle)^\dagger=\langle v\rvert U^\dagger.
  3. Step 3:
    =⟨ψ,ψ,eψ∣ϕ,ϕ,eϕ⟩=⟨ψ∣ϕ⟩2 ⟨eψ∣eϕ⟩=\big\langle\psi,\psi,e_\psi\big\vert\phi,\phi,e_\phi\big\rangle=\langle\psi\vert\phi\rangle^2\,\langle e_\psi\vert e_\phi\rangle
    Substitute the two cloning equations and factorise again.
  4. Step 4:
    x:=⟨ψ∣ϕ⟩ ⟹ ∣x∣=∣x∣2 ∣⟨eψ∣eϕ⟩∣≤∣x∣2x:=\langle\psi\vert\phi\rangle\ \Longrightarrow\ \lvert x\rvert=\lvert x\rvert^2\,\big\lvert\langle e_\psi\vert e_\phi\rangle\big\rvert\le\lvert x\rvert^2
    Take moduli of x=x2⟨eψ∣eϕ⟩x=x^2\langle e_\psi\vert e_\phi\rangle; by Cauchy–Schwarz ∣⟨eψ∣eϕ⟩∣≤∥eψ∥ ∥eϕ∥=1\lvert\langle e_\psi\vert e_\phi\rangle\rvert\le\lVert e_\psi\rVert\,\lVert e_\phi\rVert=1.
  5. Step 5:
    ∣x∣≤∣x∣2 ⟹ ∣x∣(1−∣x∣)≤0\lvert x\rvert\le\lvert x\rvert^2\ \Longrightarrow\ \lvert x\rvert\big(1-\lvert x\rvert\big)\le0
    Subtract ∣x∣2\lvert x\rvert^2 from both sides.
  6. Step 6:
    0≤∣x∣≤1 ⟹ ∣x∣(1−∣x∣)≥0 ⟹ ∣x∣(1−∣x∣)=0 ⟹ ∣x∣∈{0,1}0\le\lvert x\rvert\le1\ \Longrightarrow\ \lvert x\rvert\big(1-\lvert x\rvert\big)\ge0\ \Longrightarrow\ \lvert x\rvert\big(1-\lvert x\rvert\big)=0\ \Longrightarrow\ \lvert x\rvert\in\{0,1\}
    Cauchy–Schwarz for the normalised inputs gives ∣x∣≤1\lvert x\rvert\le1, so both factors are non-negative; together with the previous line the product is exactly 00.
  7. Step 7:
    ∣x∣=1 ⟹ ∣ϕ⟩=eiγ∣ψ⟩\lvert x\rvert=1\ \Longrightarrow\ \lvert\phi\rangle=e^{i\gamma}\lvert\psi\rangle
    Equality in Cauchy–Schwarz holds only for parallel vectors; two normalised parallel vectors differ by a phase eiγe^{i\gamma}, i.e. they are the same physical state.
  8. Step 8:
    U∣ψ,s,e⟩=eiγψ∣ψ,ψ⟩∣eψ⟩=∣ψ,ψ⟩(eiγψ∣eψ⟩)U\lvert\psi,s,e\rangle=e^{i\gamma_\psi}\lvert\psi,\psi\rangle\lvert e_\psi\rangle=\lvert\psi,\psi\rangle\big(e^{i\gamma_\psi}\lvert e_\psi\rangle\big)
    A cloner that is only required to copy up to a global phase is covered too: absorb the phase into the (still normalised) environment state eiγψ∣eψ⟩e^{i\gamma_\psi}\lvert e_\psi\rangle and apply the theorem.

∎

Derivation

Every physical cloner has this form

  1. Step 1:
    C(ρ)=TrE[V(ρ⊗∣s⟩⟨s∣⊗∣e⟩⟨e∣)V†]\mathcal C(\rho)=\mathrm{Tr}_E\Big[V\big(\rho\otimes\lvert s\rangle\langle s\rvert\otimes\lvert e\rangle\langle e\rvert\big)V^\dagger\Big]
    Any quantum operation from the input qubit (with blank) to the two output qubits can be written as a unitary VV on system plus environment followed by discarding the environment (Stinespring dilation).
  2. Step 2:
    C(∣ψ⟩⟨ψ∣)=∣ψψ⟩⟨ψψ∣andV∣ψ,s,e⟩=∑kλk∣ak⟩∣bk⟩\mathcal C\big(\lvert\psi\rangle\langle\psi\rvert\big)=\lvert\psi\psi\rangle\langle\psi\psi\rvert\quad\text{and}\quad V\lvert\psi,s,e\rangle=\textstyle\sum_k\lambda_k\lvert a_k\rangle\lvert b_k\rangle
    Perfect cloning means the two-qubit output is the pure state ∣ψψ⟩\lvert\psi\psi\rangle. Write the global pure state V∣ψ,s,e⟩V\lvert\psi,s,e\rangle in Schmidt form across the cut (two qubits ∣\vert environment), λk>0\lambda_k>0, ∑kλk2=1\sum_k\lambda_k^2=1.
  3. Step 3:
    TrE(∑k,lλkλl∣ak⟩⟨al∣⊗∣bk⟩⟨bl∣)=∑kλk2∣ak⟩⟨ak∣\mathrm{Tr}_E\Big(\textstyle\sum_{k,l}\lambda_k\lambda_l\lvert a_k\rangle\langle a_l\rvert\otimes\lvert b_k\rangle\langle b_l\rvert\Big)=\sum_k\lambda_k^2\lvert a_k\rangle\langle a_k\rvert
    The partial trace uses Tr∣bk⟩⟨bl∣=⟨bl∣bk⟩=δkl\mathrm{Tr}\lvert b_k\rangle\langle b_l\rvert=\langle b_l\vert b_k\rangle=\delta_{kl} for the orthonormal Schmidt vectors.
  4. Step 4:
    ∑kλk2∣ak⟩⟨ak∣=∣ψψ⟩⟨ψψ∣ ⟹ one term, λ1=1, ∣a1⟩=eiγ∣ψψ⟩\textstyle\sum_k\lambda_k^2\lvert a_k\rangle\langle a_k\rvert=\lvert\psi\psi\rangle\langle\psi\psi\rvert\ \Longrightarrow\ \text{one term},\ \lambda_1=1,\ \lvert a_1\rangle=e^{i\gamma}\lvert\psi\psi\rangle
    The left side is a spectral decomposition with eigenvalues λk2\lambda_k^2; the right side has rank one with eigenvalue 11, so exactly one Schmidt coefficient is non-zero and its vector spans the same ray as ∣ψψ⟩\lvert\psi\psi\rangle.
  5. Step 5:
    V∣ψ,s,e⟩=∣ψψ⟩⊗∣eψ⟩,∣eψ⟩:=eiγ∣b1⟩V\lvert\psi,s,e\rangle=\lvert\psi\psi\rangle\otimes\lvert e_\psi\rangle,\qquad\lvert e_\psi\rangle:=e^{i\gamma}\lvert b_1\rangle
    So a perfect cloner of pure states is exactly a unitary with an environment as in the theorem, and the theorem rules it out for any two non-orthogonal, distinct inputs.

Result

no quantum operation C satisfies C(∣ψ⟩⟨ψ∣)=∣ψψ⟩⟨ψψ∣ for two inputs with 0<∣⟨ψ∣ϕ⟩∣<1\boxed{\text{no quantum operation }\mathcal C\text{ satisfies }\mathcal C\big(\lvert\psi\rangle\langle\psi\rvert\big)=\lvert\psi\psi\rangle\langle\psi\psi\rvert\text{ for two inputs with }0<\lvert\langle\psi\vert\phi\rangle\rvert<1}

Quick check

Can a large environment that ends in input-dependent states ∣eψ⟩\lvert e_\psi\rangle make perfect cloning of ∣0⟩\lvert0\rangle and ∣+⟩\lvert+\rangle possible?

Select one answer

Cloning would allow faster-than-light signalling#

Alice and Bob share the Bell state ∣Φ+⟩AB=12(∣0⟩A∣0⟩B+∣1⟩A∣1⟩B)\lvert\Phi^+\rangle_{AB}=\tfrac1{\sqrt2}(\lvert0\rangle_A\lvert0\rangle_B+\lvert1\rangle_A\lvert1\rangle_B) and are far apart. Alice decides, at a time of her choosing, to measure her qubit either in the ZZ basis {∣0⟩,∣1⟩}\{\lvert0\rangle,\lvert1\rangle\} or in the XX basis {∣+⟩,∣−⟩}\{\lvert+\rangle,\lvert-\rangle\}. Her choice is one bit of information. Quantum mechanics guarantees that Bob cannot read it without a classical message, and a cloner would break that guarantee. This was the content of N. Herbert's 1982 "FLASH" proposal, which prompted the no-cloning papers of Wootters–Zurek and Dieks.

Derivation

Bob's qubit after Alice's measurement

  1. Step 1:
    (∣a⟩⟨a∣⊗I)∣Φ+⟩=12∣a⟩∣a⟩,P(a)=12,a∈{0,1}\big(\lvert a\rangle\langle a\rvert\otimes I\big)\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\lvert a\rangle\lvert a\rangle,\qquad P(a)=\tfrac12,\qquad a\in\{0,1\}
    Alice measures ZZ: the projector keeps one term of ∣Φ+⟩\lvert\Phi^+\rangle, whose squared norm is 12\tfrac12 (Born rule). After renormalising, Bob holds ∣a⟩\lvert a\rangle.
  2. Step 2:
    ∣00⟩+∣11⟩=12[(∣+⟩+∣−⟩)(∣+⟩+∣−⟩)+(∣+⟩−∣−⟩)(∣+⟩−∣−⟩)]=∣++⟩+∣−−⟩\lvert00\rangle+\lvert11\rangle=\tfrac12\Big[\big(\lvert+\rangle+\lvert-\rangle\big)\big(\lvert+\rangle+\lvert-\rangle\big)+\big(\lvert+\rangle-\lvert-\rangle\big)\big(\lvert+\rangle-\lvert-\rangle\big)\Big]=\lvert{+}{+}\rangle+\lvert{-}{-}\rangle
    Substitute ∣0⟩=12(∣+⟩+∣−⟩)\lvert0\rangle=\tfrac1{\sqrt2}(\lvert+\rangle+\lvert-\rangle) and ∣1⟩=12(∣+⟩−∣−⟩)\lvert1\rangle=\tfrac1{\sqrt2}(\lvert+\rangle-\lvert-\rangle) in both factors; the mixed terms ∣+−⟩\lvert{+}{-}\rangle and ∣−+⟩\lvert{-}{+}\rangle cancel and ∣++⟩\lvert{+}{+}\rangle, ∣−−⟩\lvert{-}{-}\rangle each appear twice.
  3. Step 3:
    ∣Φ+⟩=12(∣++⟩+∣−−⟩) ⟹ P(±)=12, Bob holds ∣±⟩\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\big(\lvert{+}{+}\rangle+\lvert{-}{-}\rangle\big)\ \Longrightarrow\ P(\pm)=\tfrac12,\ \text{Bob holds }\lvert\pm\rangle
    Alice measures XX: the same Born-rule computation in the XX basis.
  4. Step 4:
    ρB(Z)=12∣0⟩⟨0∣+12∣1⟩⟨1∣=12(1001)=I2\rho_B^{(Z)}=\tfrac12\lvert0\rangle\langle0\rvert+\tfrac12\lvert1\rangle\langle1\rvert=\tfrac12\begin{pmatrix}1&0\\0&1\end{pmatrix}=\tfrac I2
    Bob does not know Alice's outcome, so his state is the equal mixture of the two possible post-measurement states.
  5. Step 5:
    ρB(X)=12∣+⟩⟨+∣+12∣−⟩⟨−∣=12⋅12(1111)+12⋅12(1−1−11)=I2\rho_B^{(X)}=\tfrac12\lvert+\rangle\langle+\rvert+\tfrac12\lvert-\rangle\langle-\rvert=\tfrac12\cdot\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}+\tfrac12\cdot\tfrac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}=\tfrac I2
    ∣±⟩⟨±∣=12(1±1±11)\lvert\pm\rangle\langle\pm\rvert=\tfrac12\begin{pmatrix}1&\pm1\\\pm1&1\end{pmatrix}; the off-diagonal entries cancel.
  6. Step 6:
    ρB=TrA∣Φ+⟩⟨Φ+∣=I2\rho_B=\mathrm{Tr}_A\lvert\Phi^+\rangle\langle\Phi^+\rvert=\tfrac I2
    Before (or without) any measurement by Alice, the reduced state of a Bell pair is maximally mixed (lesson on Bell states). All three descriptions coincide.
  7. Step 7:
    PBob(m)=Tr(Mm ρB)=12 Tr Mmfor every measurement {Mm}P_{\text{Bob}}(m)=\mathrm{Tr}\big(M_m\,\rho_B\big)=\tfrac12\,\mathrm{Tr}\,M_m\quad\text{for every measurement }\{M_m\}
    Every statistic Bob can collect depends on his qubit only through ρB\rho_B, which is the same whatever Alice chose: no signal.

Result

ρB(Z)=ρB(X)=TrA∣Φ+⟩⟨Φ+∣=I2\boxed{\rho_B^{(Z)}=\rho_B^{(X)}=\mathrm{Tr}_A\lvert\Phi^+\rangle\langle\Phi^+\rvert=\tfrac I2}
Derivation

A cloner turns the Bell pair into a telegraph

  1. Step 1:
    ∣χ⟩∣s⟩⊗(N−1) ⟼ ∣χ⟩⊗N,∣χ⟩∈{∣0⟩,∣1⟩,∣+⟩,∣−⟩}\lvert\chi\rangle\lvert s\rangle^{\otimes(N-1)}\ \longmapsto\ \lvert\chi\rangle^{\otimes N},\qquad\lvert\chi\rangle\in\{\lvert0\rangle,\lvert1\rangle,\lvert+\rangle,\lvert-\rangle\}
    Assume Bob owns a machine that makes NN perfect copies of these four states (apply a two-copy cloner N−1N-1 times).
  2. Step 2:
    Alice Z: ∣a⟩⊗N ⇒ all N outcomes of Z equal a,  P(all equal)=1\text{Alice }Z:\ \lvert a\rangle^{\otimes N}\ \Rightarrow\ \text{all }N\text{ outcomes of }Z\text{ equal }a,\ \ P(\text{all equal})=1
    Bob clones his qubit and measures every copy in the ZZ basis; a basis state gives the same outcome every time.
  3. Step 3:
    Alice X: ∣±⟩⊗N,  P(b)=∣⟨b∣±⟩∣2=12 ⇒ P(all equal)=2⋅2−N=21−N\text{Alice }X:\ \lvert\pm\rangle^{\otimes N},\ \ P(b)=\lvert\langle b\vert\pm\rangle\rvert^2=\tfrac12\ \Rightarrow\ P(\text{all equal})=2\cdot2^{-N}=2^{1-N}
    The copies form a product state, so the NN outcomes are independent fair coins; "all equal" means all 00 or all 11, each with probability 2−N2^{-N}.
  4. Step 4:
    P(error)=12⋅0+12⋅21−N=2−NP(\text{error})=\tfrac12\cdot0+\tfrac12\cdot2^{1-N}=2^{-N}
    Bob decodes "Alice chose ZZ" exactly when all outcomes agree. With equally likely choices he errs only when Alice chose XX and the coins happen to agree; for N=20N=20 the error is below 10−610^{-6}.
  5. Step 5:
    12(∣00⟩⟨00∣+∣11⟩⟨11∣) ≠ 12(∣++⟩⟨++∣+∣−−⟩⟨−−∣)\tfrac12\big(\lvert00\rangle\langle00\rvert+\lvert11\rangle\langle11\rvert\big)\ \neq\ \tfrac12\big(\lvert{+}{+}\rangle\langle{+}{+}\rvert+\lvert{-}{-}\rangle\langle{-}{-}\rvert\big)
    The contradiction in density-matrix form (N=2N=2): the ∣01⟩⟨01∣\lvert01\rangle\langle01\rvert entry is 00 on the left and 12⋅14+12⋅14=14\tfrac12\cdot\tfrac14+\tfrac12\cdot\tfrac14=\tfrac14 on the right, since ∣⟨01∣±±⟩∣2=14\lvert\langle01\vert{\pm}{\pm}\rangle\rvert^2=\tfrac14.
  6. Step 6:
    C(I2)=12C(∣0⟩⟨0∣)+12C(∣1⟩⟨1∣)=12C(∣+⟩⟨+∣)+12C(∣−⟩⟨−∣)\mathcal C\big(\tfrac I2\big)=\tfrac12\mathcal C\big(\lvert0\rangle\langle0\rvert\big)+\tfrac12\mathcal C\big(\lvert1\rangle\langle1\rvert\big)=\tfrac12\mathcal C\big(\lvert+\rangle\langle+\rvert\big)+\tfrac12\mathcal C\big(\lvert-\rangle\langle-\rvert\big)
    Any physical map C\mathcal C is linear on density operators, and both mixtures equal I2\tfrac I2, so a physical cloner must give the same output in both cases. A perfect cloner would give the two different outputs of the previous line: it would be non-linear, and it would send Alice's bit to Bob instantly, contradicting relativistic causality. Hence no such cloner exists.

Result

perfect cloning ⟹ Bob reads Alice’s basis choice with error 2−N ⟹ superluminal signalling\boxed{\text{perfect cloning}\ \Longrightarrow\ \text{Bob reads Alice's basis choice with error }2^{-N}\ \Longrightarrow\ \text{superluminal signalling}}

Quick check

Without a cloner, why can Bob not tell whether Alice measured in the ZZ or the XX basis?

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No-cloning and quantum key distribution#

In the BB84 protocol (Bennett and Brassard, 1984) Alice sends single qubits prepared at random in one of the four states ∣0⟩,∣1⟩,∣+⟩,∣−⟩\lvert0\rangle,\lvert1\rangle,\lvert+\rangle,\lvert-\rangle, and Bob measures each in a randomly chosen basis ZZ or XX. An eavesdropper Eve would like to keep a copy of every qubit and forward the original untouched. No-cloning forbids exactly this, and the inner-product argument gives a sharper statement: any interaction that leaves two non-orthogonal signals undisturbed teaches Eve nothing about them.

Derivation

Information gain implies disturbance

  1. Step 1:
    W∣ψ⟩∣e⟩=∣ψ⟩∣eψ⟩,W∣ϕ⟩∣e⟩=∣ϕ⟩∣eϕ⟩W\lvert\psi\rangle\lvert e\rangle=\lvert\psi\rangle\lvert e_\psi\rangle,\qquad W\lvert\phi\rangle\lvert e\rangle=\lvert\phi\rangle\lvert e_\phi\rangle
    Eve couples her probe, initially ∣e⟩\lvert e\rangle, to the signal with a unitary WW and leaves both possible signals ∣ψ⟩,∣ϕ⟩\lvert\psi\rangle,\lvert\phi\rangle unchanged.
  2. Step 2:
    ⟨ψ∣ϕ⟩⟨e∣e⟩=⟨ψ∣ϕ⟩⟨eψ∣eϕ⟩ ⟹ ⟨ψ∣ϕ⟩(1−⟨eψ∣eϕ⟩)=0\langle\psi\vert\phi\rangle\langle e\vert e\rangle=\langle\psi\vert\phi\rangle\langle e_\psi\vert e_\phi\rangle\ \Longrightarrow\ \langle\psi\vert\phi\rangle\big(1-\langle e_\psi\vert e_\phi\rangle\big)=0
    Take the inner product of the two equations, as in the no-cloning proof, using W†W=IW^\dagger W=I and ⟨e∣e⟩=1\langle e\vert e\rangle=1.
  3. Step 3:
    ⟨ψ∣ϕ⟩≠0 ⟹ ⟨eψ∣eϕ⟩=1 ⟹ ∣eψ⟩=∣eϕ⟩\langle\psi\vert\phi\rangle\neq0\ \Longrightarrow\ \langle e_\psi\vert e_\phi\rangle=1\ \Longrightarrow\ \lvert e_\psi\rangle=\lvert e_\phi\rangle
    For non-orthogonal signals the second factor must vanish; equal unit vectors follow from ∥eψ−eϕ∥2=2−2 Re⟨eψ∣eϕ⟩=0\lVert e_\psi-e_\phi\rVert^2=2-2\,\mathrm{Re}\langle e_\psi\vert e_\phi\rangle=0.
  4. Step 4:
    ⟨0∣+⟩=⟨0∣−⟩=⟨1∣+⟩=12,⟨1∣−⟩=−12\langle0\vert+\rangle=\langle0\vert-\rangle=\langle1\vert+\rangle=\tfrac1{\sqrt2},\qquad\langle1\vert-\rangle=-\tfrac1{\sqrt2}
    Every BB84 state from one basis overlaps every state from the other basis, so an undisturbing probe ends in the same state for all four signals and carries no information.
  5. Step 5:
    P(error)=12⏟Eve’s basis wrong⋅12⏟Bob’s outcome random=14P(\text{error})=\underbrace{\tfrac12}_{\text{Eve's basis wrong}}\cdot\underbrace{\tfrac12}_{\text{Bob's outcome random}}=\tfrac14
    Intercept-resend attack, on the bits where Alice and Bob used the same basis: Eve measures in a random basis and resends what she found. If she guessed Alice's basis she resends the correct state; otherwise she resends an eigenstate of the wrong basis, and Bob's outcome is then correct with probability ∣12∣2=12\lvert\tfrac1{\sqrt2}\rvert^2=\tfrac12. Comparing a random sample of the key reveals this 25%25\% error rate.

Result

undisturbed non-orthogonal signals ⟹ ∣eψ⟩=∣eϕ⟩ ⟹ Eve learns nothing\boxed{\text{undisturbed non-orthogonal signals}\ \Longrightarrow\ \lvert e_\psi\rangle=\lvert e_\phi\rangle\ \Longrightarrow\ \text{Eve learns nothing}}

Numerical verification and lab#

The script below runs the CNOT copier on six input states with the simulator's statevector, computes the fidelity with ∣ψψ⟩\lvert\psi\psi\rangle, and compares it with the closed form c6+s6+2c3s3cos⁡ϕc^6+s^6+2c^3s^3\cos\phi. It then averages FF and the single-copy fidelity c4+s4c^4+s^4 over the Bloch sphere with a 200×200200\times200 midpoint rule, reproducing F‾=12\overline F=\tfrac12 and F‾single=23\overline F_{\text{single}}=\tfrac23.

TypeScriptrunnable
import { PAULI_X, Statevector } from '@/lib/quantum-simulator';

// The CNOT copier: |psi>|0> -> CNOT(0,1) -> compare with |psi>|psi>
function copier(theta, phi) {
  const c = Math.cos(theta / 2);
  const s = Math.sin(theta / 2);
  const psi = Statevector.fromAmplitudes([{ re: c, im: 0 }, { re: s * Math.cos(phi), im: s * Math.sin(phi) }]);
  const out = Statevector.product(psi, Statevector.zero(1)).applyMatrix(PAULI_X, [1], [0]);
  const F = Statevector.product(psi, psi).fidelity(out);
  const formula = c ** 6 + s ** 6 + 2 * c ** 3 * s ** 3 * Math.cos(phi);
  return { F, formula };
}

const inputs = [
  { name: '|0>  ', theta: 0, phi: 0 },
  { name: '|1>  ', theta: Math.PI, phi: 0 },
  { name: '|+>  ', theta: Math.PI / 2, phi: 0 },
  { name: '|+i> ', theta: Math.PI / 2, phi: Math.PI / 2 },
  { name: '|->  ', theta: Math.PI / 2, phi: Math.PI },
  { name: 'lab 2', theta: Math.PI / 3, phi: 0 },
];
for (const { name, theta, phi } of inputs) {
  const { F, formula } = copier(theta, phi);
  console.log(name, 'F =', F.toFixed(6), '| matches c^6+s^6+2c^3s^3cos(phi):', Math.abs(F - formula) < 1e-12);
}

// Averages over the Bloch sphere, area element sin(theta) dtheta dphi / (4 pi)
const n = 200;
let avgF = 0;
let avgSingle = 0;
for (let i = 0; i < n; i++) {
  const theta = ((i + 0.5) * Math.PI) / n;
  for (let j = 0; j < n; j++) {
    const phi = ((j + 0.5) * 2 * Math.PI) / n;
    const w = (Math.sin(theta) * (Math.PI / n) * ((2 * Math.PI) / n)) / (4 * Math.PI);
    avgF += w * copier(theta, phi).F;
    avgSingle += w * (Math.cos(theta / 2) ** 4 + Math.sin(theta / 2) ** 4);
  }
}
console.log('sphere average of F        =', avgF.toFixed(4));
console.log('sphere average of F_single =', avgSingle.toFixed(4));

Expected output

|0>   F = 1.000000 | matches c^6+s^6+2c^3s^3cos(phi): true
|1>   F = 1.000000 | matches c^6+s^6+2c^3s^3cos(phi): true
|+>   F = 0.500000 | matches c^6+s^6+2c^3s^3cos(phi): true
|+i>  F = 0.250000 | matches c^6+s^6+2c^3s^3cos(phi): true
|->   F = 0.000000 | matches c^6+s^6+2c^3s^3cos(phi): true
lab 2 F = 0.599880 | matches c^6+s^6+2c^3s^3cos(phi): true
sphere average of F        = 0.5000
sphere average of F_single = 0.6667

References

  1. W. K. Wootters and W. H. Zurek, A single quantum cannot be cloned, Nature 299, 802–803 (1982).doi:10.1038/299802a0
  2. D. Dieks, Communication by EPR devices, Physics Letters A 92, 271–272 (1982).doi:10.1016/0375-9601(82)90084-6
  3. V. Bužek and M. Hillery, Quantum copying: Beyond the no-cloning theorem, Physical Review A 54, 1844–1852 (1996).doi:10.1103/PhysRevA.54.1844
  4. M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), Box 12.1 (the no-cloning theorem) and §12.6 (quantum cryptography).doi:10.1017/CBO9780511976667

Hands-on labs

  • How good is the CNOT copier?

    2 steps · auto-graded

    The CNOT gate maps ∣0⟩∣0⟩↦∣00⟩\lvert0\rangle\lvert0\rangle\mapsto\lvert00\rangle and ∣1⟩∣0⟩↦∣11⟩\lvert1\rangle\lvert0\rangle\mapsto\lvert11\rangle, so it copies basis states perfectly. Compute its copy fidelity F=∣⟨ψψ∣ CNOT ∣ψ0⟩∣2F=\lvert\langle\psi\psi\vert\,\mathrm{CNOT}\,\vert\psi0\rangle\rvert^2 for the superpositions ∣+⟩\lvert+\rangle and cos⁡π6∣0⟩+sin⁡π6∣1⟩\cos\tfrac\pi6\lvert0\rangle+\sin\tfrac\pi6\lvert1\rangle.

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