Skip to article frontmatterSkip to article content
Tutorial

No Cloning Theorem

Abstract

Proof that there exists no universal unitary operator that can copy quantum information.

Copying Classical Information

Given some bit register ∣i⟩\ket{i} such that i∈{0,1}×1 i \in \{0,1\}^{\times 1}, a cloner CC is a joint operation on ∣i⟩\ket{i} and some empty register, initialised in ∣0⟩\ket{0}, that copies the information in the first register into the second

C(∣i⟩⊗∣0⟩)=∣i⟩⊗∣i⟩.C (\ket{i} \otimes \ket{0}) = \ket{i} \otimes \ket{i}.
(1)

It can easily be seen that if CC is a CNOT gate then the above operation is achievable.

Copying Quantum Information

The analogous operation in the quantum case would be some unitary UcU_c such that

Uc(∣ψ⟩⊗∣0⟩)=∣ψ⟩⊗∣ψ⟩,U_c (\ket{\psi} \otimes \ket{0}) = \ket{\psi} \otimes \ket{\psi},
(2)

where ∣ψ⟩\ket{\psi} is some arbitrary quantum state.

However, it can be shown that no such unitary exists Scarani et al. (2005). Below is three proofs of this fact.

Proof 1: Linearity

The first proof is by linearity. Consider the state

∣Θ⟩=12(∣ψ⟩+∣ϕ⟩).\ket{\Theta} = \frac{1}{\sqrt{2}} \big( \ket{\psi} + \ket{\phi} \big).
(3)

The universal cloner acting on this state should give

Uc(∣Θ⟩⊗∣0⟩)=∣Θ⟩⊗∣Θ⟩.U_c (\ket{\Theta} \otimes \ket{0}) = \ket{\Theta} \otimes \ket{\Theta}.
(4)

However, due to linearity it should also give

Uc(∣Θ⟩⊗∣0⟩)=12(Uc(∣ψ⟩⊗∣0⟩)+Uc(∣ϕ⟩⊗∣0⟩))=12(∣ψ⟩⊗∣ψ⟩+∣ϕ⟩⊗∣ϕ⟩)≠∣Θ⟩⊗∣Θ⟩.\begin{align*} U_c (\ket{\Theta} \otimes \ket{0}) &= \frac{1}{\sqrt{2}} \big( U_c(\ket{\psi} \otimes \ket{0}) + U_c (\ket{\phi} \otimes \ket{0}) \big) \\ &= \frac{1}{\sqrt{2}} \big( \ket{\psi} \otimes \ket{\psi} + \ket{\phi} \otimes \ket{\phi}) \\ &\neq \ket{\Theta} \otimes \ket{\Theta}. \end{align*}
(5)

The output state is not the desired product state and is instead some entangled state.

Proof 2: Distinguishability

The second proof concerns distinguishability. Let ∣ψ⟩\ket{\psi} and ∣ϕ⟩\ket{\phi} be two arbitary quantum states such that

0≤∣⟨ϕ∣ψ⟩∣≤1.0 \leq \vert \braket{\phi | \psi} \vert \leq 1.
(6)

As we are requiring our cloner to be unitary, it should preserve the inner product. Therefore,

(⟨ψ∣⊗⟨0∣)Uc†Uc(∣ϕ⟩⊗∣0⟩)=(⟨ψ∣⊗⟨0∣)(∣ϕ⟩⊗∣0⟩).\begin{align*} ( \bra{\psi} \otimes \bra{0} )U^\dagger_c U_c ( \ket{\phi} \otimes \ket{0} ) = ( \bra{\psi} \otimes \bra{0} )( \ket{\phi} \otimes \ket{0} ). \end{align*}
(7)

If one considers the left hand side of the above equation it becomes

(⟨ψ∣⊗⟨0∣)Uc†Uc(∣ϕ⟩⊗∣0⟩)=(⟨ψ∣⊗⟨ψ∣)(∣ϕ⟩⊗∣ϕ⟩)=⟨ψ∣ϕ⟩2.\begin{align*} ( \bra{\psi} \otimes \bra{0} )U^\dagger_c U_c ( \ket{\phi} \otimes \ket{0} ) &= (\bra{\psi} \otimes \bra{\psi})(\ket{\phi} \otimes \ket{\phi}) \\ &= \braket{\psi | \phi}^2. \end{align*}
(8)

The right hand side is then

(⟨ψ∣⊗⟨0∣)(∣ϕ⟩⊗∣0⟩)=⟨ψ∣ϕ⟩⟨0∣0⟩=⟨ψ∣ϕ⟩,\begin{align*} ( \bra{\psi} \otimes \bra{0} )( \ket{\phi} \otimes \ket{0} ) &= \braket{\psi \vert \phi} \braket{0 \vert 0} \\ &= \braket{\psi \vert \phi}, \end{align*}
(9)

meaning one has

⟨ψ∣ϕ⟩2=⟨ψ∣ϕ⟩.\braket{\psi | \phi}^2 = \braket{\psi | \phi}.
(10)

This only has a solution if ⟨ψ∣ϕ⟩=1\braket{\psi | \phi} = 1, such that ∣ψ⟩=∣ϕ⟩\ket{\psi}=\ket{\phi}, or ⟨ψ∣ϕ⟩=0\braket{\psi | \phi}=0, such that ∣ψ⟩\ket{\psi} is orthogonal to ∣ϕ⟩\ket{\phi}. This tells us that a universal cloner does exists for states that are othronormal, such as in the example of classical bits above.

Proof 3: No-signalling

The third proof concerns no-signalling. Let Alice and Bob be two spatially separated parties who share the Bell state

∣Φ+⟩AB=∣00⟩AB+∣11⟩AB2,\ket{\Phi^+}_{AB} = \frac{\ket{00}_{AB} + \ket{11}_{AB}}{\sqrt{2}},
(11)

and assume a universal cloner exists. It is assumed that Alice can measure either ZZ or XX on her system. If Bob is able to determine which measurement Alice made without some subluminal communication, then superluminal signalling would be possible.

If Alice measures ZZ and gets the outcome +1 then Bob has the state ∣0⟩\ket{0}, if she gets the outcome -1 then Bob has the state ∣1⟩\ket{1}. As Bob does not know Alice’s measurement outcome he has the local state

ρB=12∣0⟩⟨0∣+12∣1⟩⟨1∣=12I.\begin{align*} \rho_B &= \frac{1}{2} \vert 0 \rangle \langle 0 \vert + \frac{1}{2} \vert 1 \rangle \langle 1 \vert \\ &= \frac{1}{2} \mathbb{I}. \end{align*}
(12)

This is the same as the state he had before Alice made her measurement.

If Alice measures XX and gets the outcome +1 then Bob has the state ∣+⟩\ket{+}, if she gets the outcome -1 then Bob has the state ∣−⟩\ket{-}. As Bob does not know Alice’s measurement outcome he has the local state

ρB=12∣+⟩⟨+∣+12∣−⟩⟨−∣=12I.\begin{align*} \rho_B &= \frac{1}{2} \vert + \rangle \langle + \vert + \frac{1}{2} \vert - \rangle \langle - \vert \\ &= \frac{1}{2} \mathbb{I}. \end{align*}
(13)

Which is also the same as the state he had before Alice made her measurement.

Hence, Bob cannot tell if Alice has made a measurement, let alone which measurement she has made.

Now assume that Bob waits for some amount of time so that he knows Alice has made her measurement. This solve the issue of Bob not knowing if Alice has made a measurement. He then puts his state through the cloner. If Alice measured ZZ, Bob will have the state

ρB=12∣00⟩⟨00∣+12∣11⟩⟨11∣.\begin{align*} \rho_B &= \frac{1}{2} \vert 00 \rangle \langle 00 \vert + \frac{1}{2} \vert 11 \rangle \langle 11 \vert. \end{align*}
(14)

If Alice measured XX, Bob will have the state

ρB=12∣++⟩⟨++∣+12∣−−⟩⟨−−∣.\begin{align*} \rho_B &= \frac{1}{2} \vert ++ \rangle \langle ++ \vert + \frac{1}{2} \vert -- \rangle \langle -- \vert. \end{align*}
(15)

As these two state have different probabilities of measurement outcomes, one will at least probabilistically be able to perform superluminal communication if a cloner exists. By making repeat measurement on copies of the state Bob will be able to determine which of the two measurements Alice made, allowing her to communicate one bit superluminally.

Broadcasting Quantum Information

The above definition of a cloner was given in terms of pure states but can be generalised to mixed states and subsets of the states as follows: let {ρi}i\{\rho^i\}_{i} be a set of quantum states, then there exists no unitary UcU_c such that

Uc(ρAi⊗∣0⟩⟨0∣B)Uc=ρAi⊗ρBi,U_c (\rho^i_A \otimes \vert 0 \rangle \langle 0 \vert_B) U_c = \rho^i_A \otimes \rho^i_B,
(16)

unless all elements of the set {ρi}i\{\rho^i\}_{i} are mutually orthogonal, tr[ρjρi]=δij ∀ i,j{\rm tr}\big[ \rho^j\rho^i \big]= \delta_{ij}~\forall~i,j.

This task can be generalised further to the notion of broadcasting, or local cloning, where ones asks if there exists a device such that

(ρAi⊗∣0⟩⟨0∣B)→σAB,(\rho^i_A \otimes \vert 0 \rangle \langle 0 \vert_B) \rightarrow \sigma_{AB},
(17)

where

trA[σAB]=ρi,  trB[σAB]=ρi.{\rm tr}_{A} \big[\sigma_{AB} \big] = \rho^i, ~ ~ {\rm tr}_{B} \big[\sigma_{AB} \big] = \rho^i.
(18)

It can be shown that broadcasting quantum information is also impossible Barnum et al. (1996).

References
  1. Scarani, V., Iblisdir, S., Gisin, N., & Acín, A. (2005). Quantum cloning. Reviews of Modern Physics, 77(4), 1225–1256. 10.1103/revmodphys.77.1225
  2. Barnum, H., Caves, C. M., Fuchs, C. A., Jozsa, R., & Schumacher, B. (1996). Noncommuting Mixed States Cannot Be Broadcast. Physical Review Letters, 76(15), 2818–2821. 10.1103/physrevlett.76.2818
Quantum Information 1
Super-dense Coding
Quantum Information 1
State Exclusion