Quantum cloning

A cloning operation for a family of pure states would mapfor every state in that family. A single physical operation cannot clone every unknown quantum state.

The no-cloning theorem forbids a universal unknown-state cloning operation, while orthogonal families can be cloned.

A two-qubit unitary gate can approximately clone the known pair $|0⟩,|+⟩$ with global cloning fidelity ${\cos }^2(\pi /24)$ for each input, using no additional workspace. To construct it, put $a=|00⟩$, $b=|++⟩$, $u=(a+b)/\sqrt{3}$ and $v=a-b$. The vectors $u,v$ are orthonormal, and $a,b$ have coefficients $(\cos (\pi /6),\pm \sin (\pi /6))$ in that basis. Choose output vectors with coefficients $(\cos (\pi /8),\pm \sin (\pi /8))$. Their inner product is $\cos (\pi /4)=1/\sqrt{2}$, exactly the overlap of the inputs $|00⟩,|+0⟩$. Their linear matching therefore defines a linear isometry on the input span and extends to a two-qubit unitary operator by unitary extension of a finite-dimensional isometry. Each output's overlap with its ideal copy is $\cos (\pi /6-\pi /8)=\cos (\pi /24)$, proving the stated squared quantum fidelity.

For a quantum cloning device with pure state output $U|\varphi ,0⟩$, the global cloning fidelity is its squared quantum fidelity with the desired joint two-copy pure state. It is the probability that a joint projector test for two ideal copies accepts the actual output. It differs from a one-clone marginal fidelity: correlations between the output qubits affect the joint overlap. For a controlled-NOT gate with input $|+⟩|0⟩$, the output is a Bell state and its global cloning fidelity with $|++⟩$ is $1/2$; for input $|0⟩|0⟩$ it is one.

For two pure states of overlap $0<c<1$, a quantum channel attempts to turn $M$ copies into $N>M$ copies. Let $e_i$ be the failure probability of the all-copies projector test on input $i$. The pure-state trace distance and fidelity identity, trace-distance contraction under quantum channels, pure-target upper bound on trace distance and triangle inequality give the displayed strictly positive gap $\Delta$. Hence ${\max }_i{e}_i\ge {\Delta }^2/4$, and for positive priors ${\pi }_0+{\pi }_1=1$, weighted Cauchy-Schwarz inequality gives ${\pi }_0{e}_0+{\pi }_1{e}_1\ge {\pi }_0{\pi }_1{\Delta }^2$. No positive state-by-state bound holds: preparing one candidate perfectly makes its own conditional failure probability zero.

Let distinct nonorthogonal pure states have overlap magnitude $0<c<1$, and let one quantum channel take $n$ identical inputs to $m$ outputs. Let ${ϵ}_u,{ϵ}_v$ be the respective probabilities that a test for the ideal $m$-copy pure state fails. Trace distance contraction, its triangle inequality and the pure-target upper bound on trace distance give $\sqrt{1-c^{2m}}\le \sqrt{{ϵ}_u}+\sqrt{1-c^{2n}}+\sqrt{{ϵ}_v}$. Squaring with the Cauchy-Schwarz inequality gives the displayed strategy-independent positive bound. It establishes both a worst-case bound and an equal-prior average bound. Averaging rotated pairs establishes a positive Haar-average bound for unknown qubit inputs, without assuming independent clone errors.

No deterministic quantum operation can make two perfect copies of every unknown pure state. For a proposed unitary operator acting with a fixed blank and environment, preservation of inner products gives $c=c^{2}e$, where $c$ is the input overlap and $|e|\le 1$ is the overlap of final environment states. Nonorthogonal distinct states have $0<|c|<1$, contradicting this equality. The no-cloning theorem for two pure states is its elementary two-state form.

Relative to a fixed orthonormal basis, the displayed linear map is an isometry because its basis images are orthonormal. Its adjoint obeys $C^{\ast }|k⟩|\ell ⟩={\delta }_{k\ell }|k⟩$. If vectors are described by coefficient functions, this adjoint maps $|f⟩|g⟩$ to $|fg⟩$, where the product is pointwise. Although $C$ copies every basis quantum state, it does not copy arbitrary superpositions. For example it sends $(|0⟩+|1⟩)/\sqrt{2}$ to $(|00⟩+|11⟩)/\sqrt{2}$, whereas two copies would also contain the cross terms $|01⟩$ and $|10⟩$.

For the normalized discrete Fourier transform basis over ${\mathbb{Z}}_N$, products of coefficient functions satisfy $f_m{f}_n=N^{-1/2}{f}_{m+n}$. Taking adjoints of the multiplication rule for the basis-copying isometry gives the displayed identity, with addition modulo $N$. Its $N$ output terms are orthonormal, so the prefactor is required by norm preservation. Each reduced quantum state is $I/N$, making the output a maximally entangled state for $N\ge 2$. This explains concretely why copying one basis is incompatible with copying its Fourier basis.

No unitary operator $U$ can satisfyfor two distinct nonorthogonal pure states. Taking the inner product of these two equations would requirewhich is impossible when $0<|⟨c_0|c_1⟩|<1$.

If a device produces arbitrarily many copies of either of two distinct pure-state rays, their $N$-copy overlap isThe Helstrom measurement on the copies therefore distinguishes the alternatives with success probability tending to one.

If a device perfectly identifies which member of a known finite state family was supplied, its classical output can control state-preparation unitaries that prepare any requested number of fresh copies. The discrimination may destroy the supplied system.