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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.

Construct an economical two-state quantum cloner by moving both actual outputs slightly toward one another relative to the two desired copies. Set $|a⟩=|00⟩$ and $|b⟩=|++⟩$. Their overlap is $1/2$, so their sum and difference yield the orthonormal vectorsThe desired outputs have the symmetric expressionsChoose the actual outputs to beThey are normalized and satisfyConsequently the prescription $|00⟩\mapsto |c_0⟩$, $|+0⟩\mapsto |c_{+}⟩$ preserves the Gram matrix of the two inputs and defines a linear isometry on their span. The unitary extension of a finite-dimensional isometry gives a unitary operator on the same four-dimensional, two-qubit Hilbert space; no extra workspace is required.

For an explicit full-space construction, define $c=\cos (\pi /8)$, $s=\sin (\pi /8)$ and the orthonormal basisChoose two further orthonormal vectors perpendicular to $u,v$,Thenmaps one orthonormal basis to another and is therefore unitary. Since $|00⟩=c|p⟩+s|m⟩$ and $|+0⟩=c|p⟩-s|m⟩$, it has exactly the required two outputs.

Finally,Both inputs therefore have the same global cloning fidelity, and so does their average:This proves the requested existence with a two-qubit unitary gate.

For input $|0{⟩}_S$, the controlled-NOT gate fixes $|00{⟩}_{SC}$, which is the ideal two-copy pure state. Thus its global cloning fidelity is $F_0=1$. For input $|+{⟩}_S$, the actual output is the Bell state $|{\varphi }_{+}⟩$, whereas the desired output is $|++⟩$. Their overlap isWith the given equal input probabilities,These values use the specified squared quantum fidelity for the full output, rather than a one-copy marginal fidelity.