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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.

Use the Dirac notation convention that the inner product is linear in the ket. Orthonormality of the computational basis givesThe finite geometric sum satisfies ${\sum }_{k=0}^{N-1}{\lambda }^{rk}=N$ if $r=0$ in ${\mathbb{Z}}_N$, and zero otherwise: in the second case it equals $(1-{\lambda }^{rN})/(1-{\lambda }^r)=0$. ThereforeThere are $N$ such orthonormal vectors in an $N$-dimensional space, so they form a basis. The inverse expansion is verified by the same sum:This is the unitary discrete Fourier transform and its inverse.

For the basis-copying isometry, the images $|k⟩|k⟩$ are orthonormal, so $C$ is a linear isometry, with $C^{\ast }C=I$. Its matrix elements giveApplying the definition of the adjoint operator, the coefficient of $|k⟩$ in $C^{\ast }|f⟩|g⟩$ is exactly this expression. ThusEquivalently $C^{\ast }|k⟩|\ell ⟩={\delta }_{k\ell }|k⟩$. The adjoint of basis copying acts as pointwise multiplication of coefficient functions.

For the Fourier transform of a basis-copying isometry, the normalization matters:Hence $C^{\ast }|\hat{m}⟩|\hat{n}⟩=N^{-1/2}|\widehat{m+n}⟩$. Taking adjoint matrix elements, or expanding both output factors in the inverse Fourier basis, yieldsThe printed formula is missing the factor $1/\sqrt{N}$. There are $N$ mutually orthogonal terms in its sum, so without that factor its right side has norm $\sqrt{N}$, whereas its left side has norm one because $C$ is an isometry. The formula above is the normalized identity consistent with all the preceding definitions. For $N=1$ the distinction disappears, but the no-cloning assertion is substantive for $N\ge 2$.

The normalized output is not $|\hat{j}⟩|\hat{j}⟩$. Its coefficient matrix pairs every Fourier label $m$ with $j-m$; tracing out either factor gives $I/N$. Thus it is a maximally entangled state, while two clones of a pure Fourier quantum state would have pure reduced quantum states. A copier for the computational basis therefore does not copy the Fourier basis.

More generally, suppose a linear isometry with a fixed blank input could clone every normalized quantum state. Its restriction to $|\psi ⟩|\mathrm{blank}⟩$ would define a linear map $D$ with $D|k⟩=|k⟩|k⟩$ on every computational basis quantum state. Linearity forces $D=C$, which has just been shown to fail on Fourier quantum states. This proves the no-cloning theorem in the pure-quantum state isometric setting.

The inner-product proof makes the obstruction independent of the chosen basis. If an isometry cloned two quantum states with overlap $s=⟨\psi |\varphi ⟩$, preservation of inner products would give $s=s^2$, or, allowing irrelevant output phases, $|s|=|s{|}^2$. Distinct nonorthogonal rays have $0<|s|<1$, which is impossible. Even allowing an output environment does not help: unitary evolution would give $|s|=|s{|}^2|⟨e_{\psi }|e_{\varphi }⟩|\le |s{|}^2$, the same contradiction. Arbitrary unknown quantum states cannot be cloned, although any fixed orthonormal basis can be copied and a one-dimensional quantum state space is trivial.

Use the Dirac notation convention that the inner product is linear in the ket. Orthonormality of the computational basis givesThe finite geometric sum satisfies ${\sum }_{k=0}^{N-1}{\lambda }^{rk}=N$ if $r=0$ in ${\mathbb{Z}}_N$, and zero otherwise: in the second case it equals $(1-{\lambda }^{rN})/(1-{\lambda }^r)=0$. ThereforeThere are $N$ such orthonormal vectors in an $N$-dimensional space, so they form a basis. The inverse expansion is verified by the same sum:This is the unitary discrete Fourier transform and its inverse.

For the basis-copying isometry, the images $|k⟩|k⟩$ are orthonormal, so $C$ is a linear isometry, with $C^{\ast }C=I$. Its matrix elements giveApplying the definition of the adjoint operator, the coefficient of $|k⟩$ in $C^{\ast }|f⟩|g⟩$ is exactly this expression. ThusEquivalently $C^{\ast }|k⟩|\ell ⟩={\delta }_{k\ell }|k⟩$. The adjoint of basis copying acts as pointwise multiplication of coefficient functions.

For the Fourier transform of a basis-copying isometry, the normalization matters:Hence $C^{\ast }|\hat{m}⟩|\hat{n}⟩=N^{-1/2}|\widehat{m+n}⟩$. Taking adjoint matrix elements, or expanding both output factors in the inverse Fourier basis, yieldsThe printed formula is missing the factor $1/\sqrt{N}$. There are $N$ mutually orthogonal terms in its sum, so without that factor its right side has norm $\sqrt{N}$, whereas its left side has norm one because $C$ is an isometry. The formula above is the normalized identity consistent with all the preceding definitions. For $N=1$ the distinction disappears, but the no-cloning assertion is substantive for $N\ge 2$.

The normalized output is not $|\hat{j}⟩|\hat{j}⟩$. Its coefficient matrix pairs every Fourier label $m$ with $j-m$; tracing out either factor gives $I/N$. Thus it is a maximally entangled state, while two clones of a pure Fourier quantum state would have pure reduced quantum states. A copier for the computational basis therefore does not copy the Fourier basis.

More generally, suppose a linear isometry with a fixed blank input could clone every normalized quantum state. Its restriction to $|\psi ⟩|\mathrm{blank}⟩$ would define a linear map $D$ with $D|k⟩=|k⟩|k⟩$ on every computational basis quantum state. Linearity forces $D=C$, which has just been shown to fail on Fourier quantum states. This proves the no-cloning theorem in the pure-quantum state isometric setting.

The inner-product proof makes the obstruction independent of the chosen basis. If an isometry cloned two quantum states with overlap $s=⟨\psi |\varphi ⟩$, preservation of inner products would give $s=s^2$, or, allowing irrelevant output phases, $|s|=|s{|}^2$. Distinct nonorthogonal rays have $0<|s|<1$, which is impossible. Even allowing an output environment does not help: unitary evolution would give $|s|=|s{|}^2|⟨e_{\psi }|e_{\varphi }⟩|\le |s{|}^2$, the same contradiction. Arbitrary unknown quantum states cannot be cloned, although any fixed orthonormal basis can be copied and a one-dimensional quantum state space is trivial.