What is the proof behind this identity used in proving the No-Cloning Theorem

Question

I am trying to follow an elementary proof of the No-Cloning theorem. The source I am using implies that $\langle \space(\lvert\psi\rangle \otimes |\psi\rangle) \space | \space (|\phi\rangle\otimes |\phi\rangle)\space \rangle = \langle\psi\rvert\phi\rangle^2$. I have limited experience in tensor calculus and so do not understand how this step is made. Why is this true? Alternatively, if anyone could direct me to a source that explains it that would also be appreciated. N.B. This identity appears to be implied by the source rather than explicitly stated. If this is incorrect please do say.

Answer 1

This simply boils down to the induced definition of the scalar product on a tensor product space^† of inner-product spaces $\mathcal{G}$ and $\mathcal{H}$: if $χ,ξ\in \mathcal{G}$ and $φ,ψ\in\mathcal{H}$, then $$ \langle χ\otimes φ, ξ\otimes ψ\rangle_{\mathcal{G}\otimes\mathcal{H}} := \langle χ,ξ\rangle_{\mathcal{G}} \cdot \langle φ,ψ\rangle_{\mathcal{H}}. $$ Why is that the definition? Well, you want this scalar product to have certain properties, which are easily fulfilled by that definition. In particular, both the tensor product and scalar product must be bilinear (or sesquilinear for a complex scalar product). This already pins the definition largely down to the one given above, but I don't think it can uniquely determine the prefactor.

You can easily derive that exact definition though if you consider baseis $\{g_i|i\in\mathscr{I}_\mathcal{G}\}\subset \mathcal{G}$ and $\{h_k|k\in\mathscr{I}_\mathcal{H}\} \subset \mathcal{H}$. These straightforwardly induce a basis of $\mathcal{G}\otimes\mathcal{H}$ as $$ \{g_i\otimes h_k | i\in\mathscr{I}_\mathcal{G},k\in\mathscr{I}_\mathcal{H}\}. $$ Now, we usually want all baseis to be orthonormal with respect to the inner products on the corresponding space. For the tensor product, this means: $$ \langle g_i\otimes h_k,\, g_j\otimes h_l\rangle \stackrel!= δ_{ij}⋅δ_{kl}. $$ Therefore (written in summation convention) $$\begin{align} \langle χ\otimes φ, ξ\otimes ψ\rangle_{\mathcal{G}\otimes\mathcal{H}} =& \langle χ_i g_i\otimes φ_k h_k, ξ_jg_j\otimes ψ_lh_l\rangle \\=& χ_iφ_k\bar ξ_j\bar ψ_l \cdot \langle g_i\otimes h_k, g_j\otimes h_l\rangle \\=& χ_i\bar ξ_j \cdot φ_k\bar ψ_l \cdot δ_{ij}\cdot δ_{kl} \\=& χ_i\bar ξ_jδ_{ij}\cdot φ_k\bar ψ_l δ_{kl} \\=& \langle χ_i g_i, ξ_jg_j\rangle\cdot \langle φ_kh_k,ψ_l h_l\rangle \\=& \langle χ, ξ\rangle_{\mathcal{G}} \cdot \langle φ,ψ\rangle_{\mathcal{H}}. \end{align}$$

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^†_{Of course, this definition doesn't yet cover the entire space: a general vector in a tensor product can only be written as a sum over such $χ\otimes φ$-pairs, but the extension follows readily from the sesquilinearity property.}