Teleportation preserves a finite reference

For any instance of the theorem, choose an ordered orthonormal basis ${\mathcal{E}}_R=(e_j{)}_{j<d}$ of $R$ using [F11]. This supplies the reference-factor basis required by [F1]; together with the qubit computational bases, it gives product bases for the initial order $A,R,B,C$ and the stated global order $A,B,C,R$. Write $\mathrm{\mid }\mathrm{\Psi }{⟩}_{AR}={\sum }_{j<d}\mathrm{\mid }{\psi }_j{⟩}_A\otimes e_j$, where $\mathrm{\mid }{\psi }_j⟩={\alpha }_j\mathrm{\mid }0⟩+{\beta }_j\mathrm{\mid }1⟩$. The canonical wire permutation in [F1] sends the initial vector to ${\sum }_j\mathrm{\mid }{\psi }_j{⟩}_A\otimes \mathrm{\mid }{\mathrm{\Phi }}^{+}{⟩}_{BC}\otimes e_j$. By [F3], for each (not necessarily normalized) qubit vector $\mathrm{\mid }{\psi }_j⟩$, the pre-measurement gates give $\frac{1}{2}{\sum }_{a,b}\mathrm{\mid }ab{⟩}_{AB}\otimes X^b{Z}^a\mathrm{\mid }{\psi }_j{⟩}_C$. Tensoring these identities with $e_j$ using [F2] and summing yields $\mathrm{\mid }\mathrm{\Omega }{⟩}_{ABCR}=\frac{1}{2}{\sum }_{a,b}\mathrm{\mid }ab{⟩}_{AB}\otimes (X^b{Z}^a\otimes I_R)\mathrm{\mid }\mathrm{\Psi }{⟩}_{CR}\mathrm{.}$

Define $P_{ab}$ on $A,B,C,R$ as the diagonal projection onto product-basis vectors whose first two bits are $ab$. These four projections form a PVM because they partition the product basis from step 1.1; that product basis is orthonormal by [F1] and [F10]. Write $\mathrm{\mid }\mathrm{\Omega }⟩={\sum }_i{c}_i{e}_i$ in this basis and define ${\rho }_{ij}=c_i\overline{c_j}$. This rank-one matrix is self-adjoint and has nonnegative quadratic form $⟨\rho z,z⟩=\mathrm{\mid }{\sum }_j{z}_j\overline{c_j}{\mathrm{\mid }}^2$. For each label, write $\mathrm{\mid }{\chi }_{ab}⟩=(X^b{Z}^a\otimes I_R)\mathrm{\mid }\mathrm{\Psi }⟩$. The product inner product and the chosen basis of $R$ give $\mathrm{\parallel }{\chi }_{ab}{\mathrm{\parallel }}^2={\sum }_{j<d}\mathrm{\parallel }{X}^b{Z}^a\mathrm{\mid }{\psi }_j⟩{\mathrm{\parallel }}^2={\sum }_{j<d}\mathrm{\parallel }{\psi }_j{\mathrm{\parallel }}^2=\mathrm{\parallel }\mathrm{\Psi }{\mathrm{\parallel }}^2=1,$ since [F9] shows that $X$ and $Z$ each preserve the qubit norm. The four $AB$ labels in step 1.1 are orthogonal, so $\mathrm{\parallel }\mathrm{\Omega }{\mathrm{\parallel }}^2=\frac{1}{4}{\sum }_{a,b}\mathrm{\parallel }{\chi }_{ab}{\mathrm{\parallel }}^2=1$. Thus [F8] gives $\mathrm{tr}(\rho )={\sum }_i\mathrm{\mid }{c}_i{\mathrm{\mid }}^2=1$, and $\rho$ is a density operator by [F12]. The weight $\mathrm{tr}(\rho P_{ab})$ is the sum of the diagonal entries selected by $P_{ab}$, equal to $\mathrm{\parallel }{P}_{ab}\mathrm{\mid }\mathrm{\Omega }⟩{\mathrm{\parallel }}^2=\frac{1}{4}$; by [F6] these weights are probabilities.

Applying $P_{ab}$ to step 1.1 gives $\frac{1}{2}\mathrm{\mid }ab⟩\otimes \mathrm{\mid }{\chi }_{ab}⟩$. Since $\mathrm{\parallel }{\chi }_{ab}\mathrm{\parallel }=1$ by step 2.1, this projected vector has norm $1\mathrm{/}2$ and its normalized branch vector is exactly $\mathrm{\mid }ab⟩\otimes \mathrm{\mid }{\chi }_{ab}⟩$. Equivalently, [F7] gives conditional density $\mathrm{\mid }ab⟩⟨ab{\mathrm{\mid }}_{AB}\otimes \mathrm{\mid }{\chi }_{ab}⟩⟨{\chi }_{ab}{\mathrm{\mid }}_{CR}$ after division by the outcome weight $1\mathrm{/}4$. Thus the conditional vector on $C\otimes R$ before Bob's correction is $\mathrm{\mid }{\chi }_{ab}⟩$.

Bob's correction acts on the $C$ factor as $Z^a{X}^b$. By [F2] and [F4], on every elementary tensor $x_C\otimes r_R$ its composition with $X^b{Z}^a\otimes I_R$ is $x_C\otimes r_R$; elementary tensors span by [F1], so it fixes the whole joint vector $\mathrm{\mid }\mathrm{\Psi }{⟩}_{CR}$. The corrected normalized branch vector is therefore $\mathrm{\mid }ab{⟩}_{AB}\otimes \mathrm{\mid }\mathrm{\Psi }{⟩}_{CR}$, whose $C\otimes R$ factor is exactly the original input vector for every label.