Bell States, Teleportation, and Dense Coding

The matrix in [F2] gives $H\mathrm{\mid }a⟩=(\mathrm{\mid }0⟩+(-1{)}^a\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$ for each bit $a$. By [F6], $(H\otimes I)\mathrm{\mid }ab⟩=(\mathrm{\mid }0,b⟩+(-1{)}^a\mathrm{\mid }1,b⟩)\mathrm{/}\sqrt{2}$.

From [F3], applying CNOT twice fixes every basis vector because $(t\oplus a)\oplus a=t$. Its matrix permutes the orthonormal basis [F4], so its inverse is its transpose and equals itself. The matrix of $H\otimes I$ in [F4] is real symmetric and its square is the identity by [F2]. Thus both factors are self-adjoint involutions, and direct composition gives $(H\otimes I){\mathrm{CNOT}}_{1\to 2}B=I=B(H\otimes I){\mathrm{CNOT}}_{1\to 2}$.

Apply the control-first action in [F3] to the two terms of step 1.1. The control-zero term is fixed, while the control-one term changes its target from $b$ to $b\oplus 1$. Therefore $B\mathrm{\mid }ab⟩=(\mathrm{\mid }0,b⟩+(-1{)}^a\mathrm{\mid }1,b\oplus 1⟩)\mathrm{/}\sqrt{2}$.

Substituting $00,01,10,11$ into step 2.1 gives ${\mathrm{\Phi }}^{+},{\mathrm{\Psi }}^{+},{\mathrm{\Phi }}^{-},{\mathrm{\Psi }}^{-}$ in that order, exactly as defined in [F1]. Each has norm one because it is a signed sum of two distinct orthonormal computational basis vectors from [F5], divided by $\sqrt{2}$. The inner product of ${\mathrm{\Phi }}^{+}$ with ${\mathrm{\Phi }}^{-}$ and of ${\mathrm{\Psi }}^{+}$ with ${\mathrm{\Psi }}^{-}$ is $(1-1)\mathrm{/}2=0$; cross-pair inner products are zero because their computational supports are disjoint.

The four vectors in step 3.1 are an orthonormal set of four vectors in the four-dimensional space whose product basis is listed in [F4]; hence they are an orthonormal basis.

In matrices, $(AB{)}_{ij}={\sum }_k{A}_{ik}{B}_{kj}$, so conjugating and transposing this finite sum gives $(AB{)}^{†}=B^{†}{A}^{†}$. By [F7]–[F8] this is the adjoint product rule in the register pairing. Since both factors in [F6] and [F3] are self-adjoint by step 1.2, $B^{†}=(H\otimes I){\mathrm{CNOT}}_{1\to 2}$. Applying CNOT sends the four Bell vectors to $\frac{\mathrm{\mid }0⟩+\mathrm{\mid }1⟩}{\sqrt{2}}\otimes \mathrm{\mid }0⟩$, $\frac{\mathrm{\mid }0⟩+\mathrm{\mid }1⟩}{\sqrt{2}}\otimes \mathrm{\mid }1⟩$, $\frac{\mathrm{\mid }0⟩-\mathrm{\mid }1⟩}{\sqrt{2}}\otimes \mathrm{\mid }0⟩$, and $\frac{\mathrm{\mid }0⟩-\mathrm{\mid }1⟩}{\sqrt{2}}\otimes \mathrm{\mid }1⟩$, respectively. By the matrix in [F2], $H$ sends the first factors in these expressions to $\mathrm{\mid }0⟩,\mathrm{\mid }0⟩,\mathrm{\mid }1⟩,\mathrm{\mid }1⟩$, respectively, so the reverse sequence recovers $00,01,10,11$.

By [F2], the initial vector is $2^{-1\mathrm{/}2}(\alpha \mathrm{\mid }000⟩+\alpha \mathrm{\mid }011⟩+\beta \mathrm{\mid }100⟩+\beta \mathrm{\mid }111⟩)$ in wire order $A,B,C$.

Applying [F4] to the four basis vectors gives $2^{-1\mathrm{/}2}(\alpha \mathrm{\mid }000⟩+\alpha \mathrm{\mid }011⟩+\beta \mathrm{\mid }110⟩+\beta \mathrm{\mid }101⟩)$ after CNOT: the first two terms have control bit $0$ and are fixed, while the last two have control bit $1$ and their target bits flip.

By [F3] and [F6], $H_A$ sends each first-wire $\mathrm{\mid }0⟩$ to $(\mathrm{\mid }0⟩+\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$ and each $\mathrm{\mid }1⟩$ to $(\mathrm{\mid }0⟩-\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. Regrouping the resulting eight terms by the first two wires gives $\frac{1}{2}(\mathrm{\mid }00⟩(\alpha \mathrm{\mid }0⟩+\beta \mathrm{\mid }1⟩)+\mathrm{\mid }01⟩(\alpha \mathrm{\mid }1⟩+\beta \mathrm{\mid }0⟩)+\mathrm{\mid }10⟩(\alpha \mathrm{\mid }0⟩-\beta \mathrm{\mid }1⟩)+\mathrm{\mid }11⟩(\alpha \mathrm{\mid }1⟩-\beta \mathrm{\mid }0⟩))$.

The four receiver vectors in step 3.1 are respectively $\mathrm{\mid }\psi ⟩$, $X\mathrm{\mid }\psi ⟩$, $Z\mathrm{\mid }\psi ⟩$, and $XZ\mathrm{\mid }\psi ⟩$ by the actions in [F5]. These are exactly $X^b{Z}^a\mathrm{\mid }\psi ⟩$ for labels $00,01,10,11$, so step 3.1 is the sum in the Statement.

For each $(a,b)$, let $P_{ab}$ be the diagonal projection onto the two basis vectors $\mathrm{\mid }ab0⟩,\mathrm{\mid }ab1⟩$ of the ordered three-wire register. These four projections are self-adjoint idempotents, their ranges are pairwise orthogonal, and their sum is the identity because the eight orthonormal computational basis vectors from [F8] are partitioned by their first two bits. Hence they form a projective measurement as in [F2].

Write the post-gate vector in [F1] as $\mathrm{\Omega }={\sum }_i{c}_i{e}_i$ in the orthonormal basis [F4]. For the matrix $\rho$ with entries ${\rho }_{ij}=c_i\overline{c_j}$, self-adjointness is immediate; for any coordinate vector $z$, $⟨\rho z,z⟩=\mathrm{\mid }{\sum }_j{z}_j\overline{c_j}{\mathrm{\mid }}^2\ge 0$. The matrix [F6] sends $(\alpha ,\beta )$ to $(\beta ,\alpha )$ under $X$ and to $(\alpha ,-\beta )$ under $Z$, so each preserves $\mathrm{\mid }\alpha {\mathrm{\mid }}^2+\mathrm{\mid }\beta {\mathrm{\mid }}^2$ and every $X^b{Z}^a$ preserves norm. The branch vectors in [F1] have orthogonal $AB$ labels, hence $\mathrm{tr}(\rho )={\sum }_i\mathrm{\mid }{c}_i{\mathrm{\mid }}^2=\mathrm{\parallel }\mathrm{\Omega }{\mathrm{\parallel }}^2=\mathrm{\parallel }\psi {\mathrm{\parallel }}^2=1$ by [A1] and [F7]. Thus $\rho$ is a density operator by [F9], since the preceding calculation establishes nonnegativity and trace one.

By [F2] and [F7], the probability weight of label $(a,b)$ is $\mathrm{tr}(\rho P_{ab})=\mathrm{\mid }{c}_{ab0}{\mathrm{\mid }}^2+\mathrm{\mid }{c}_{ab1}{\mathrm{\mid }}^2$, the sum of the two diagonal entries selected by $P_{ab}$. The branch vector in [F1] has these two coefficients equal to those of $\frac{1}{2}{X}^b{Z}^a\mathrm{\mid }\psi ⟩$, so this sum is $\frac{1}{4}\mathrm{\parallel }{X}^b{Z}^a\psi {\mathrm{\parallel }}^2$.

The matrices in [F6] are unitary: direct multiplication gives $X^{\ast }X=I$ and $Z^{\ast }Z=I$. Hence $\mathrm{\parallel }{X}^b{Z}^a\psi {\mathrm{\parallel }}^2=\mathrm{\parallel }\psi {\mathrm{\parallel }}^2=1$, so every weight in step 2.1 is $1\mathrm{/}4$. By [F3] these are the outcome probabilities, and the four weights sum to one.

From the matrices in [F2], direct multiplication gives $X^2=I$ and $Z^2=I$. Therefore $X^b{X}^b=I$ and $Z^a{Z}^a=I$ for the two possible values of each bit.

Keeping the operator order fixed, $Z^a{X}^b{X}^b{Z}^a=Z^a(X^b{X}^b)Z^a=Z^{a}I{Z}^a=Z^a{Z}^a=I$ by step 1.1. This also covers $a=b=1$ without using an anticommutation swap.

Applying the identity in step 2.1 to the byproduct vector from [F1] yields $Z^a{X}^b(X^b{Z}^a\mathrm{\mid }\psi ⟩)=\mathrm{\mid }\psi ⟩$ for every qubit vector.

Applying $Z_A^a{X}_A^b$ to $\mathrm{\mid }{\mathrm{\Phi }}^{+}⟩=(\mathrm{\mid }00⟩+\mathrm{\mid }11⟩)\mathrm{/}\sqrt{2}$ and using [F4] gives the four cases: $00\mapsto {\mathrm{\Phi }}^{+}$, $01\mapsto {\mathrm{\Psi }}^{+}$, $10\mapsto {\mathrm{\Phi }}^{-}$, and $11\mapsto {\mathrm{\Psi }}^{-}$. In the last case, $Z_A{X}_A\mathrm{\mid }00⟩=-\mathrm{\mid }10⟩$ and $Z_A{X}_A\mathrm{\mid }11⟩=\mathrm{\mid }01⟩$, so the state is exactly ${\mathrm{\Psi }}^{-}$ in the chosen convention.

For $x\in \{0,1\}$, [F4] gives $Z^a\mathrm{\mid }x⟩=(-1{)}^{ax}\mathrm{\mid }x⟩$. Expanding the left side and then substituting $y=x\oplus b$ gives $(Z^a{X}^b\otimes I)\mathrm{\mid }{\mathrm{\Phi }}^{+}⟩=2^{-1\mathrm{/}2}{\sum }_x(-1{)}^{a(x\oplus b)}\mathrm{\mid }x\oplus b,x⟩=2^{-1\mathrm{/}2}{\sum }_y(-1{)}^{ay}\mathrm{\mid }y,y\oplus b⟩=(I\otimes X^b{Z}^a)\mathrm{\mid }{\mathrm{\Phi }}^{+}⟩$. The same matrices give $XZ=-ZX$, hence $X^b{Z}^a=(-1{)}^{ab}{Z}^a{X}^b$. The scalar $\lambda =(-1{)}^{ab}$ leaves the rank-one projector unchanged because $(\lambda v)(\lambda v{)}^{\ast }=\mathrm{\mid }\lambda {\mathrm{\mid }}^{2}v{v}^{\ast }=v{v}^{\ast }$; it changes no projective outcome weight, while the direct four-case vectors in step 1.1 fix the exact label phase.

By [F3], the four encoded vectors in step 1.1 are $B\mathrm{\mid }ab⟩$, and $B^{†}$ sends each to $\mathrm{\mid }ab⟩$. The output projectors are the orthogonal basis projections in [F8]. The rank-one matrix of $\mathrm{\mid }ab⟩$ has a single diagonal entry $1$ at label $(a,b)$, so its quadratic form is nonnegative and its trace is one by [F7]; it is a density operator by [F9]. Then [F5] and [F7] give weight one at $(a,b)$ and zero at the other labels. By [F6] this is a deterministic output law.

The procedure [F1] begins with one shared pair, sends only wire $A$ once, and outputs the two input bits at Bob's measurement. Its instructions contain at most two conditional Pauli gate occurrences for Alice's encoding, one CNOT and one Hadamard for Bob, and two final measured wires. Thus it achieves the stated resource exchange for this procedure; no optimality statement is used.

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.

By [F2], each of the four possible labels has probability $1\mathrm{/}4$, so every branch has positive probability and the conditional branch is defined. The Bell expansion [F6] identifies its receiver vector as $X^b{Z}^a\mathrm{\mid }\psi ⟩$.

Applying the correction from [F1] to the byproduct in step 1.1 gives $Z^a{X}^b{X}^b{Z}^a\mathrm{\mid }\psi ⟩=\mathrm{\mid }\psi ⟩$ by [F3]. Hence every branch has the exact corrected output.

For a joint input with any finite reference, the same branchwise output statement is the conclusion of [F4]. This verifies the protocol preserves the full input-reference vector, not only its reduced input state.

The resource record [F5] separates one preshared Bell-pair use, the two classical bits sent from Alice to Bob, zero qubits sent after the pair is available, the two Alice gates, the two-bit Alice measurement record, and up to two conditional Bob Pauli gates. These are counts of the stated procedure only; no lower bound or physical preparation claim follows.