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The Bell basis diagonalizes the commuting Pauli operators $Z\otimes Z$ and $X\otimes X$. Two local anticommutations make the joint operators commute. Their signs encode parity and relative phase. The displayed projectors turn the signed eigenvalues into literal zero-or-one bit eigenvalues: even parity and plus phase have bit zero.

Original 1997 paper introducing the magic-basis concurrence description used here. Primary preprint: arxiv.org/abs/quant-ph/9703041 . Full text: arxiv.org/pdf/quant-ph/9703041 . Eqs.(3)–(6) specify the phased Bell basis and its pure-state entropy relation.

A Bell basis with phases chosen so that local special-unitary transformations of the two qubits are represented by real four-dimensional rotations. For one standard ordering it is ${\Phi }_{+},i{\Phi }_{-},i{\Psi }_{+},{\Psi }_{-}$. Permuting these vectors changes generator conventions but preserves the invariant concurrence $|{\sum }_j{b}_j^2|$.^[ref]

Normalization requires $2p+2q=1$, so $q=1/2-p$ and $0\le p\le 1/2$. With logarithms to base two, the Shannon entropy isThe convention $0\log 0=0$ gives $H=1$ at both endpoints. In the interior,Hence the unique maximum is $H=2$ bits at $p=q=1/4$. The curve is concave and symmetric about $p=1/4$.

Choose the message assignment $x_1\leftrightarrow {\Psi }^{+}$, $x_2\leftrightarrow {\Psi }^{-}$, $x_3\leftrightarrow {\Phi }^{+}$ and $x_4\leftrightarrow {\Phi }^{-}$. Starting from $|{\Psi }^{+}⟩$, apply respectively $I,Z,X,XZ$ to the first qubit. This produces those four orthonormal Bell states, up to a physically irrelevant overall sign in the last case. The ensemble density operator isin computational order $00,01,10,11$, where ${\Pi }_B=|B⟩⟨B|$. The cancellation of off-diagonal terms follows from the equal probabilities within each sign pair; the individual encoded states remain entangled.

A Bell basis measurement reads the label perfectly because the four code states are orthogonal. Thus this is equivalent to a classical error-free encoding when unrestricted joint measurement is allowed. Each conditional encoded state is pure, soThe second formula is the classical-quantum mutual information, here equal to attainable classical information. The first says there is no uncertainty about the joint pure code state once the message is known; it does not say that measuring its individual qubits has deterministic outcomes.

In superdense coding, Alice and Bob initially share $|{\Psi }^{+}{⟩}_{AB}$. Alice selects the same one-qubit Pauli gate $I,Z,X,XZ$ according to her message and sends her single encoded qubit to Bob. He now has both halves and measures in the Bell basis. For the chosen assignment,which reaches two bits for the uniform distribution. A circuit implementation first converts the resource convention with $X_B$, then applies ${\mathrm{CNOT}}_{A\to B}$ and $H_A$, and measures both in the computational basis. The four outcomes are distinct; relabeling them restores the selected message order.

Before Alice's transmission, Bob's reduced density operator is $I/2$ for every choice of her local operation. Thus distributing the message-independent entangled resource has transmitted no information about the subsequently chosen message. It is nevertheless a real physical resource: the two-bit rate counts the one additional transmitted qubit after entanglement has been shared, not a free initial distribution. The deterministic conditional probabilities above establish perfect correlation rather than merely matching output and input entropies.

Let $Q$ be Alice's unknown qubit, with pure state $|\varphi {⟩}_Q=\alpha |0⟩+\beta |1⟩$. Alice and Bob initially share the Bell state $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Alice holds $Q,A$ and Bob holds $B$. Define the Bell basisExpanding in this orthonormal basis gives the Bell-basis teleportation identityFor example, the last branch on $B$ is $\alpha |1⟩-\beta |0⟩$, which is $XZ|\varphi ⟩$; the ordering of the Pauli operators therefore matters.

Alice performs a projective measurement in the Bell basis on her two local qubits. Each outcome has probability $1/4$, independent of the unknown amplitudes. She encodes the four possible outcomes as two classical bits and sends them to Bob. For outcomes ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$, respectively, Bob applies the local unitary operationsThese are inverses of the corresponding branch Pauli operators, so Bob obtains exactly the unknown input qubit in every branch. Using $XZ$ instead of $ZX$ as the last correction also succeeds up to an irrelevant overall phase. Alice need not know $\alpha$ or $\beta$. Every quantum operation is local to one of the parties, and the only transmitted information is the two-bit classical outcome. The shared Bell pair is consumed.

Before the message arrives, Bob's density operator isso no unknown-state information is available without the classical communication. Alice's original qubit has been included in the destructive Bell-basis measurement; the protocol creates no extra copy. Since the corrected transformation is the identity on every input vector and is linear, it also preserves entanglement with an external reference and transfers mixed inputs. Thus this is deterministic quantum teleportation, using precisely one shared maximally entangled Bell pair, two classical bits, and local operations.

Apply a Hadamard gate to qubit $A$, followed by a controlled-NOT gate with $A$ controlling $B$. The resulting Bell-basis conversion circuit is $U=\mathrm{CNOT}(H\otimes I)$ and givesThus $00,01,10,11$ map respectively to ${\Phi }^{+},{\Psi }^{+},{\Phi }^{-},{\Psi }^{-}$. The first input bit becomes the phase bit and the second becomes the parity bit.

Both gates are self-inverse, but inversion of their product reverses the order:So the same two gates convert the Bell basis back to the computational basis when applied in reverse order: first the controlled-NOT, then the Hadamard. Using the forward order twice does not generally work, since these gates do not commute. For example, $U|{\Phi }^{+}⟩=(|00⟩+|01⟩-|10⟩+|11⟩)/2$, which is not a computational-basis state.

The Bell basis consists ofLet $Z={\sigma }_z$, $X={\sigma }_x$. The Pauli operators $Z\otimes Z$ and $X\otimes X$ commute: each of the two local anticommutations contributes a minus sign, so they cancel. Their eigenvalue table isTo make the bit values literally zero or one rather than signed eigenvalues, the commuting Bell parity and phase observables areThey distinguish all four states jointly. The parity bit records whether the computational bits agree; the phase bit records the relative sign.

Write the arbitrary joint pure state of the input $A$ and an external reference $R$ asThe reference vectors need not be normalized or orthogonal. Share $|{\Phi }^{+}{⟩}_{aB}$. Label Alice's Bell states byProjecting onto this Bell basis gives the unnormalized stateEach outcome has probability $1/4$. Alice sends $(r,s)$ and Bob applies $Z_B^r{X}_B^s$, reversing the two Pauli gates. The final state is exactly $|\Omega {⟩}_{BR}$ for every outcome. In particular all input entanglement and other correlations with $R$ now belong to $B$, while the reference marginal is unchanged. This is teleportation as an identity channel on a reference.

Conditioned on the record, Alice's two measured carriers are in $|{\beta }_{rs}⟩$ and factor from $BR$. Thus $A$ no longer retains its original correlations with $R$. There is no second copy, even when the teleported qubit was initially entangled with a larger system.

In quantum teleportation, Alice transfers an arbitrary unknown qubit $|\chi ⟩=\alpha |0⟩+\beta |1⟩$ using one shared Bell pair and two bits of classical communication. Define $|{\beta }_{mn}⟩=(I\otimes X^n{Z}^m)|{\Phi }^{+}⟩$, with $m,n\in \{0,1\}$. Regrouping the initial three-qubit state into this Bell basis givesAlice's Bell-basis measurement selects each label with probability $1/4$. She sends $(m,n)$ to Bob, who applies the inverse $Z^m{X}^n$, restoring $|\chi ⟩$. The correction depends on the measurement label, not on the unknown amplitudes. Before the message, Bob's average state is $I/2$, independent of the input; quantum no-signalling therefore prevents superluminal communication. The shared pair is consumed and Alice does not retain an independent copy, in agreement with the no-cloning theorem.

The Bouwmeester quantum teleportation experiment used photon polarization, a down-conversion entangled pair, a beam-splitter Bell-state analyzer and coincidence-conditioned output polarization measurements. Temporal indistinguishability of the two analyzer photons was essential. Its successful selected Bell outcome demonstrated the state transfer, rather than a deterministic implementation of all four outcomes.^[ref]

The general experimental tasks are to distribute entanglement, prepare and synchronize the input, perform a sufficiently discriminating joint measurement, communicate the label and apply or account for the correction. A demonstration must test nonorthogonal input states, not merely transfer a known classical basis label. Finite efficiency, accidental coincidences, distinguishability and phase drift reduce success probability or teleportation fidelity. Postselected verification, a heralded usable output and unconditional deterministic teleportation are distinct experimental claims.

The specified Bell-state phases giveThereforeThe sum contains complex squares, not squared moduli. Define $q={\sum }_j{b}_j^2$ and $\mathcal{C}=|q|$. Normalization gives ${\sum }_j|b_j{|}^2=1$, so the triangle inequality ensures $0\le \mathcal{C}\le 1$.

The characteristic polynomial is ${\lambda }^2-\lambda +{\mathcal{C}}^2/4$. HenceHere logarithms are natural and $0\ln 0$ means its limiting value zero. Using logarithms to base two gives entropy in bits instead. Product states have $\mathcal{C}=0$, spectrum $(1,0)$ and entropy zero; maximally entangled states have $\mathcal{C}=1$, spectrum $(1/2,1/2)$ and entropy $\ln 2$.

This phased Bell basis is a magic basis, and $\mathcal{C}$ is the pure-state concurrence. The paper's ordering is a permutation of the usual magic-basis ordering, which leaves $|{\sum }_j{b}_j^2|$ unchanged.^[ref]

In the printed phased Bell basis, the actions on its four kets areEach matrix is purely imaginary and antisymmetric. Let $T$ have these four basis kets as columns in the computational basis. The infinitesimal generators of the standard rotation $U_A=\exp [-i\theta \mathbf{n}\cdot {\mathbf{\sigma }}_A/2]$ areUsing the displayed actions gives the phase-convention-specific identitiesThey are real antisymmetric matrices. A finite $SU(2)$ rotation on $A$ therefore acts on the magic-basis coefficient vector as $b\mapsto Ob$, with $O\in SO(4)$. ConsequentlyIt preserves concurrence and hence preserves the reduced eigenvalues and entanglement entropy. A general $U(2)$ operation adds a common phase to $b$; that changes $q$ by a phase of twice the angle but leaves $|q|$ unchanged. The real-rotation description of local transformations is a characteristic property of the magic basis.^[ref]

As an independent direct check, a local unitary sends ${\rho }_A$ to $U_A{\rho }_A{U}_A^{†}$, which is a unitary similarity transformation and cannot change its spectrum. This local-unitary invariance of entanglement entropy also explains why merely rotating one subsystem cannot create entanglement.