Quantum Teleportation

Derivation

Quantum Teleportation

D-407 Home PU-404 Threads chance · matter · symmetry Depends on The No-Cloning Theorem, born-rule, Reduced States and the Partial Trace

Statement

Given one qubit in an unknown pure state \( |\psi\rangle = \alpha|0\rangle + \beta|1\rangle \) held by Alice, together with a shared maximally entangled Bell pair \( |\Phi^+\rangle_{AB} = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle) \) split between Alice and Bob, we derive that Alice can transfer \( |\psi\rangle \) to Bob exactly by performing one Bell-basis measurement on her two qubits and sending Bob the two resulting classical bits, upon which Bob applies a Pauli correction. No copy of \( |\psi\rangle \) remains with Alice and the amplitudes \( \alpha,\beta \) are never learned, so the protocol is fully consistent with no-cloning.

Why it matters

Teleportation is the operational cornerstone of quantum information: it shows that shared entanglement plus classical communication can substitute for physically sending a quantum system through space. It underlies quantum repeaters, gate teleportation in fault-tolerant computing, and the transfer of quantum states between distant nodes of a network.

Conceptually it sharpens what "unknown quantum information" means. The state moves without \( \alpha \) or \( \beta \) ever being measured or duplicated, and without the amplitudes travelling faster than light: the two classical bits are the causal bottleneck. It is the clean case study for how no-cloning, the Born rule, and entanglement coexist.

Assumptions

The shared pair is a pure, maximally entangled Bell state \( |\Phi^+\rangle \).If the pair is mixed or only partially entangled, the corrected output is a mixed state and fidelity falls below one; perfect transfer requires one full ebit of entanglement.

Alice can perform a projective measurement in the two-qubit Bell basis.Without a joint entangling measurement (e.g. only local product-basis measurements) the four outcomes cannot be distinguished, Bob's conditional states cannot be sorted, and no deterministic correction exists.

A noiseless classical channel delivers the two bits, and all gates and measurements are ideal.If the two classical bits are lost, Bob holds \( \tfrac{1}{2}\mathbf{1} \) (maximally mixed) and recovers nothing; gate or measurement noise degrades fidelity below one.

The unknown qubit enters uncorrelated with any third system, in a definite if unknown pure state.If \( |\psi\rangle \) is itself entangled with a reference, the same algebra still teleports it faithfully (entanglement swapping), but the transported "state" must then be described by the joint reference correlations, not a lone Bloch vector.

Derivation

\[ |\Psi_0\rangle = |\psi\rangle_1 \otimes |\Phi^+\rangle_{AB} = \big(\alpha|0\rangle_1 + \beta|1\rangle_1\big) \otimes \tfrac{1}{\sqrt{2}}\big(|00\rangle_{AB} + |11\rangle_{AB}\big) \]

Write the full three-qubit input as a tensor product of the unknown qubit (label 1) and the shared Bell pair (Alice's half \(A\), Bob's half \(B\)); product form because the systems are prepared independently. A

\[ |\Psi_0\rangle = \tfrac{1}{\sqrt{2}}\Big( \alpha|0\rangle_1|0\rangle_A|0\rangle_B + \alpha|0\rangle_1|1\rangle_A|1\rangle_B + \beta|1\rangle_1|0\rangle_A|0\rangle_B + \beta|1\rangle_1|1\rangle_A|1\rangle_B \Big) \]

Expand the outer product by distributing over the four basis terms; pure bookkeeping of the tensor algebra. A

\[ |\Phi^\pm\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle), \qquad |\Psi^\pm\rangle = \tfrac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle) \]

Introduce the Bell basis for Alice's two qubits (labels \(1\) and \(A\)); these four orthonormal states span the two-qubit space and are the natural basis for her joint measurement. A

\[ |00\rangle = \tfrac{1}{\sqrt{2}}(|\Phi^+\rangle + |\Phi^-\rangle),\quad |11\rangle = \tfrac{1}{\sqrt{2}}(|\Phi^+\rangle - |\Phi^-\rangle),\quad |01\rangle = \tfrac{1}{\sqrt{2}}(|\Psi^+\rangle + |\Psi^-\rangle),\quad |10\rangle = \tfrac{1}{\sqrt{2}}(|\Psi^+\rangle - |\Psi^-\rangle) \]

Invert the Bell definitions to express the computational basis of qubits \(1,A\) in the Bell basis. Legal because the Bell states are a fixed unitary rotation of the product basis. B

\[ |\Psi_0\rangle = \tfrac{1}{2}\Big[\, |\Phi^+\rangle_{1A}\big(\alpha|0\rangle_B + \beta|1\rangle_B\big) + |\Phi^-\rangle_{1A}\big(\alpha|0\rangle_B - \beta|1\rangle_B\big) + |\Psi^+\rangle_{1A}\big(\alpha|1\rangle_B + \beta|0\rangle_B\big) + |\Psi^-\rangle_{1A}\big(\alpha|1\rangle_B - \beta|0\rangle_B\big) \,\Big] \]

Substitute Step 4 into Step 2 and collect terms by Alice's Bell state; the two \(\tfrac{1}{\sqrt{2}}\) prefactors combine to \(\tfrac{1}{2}\). This regrouping of the same superposition is the algebraic heart of the protocol. B

\[ \alpha|0\rangle+\beta|1\rangle = \mathbf{1}|\psi\rangle,\quad \alpha|0\rangle-\beta|1\rangle = \hat{Z}|\psi\rangle,\quad \alpha|1\rangle+\beta|0\rangle = \hat{X}|\psi\rangle,\quad \alpha|1\rangle-\beta|0\rangle = \hat{X}\hat{Z}|\psi\rangle \]

Recognise each of Bob's four conditional kets as a Pauli acting on the original \( |\psi\rangle \), using \( \hat{Z}|0\rangle=|0\rangle,\ \hat{Z}|1\rangle=-|1\rangle,\ \hat{X}|0\rangle=|1\rangle,\ \hat{X}|1\rangle=|0\rangle \). Definitional identification. A

\[ |\Psi_0\rangle = \tfrac{1}{2}\Big[\, |\Phi^+\rangle_{1A}\,\mathbf{1}|\psi\rangle_B + |\Phi^-\rangle_{1A}\,\hat{Z}|\psi\rangle_B + |\Psi^+\rangle_{1A}\,\hat{X}|\psi\rangle_B + |\Psi^-\rangle_{1A}\,\hat{X}\hat{Z}|\psi\rangle_B \,\Big] \]

Rewrite Step 5 with the identifications of Step 6. Bob's half now carries \( |\psi\rangle \) up to a known Pauli, correlated one-to-one with Alice's Bell state. A

\[ P(m) = \big\| \langle m|_{1A}\,\Psi_0\rangle \big\|^2 = \Big\| \tfrac{1}{2}\hat{U}_m|\psi\rangle \Big\|^2 = \tfrac{1}{4}\langle\psi|\hat{U}_m^\dagger\hat{U}_m|\psi\rangle = \tfrac{1}{4}, \qquad m\in\{\Phi^+,\Phi^-,\Psi^+,\Psi^-\} \]

Apply the Born rule: the probability of Bell outcome \(m\) is the squared norm of the projected state. Each \( \hat{U}_m\in\{\mathbf{1},\hat Z,\hat X,\hat X\hat Z\} \) is unitary, so \( \hat U_m^\dagger \hat U_m=\mathbf 1 \), \( \langle\psi|\psi\rangle=1 \), and all four outcomes are equally likely, independent of \( \alpha,\beta \). C

\[ \rho_B = \mathrm{Tr}_{1A}\,|\Psi_0\rangle\langle\Psi_0| = \tfrac{1}{4}\sum_{m} \hat{U}_m |\psi\rangle\langle\psi| \hat{U}_m^\dagger = \tfrac{1}{2}\mathbf{1} \]

Trace out Alice's qubits (reduced density matrix via partial trace) before any classical bits arrive. The four Pauli-conjugated projectors sum to the maximally mixed state \( \tfrac12\mathbf1 \), so Bob learns nothing until he receives the bits: this is what forbids superluminal signalling. C

\[ \text{Alice measures } m \ \Rightarrow\ \text{Bob's collapsed state } \hat{U}_m|\psi\rangle; \qquad \text{encode } m \to (c_1 c_0)\in\{00,01,10,11\},\ \text{send 2 bits.} \]

Alice's Bell measurement collapses the joint state onto one term of Step 7, leaving Bob's qubit in \( \hat U_m|\psi\rangle \). Assign two classical bits to label which \(m\) occurred and transmit them. A

\[ \hat{U}_m^{-1}\,\hat{U}_m|\psi\rangle = |\psi\rangle, \qquad \hat{U}_m^{-1}\in\{\mathbf{1},\ \hat{Z},\ \hat{X},\ \hat{Z}\hat{X}\}\ \text{keyed by } (c_1 c_0), \quad c_1 \to \hat X,\ c_0 \to \hat Z \]

Bob applies the inverse Pauli selected by the two bits. Since each Pauli is its own inverse (\( \hat X^2=\hat Z^2=\mathbf1 \)), the correction exactly undoes \( \hat U_m \), regardless of the unknown \( \alpha,\beta \). A

\[ |\psi\rangle_B = \alpha|0\rangle_B + \beta|1\rangle_B, \qquad \rho_1^{\text{final}} = \tfrac{1}{2}\mathbf{1} \]

After correction Bob's qubit is exactly the original state, while Alice's post-measurement qubit is left maximally mixed: the input is destroyed at the source. One instance, not two, so no-cloning is respected. A

Result

\[ |\psi\rangle_1 \otimes |\Phi^+\rangle_{AB} \ \xrightarrow{\ \text{Bell meas.} + 2\ \text{bits} + \hat{U}_m^{-1}\ }\ |\psi\rangle_B, \qquad P(m)=\tfrac14,\quad F=1 \]

Reading. An unknown qubit is transported perfectly (fidelity \(F=1\)) using exactly one shared ebit and two classical bits. Each of the four measurement outcomes is equally likely with probability \(\tfrac14\) and carries no information about \( \alpha,\beta \); the state is reconstructed at Bob only after his Pauli correction. The original at Alice is left maximally mixed, so exactly one copy exists at every instant \(-\) consistent with no-cloning \(-\) and because the classical bits travel no faster than light, no signalling is possible.

Units check. All quantities are dimensionless. Amplitudes satisfy \( |\alpha|^2+|\beta|^2 = 1 \) (probability, dimensionless). Probabilities \( P(m)=\tfrac14 \) sum to \( \sum_m P(m)=1 \). The resource ledger balances: 1 qubit transported \(=\) 1 ebit \(+\) 2 classical bits (each an information count, no physical dimension). Fidelity \( F=|\langle\psi|\psi\rangle|^2=1 \) is a pure number.

Limiting cases

Outcome \( m=\Phi^+ \) (bits \(00\)): \( \hat U_m=\mathbf1 \), Bob's qubit is already \( |\psi\rangle \) and no correction is needed \(-\) happens \(\tfrac14\) of the time.
Basis states \( \alpha=1,\beta=0 \): \( |\psi\rangle=|0\rangle \) still teleports; the four conditional states become \( |0\rangle,|0\rangle,|1\rangle,|1\rangle \), corrected identically. The protocol never uses the values of \( \alpha,\beta \).
Reference-entangled input (entanglement swapping): if qubit 1 is half of a Bell pair with a reference \(R\), the same steps leave \(R\) entangled with Bob instead, faithfully "moving" one end of the entanglement.
Discarding the bits: averaging over unknown \(m\) returns \( \rho_B=\tfrac12\mathbf1 \), recovering the no-signalling reduced state of Step 9.

Breaks when

Partial or mixed entanglement. If the shared pair is \( \sqrt{p}\,|00\rangle+\sqrt{1-p}\,|11\rangle \) with \( p\neq\tfrac12 \), or a Werner state, the four branches no longer differ by a unitary alone; Bob's corrected state is mixed and \( F<1 \). Perfect teleportation strictly requires one maximally entangled ebit.
Lost or corrupted classical bits. Without the two bits Bob cannot choose \( \hat U_m^{-1} \); his best description is \( \tfrac12\mathbf1 \), fidelity \( \tfrac12 \) on average \(-\) no better than a guess. The classical channel is indispensable and enforces the light-speed limit.
No joint Bell measurement available. If Alice can only measure qubits \(1\) and \(A\) separately in product bases, she cannot resolve \( \Phi^\pm \) from \( \Psi^\pm \) coherently; the outcome labels become ambiguous and no single deterministic correction restores \( |\psi\rangle \).
Decoherence or gate noise. Dephasing/depolarising of the stored qubits or imperfect CNOT/Hadamard in the Bell measurement injects errors so that the transported state is \( \mathcal{E}(|\psi\rangle\langle\psi|) \) with \( F<1 \).

Failure modes

"Teleportation is faster than light." The corrected state only exists after the two classical bits arrive; before that Bob has \( \tfrac12\mathbf1 \). The protocol is bounded by \(c\).
"The original is copied." Alice's qubit collapses to a Pauli-random mixed state under the Bell measurement; it is not a second copy of \( |\psi\rangle \). Believing otherwise contradicts no-cloning.
Wrong correction order or Pauli assignment. Swapping which bit triggers \( \hat X \) versus \( \hat Z \), or applying \( \hat X\hat Z \) instead of \( \hat Z\hat X=(\hat X\hat Z)^{-1} \), introduces a stray phase or bit flip. \( \hat Z\hat X = -\hat X\hat Z \), but the global phase is unobservable only if both are handled consistently.
Assuming Alice must know \( |\psi\rangle \). She never measures it in the computational basis; doing so would collapse it and destroy the amplitudes. The Bell measurement is joint and reveals only \(m\).
Forgetting to normalise the Bell substitution. Dropping a \( \tfrac{1}{\sqrt2} \) in Step 4 gives probabilities that fail to sum to one; every branch must carry weight \( \tfrac14 \).
Confusing \( |\Psi^+\rangle \) and \( |\Phi^+\rangle \) branches. Mislabelling swaps the bit-flip and identity corrections, sending \( |\psi\rangle\to\hat X|\psi\rangle \).

Discussion

The physical content lies entirely in Step 5: the fixed three-qubit state, when re-expanded in Alice's Bell basis, factorises so that each of her four measurement outcomes leaves Bob's qubit in a definite Pauli image of \( |\psi\rangle \). Nothing about \( \alpha,\beta \) is ever extracted; the amplitudes ride along inside the entanglement and are decoded only by the correction. Teleportation thus separates the "quantum" part of the information (carried by the pre-shared ebit) from the "classical" part (the two bits), which is why the resource identity 1 qubit \(=\) 1 ebit \(+\) 2 cbits is exact and cannot be beaten \(-\) two bits are both necessary and sufficient.

The Born-rule calculation (Step 8) and the partial trace (Step 9) together enforce consistency with relativity. Because \( \rho_B=\tfrac12\mathbf1 \) is independent of \( \alpha,\beta \) and of which measurement Alice performs, Bob's local statistics cannot change until the classical message arrives. This is the no-signalling theorem realised concretely: entanglement alone transmits no information, and only the light-speed-limited bits complete the transfer.

No-cloning is respected by an exact accounting. The linear map that would output \( |\psi\rangle|\psi\rangle \) for arbitrary \( |\psi\rangle \) is impossible; teleportation instead moves the single instance, annihilating it at the source. The measurement is the destructive step, and its randomness (four equal outcomes) is precisely what hides the amplitudes and prevents duplication. Teleportation and no-cloning are two faces of the same linearity constraint.

Teleportation is dual to superdense coding under the exchange of quantum and classical resources: there, one ebit plus one transmitted qubit conveys two classical bits; here, one ebit plus two classical bits conveys one qubit. Both saturate the same resource inequalities. Gate teleportation generalises the protocol by pre-applying a unitary to the entangled resource, so that measuring in a rotated Bell basis implements a logical gate on the teleported qubit \(-\) the basis of magic-state injection and fault-tolerant \(T\)-gates in surface codes. The entanglement-swapping limit turns the same algebra into the elementary link of a quantum repeater.

Common misconceptions. Teleportation does not move matter or energy \(-\) only the quantum state; Bob's qubit was already present. It does not clone \(-\) the source is destroyed. It is not instantaneous \(-\) it waits for two classical bits. And Alice never learns the state she sends.

Worked examples

\[ |\psi\rangle = \cos\tfrac{\theta}{2}\,|0\rangle + \sin\tfrac{\theta}{2}\,|1\rangle,\quad \theta=\tfrac{\pi}{3} \ \Rightarrow\ \alpha=\cos 30^\circ=\tfrac{\sqrt3}{2},\ \ \beta=\sin 30^\circ=\tfrac12 \]

Take a definite unknown-to-Alice state on the Bloch sphere; \( |\alpha|^2+|\beta|^2 = \tfrac34+\tfrac14 = 1 \). A

\[ \text{Suppose Alice's Bell outcome is } m=\Psi^-\ \Rightarrow\ (c_1 c_0)=11,\quad \text{Bob holds } \hat X\hat Z|\psi\rangle = \tfrac{\sqrt3}{2}|1\rangle - \tfrac12|0\rangle \]

From Step 7, outcome \( \Psi^- \) leaves Bob in \( \hat X\hat Z|\psi\rangle \): \( \hat Z \) flips the sign of \( \beta \), then \( \hat X \) swaps the labels. B

\[ \hat U_m^{-1}=\hat Z\hat X:\quad \hat X\big(\tfrac{\sqrt3}{2}|1\rangle-\tfrac12|0\rangle\big)=\tfrac{\sqrt3}{2}|0\rangle-\tfrac12|1\rangle,\quad \hat Z\big(\tfrac{\sqrt3}{2}|0\rangle-\tfrac12|1\rangle\big)=\tfrac{\sqrt3}{2}|0\rangle+\tfrac12|1\rangle \]

Apply the inverse correction keyed by \(11\): first \( \hat X \), then \( \hat Z \). Each is its own inverse, so the composition undoes \( \hat X\hat Z \). B

\[ |\psi\rangle_B = \tfrac{\sqrt3}{2}|0\rangle + \tfrac12|1\rangle = |\psi\rangle \quad\checkmark \]

Reading. Bob recovers the exact input, fidelity \( F=|\langle\psi|\psi\rangle|^2=1 \). Units: amplitudes dimensionless, \( \tfrac34+\tfrac14=1 \).

\[ |\psi\rangle = \tfrac{1}{\sqrt2}\big(|0\rangle + i|1\rangle\big)\quad(\alpha=\tfrac{1}{\sqrt2},\ \beta=\tfrac{i}{\sqrt2},\ \text{the }+y\text{ eigenstate}) \]

A complex-amplitude state tests that phases survive teleportation; \( |\alpha|^2+|\beta|^2=\tfrac12+\tfrac12=1 \). A

\[ \text{Suppose } m=\Phi^-\ \Rightarrow\ (c_1 c_0)=01,\quad \text{Bob holds } \hat Z|\psi\rangle = \tfrac{1}{\sqrt2}\big(|0\rangle - i|1\rangle\big) \]

Outcome \( \Phi^- \) applies \( \hat Z \), flipping the sign of the \( |1\rangle \) amplitude (the \(+y\) state becomes \(-y\)). B

\[ \hat U_m^{-1}=\hat Z:\quad \hat Z\cdot\tfrac{1}{\sqrt2}\big(|0\rangle - i|1\rangle\big) = \tfrac{1}{\sqrt2}\big(|0\rangle + i|1\rangle\big) \]

Bit \( c_0=1 \) triggers a single \( \hat Z \); since \( \hat Z^2=\mathbf1 \) the phase is restored exactly. No \( \hat X \) because \( c_1=0 \). A

\[ |\psi\rangle_B = \tfrac{1}{\sqrt2}\big(|0\rangle + i|1\rangle\big) = |\psi\rangle \quad\checkmark \]

Reading. The complex phase \(i\) is transported faithfully; had Bob skipped the \( \hat Z \), he would hold the orthogonal \(-y\) state, \( F=|\langle +y|-y\rangle|^2=0 \). Units: dimensionless, norm \(=1\).

Problems

Show that for an arbitrary \( |\psi\rangle=\alpha|0\rangle+\beta|1\rangle \), the probability of Alice's outcome \( \Phi^+ \) is exactly \( \tfrac14 \), independent of \( \alpha,\beta \).
Solution
From Step 7 the \( \Phi^+ \) branch is \( \tfrac12|\Phi^+\rangle_{1A}\,\mathbf1|\psi\rangle_B \). By the Born rule \( P(\Phi^+)=\|\tfrac12\mathbf1|\psi\rangle\|^2=\tfrac14\langle\psi|\psi\rangle=\tfrac14(|\alpha|^2+|\beta|^2)=\tfrac14 \). Independent of the amplitudes because \( \mathbf1 \) is unitary and the state is normalised.
Alice's outcome is \( \Psi^+ \) (bits \(10\)). Write Bob's uncorrected state and the correction he must apply, then verify recovery for \( |\psi\rangle=\tfrac{\sqrt3}{2}|0\rangle+\tfrac12|1\rangle \).
Solution
\( \Psi^+ \) leaves Bob in \( \hat X|\psi\rangle=\tfrac{\sqrt3}{2}|1\rangle+\tfrac12|0\rangle=\tfrac12|0\rangle+\tfrac{\sqrt3}{2}|1\rangle \). Correction \( \hat U_m^{-1}=\hat X \) (bit \(c_1=1\) only): \( \hat X(\tfrac12|0\rangle+\tfrac{\sqrt3}{2}|1\rangle)=\tfrac12|1\rangle+\tfrac{\sqrt3}{2}|0\rangle=\tfrac{\sqrt3}{2}|0\rangle+\tfrac12|1\rangle=|\psi\rangle \). Recovered, \( F=1 \).
Compute the reduced state \( \rho_B \) that Bob holds before receiving the classical bits, and use it to argue teleportation cannot signal faster than light.
Solution
Averaging over the four equally likely outcomes, \( \rho_B=\tfrac14\sum_m \hat U_m|\psi\rangle\langle\psi|\hat U_m^\dagger \). Writing \( |\psi\rangle\langle\psi|=\tfrac12(\mathbf1+\vec r\cdot\vec\sigma) \), the four Paulis map \( \vec r\to(\vec r,\ (r_x,-r_y,-r_z),\ (-r_x,r_y,-r_z)\ \text{etc.}) \) and the Bloch vectors cancel, leaving \( \rho_B=\tfrac12\mathbf1 \). Since this is independent of \( \alpha,\beta \) and of Alice's actions, Bob's local measurement statistics are unchanged until the light-speed-limited bits arrive: no signalling.
A student receives bits \(11\) but applies the correction in the order \( \hat X \) then... forgets \( \hat Z \), applying only \( \hat X \). For \( |\psi\rangle=\cos\tfrac{\theta}{2}|0\rangle+\sin\tfrac{\theta}{2}|1\rangle \), what fidelity does the student obtain?
Solution
Correct state after full \( \hat Z\hat X \) is \( |\psi\rangle \). Applying only \( \hat X \) to Bob's raw \( \hat X\hat Z|\psi\rangle \) gives \( \hat X\hat X\hat Z|\psi\rangle=\hat Z|\psi\rangle=\cos\tfrac{\theta}{2}|0\rangle-\sin\tfrac{\theta}{2}|1\rangle \). Fidelity \( F=|\langle\psi|\hat Z|\psi\rangle|^2=(\cos^2\tfrac{\theta}{2}-\sin^2\tfrac{\theta}{2})^2=\cos^2\theta \). For \( \theta=\tfrac{\pi}{3} \), \( F=\cos^2 60^\circ=\tfrac14 \). Only \( \theta=0,\pi \) (basis states) escape the error.
Show that the resource cost is optimal: prove two classical bits are necessary, i.e. one bit cannot suffice for deterministic perfect teleportation of an arbitrary qubit.
Solution
One classical bit selects among only two correction unitaries. Alice's Bell measurement yields four equally likely outcomes (Problem 1), each requiring a distinct correction from \( \{\mathbf1,\hat Z,\hat X,\hat X\hat Z\} \) to undo a different \( \hat U_m \). With only two labels, at least two of the four outcomes must share a correction; for those, one branch is left with a residual non-identity Pauli \( \hat U_m\hat U_{m'}^{-1}\neq\mathbf1 \) acting on \( |\psi\rangle \), giving \( F<1 \) for generic \( |\psi\rangle \). Hence two bits (four labels) are necessary; Step 11 shows they are sufficient. This also matches the superdense-coding dual, where one qubit carries at most two bits.