In short

Quantum teleportation moves an unknown qubit state |\psi\rangle = \alpha|0\rangle + \beta|1\rangle from Alice to Bob without ever sending the qubit itself. The recipe: Alice and Bob share an entangled Bell pair |\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle) prepared earlier. Alice holds the unknown qubit S plus her half A of the Bell pair; Bob holds the other half B, possibly thousands of kilometres away. Alice performs a Bell-basis measurement on S and A — implemented as CNOT, then Hadamard, then measure both in the computational basis — and reads two classical bits (m_1, m_2). She phones, emails, or radio-transmits those two bits to Bob. Bob applies a Pauli correction X^{m_2} Z^{m_1} to B and his qubit is now exactly |\psi\rangle. The original in Alice's hand was destroyed at measurement, so no-cloning is respected. The two classical bits travel at the speed of light or slower, so no-faster-than-light signalling happens either. Teleportation transfers state, not matter; it does not beam atoms, and it consumes one Bell pair per qubit moved. It is the foundational routine for quantum networks, distributed quantum computing, and the quantum internet.

You have a qubit. Its state is |\psi\rangle = \alpha|0\rangle + \beta|1\rangle — some specific \alpha, \beta that together carry the information you care about. A friend in another city needs this exact quantum state. Three things are true at once.

First, you cannot measure |\psi\rangle and tell your friend what \alpha and \beta are. A measurement in the computational basis gives you either 0 (with probability |\alpha|^2) or 1 (with probability |\beta|^2) and nothing else — just one classical bit, with all of the phase and amplitude structure of the original state gone. The state has collapsed; the numbers \alpha and \beta are lost.

Second, you cannot copy |\psi\rangle and post one of the copies. The no-cloning theorem rules it out: no unitary on any ancilla can take |\psi\rangle |0\rangle to |\psi\rangle |\psi\rangle for arbitrary |\psi\rangle. The proof is three lines of linearity, and it leaves no escape.

Third, you cannot put the physical qubit on a plane and fly it over. Qubits live in superconducting circuits at 15 millikelvin, or in trapped ions held by electromagnetic fields, or as polarisation states of individual photons in optical fibre. "Putting one in an envelope" is not a supported operation — the qubit decoheres at room temperature in nanoseconds.

Teleportation solves this. Not by getting around any of the three constraints — each remains in force — but by routing around them, using entanglement as a pre-laid rail. The 1993 protocol by Bennett, Brassard, Crépeau, Jozsa, Peres, and Wootters [1] is a five-step recipe that moves the state of |\psi\rangle from Alice to Bob with no qubit in transit between them. By the end of this chapter you will know the recipe, you will have traced it line by line, you will know which physical experiments have demonstrated it, and you will know — precisely — what teleportation does and does not do.

The three ingredients

Teleportation needs three qubits, labelled S, A, B.

The Bell pair A, B is prepared in the state |\Phi^+\rangle_{AB} = \tfrac{1}{\sqrt{2}}(|00\rangle_{AB} + |11\rangle_{AB}) ahead of time, typically by someone at a central station, and the two qubits are then sent — one to Alice, one to Bob — by whatever physical means (optical fibre, free-space photon, trapped-ion transport). Once shared, the pair stays on the shelf until Alice wants to teleport.

Teleportation setupA schematic showing Alice on the left with two qubits S and A, Bob on the right with one qubit B. A wavy line labelled shared entanglement connects qubit A on Alice's side to qubit B on Bob's side. A dashed arrow labelled classical channel connects Alice to Bob, indicating the eventual transmission of two classical bits.AliceSunknown |ψ⟩Ahalf of Bell pairholds S and ABobBother half of Bell pairholds Bshared entanglement |Φ⁺⟩2 classical bits (phone, email, radio)Resources: 1 Bell pair (pre-shared) + 2 classical bits (sent during protocol).
The setup. Alice holds $S$ (unknown) and $A$ (half of a Bell pair). Bob holds $B$ (the other half). A pre-shared entangled pair links $A$ and $B$; a classical channel carries two bits from Alice to Bob during the protocol.

Write out the full three-qubit state at the starting moment. S is in |\psi\rangle; A and B are in |\Phi^+\rangle. The joint state is

|\psi\rangle_S \otimes |\Phi^+\rangle_{AB} \;=\; \bigl(\alpha|0\rangle + \beta|1\rangle\bigr)_S \otimes \tfrac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr)_{AB}.

Distribute the tensor product:

= \tfrac{1}{\sqrt{2}}\bigl(\alpha|000\rangle + \alpha|011\rangle + \beta|100\rangle + \beta|111\rangle\bigr)_{SAB}.

Why this expansion: tensor product distributes over sums. (\alpha|0\rangle + \beta|1\rangle) \otimes (|00\rangle + |11\rangle) = \alpha|0\rangle|00\rangle + \alpha|0\rangle|11\rangle + \beta|1\rangle|00\rangle + \beta|1\rangle|11\rangle, which you then write as 3-qubit kets in SAB order.

This is the state Alice will now operate on. Importantly, it is a product between S and AB — the unknown state |\psi\rangle has not yet touched the Bell pair. Teleportation works by entangling them.

The protocol — five steps

Here is the complete recipe. The circuit diagram appears below; walk through it while reading.

  1. Alice applies CNOT with S as control and A as target.
  2. Alice applies Hadamard to S.
  3. Alice measures S and A in the computational basis, recording two classical bits (m_1, m_2). Here m_1 is the outcome from measuring S and m_2 is the outcome from measuring A.
  4. Alice sends (m_1, m_2) to Bob by classical means — phone, email, radio, pigeon carrier, whatever works. The channel can be as slow as physics requires; it just has to arrive.
  5. Bob applies a Pauli correction to B: first X^{m_2} (apply X if m_2 = 1, otherwise do nothing), then Z^{m_1} (apply Z if m_1 = 1, otherwise do nothing).

After step 5, Bob's qubit B is in state |\psi\rangle — exactly the original Alice had. Alice's qubits S and A are each in some computational-basis state (|0\rangle or |1\rangle) determined by the measurement; they no longer hold any trace of |\psi\rangle.

Full teleportation circuitA three-qubit quantum circuit. Top wire S starts in state psi. Middle wire A starts in ket 0. Bottom wire B starts in ket 0. A Hadamard on A followed by CNOT from A to B prepares the Bell pair on the right of the start. Then a CNOT from S to A fires, then a Hadamard on S, then measurement meters on S and A produce classical bits m1 and m2. Two double-line classical wires carry m1 and m2 to Bob's side, where they control a Z gate and an X gate on wire B. Output wire B is labelled psi.|ψ⟩_S|0⟩_A|0⟩_BHprepare Bell pair (A,B)HBell measurement on (S, A)m₁m₂ZXif m₁=1if m₂=1|ψ⟩(pre-shared)Alice: entangle S into the pair, measureBob: Pauli correctionBob's qubit ends up in |ψ⟩ regardless of which (m₁, m₂) outcome fires.
The full teleportation circuit. The leftmost $H$ + CNOT prepares the Bell pair (done ahead of time). Then Alice's CNOT + $H$ rotates $|S, A\rangle$ into the Bell basis, followed by a standard computational-basis measurement that yields two classical bits. The bits travel as double-line classical wires to Bob's side, controlling $Z$ and $X$ corrections. Bob's output qubit is $|\psi\rangle$.

The rest of the chapter unpacks why this works — step by step, with every tensor product written out and the four measurement branches traced individually.

The derivation — one expansion at a time

Start from the three-qubit state established earlier:

|\Omega_0\rangle \;=\; \tfrac{1}{\sqrt{2}}\bigl(\alpha|000\rangle + \alpha|011\rangle + \beta|100\rangle + \beta|111\rangle\bigr)_{SAB}.

Step 1: Alice applies CNOT_{S \to A}. This flips qubit A whenever qubit S is |1\rangle. Apply term by term:

Why CNOT here: Alice wants the unknown state |\psi\rangle to entangle with her half of the Bell pair. CNOT is the simplest entangling gate. With S as the control, it begins correlating "S-value" with "A-value" — the first step toward a Bell-basis measurement on (S, A).

The state after Step 1:

|\Omega_1\rangle \;=\; \tfrac{1}{\sqrt{2}}\bigl(\alpha|000\rangle + \alpha|011\rangle + \beta|110\rangle + \beta|101\rangle\bigr)_{SAB}.

Step 2: Alice applies H to S. The Hadamard takes |0\rangle \to \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle) and |1\rangle \to \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle). Apply to each term (acting only on the first ket — the S qubit):

Collecting everything under the shared \tfrac{1}{\sqrt{2}} from |\Omega_1\rangle:

|\Omega_2\rangle \;=\; \tfrac{1}{2}\bigl[\alpha|000\rangle + \alpha|100\rangle + \alpha|011\rangle + \alpha|111\rangle + \beta|010\rangle - \beta|110\rangle + \beta|001\rangle - \beta|101\rangle\bigr].

Why the Hadamard next: CNOT alone does not produce a Bell-basis measurement. The combination CNOT-then-H is the inverse of the Bell preparation circuit (H-then-CNOT) that prepares |\Phi^+\rangle from |00\rangle. Running the inverse circuit takes the four Bell states back to the four computational-basis states of (S, A), so measuring (S, A) in the computational basis after CNOT-then-H is exactly a Bell-basis measurement on the original (S, A).

Step 3: regroup by (S, A) outcomes. Collect terms that share the same |S A\rangle value, and factor out the corresponding state on B:

|\Omega_2\rangle \;=\; \tfrac{1}{2}\Bigl[|00\rangle_{SA}\bigl(\alpha|0\rangle + \beta|1\rangle\bigr)_B + |01\rangle_{SA}\bigl(\alpha|1\rangle + \beta|0\rangle\bigr)_B + |10\rangle_{SA}\bigl(\alpha|0\rangle - \beta|1\rangle\bigr)_B + |11\rangle_{SA}\bigl(\alpha|1\rangle - \beta|0\rangle\bigr)_B\Bigr].

This is the key line of the whole derivation. Read each of the four pieces:

Why grouping works: after CNOT-then-H, the (S, A) computational basis correlates one-to-one with the pre-measurement Bell-basis labels. Each of the four (S, A) outcomes is paired with exactly one of four Pauli-transformed versions of |\psi\rangle on B. This is the structural heart of teleportation.

Step 4: measurement. Alice measures (S, A) in the computational basis. The four outcomes (m_1, m_2) = (0, 0), (0, 1), (1, 0), (1, 1) each occur with probability \tfrac{1}{4}, because each |m_1 m_2\rangle_{SA} appears with amplitude \tfrac{1}{2} in |\Omega_2\rangle, and |\tfrac{1}{2}|^2 = \tfrac{1}{4}.

Why each outcome is equally likely: the norm-squared of each branch's amplitude is \tfrac{1}{4} summed over the B-side state (which is already normalised), and this gives the measurement probability. The probabilities do not depend on \alpha or \beta — the measurement outcome is fully random and carries no information about |\psi\rangle. That is crucial: if the outcome statistics leaked anything about \alpha and \beta, Alice could send |\psi\rangle to Bob without any quantum channel, which would contradict the no-cloning theorem and destroy security of QKD.

After measurement, Alice's qubits are in the specific computational-basis state |m_1 m_2\rangle_{SA}. Bob's qubit is left in the corresponding member of the four Pauli-transformed states listed above.

Step 5: the Pauli correction. Alice sends (m_1, m_2) by classical channel. Bob receives the two bits and applies X^{m_2} Z^{m_1} to his qubit B — meaning: first apply X if m_2 = 1, then apply Z if m_1 = 1. Run through the four cases:

(m_1, m_2) Bob's state before Correction Bob's state after
(0, 0) \alpha|0\rangle + \beta|1\rangle = |\psi\rangle I |\psi\rangle
(0, 1) \alpha|1\rangle + \beta|0\rangle = X|\psi\rangle X X \cdot X|\psi\rangle = |\psi\rangle
(1, 0) \alpha|0\rangle - \beta|1\rangle = Z|\psi\rangle Z Z \cdot Z|\psi\rangle = |\psi\rangle
(1, 1) \alpha|1\rangle - \beta|0\rangle = XZ|\psi\rangle (up to sign) ZX |\psi\rangle (up to sign — a global phase)

Why the corrections always give |\psi\rangle: each Pauli is its own inverse (X^2 = I, Z^2 = I), so X applied to an X-distorted state undoes the distortion, and similarly for Z. The (1, 1) branch accumulates a global phase of -1 which is physically unobservable — Bob's qubit is |\psi\rangle in every measurable sense.

In all four branches, Bob's qubit ends up in |\psi\rangle. Teleportation complete.

Four teleportation branchesA table-like diagram. Four rows, each showing a possible measurement outcome (m1, m2) on the left, Bob's qubit state before correction in the middle, and the Pauli correction plus final state on the right. Rows show (0,0) leading to psi directly, (0,1) leading to X correction, (1,0) leading to Z correction, (1,1) leading to ZX correction. All four final states are labelled psi.(m₁, m₂)Bob's qubit B (before)applyBob's qubit B (after)(0, 0)α|0⟩ + β|1⟩ = |ψ⟩I (nothing)|ψ⟩(0, 1)α|1⟩ + β|0⟩ = X|ψ⟩X|ψ⟩(1, 0)α|0⟩ − β|1⟩ = Z|ψ⟩Z|ψ⟩(1, 1)α|1⟩ − β|0⟩ = XZ|ψ⟩Z then X|ψ⟩ (up to ±)each outcome has probability 1/4; each correction fires only when the corresponding bit is 1
The four teleportation branches. Whatever $(m_1, m_2)$ Alice reads, Bob's correction rule $X^{m_2} Z^{m_1}$ maps the distorted qubit back to the original $|\psi\rangle$.

What just happened — the bookkeeping

The three-qubit state started in |\psi\rangle_S \otimes |\Phi^+\rangle_{AB}. After the CNOT + H + measure cycle, it ended with |m_1 m_2\rangle_{SA} \otimes |\psi\rangle_B (once Bob's correction is applied). The state |\psi\rangle has moved from qubit S to qubit B. Qubit S is now in one of |0\rangle, |1\rangle — the information \alpha, \beta is not on S any more.

Resources consumed: one Bell pair (entanglement) + two classical bits (communication) = one qubit-state transfer. This is the famous resource equality known as the teleportation inequality:

\text{1 ebit} + \text{2 cbits} \;\geq\; \text{1 qubit transmission}.

"Ebit" is the standard name for one Bell pair worth of entanglement; "cbit" is one classical bit; the inequality says these resources suffice to transmit one qubit. Bennett, Brassard, and collaborators formalised this accounting in 1996 as part of the broader resource theory of entanglement.

Teleportation resource equalityA triangular diagram. Left node labelled 1 Bell pair (1 ebit). Middle node labelled 2 classical bits (2 cbits). Arrow from the combination of left and middle to the right node labelled 1 qubit transmitted. Below, a caption explains that pre-shared entanglement plus a classical channel equals a quantum channel for state transfer.1 Bell pair= 1 ebit of entanglement+2 classical bits= 2 cbits of communication1 qubitstate transferredentanglement + classical channel = quantum channelBennett–Brassard–Crépeau–Jozsa–Peres–Wootters, 1993
The resource accounting. One Bell pair (pre-shared) plus two classical bits (sent during the protocol) equals one qubit transmitted. This is the foundational equation of quantum Shannon theory.

Hype check

Teleportation is the most misreported topic in quantum computing. Pop-science has it either as "Star Trek transporters" or as "faster-than-light communication," and both are wrong. The actual story is more elegant, and more constrained, than either.

Hype check.

It does not transmit information faster than light. The two classical bits (m_1, m_2) must be sent by a normal communication channel — by radio, by fibre, by post, whatever — and none of those beats the speed of light. Until Bob receives the bits, his qubit B is, from his perspective, in a maximally-mixed state that contains no information about |\psi\rangle. The classical bits are essential, and they enforce the relativistic speed limit.

It does not violate no-cloning. At the moment of Alice's measurement, |\psi\rangle is destroyed on S — the original is gone. Bob ends up with the only remaining copy. At no instant do two copies of |\psi\rangle exist simultaneously, which is exactly what no-cloning requires.

It does not transmit matter. Only the quantum state moves. The qubit S is still in Alice's lab after teleportation (now in state |0\rangle or |1\rangle), and the qubit B is still in Bob's lab. No atom or photon travelled from Alice to Bob during the protocol. "Teleportation" is a slightly misleading name that has stuck since 1993.

It does not produce a second |\psi\rangle anywhere. One |\psi\rangle went in, one |\psi\rangle came out, the input was destroyed. The protocol is a state-swap with collateral destruction, not a state-copy.

Understanding teleportation as "entanglement as a pre-laid rail for a state transfer that relies on classical communication" is the only framing that survives close inspection. Every pop-science framing that hints at magical instantaneity has quietly dropped the two classical bits, and with them the relativistic safety mechanism.

Worked examples

Example 1: Trace all four branches for $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$

Follow the protocol symbolically all the way through, branch by branch, and confirm each branch ends in |\psi\rangle after correction.

Setup. Alice: qubits S (in |\psi\rangle) and A (half of |\Phi^+\rangle). Bob: qubit B (other half). Initial state:

|\Omega_0\rangle = \tfrac{1}{\sqrt{2}}\bigl(\alpha|000\rangle + \alpha|011\rangle + \beta|100\rangle + \beta|111\rangle\bigr)_{SAB}.

Step 1 — CNOT_{S \to A}. Flip A when S is 1:

|\Omega_1\rangle = \tfrac{1}{\sqrt{2}}\bigl(\alpha|000\rangle + \alpha|011\rangle + \beta|110\rangle + \beta|101\rangle\bigr)_{SAB}.

Why: CNOT acts term-by-term on each basis state in the superposition. Terms with S = 0 are untouched; terms with S = 1 get the A-bit flipped. The amplitudes \alpha, \beta ride along.

Step 2 — H on S. Substitute |0\rangle_S \to \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)_S and |1\rangle_S \to \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle)_S:

|\Omega_2\rangle = \tfrac{1}{2}\bigl[\alpha(|000\rangle + |100\rangle) + \alpha(|011\rangle + |111\rangle) + \beta(|010\rangle - |110\rangle) + \beta(|001\rangle - |101\rangle)\bigr].

Why the minus signs appear only on the |1\rangle-branch of S: H applied to |1\rangle gives |-\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle), so whenever the original state had S = 1 (the \beta terms), the Hadamard produces a relative minus sign between the two new S-outcomes.

Step 3 — regroup by (S, A). Collect all eight terms by their |SA\rangle value:

|\Omega_2\rangle = \tfrac{1}{2}\Bigl[|00\rangle_{SA}\bigl(\alpha|0\rangle + \beta|1\rangle\bigr) + |01\rangle_{SA}\bigl(\alpha|1\rangle + \beta|0\rangle\bigr) + |10\rangle_{SA}\bigl(\alpha|0\rangle - \beta|1\rangle\bigr) + |11\rangle_{SA}\bigl(\alpha|1\rangle - \beta|0\rangle\bigr)\Bigr].

The B-qubit factors in each bracket are, respectively, |\psi\rangle, X|\psi\rangle, Z|\psi\rangle, and XZ|\psi\rangle (the last up to a global sign).

Step 4 — measure (S, A) and apply correction. Alice's measurement picks one of the four (m_1, m_2) outcomes with probability \tfrac{1}{4} each; Bob receives the bits and applies X^{m_2} Z^{m_1}. Case by case:

  • (0, 0): Bob's state is |\psi\rangle; no correction; final |\psi\rangle. ✓
  • (0, 1): Bob's state is X|\psi\rangle; apply X; final X \cdot X|\psi\rangle = |\psi\rangle. ✓
  • (1, 0): Bob's state is Z|\psi\rangle; apply Z; final Z \cdot Z|\psi\rangle = |\psi\rangle. ✓
  • (1, 1): Bob's state is XZ|\psi\rangle; apply Z then X; final X Z \cdot X Z |\psi\rangle = -|\psi\rangle, which is |\psi\rangle up to a global phase. ✓

Result. In every branch, Bob's qubit after correction is |\psi\rangle (up to a physically-irrelevant global phase). The protocol succeeds deterministically — not probabilistically. The randomness of the measurement outcome is exactly compensated by the Pauli correction keyed to that outcome.

The four branches meet at |ψ⟩A branching flow diagram. A single source node labelled |psi> fans out to four branches corresponding to measurement outcomes (0,0), (0,1), (1,0), (1,1). Each branch passes through a correction box labelled I, X, Z, or ZX, and all four branches converge to a single target node labelled |psi>.|ψ⟩start(0,0) → I(0,1) → X(1,0) → Z(1,1) → ZX|ψ⟩on Bob's qubitFour measurement branches — one input state — one output state. All four paths arrive.
The measurement outcome randomises which path is taken, but every path recovers $|\psi\rangle$ on Bob's side after the correction keyed to that outcome.

Example 2: Teleport $|+\rangle$ — concrete numbers

Pick |\psi\rangle = |+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle). So \alpha = \beta = \tfrac{1}{\sqrt{2}}. Run through the protocol, watching the amplitudes evolve concretely.

Setup. Initial three-qubit state:

|\Omega_0\rangle = \tfrac{1}{\sqrt{2}}|+\rangle_S \otimes \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)_{AB} = \tfrac{1}{2}\bigl(|000\rangle + |011\rangle + |100\rangle + |111\rangle\bigr)_{SAB}.

Why the overall factor is \tfrac{1}{2}: |+\rangle contributes \tfrac{1}{\sqrt{2}} and |\Phi^+\rangle contributes \tfrac{1}{\sqrt{2}}, so \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{\sqrt{2}} = \tfrac{1}{2}.

Step 1 — CNOT_{S \to A}. Flip A when S is 1:

|\Omega_1\rangle = \tfrac{1}{2}\bigl(|000\rangle + |011\rangle + |110\rangle + |101\rangle\bigr).

Step 2 — H on S. Apply to each S-ket:

  • |000\rangle \to \tfrac{1}{\sqrt{2}}(|000\rangle + |100\rangle)
  • |011\rangle \to \tfrac{1}{\sqrt{2}}(|011\rangle + |111\rangle)
  • |110\rangle \to \tfrac{1}{\sqrt{2}}(|010\rangle - |110\rangle)
  • |101\rangle \to \tfrac{1}{\sqrt{2}}(|001\rangle - |101\rangle)

Combining (with the \tfrac{1}{2} prefactor \times \tfrac{1}{\sqrt{2}} = \tfrac{1}{2\sqrt{2}}):

|\Omega_2\rangle = \tfrac{1}{2\sqrt{2}}\bigl[|000\rangle + |100\rangle + |011\rangle + |111\rangle + |010\rangle - |110\rangle + |001\rangle - |101\rangle\bigr].

Step 3 — regroup by (S, A). Collect:

|\Omega_2\rangle = \tfrac{1}{2\sqrt{2}}\Bigl[|00\rangle_{SA}(|0\rangle + |1\rangle)_B + |01\rangle_{SA}(|1\rangle + |0\rangle)_B + |10\rangle_{SA}(|0\rangle - |1\rangle)_B + |11\rangle_{SA}(|1\rangle - |0\rangle)_B\Bigr].

Normalise each B-bracket to unit length — each has norm \sqrt{2} before the extra \tfrac{1}{\sqrt{2}} is absorbed — to get the standard plus/minus states:

|\Omega_2\rangle = \tfrac{1}{2}\Bigl[|00\rangle_{SA}|+\rangle_B + |01\rangle_{SA}|+\rangle_B + |10\rangle_{SA}|-\rangle_B - |11\rangle_{SA}|-\rangle_B\Bigr].

Why the B-brackets simplify: |0\rangle + |1\rangle = \sqrt{2}|+\rangle and |0\rangle - |1\rangle = \sqrt{2}|-\rangle. The extra \sqrt{2} absorbs into the \tfrac{1}{2\sqrt{2}} to give \tfrac{1}{2}. Note that in the (0,1) branch, |1\rangle + |0\rangle = |0\rangle + |1\rangle = \sqrt{2}|+\rangle, so Bob's qubit is still |+\rangle = X|+\rangle (since X|+\rangle = |+\rangle). In the (1,1) branch, |1\rangle - |0\rangle = -(|0\rangle - |1\rangle) = -\sqrt{2}|-\rangle — the extra minus is a branch-dependent global phase that Bob's correction will not affect.

Step 4 — measurement probabilities. Each of the four (m_1, m_2) outcomes has amplitude \tfrac{1}{2} (with a possible overall sign), so probability \tfrac{1}{4}. Take the (m_1, m_2) = (1, 0) branch as a concrete case.

Step 5 — Bob's correction for (m_1, m_2) = (1, 0). Before correction, Bob's qubit is |-\rangle. Bob applies X^{m_2} Z^{m_1} = X^0 Z^1 = Z. Compute:

Z|-\rangle = Z \cdot \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = |+\rangle = |\psi\rangle.

Why Z|-\rangle = |+\rangle: Z|0\rangle = |0\rangle and Z|1\rangle = -|1\rangle, so Z(|0\rangle - |1\rangle) = |0\rangle + |1\rangle, which is \sqrt{2}|+\rangle. The \tfrac{1}{\sqrt{2}} normalisation makes the result exactly |+\rangle.

Result. For input |\psi\rangle = |+\rangle and measurement outcome (1, 0), Bob's final state is |+\rangle — matching the input. The other three branches work out identically by the same logic: in each branch, Bob applies the correction keyed to (m_1, m_2) and recovers |+\rangle. The teleportation succeeds deterministically regardless of which branch the random measurement selects.

Teleporting |+⟩ — outcome (1, 0) branchBefore-and-after diagram for the (1, 0) branch of teleporting |+⟩. On the left, a qubit B is shown in state |-⟩. Arrow labelled apply Z. On the right, qubit B is in state |+⟩. Below, a small caption says this matches the input state |ψ⟩ = |+⟩ that Alice started with.B before correction|−⟩outcome (m₁, m₂) = (1, 0)apply Zm₁ = 1 fires ZB after correction|+⟩matches Alice's |ψ⟩ = |+⟩Deterministic recovery in every branch — the correction is keyed to the measurement outcome.
Concrete branch for $|\psi\rangle = |+\rangle$ and outcome $(1, 0)$: Bob's qubit is $|-\rangle$ before correction, becomes $|+\rangle = |\psi\rangle$ after the $Z$ correction fires.

Common confusions

Going deeper

If you are here to understand what teleportation does and why it works, you have it. The rest of this article covers the relationship to superdense coding, experimental demonstrations, entanglement as a resource, deferred-measurement rewriting, and the Indian contributions to satellite-based quantum communication.

The duality with superdense coding

Teleportation consumes one Bell pair + 2 classical bits to transmit 1 qubit. Superdense coding (ch.51) consumes one Bell pair + 1 qubit to transmit 2 classical bits. These two protocols are duals in the resource theory of quantum Shannon theory. Writing them as inequalities:

\text{Teleportation:}\quad 1\text{ ebit} + 2\text{ cbits} \;\geq\; 1\text{ qubit channel}.
\text{Superdense coding:}\quad 1\text{ ebit} + 1\text{ qubit channel} \;\geq\; 2\text{ cbits}.

Neither direction gives you something for nothing — entanglement is consumed, and each protocol requires the resource that its dual produces. But together they tell you how the three currencies of quantum information — ebits, cbits, qubits — exchange for each other.

Experimental demonstrations

The first experimental teleportation of a qubit was performed by Anton Zeilinger's group in Innsbruck in 1997, using polarisation-entangled photon pairs. A single-photon state was teleported over a distance of about one metre inside the lab. Zeilinger received the 2022 Nobel Prize in Physics, jointly with Alain Aspect and John Clauser, for this and related foundational experiments with entangled photons.

Subsequent milestones:

The fidelity of typical experiments is 85-95%, limited by detector losses, gate imperfections, and decoherence. The theoretical fidelity is 100%; the gap between experiment and theory is a direct measure of how far the hardware is from the idealised protocol.

ISRO and Indian quantum communication

India's own quantum-communication work sits at the intersection of teleportation theory and satellite-based quantum networking. In 2022, ISRO (in partnership with the Raman Research Institute, Bengaluru, and the Physical Research Laboratory, Ahmedabad) demonstrated satellite-based quantum key distribution over a free-space link — one of the stepping-stones to full entanglement-based networks. The protocol they demonstrated (a polarisation-based BB84 variant) is not yet a full teleportation protocol, but it exercises the same underlying technology: Bell-pair generation, entanglement distribution, and single-photon detection at kilometre scales.

The National Quantum Mission (announced 2023, ₹6003 crore over 8 years) explicitly funds work on quantum repeaters, entanglement distribution, and the beginnings of a quantum internet backbone between Indian metros. Active academic groups: TIFR Mumbai (superconducting qubits and entanglement), IIT Madras (trapped-ion QC), IISc Bengaluru (photonic quantum networks), and IIT Delhi (quantum algorithms and communication).

When a full teleportation-based quantum network lands in India — plausibly a satellite-ground link in the late 2020s or early 2030s — the protocol at its heart will be exactly the one derived above, implemented over a Bell pair with one end in Bengaluru and the other on an orbiting satellite.

Deferred-measurement rewriting

The deferred-measurement principle says that any mid-circuit measurement plus classical-controlled operation can be rewritten as a quantum-controlled operation plus a measurement at the end, with identical output statistics. Apply this to teleportation:

The rewritten circuit is purely quantum, no classical feedback, no double-line wires. It is mathematically equivalent to the original — same output distribution on every qubit — but it costs two CZ + CNOT gates across Alice-Bob instead of two classical bits communicated across Alice-Bob. On real hardware with fast classical feedforward (IBM's dynamic circuits, Quantinuum's adaptive ion traps), the original version with classical control is preferred; on pen-and-paper analyses and some older simulators, the deferred version is cleaner.

The rewrite is a reminder that teleportation is not about classical information being fundamental to the protocol — it is about the resource trade-off between classical and quantum communication. Classical control is just the cheapest implementation.

Teleportation in quantum networks

Single-hop teleportation (one Bell pair, one transfer) is limited by the maximum fibre distance over which a Bell pair can be distributed — roughly 100-200 km in current optical fibre, limited by photon loss. To go farther, you chain teleportations: prepare Bell pairs over short hops, then use entanglement swapping (a close cousin of teleportation, covered in the next chapter) to stitch short-hop Bell pairs into long-hop ones. The resulting architecture is called a quantum repeater network, and it is the core building block of the envisioned quantum internet.

The quantum internet is not a replacement for the classical internet — it is an overlay that supports tasks classical networks cannot: blind quantum computing, quantum key distribution with information-theoretic security, distributed quantum sensing, and networked quantum computers running algorithms larger than any single chip can host. Teleportation is the fundamental moving-parts primitive in that architecture.

Where this leads next

References

  1. Bennett, Brassard, Crépeau, Jozsa, Peres, Wootters, Teleporting an unknown quantum state via dual classical and Einstein-Podolsky-Rosen channels (1993) — the original protocol paper. DOI:10.1103/PhysRevLett.70.1895.
  2. Wikipedia, Quantum teleportation — protocol, history, and experimental demonstrations.
  3. John Preskill, Lecture Notes on Quantum Computation, Ch. 4 — theory.caltech.edu/~preskill/ph229.
  4. Nielsen and Chuang, Quantum Computation and Quantum Information (2010), §1.3.7 — Cambridge University Press.
  5. Qiskit Textbook, Quantum Teleportation — Python code and runnable examples on IBM hardware.
  6. Ren et al., Ground-to-satellite quantum teleportation (2017) — Micius satellite experiment. arXiv:1707.00934.