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Lesson 8: Teleportation

Move an unknown qubit with one Bell pair and two classical bits

about 40 min15 partsbest after lesson 5 and lesson 7

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The interactive version checks every answer, gives hints and saves your progress in this browser. This page is the same lesson on one page, with the answers folded away.

1 / 15 · Learn

Before you start

Lesson 7 showed that an unknown state can’t be copied, and that a Bell pair’s correlations carry no message by themselves. Teleportation combines a Bell pair with two ordinary bits to move an unknown state anyway.

About 40 minutes in two parts. Part A follows the state through Alice’s gates. Part B, from “Alice measures”, covers the measurement, Bob’s corrections and what teleportation cannot do; that is a good place for a break.

By the end of this lesson you can:

  • compute the states ∣ψ0⟩\ket{\psi_0}, ∣ψ1⟩\ket{\psi_1} and ∣ψ2⟩\ket{\psi_2} of the teleportation circuit;
  • regroup ∣ψ2⟩\ket{\psi_2} by Alice’s two measurement results;
  • choose Bob’s correction ZM1XM2Z^{M_1}X^{M_2} for each result;
  • explain why teleportation sends nothing faster than light and makes no copy;
  • count the cost: one Bell pair plus two classical bits move one qubit’s state.

Check yourself

A warm-up: make a guess before reading on. The lesson comes back to it.

Alice and Bob already share a Bell pair. Alice now wants Bob to end up with a qubit in an unknown state ∣ψ⟩\ket\psi that she holds. What must she send him?

  1. AHer qubit itself, through a quantum channel
  2. BTwo classical bits
  3. CThe numbers α\alpha and β\beta, written out
  4. DNothing more: the shared pair carries ∣ψ⟩\ket\psi over
Show the answer

Answer: B. Two classical bits

Alice sends two classical bits, her measurement results M1M_1 and M2M_2. They tell Bob which of four fixed corrections to apply.

Why not the others

  • A. Sending the qubit would work, but the protocol avoids it. Once the pair is shared, only classical bits go from Alice to Bob, and her qubit stays with her, ending in ∣0⟩\ket0 or ∣1⟩\ket1. It seems something quantum must travel; it did, in advance, as Bob’s half of the pair.
  • C. It feels as if a measurement would let Alice read off α\alpha and β\beta. One measurement of one copy, in any basis, gives one bit and replaces the state, so she can never learn them; and an exact complex number would need unlimited digits anyway. The protocol works without anyone learning α\alpha or β\beta.
  • D. A common picture is that entanglement does the carrying. But nothing Alice does to her half changes what Bob can measure: his qubit stays 50/50 in every basis until a message arrives. He needs her two measurement results to finish, and they travel no faster than light. The name “teleportation” makes the instant picture tempting.

Check yourself

A warm-up: make a guess before reading on. The lesson comes back to it.

Alice and Bob share a Bell pair ∣β00⟩\ket{\beta_{00}} and nothing else: no phone, no radio. Can Alice use the pair alone to send Bob one bit instantly, for example by choosing whether or not to measure her half?

  1. AYes, if they agreed in advance which basis Alice would measure in
  2. BNo, because entanglement fades with distance
  3. CYes: measuring her half changes Bob’s qubit at once, and Bob can detect the change
  4. DNo: whatever Alice does to her half, Bob’s results have the same statistics
Show the answer

Answer: D. No: whatever Alice does to her half, Bob’s results have the same statistics

Bob’s half of ∣β00⟩\ket{\beta_{00}} gives 0 or 1 with probability 12\tfrac12 in every basis, and nothing Alice does to her half changes that. A signal would need Bob’s statistics to depend on Alice’s choice, so the pair alone sends nothing.

Why not the others

  • A. Agreeing on a basis doesn’t help. Whatever basis Alice measures in, and whether she measures at all, Bob’s own results are 50/50 in every basis he picks. The correlations are real, but Bob can see them only by comparing his list with Alice’s, and that takes ordinary communication.
  • B. Distance by itself doesn’t weaken the pair: in the ideal case the qubits stay entangled however far apart they are. The real reason is that Bob’s statistics on his own qubit never depend on what Alice does. Real entanglement is fragile, which makes this reason tempting, but that is noise, not distance.
  • C. It seems Bob should notice the change. Check it: if Alice measures in the 0/1 basis, Bob’s 0/1 measurement gives 0 or 1 with probability 12\tfrac12 each; if she doesn’t measure, it still gives 12\tfrac12 each. Nothing Bob sees on his own depends on what Alice did. The correlation shows up only when they compare notes, which needs a message.

2 / 15 · Learn

Part A · The job and the protocol

Alice holds a qubit in a state ∣ψ⟩=α∣0⟩+β∣1⟩\ket\psi = \alpha\ket0 + \beta\ket1 that nobody knows. She wants Bob, far away, to end up with a qubit in exactly that state.

Two obvious routes are closed. She can’t learn α\alpha and β\beta by measuring: one measurement, in any basis, gives one bit and replaces the state (Lesson 1, Lesson 5). She can’t copy the qubit and study the copies either: that is no-cloning (Lesson 7).

What she does have: half of a Bell pair ∣β00⟩=(∣00⟩+∣11⟩)/2\ket{\beta_{00}} = (\ket{00}+\ket{11})/\sqrt2, shared with Bob earlier, and an ordinary classical line for sending bits.

Check yourself

Why can’t Alice simply measure her qubit in the 0/1 basis and phone Bob the result?

  1. AShe could, if she read α\alpha and β\beta off the result
  2. BMeasuring is fine; the problem is that Bob can’t build a state from a description
  3. CThe result is one bit, and the measurement leaves her qubit in ∣0⟩\ket0 or ∣1⟩\ket1: α\alpha and β\beta are gone after one try
  4. DShe could, but a phone call is slower than teleportation
Show the answer

Answer: C. The result is one bit, and the measurement leaves her qubit in ∣0⟩\ket0 or ∣1⟩\ket1: α\alpha and β\beta are gone after one try

One measurement of one copy yields one bit and destroys the superposition. Estimating α\alpha and β\beta would take many identical copies, and Alice has one.

Why not the others

  • A. A measurement doesn’t display amplitudes. It returns one outcome, 0 with probability ∣α∣2\abs\alpha^2 or 1 with probability ∣β∣2\abs\beta^2, and leaves ∣0⟩\ket0 or ∣1⟩\ket1 behind; measuring again in the same basis only repeats that outcome. The idea is tempting because a classical measurement reads off a value that was already there.
  • B. It is natural to suspect the receiving end, but Bob could prepare any state whose α\alpha and β\beta he knew: that is just a choice of gates. The obstacle is on Alice’s side. She has one copy of an unknown state, and one measurement can’t reveal its amplitudes.
  • D. The name suggests teleportation beats a phone call, but it needs one: Alice’s two bits must reach Bob before he can finish, so it is never faster than ordinary communication. The measuring route fails for another reason: one 0/1 result can’t tell Bob α\alpha and β\beta.

After the pair is shared, the protocol has four moves:

  1. Alice applies CNOT, with her ∣ψ⟩\ket\psi qubit (top wire) as control and her half of the pair (middle wire) as target.
  2. She applies HH to the top wire.
  3. She measures both of her qubits in the 0/1 basis: M1M_1 from the top wire, M2M_2 from the middle wire.
  4. She sends the two bits to Bob. He applies XX if M2=1M_2 = 1, then ZZ if M1=1M_1 = 1.

Bob’s qubit then holds ∣ψ⟩\ket\psi. The rest of the lesson checks every step.

Check yourself

Put the protocol in time order.

Put these in order.

  • Bob applies XX if M2=1M_2 = 1
  • Alice: HH on her ∣ψ⟩\ket\psi qubit
  • Alice: CNOT from her ∣ψ⟩\ket\psi qubit onto her half of the pair
  • Bob applies ZZ if M1=1M_1 = 1
  • Alice measures both of her qubits in the 0/1 basis
  • Alice sends the bits M1M2M_1M_2 to Bob over the classical line
  • Alice and Bob share ∣β00⟩\ket{\beta_{00}} (done in advance)
Hints
  1. Bob can’t do anything useful until he learns something. When does he learn it?
  2. Alice’s two gates come before her measurements. For Bob, XX comes before ZZ.
Show the answer
  1. Alice and Bob share ∣β00⟩\ket{\beta_{00}} (done in advance)
  2. Alice: CNOT from her ∣ψ⟩\ket\psi qubit onto her half of the pair
  3. Alice: HH on her ∣ψ⟩\ket\psi qubit
  4. Alice measures both of her qubits in the 0/1 basis
  5. Alice sends the bits M1M2M_1M_2 to Bob over the classical line
  6. Bob applies XX if M2=1M_2 = 1
  7. Bob applies ZZ if M1=1M_1 = 1

CNOT, then HH, then the two 0/1-basis measurements; the bits travel; Bob applies ZM1XM2Z^{M_1}X^{M_2}, which means XM2X^{M_2} first, because the operator written next to the ket acts first.

3 / 15 · Learn

Pass 1 · Before and after Alice’s CNOT

Part A is a worked example in passes. Pass 1 follows the state with the letters α\alpha and β\beta; pass 2 puts in numbers; pass 3, in Part B, is mostly yours.

Before Alice acts, the three qubits are in ∣ψ0⟩=∣ψ⟩∣β00⟩\ket{\psi_0} = \ket\psi\ket{\beta_{00}}. Kets list the wires top to bottom: the first label is Alice’s ∣ψ⟩\ket\psi qubit, the second her half of the pair, the third Bob’s. Multiply out term by term:

∣ψ0⟩=12[α∣0⟩(∣00⟩+∣11⟩)+β∣1⟩(∣00⟩+∣11⟩)]=12(α∣000⟩+α∣011⟩+β∣100⟩+β∣111⟩)\begin{aligned}\ket{\psi_0} &= \tfrac{1}{\sqrt2}\big[\alpha\ket0(\ket{00}+\ket{11}) \\ &\qquad + \beta\ket1(\ket{00}+\ket{11})\big] \\ &= \tfrac{1}{\sqrt2}\big(\alpha\ket{000} + \alpha\ket{011} \\ &\qquad + \beta\ket{100} + \beta\ket{111}\big)\end{aligned}

Four of the eight amplitudes are nonzero. Nothing is new yet: ∣ψ⟩\ket\psi sits on its own, and the other two qubits share a Bell pair.

Refresher: Tensor products Multiplying out ∣ψ⟩∣β00⟩\ket\psi\ket{\beta_{00}} is a tensor product: each amplitude of the first factor times each amplitude of the second.

Check yourself

Take ∣ψ⟩=∣+⟩\ket\psi = \ket+, so α=β=1/2\alpha = \beta = 1/\sqrt2. Enter ∣ψ0⟩\ket{\psi_0} as eight amplitudes, ∣000⟩\ket{000} to ∣111⟩\ket{111} (a box left empty counts as 0).

Hints
  1. Each term is (an amplitude of ∣ψ⟩\ket\psi) times (an amplitude of ∣β00⟩\ket{\beta_{00}}).
  2. Each product is 12⋅12=12\tfrac{1}{\sqrt2}\cdot\tfrac{1}{\sqrt2} = \tfrac12, and only labels whose last two bits are 0000 or 1111 appear.
Show the answer

Answer: 12∣000⟩+12∣011⟩+12∣100⟩+12∣111⟩\frac{1}{2}|000\rangle + \frac{1}{2}|011\rangle + \frac{1}{2}|100\rangle + \frac{1}{2}|111\rangle

∣ψ0⟩=12(∣000⟩+∣011⟩+∣100⟩+∣111⟩)\ket{\psi_0} = \tfrac12(\ket{000}+\ket{011}+\ket{100}+\ket{111}): every nonzero amplitude is 12×12\tfrac{1}{\sqrt2}\times\tfrac{1}{\sqrt2}.

Check yourself

Alice’s CNOT has the top wire (her ∣ψ⟩\ket\psi qubit) as control and the middle wire (her half of the pair) as target. Which terms of ∣ψ0⟩\ket{\psi_0} does it change?

  1. AOnly β∣100⟩\beta\ket{100} and β∣111⟩\beta\ket{111}, the terms whose first label is 1
  2. BAll four: a CNOT changes every term of a superposition
  3. COnly α∣011⟩\alpha\ket{011} and β∣111⟩\beta\ket{111}, the terms whose middle label is 1
  4. DNone: the middle qubit is entangled with Bob’s, so a gate on it alone can’t act
Hint
  1. A CNOT works term by term. Which label decides whether a term changes?
Show the answer

Answer: A. Only β∣100⟩\beta\ket{100} and β∣111⟩\beta\ket{111}, the terms whose first label is 1

CNOT flips the middle label exactly where the first label is 1: ∣100⟩→∣110⟩\ket{100}\to\ket{110} and ∣111⟩→∣101⟩\ket{111}\to\ket{101}. The α\alpha terms are untouched.

Why not the others

  • B. A gate acting on a superposition can feel like it must touch every term. But CNOT acts term by term and flips the target only where the control label is 1: the two α\alpha terms start with 0 and pass through unchanged.
  • C. That treats the middle wire as the control, an easy slip since the middle label is the one that changes. Here the control is the top wire, the one holding ∣ψ⟩\ket\psi, so CNOT flips the middle label only in terms whose first label is 1.
  • D. Gates act on wires, whether or not a wire has a state of its own. Apply CNOT term by term: β∣100⟩\beta\ket{100} becomes β∣110⟩\beta\ket{110}. The worry is natural after Lesson 7, but entangled only means the joint state doesn’t factor; every basis term still transforms in the usual way.

So the two α\alpha terms pass unchanged, the two β\beta terms become ∣110⟩\ket{110} and ∣101⟩\ket{101}, and Bob’s label never changes:

∣ψ1⟩=12[α∣0⟩(∣00⟩+∣11⟩)+β∣1⟩(∣10⟩+∣01⟩)]\begin{aligned}\ket{\psi_1} = \tfrac{1}{\sqrt2}\big[&\alpha\ket0(\ket{00}+\ket{11}) \\ + &\beta\ket1(\ket{10}+\ket{01})\big]\end{aligned}

Now the ∣ψ⟩\ket\psi qubit is entangled with the pair: the α\alpha branch carries ∣β00⟩\ket{\beta_{00}}, the β\beta branch carries ∣β01⟩=(∣01⟩+∣10⟩)/2\ket{\beta_{01}} = (\ket{01}+\ket{10})/\sqrt2.

4 / 15 · Learn

Pass 1 · Alice’s Hadamard

Next Alice applies HH to the top wire. In each term, replace the first label using H∣0⟩=(∣0⟩+∣1⟩)/2H\ket0 = (\ket0+\ket1)/\sqrt2 and H∣1⟩=(∣0⟩−∣1⟩)/2H\ket1 = (\ket0-\ket1)/\sqrt2, and leave the other two labels alone:

∣ψ2⟩=12[α(∣0⟩+∣1⟩)(∣00⟩+∣11⟩)+β(∣0⟩−∣1⟩)(∣10⟩+∣01⟩)]\begin{aligned}\ket{\psi_2} = \tfrac12\big[&\alpha(\ket0+\ket1)(\ket{00}+\ket{11}) \\ + &\beta(\ket0-\ket1)(\ket{10}+\ket{01})\big]\end{aligned}

The two factors of 1/21/\sqrt2, one from the pair and one from HH, make 12\tfrac12. Multiplied out there are eight terms, each with amplitude ±α/2\pm\alpha/2 or ±β/2\pm\beta/2. The minus signs all come from H∣1⟩H\ket1, so they sit on β\beta terms whose first label is 1.

Check yourself

What is the amplitude of ∣110⟩\ket{110} in ∣ψ2⟩\ket{\psi_2}?

  1. Aα/2\alpha/2
  2. B−β/2-\beta/2
  3. Cβ/2\beta/2
  4. D−β/2-\beta/\sqrt2
Show the answer

Answer: B. −β/2-\beta/2

∣110⟩\ket{110} comes from 12β(∣0⟩−∣1⟩)∣10⟩\tfrac12\beta(\ket0-\ket1)\ket{10}: its ∣1⟩∣10⟩\ket1\ket{10} part carries −β/2-\beta/2.

Why not the others

  • A. α\alpha comes with the pair state ∣00⟩+∣11⟩\ket{00}+\ket{11}, so its terms end in 0000 or 1111. A term ending in 1010 belongs to the β\beta branch, whose pair the CNOT turned into ∣10⟩+∣01⟩\ket{10}+\ket{01}.
  • C. The sign is easy to drop because β\beta itself carries none. But the term comes from β∣1⟩∣10⟩\beta\ket1\ket{10}, and H∣1⟩=(∣0⟩−∣1⟩)/2H\ket1 = (\ket0 - \ket1)/\sqrt2 puts a minus sign on its new ∣1⟩\ket1 part. So ∣110⟩\ket{110} gets −β/2-\beta/2, while ∣010⟩\ket{010} gets +β/2+\beta/2.
  • D. Two factors of 1/21/\sqrt2 multiply here, one from the Bell pair and one from HH, giving 12\tfrac12. Check: eight amplitudes of size ∣α∣/2\abs\alpha/2 or ∣β∣/2\abs\beta/2 give total probability 4∣α∣2/4+4∣β∣2/4=14\abs\alpha^2/4 + 4\abs\beta^2/4 = 1.

5 / 15 · Learn

Pass 1 · Regrouping by Alice’s results

This is the step most readers find hardest. Multiply out ∣ψ2⟩\ket{\psi_2}:

∣ψ2⟩=12[α∣000⟩+α∣011⟩+α∣100⟩+α∣111⟩+β∣010⟩+β∣001⟩−β∣110⟩−β∣101⟩]\begin{aligned}\ket{\psi_2} = \tfrac12\big[&\alpha\ket{000} + \alpha\ket{011} \\ + &\alpha\ket{100} + \alpha\ket{111} \\ + &\beta\ket{010} + \beta\ket{001} \\ - &\beta\ket{110} - \beta\ket{101}\big]\end{aligned}

Now collect the terms by their first two labels (Alice’s qubits) and factor Bob’s qubit out of each group. For example, the terms starting with 0101 are β∣010⟩\beta\ket{010} and α∣011⟩\alpha\ket{011}, which make ∣01⟩(β∣0⟩+α∣1⟩)\ket{01}(\beta\ket0 + \alpha\ket1). Try the 1111 group yourself.

Check yourself

Which terms of the expanded ∣ψ2⟩\ket{\psi_2} make up the ∣11⟩\ket{11} group?

  1. Aα2∣011⟩\tfrac{\alpha}{2}\ket{011} and α2∣111⟩\tfrac{\alpha}{2}\ket{111}, giving α2(∣0⟩+∣1⟩)∣11⟩\tfrac{\alpha}{2}(\ket0+\ket1)\ket{11}
  2. BEvery group contains ∣ψ⟩\ket\psi, so the ∣11⟩\ket{11} group is 12∣11⟩(α∣0⟩+β∣1⟩)\tfrac12\ket{11}(\alpha\ket0 + \beta\ket1)
  3. Cα2∣111⟩\tfrac{\alpha}{2}\ket{111} and −β2∣110⟩-\tfrac{\beta}{2}\ket{110}, giving 12∣11⟩(α∣1⟩−β∣0⟩)\tfrac12\ket{11}(\alpha\ket1 - \beta\ket0)
  4. Dα2∣111⟩\tfrac{\alpha}{2}\ket{111} and β2∣110⟩\tfrac{\beta}{2}\ket{110}, giving 12∣11⟩(α∣1⟩+β∣0⟩)\tfrac12\ket{11}(\alpha\ket1 + \beta\ket0)
Hint
  1. Which of the eight terms start with 1111? Keep each one’s sign.
Show the answer

Answer: C. α2∣111⟩\tfrac{\alpha}{2}\ket{111} and −β2∣110⟩-\tfrac{\beta}{2}\ket{110}, giving 12∣11⟩(α∣1⟩−β∣0⟩)\tfrac12\ket{11}(\alpha\ket1 - \beta\ket0)

The terms whose first two labels are 1111 are α2∣111⟩\tfrac{\alpha}{2}\ket{111} and −β2∣110⟩-\tfrac{\beta}{2}\ket{110}. Factoring out ∣11⟩\ket{11} leaves Bob’s factor α∣1⟩−β∣0⟩\alpha\ket1 - \beta\ket0.

Why not the others

  • A. That groups by the last two labels, which is how the pair was written before the CNOT. The regrouping is by Alice’s labels, the first two, because her measurement reads those; Bob’s label is what is left inside the bracket.
  • B. It is tempting to think the protocol has already moved ∣ψ⟩\ket\psi into every branch. Only the 0000 group holds α∣0⟩+β∣1⟩\alpha\ket0+\beta\ket1 as it is. The others have the amplitudes swapped, a sign flipped, or both, which is why Bob needs a correction.
  • D. The ∣110⟩\ket{110} term has a minus sign: it comes from β∣1⟩∣10⟩\beta\ket1\ket{10}, and H∣1⟩H\ket1 puts −1-1 on the new ∣1⟩\ket1 part. That sign is why result 1111 needs a ZZ as well as an XX.

All four groups together:

∣ψ2⟩=12[∣00⟩(α∣0⟩+β∣1⟩)+∣01⟩(α∣1⟩+β∣0⟩)+∣10⟩(α∣0⟩−β∣1⟩)+∣11⟩(α∣1⟩−β∣0⟩)]\begin{aligned}\ket{\psi_2} = \tfrac12\big[&\ket{00}(\alpha\ket0+\beta\ket1) \\ + &\ket{01}(\alpha\ket1+\beta\ket0) \\ + &\ket{10}(\alpha\ket0-\beta\ket1) \\ + &\ket{11}(\alpha\ket1-\beta\ket0)\big]\end{aligned}

No gate acted here. It is the same vector, written so that each possible result of Alice’s measurement stands next to the Bob state it would leave.

Check yourself

Alice has applied her CNOT and HH but has not measured yet. What is the state of Bob’s qubit on its own?

  1. AOne of the four Bob factors; we just don’t know which yet
  2. B∣ψ⟩\ket\psi: the state has already moved to Bob
  3. C∣0⟩\ket0, as at the start
  4. DIt has no ket of its own: it is entangled with Alice’s two qubits
Show the answer

Answer: D. It has no ket of its own: it is entangled with Alice’s two qubits

Each of Alice’s labels comes with a different Bob factor, so ∣ψ2⟩\ket{\psi_2} doesn’t factor into an Alice part times a Bob part. Bob’s qubit has no ket of its own until Alice measures.

Why not the others

  • A. Tempting, because the regrouped form lists four Bob states. But a sum of branches is one entangled state, not a hidden choice among four. The circuit readout above shows it: Bob’s wire is “entangled” until the meters, and gets an arrow only after a result is picked.
  • B. The name “teleportation” invites this picture. But only the 0000 group contains ∣ψ⟩\ket\psi unchanged; the other three hold altered versions. And before Alice measures, no group has been selected, so Bob can’t be said to hold any of them.
  • C. Bob’s qubit stopped being ∣0⟩\ket0 when the Bell pair was made. From then on it was entangled with Alice’s half and had no state of its own. Alice’s gates don’t change his statistics, but they don’t return his qubit to ∣0⟩\ket0 either.

6 / 15 · Worked example

Part A · Pass 2: with numbers

The same steps with numbers. Enter each state as eight amplitudes, ∣000⟩\ket{000} to ∣111⟩\ket{111}; an empty slot counts as 0. These questions want the exact vector, signs included.

Worked example

Example: α=3/2\alpha = \sqrt3/2, β=1/2\beta = 1/2

Take ∣ψ⟩=32∣0⟩+12∣1⟩\ket\psi = \tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1, so α=3/2\alpha = \sqrt3/2 and β=1/2\beta = 1/2.

  1. Before Alice acts

    Substitute into ∣ψ0⟩=12(α∣000⟩+α∣011⟩+β∣100⟩+β∣111⟩)\ket{\psi_0} = \tfrac{1}{\sqrt2}(\alpha\ket{000} + \alpha\ket{011} + \beta\ket{100} + \beta\ket{111}). Since α/2=6/4\alpha/\sqrt2 = \sqrt6/4 and β/2=2/4\beta/\sqrt2 = \sqrt2/4:

    ∣ψ0⟩=64∣000⟩+64∣011⟩+24∣100⟩+24∣111⟩\begin{aligned}\ket{\psi_0} = {}&\tfrac{\sqrt6}{4}\ket{000} + \tfrac{\sqrt6}{4}\ket{011} \\ + {}&\tfrac{\sqrt2}{4}\ket{100} + \tfrac{\sqrt2}{4}\ket{111}\end{aligned}

    Check: 616+616+216+216=1\tfrac{6}{16} + \tfrac{6}{16} + \tfrac{2}{16} + \tfrac{2}{16} = 1.

  2. After her CNOT

    Alice’s CNOT flips the middle label wherever the first label is 1.

    Your turn

    Enter ∣ψ1⟩\ket{\psi_1}.

    Hints
    1. Which two terms have a 1 in the first slot?
    2. Those two terms keep their amplitudes; only their middle label flips.
    Show the answer

    Answer: 64∣000⟩+64∣011⟩+24∣101⟩+24∣110⟩\frac{\sqrt{6}}{4}|000\rangle + \frac{\sqrt{6}}{4}|011\rangle + \frac{\sqrt{2}}{4}|101\rangle + \frac{\sqrt{2}}{4}|110\rangle

    ∣100⟩→∣110⟩\ket{100}\to\ket{110} and ∣111⟩→∣101⟩\ket{111}\to\ket{101}, so ∣ψ1⟩=64(∣000⟩+∣011⟩)+24(∣101⟩+∣110⟩)\ket{\psi_1} = \tfrac{\sqrt6}{4}(\ket{000}+\ket{011}) + \tfrac{\sqrt2}{4}(\ket{101}+\ket{110}).

  3. After her H

    Alice’s HH on the top wire turns each term c∣0xy⟩c\ket{0xy} into c2(∣0xy⟩+∣1xy⟩)\tfrac{c}{\sqrt2}(\ket{0xy} + \ket{1xy}), and each term c∣1xy⟩c\ket{1xy} into c2(∣0xy⟩−∣1xy⟩)\tfrac{c}{\sqrt2}(\ket{0xy} - \ket{1xy}). For example 64∣000⟩\tfrac{\sqrt6}{4}\ket{000} becomes 34∣000⟩+34∣100⟩\tfrac{\sqrt3}{4}\ket{000} + \tfrac{\sqrt3}{4}\ket{100}, because 6/(42)=3/4\sqrt6/(4\sqrt2) = \sqrt3/4.

    Your turn

    Enter ∣ψ2⟩\ket{\psi_2}.

    Hints
    1. Each of the four terms of ∣ψ1⟩\ket{\psi_1} splits into two: one starting with 0, one starting with 1.
    2. 64⋅12=34\tfrac{\sqrt6}{4}\cdot\tfrac{1}{\sqrt2} = \tfrac{\sqrt3}{4} and 24⋅12=14\tfrac{\sqrt2}{4}\cdot\tfrac{1}{\sqrt2} = \tfrac14. The minus signs come from the terms that started with 1.
    Show the answer

    Answer: 34∣000⟩+14∣001⟩+14∣010⟩+34∣011⟩+34∣100⟩−14∣101⟩−14∣110⟩+34∣111⟩\frac{\sqrt{3}}{4}|000\rangle + \frac{1}{4}|001\rangle + \frac{1}{4}|010\rangle + \frac{\sqrt{3}}{4}|011\rangle + \frac{\sqrt{3}}{4}|100\rangle - \frac{1}{4}|101\rangle - \frac{1}{4}|110\rangle + \frac{\sqrt{3}}{4}|111\rangle

    ∣ψ2⟩=34∣000⟩+14∣001⟩+14∣010⟩+34∣011⟩+34∣100⟩−14∣101⟩−14∣110⟩+34∣111⟩\ket{\psi_2} = \tfrac{\sqrt3}{4}\ket{000} + \tfrac14\ket{001} + \tfrac14\ket{010} + \tfrac{\sqrt3}{4}\ket{011} + \tfrac{\sqrt3}{4}\ket{100} - \tfrac14\ket{101} - \tfrac14\ket{110} + \tfrac{\sqrt3}{4}\ket{111}: α/2\alpha/2 or ±β/2\pm\beta/2 in each slot, as pass 1 said.

Here is the whole calculation once more, fully worked, with different numbers.

Another worked example

Another example: α=3/5\alpha = 3/5, β=4/5\beta = 4/5

  1. Before Alice acts

    With α=35\alpha = \tfrac35 and β=45\beta = \tfrac45:

    ∣ψ0⟩=352(∣000⟩+∣011⟩)+452(∣100⟩+∣111⟩)\begin{aligned}\ket{\psi_0} = {}&\tfrac{3}{5\sqrt2}(\ket{000}+\ket{011}) \\ + {}&\tfrac{4}{5\sqrt2}(\ket{100}+\ket{111})\end{aligned}

  2. After her CNOT

    CNOT moves only the terms that start with 1, flipping their middle label:

    ∣ψ1⟩=352(∣000⟩+∣011⟩)+452(∣110⟩+∣101⟩)\begin{aligned}\ket{\psi_1} = {}&\tfrac{3}{5\sqrt2}(\ket{000}+\ket{011}) \\ + {}&\tfrac{4}{5\sqrt2}(\ket{110}+\ket{101})\end{aligned}

  3. After her H

    HH on the top wire: each 352\tfrac{3}{5\sqrt2} term splits into two terms of 310\tfrac{3}{10}; each 452\tfrac{4}{5\sqrt2} term splits into +25+\tfrac25 (first label 0) and −25-\tfrac25 (first label 1):

    ∣ψ2⟩=310∣000⟩+25∣001⟩+25∣010⟩+310∣011⟩+310∣100⟩−25∣101⟩−25∣110⟩+310∣111⟩\begin{aligned}\ket{\psi_2} = {}&\tfrac{3}{10}\ket{000} + \tfrac25\ket{001} \\ + {}&\tfrac25\ket{010} + \tfrac{3}{10}\ket{011} \\ + {}&\tfrac{3}{10}\ket{100} - \tfrac25\ket{101} \\ - {}&\tfrac25\ket{110} + \tfrac{3}{10}\ket{111}\end{aligned}

    Check: 4⋅9100+4⋅16100=14\cdot\tfrac{9}{100} + 4\cdot\tfrac{16}{100} = 1.

Check yourself

Now without the steps. Teleport ∣ψ⟩=12∣0⟩−32∣1⟩\ket\psi = \tfrac12\ket0 - \tfrac{\sqrt3}{2}\ket1 and enter ∣ψ2⟩\ket{\psi_2}, the state right after Alice’s CNOT and HH.

Hints
  1. Use the pattern from pass 1: every amplitude is ±α/2\pm\alpha/2 or ±β/2\pm\beta/2. Which kind goes where?
  2. α\alpha sits on ∣000⟩\ket{000}, ∣011⟩\ket{011}, ∣100⟩\ket{100}, ∣111⟩\ket{111}; β\beta on the other four, with a minus on ∣101⟩\ket{101} and ∣110⟩\ket{110}.
Show the answer

Answer: 14∣000⟩−34∣001⟩−34∣010⟩+14∣011⟩+14∣100⟩+34∣101⟩+34∣110⟩+14∣111⟩\frac{1}{4}|000\rangle - \frac{\sqrt{3}}{4}|001\rangle - \frac{\sqrt{3}}{4}|010\rangle + \frac{1}{4}|011\rangle + \frac{1}{4}|100\rangle + \frac{\sqrt{3}}{4}|101\rangle + \frac{\sqrt{3}}{4}|110\rangle + \frac{1}{4}|111\rangle

α/2=14\alpha/2 = \frac{1}{4} and β/2=−34\beta/2 = -\frac{\sqrt{3}}{4}, so ∣ψ2⟩=14∣000⟩−34∣001⟩−34∣010⟩+14∣011⟩+14∣100⟩+34∣101⟩+34∣110⟩+14∣111⟩\begin{aligned}\ket{\psi_2} = {}&\frac{1}{4}|000\rangle - \frac{\sqrt{3}}{4}|001\rangle \\ - {}&\frac{\sqrt{3}}{4}|010\rangle + \frac{1}{4}|011\rangle \\ + {}&\frac{1}{4}|100\rangle + \frac{\sqrt{3}}{4}|101\rangle \\ + {}&\frac{\sqrt{3}}{4}|110\rangle + \frac{1}{4}|111\rangle\end{aligned}

7 / 15 · Learn

Part B · Alice measures

Part B: the measurement, Bob’s corrections, and what teleportation cannot do.

Alice now measures her two qubits in the 0/1 basis. In the regrouped ∣ψ2⟩\ket{\psi_2} each of her four results comes with amplitude 12\tfrac12 times a normalized Bob state, so each result has probability (12)2=14(\tfrac12)^2 = \tfrac14, whatever α\alpha and β\beta are. Getting M1M2M_1M_2 keeps just that group, renormalized:

∣ψ3⟩=∣M1M2⟩ ∣ψ3(M1M2)⟩\ket{\psi_3} = \ket{M_1M_2}\,\ket{\psi_3(M_1M_2)}

Here ∣ψ3(M1M2)⟩\ket{\psi_3(M_1M_2)} is Bob’s factor from that group, for example ∣ψ3(10)⟩=α∣0⟩−β∣1⟩\ket{\psi_3(10)} = \alpha\ket0 - \beta\ket1. Now, and only now, Bob’s qubit has a state of its own, though Bob doesn’t yet know which one.

Check yourself

Teleport ∣ψ⟩=32∣0⟩+12∣1⟩\ket\psi = \tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1. Alice measures her two qubits in the 0/1 basis. What is the probability that she gets M1M2=10M_1M_2 = 10?

P(M1M2=10)=  ?P(M_1M_2 = 10) = \;?

Hints
  1. In the regrouped ∣ψ2⟩\ket{\psi_2}, what number multiplies the ∣10⟩\ket{10} group?
  2. Its Bob factor α∣0⟩−β∣1⟩\alpha\ket0 - \beta\ket1 is normalized, so the probability is ∣12∣2\abs{\tfrac12}^2 times 1.
Show the answer

Answer: P(M1M2=10)=14P(M_1M_2 = 10) = \frac{1}{4}

The ∣10⟩\ket{10} group is 12∣10⟩(α∣0⟩−β∣1⟩)\tfrac12\ket{10}(\alpha\ket0 - \beta\ket1) and ∣α∣2+∣β∣2=1\abs\alpha^2 + \abs\beta^2 = 1, so the probability is 14\tfrac14. The same holds for every result and every ∣ψ⟩\ket\psi: Alice’s bits are uniformly random and say nothing about α\alpha or β\beta.

Common slips

  • 18\frac{1}{8}: That treats M1M_1 as a 0/1 measurement of ∣ψ⟩\ket\psi itself, P(M1=1)=∣β∣2=14P(M_1 = 1) = \abs\beta^2 = \tfrac14, times 12\tfrac12 for M2M_2. But Alice’s CNOT and HH came first. In the regrouped state every group has amplitude 12\tfrac12 times a normalized Bob state, so every result has probability 14\tfrac14.
  • 316\frac{3}{16}: That counts only the ∣100⟩\ket{100} term: ∣α/2∣2=316\abs{\alpha/2}^2 = \tfrac{3}{16}. Alice’s result says nothing about Bob’s label, so both terms of the 1010 group count, ∣100⟩\ket{100} and ∣101⟩\ket{101}: 316+116=14\tfrac{3}{16} + \tfrac{1}{16} = \tfrac14.
  • 12\frac{1}{2}: 12\tfrac12 is the amplitude in front of the group. A probability is a squared magnitude: (12)2 (∣α∣2+∣β∣2)=14(\tfrac12)^2\,(\abs\alpha^2 + \abs\beta^2) = \tfrac14.
  • 116\frac{1}{16}: That counts only the ∣101⟩\ket{101} term: ∣β/2∣2=116\abs{\beta/2}^2 = \tfrac{1}{16}. Alice’s result says nothing about Bob’s label, so both terms of the 1010 group count, ∣100⟩\ket{100} and ∣101⟩\ket{101}: 316+116=14\tfrac{3}{16} + \tfrac{1}{16} = \tfrac14.

8 / 15 · Explore

Part B · What Bob sees before the bits arrive

Predict, then see

Alice teleports a state ∣ψ⟩=α∣0⟩+β∣1⟩\ket\psi = \alpha\ket0 + \beta\ket1. She has measured her two qubits in the 0/1 basis, but her two bits have not reached Bob yet. Bob measures his qubit in the 0/1 basis. What is his chance of getting 0?

  1. A12\tfrac12, whatever ∣ψ⟩\ket\psi is
  2. BSomewhere between ∣α∣2\abs\alpha^2 and 12\tfrac12: in one run out of four Bob already holds ∣ψ⟩\ket\psi
  3. C∣α∣2\abs\alpha^2: near 1 when ∣ψ⟩\ket\psi is close to ∣0⟩\ket0, near 0 when it is close to ∣1⟩\ket1
  4. D1: Bob’s qubit started as ∣0⟩\ket0 and nobody has touched it

Bob’s P(0)P(0) read 50% at every θ\theta, even though one of Alice’s four results leaves him holding ∣ψ⟩\ket\psi. In a sentence or two, why?

Show what happens

What happens: A. 12\tfrac12, whatever ∣ψ⟩\ket\psi is

Until the bits arrive, Bob’s qubit is one of four versions of ∣ψ⟩\ket\psi, each with probability 14\tfrac14, and he can’t tell which. Their chances of giving 0 are ∣α∣2\abs\alpha^2, ∣β∣2\abs\beta^2, ∣α∣2\abs\alpha^2, ∣β∣2\abs\beta^2, which average to 12\tfrac12 for every ∣ψ⟩\ket\psi. That is why the widget draws Bob’s arrow as a dot at the centre of the sphere: no direction, no information. Only when the two bits arrive, at no more than light speed, can he turn his qubit into ∣ψ⟩\ket\psi.

Why not the others

  • B. Half right: when Alice gets 0000, Bob’s qubit is ∣ψ⟩\ket\psi. But Bob doesn’t know which runs those are, and the other versions pull the other way: X∣ψ⟩X\ket\psi and XZ∣ψ⟩XZ\ket\psi give 0 with probability ∣β∣2\abs\beta^2. Over all four, P(0)=12P(0) = \tfrac12 exactly.
  • C. This is what the name “teleportation” suggests. The widget shows otherwise: at step 4, Bob’s P(0)P(0) reads 50% for every θ\theta, near ∣0⟩\ket0 and near ∣1⟩\ket1 alike.
  • D. Bob’s qubit stopped being ∣0⟩\ket0 when the Bell pair was made (widget step 1). From then on, measured on its own in the 0/1 basis, it gives 0 or 1 with probability 12\tfrac12 each: while entangled it has no state of its own.
Why the centre of the sphere, in every basis?

On the Bloch sphere, XX takes the vector (x,y,z)(x, y, z) to (x,−y,−z)(x, -y, -z) and ZZ takes it to (−x,−y,z)(-x, -y, z). So Bob’s four possible states point along (x,y,z)(x, y, z), (x,−y,−z)(x, -y, -z), (−x,−y,z)(-x, -y, z) and (−x,y,−z)(-x, y, -z), each with weight 14\tfrac14. Their average is (0,0,0)(0, 0, 0): 50/50 along every axis, not just the 0/1 one. Before Alice measures, Bob’s qubit is half of an entangled state, and it is just as random. A more general tool, the reduced density operator, which describes one part of an entangled system on its own, makes this precise.

Check yourself

Alice teleports ∣ψ⟩=35∣0⟩+45∣1⟩\ket\psi = \tfrac35\ket0 + \tfrac45\ket1 and has measured. Before her bits arrive, Bob measures his qubit in the 0/1 basis. What is his probability of getting 0?

P(0)=  ?P(0) = \;?

Hints
  1. Bob doesn’t know M1M2M_1M_2. What are his four possible states, and how likely is each?
  2. Each has weight 14\tfrac14. Average their chances of giving 0.
Show the answer

Answer: P(0)=12P(0) = \frac{1}{2}

Bob’s four possible states give 0 with probabilities 925\tfrac{9}{25}, 1625\tfrac{16}{25}, 925\tfrac{9}{25}, 1625\tfrac{16}{25} (results 00, 01, 10, 11), each with weight 14\tfrac14. The average is 50100=12\tfrac{50}{100} = \tfrac12.

Common slips

  • 925\frac{9}{25}: That is ∣α∣2\abs\alpha^2, what ∣ψ⟩\ket\psi itself would give. Bob doesn’t hold ∣ψ⟩\ket\psi yet; he holds one of four versions of it, each with probability 14\tfrac14, and averaged over them P(0)=12P(0) = \tfrac12.
  • 1625\frac{16}{25}: That is ∣β∣2\abs\beta^2, the 0/1 statistics of X∣ψ⟩X\ket\psi, which is only one of Bob’s four possible states. He doesn’t know which one he has, so average over all four.
  • 11: That assumes Bob’s qubit is still ∣0⟩\ket0. Since the Bell pair was made it has had no state of its own, and measured alone in the 0/1 basis it gives 0 with probability 12\tfrac12.
  • 14\frac{1}{4}: 14\tfrac14 is the chance of each of Alice’s four results, not Bob’s chance of 0. Weight each of his four possible states by 14\tfrac14 and add their chances of giving 0: 14(925+1625+925+1625)=12\tfrac14\big(\tfrac{9}{25} + \tfrac{16}{25} + \tfrac{9}{25} + \tfrac{16}{25}\big) = \tfrac12.

9 / 15 · Learn

Part B · Bob’s corrections

When the bits arrive, Bob knows which group survived. His four possible states are ∣ψ⟩\ket\psi itself, or ∣ψ⟩\ket\psi with the amplitudes swapped by XX, a sign flipped by ZZ, or both:

∣ψ3(00)⟩=∣ψ⟩∣ψ3(01)⟩=X∣ψ⟩∣ψ3(10)⟩=Z∣ψ⟩∣ψ3(11)⟩=XZ∣ψ⟩\begin{aligned}\ket{\psi_3(00)} &= \ket\psi \\ \ket{\psi_3(01)} &= X\ket\psi \\ \ket{\psi_3(10)} &= Z\ket\psi \\ \ket{\psi_3(11)} &= XZ\ket\psi\end{aligned}

Each is undone by the same gates in reverse order. The whole rule fits in one line: Bob applies ZM1XM2Z^{M_1}X^{M_2}. Read it right to left, the way operators act: first XX if M2=1M_2 = 1, then ZZ if M1=1M_1 = 1. Since X2=Z2=IX^2 = Z^2 = I, this restores ∣ψ⟩\ket\psi in all four cases.

Refresher: Matrices acting on states Why ZM1XM2Z^{M_1}X^{M_2} means XX first: in a product of operators, the one next to the ket acts first.

Check yourself

Alice reports M1M2=01M_1M_2 = 01. Which gates does Bob apply, in time order?

  1. ANothing
  2. BZZ
  3. CXX
  4. DXX, then ZZ
Show the answer

Answer: C. XX

After result 0101 Bob’s qubit is in β∣0⟩+α∣1⟩\beta\ket0 + \alpha\ket1. Applying XX gives α∣0⟩+β∣1⟩\alpha\ket0 + \beta\ket1. That is ZM1XM2Z^{M_1}X^{M_2} with M1=0M_1 = 0 and M2=1M_2 = 1.

Why not the others

  • A. After result 0101 Bob’s qubit is in β∣0⟩+α∣1⟩\beta\ket0 + \alpha\ket1, which is not ∣ψ⟩\ket\psi for general α\alpha and β\beta, so Bob has to act.
  • B. Applying ZZ turns Bob’s β∣0⟩+α∣1⟩\beta\ket0 + \alpha\ket1 into β∣0⟩−α∣1⟩\beta\ket0 - \alpha\ket1, not ∣ψ⟩\ket\psi. It looks as if M1M_1 and M2M_2 were swapped: M2M_2 (from Alice’s half of the pair) controls XX, and M1M_1 (from the ∣ψ⟩\ket\psi qubit) controls ZZ.
  • D. Applying XX and then ZZ turns Bob’s β∣0⟩+α∣1⟩\beta\ket0 + \alpha\ket1 into α∣0⟩−β∣1⟩\alpha\ket0 - \beta\ket1, not ∣ψ⟩\ket\psi.

Check yourself

Alice reports M1M2=11M_1M_2 = 11. Build Bob’s correction on his wire, in time order (left to right). The checker tries your gates on every input, so the ∣0⟩\ket0 drawn at the start of the wire doesn’t matter.

Build it in the circuit builder (gates: X, Z).

Hints
  1. Bob’s state after 1111 is XZ∣ψ⟩=α∣1⟩−β∣0⟩XZ\ket\psi = \alpha\ket1 - \beta\ket0. Which gates undo XZXZ?
  2. Undo the last gate first: XX, then ZZ.
Show the answer

One circuit that works: X on q0, then Z on q0.

Z1X1Z^{1}X^{1}: XX first, then ZZ, turns XZ∣ψ⟩XZ\ket\psi back into ∣ψ⟩\ket\psi. The checker compares circuits up to a global phase, so ZZ first was accepted too: it gives −∣ψ⟩-\ket\psi, which the next question looks at.

Check yourself

On result 1111, Bob applies ZZ first and then XX, and ends with −∣ψ⟩-\ket\psi. Did the teleportation work?

  1. AYes: −∣ψ⟩-\ket\psi differs from ∣ψ⟩\ket\psi only by a global phase, which no measurement can detect
  2. BNo: −∣ψ⟩-\ket\psi gives the opposite measurement results
  3. CNo: they must start again with a fresh Bell pair
  4. DOnly after Bob also applies −I-I to remove the sign
Hint
  1. Compare the probabilities that −∣ψ⟩-\ket\psi and ∣ψ⟩\ket\psi give for any measurement (Lesson 2).
Show the answer

Answer: A. Yes: −∣ψ⟩-\ket\psi differs from ∣ψ⟩\ket\psi only by a global phase, which no measurement can detect

XZ=−ZXXZ = -ZX, so the reversed order gives −∣ψ⟩-\ket\psi. A global phase changes no probability in any basis, so Bob holds the teleported state.

Why not the others

  • B. A sign on the whole state looks like a big change, but probabilities are squared magnitudes: ∣−α∣2=∣α∣2\abs{-\alpha}^2 = \abs\alpha^2, in every basis. Only a relative sign, on one term, can change the statistics in some basis. So −∣ψ⟩-\ket\psi and ∣ψ⟩\ket\psi are the same physical state.
  • C. A wrong-looking sign suggests a failed run, but nothing needs repeating. The leftover −1-1 multiplies every amplitude, so Bob’s qubit behaves exactly like ∣ψ⟩\ket\psi in every experiment he could do.
  • D. Removing the sign seems tidy, but applying −I-I is physically the same as doing nothing: it too changes only the global phase. There is nothing to remove.

So on result 1111 the order of Bob’s two gates matters only up to a sign. Since XZ=−ZXXZ = -ZX, ZZ before XX leaves −∣ψ⟩-\ket\psi: every amplitude times the same −1-1, a global phase that no measurement in any basis can detect (Lesson 2). The order ZM1XM2Z^{M_1}X^{M_2}, XX first, gives exactly ∣ψ⟩\ket\psi, and that is the order this lesson asks for whenever a question wants the exact vector.

10 / 15 · Learn

Part B · Not a copy, and what it costs

Does teleportation copy ∣ψ⟩\ket\psi? Look at what Alice is left with. After her 0/1-basis measurement, her top qubit is in ∣M1⟩\ket{M_1} and her middle qubit in ∣M2⟩\ket{M_2}: plain basis states with no trace of α\alpha or β\beta. At the end, ∣ψ⟩\ket\psi exists in one place, Bob’s qubit, and it stopped existing on Alice’s side when she measured. That fits no-cloning (Lesson 7): the state moved; it wasn’t duplicated. And no qubit travelled during the protocol: Bob’s own qubit, delivered in advance as half of the pair, is the one that ends up in ∣ψ⟩\ket\psi.

Check yourself

A run ends with Alice’s result M1M2=10M_1M_2 = 10 and Bob’s correction done. What state is Alice’s top qubit, the one that held ∣ψ⟩\ket\psi, in now?

  1. A∣1⟩\ket1
  2. BIt is still entangled with Bob’s qubit
  3. C∣ψ⟩\ket\psi: Alice keeps the original and Bob gets a copy
  4. D∣0⟩\ket0: the protocol resets it
Show the answer

Answer: A. ∣1⟩\ket1

Alice measured her top qubit in the 0/1 basis and got 1, so it is ∣1⟩\ket1. Only Bob’s qubit ends up in ∣ψ⟩\ket\psi, so no copy was made.

Why not the others

  • B. They were entangled before the measurement, which makes this tempting. After the 0/1-basis measurement Alice’s qubits are in definite basis states, so they are entangled with nothing: the three-qubit state is ∣10⟩∣ψ3(10)⟩\ket{10}\ket{\psi_3(10)}, a product.
  • C. Many people picture teleportation as a fax: the original stays and a copy appears. Check Alice’s qubit: she measured it in the 0/1 basis and got M1=1M_1 = 1, so it is now ∣1⟩\ket1, with no α\alpha or β\beta left. That is exactly why teleportation doesn’t break no-cloning.
  • D. Qubits often start in ∣0⟩\ket0, so a reset sounds plausible. But a 0/1-basis measurement leaves a qubit in the basis state that was observed: here M1=1M_1 = 1, so it is ∣1⟩\ket1. Nothing in the protocol resets it.

Now count what was used: one Bell pair, shared in advance, plus two classical bits sent from Alice to Bob, moved one qubit’s state. Think of it as a trade between resources: a shared pair and two bits can stand in for a quantum channel that carries one qubit. Two bits are enough because they only have to say which of four known corrections to apply. The unlimited detail in α\alpha and β\beta never has to be sent.

Check yourself

Alice wants to teleport three qubits, one after another, this way. How many classical bits must she send Bob in total?

Hint
  1. How many bits does one teleported qubit need?
Show the answer

Answer: 66

Each qubit needs its own Bell pair and its own two results, so 3 pairs and 3×2=63 \times 2 = 6 bits.

Common slips

  • 33: One bit per qubit isn’t enough: Bob has to choose among four corrections (II, XX, ZZ or both), and that takes two bits for each qubit.
  • 22: Two bits cover one qubit. Each teleported qubit uses its own Bell pair and its own pair of results, so three qubits need 3×23\times 2 bits.
  • 00: Without the bits Bob’s qubits stay 50/50 in every basis and he can’t finish. The classical message is a required part of teleportation.

11 / 15 · Worked example

Part B · Pass 3: regroup and correct

Pass 3 is mostly yours: you pick out the surviving group and correct it. Enter states as eight amplitudes, ∣000⟩\ket{000} to ∣111⟩\ket{111}; these want the exact vector.

Worked example

Example: a complex amplitude, result 1111

Take ∣ψ⟩=32∣0⟩+i2∣1⟩\ket\psi = \tfrac{\sqrt3}{2}\ket0 + \tfrac{i}{2}\ket1. Every amplitude of ∣ψ2⟩\ket{\psi_2} is ±α/2\pm\alpha/2 or ±β/2\pm\beta/2, here 34\tfrac{\sqrt3}{4} or ±i4\pm\tfrac{i}{4}.

  1. After her H

    Alice’s CNOT and HH give, one line per pair of Alice’s labels:

    ∣ψ2⟩=34∣000⟩+i4∣001⟩+i4∣010⟩+34∣011⟩+34∣100⟩−i4∣101⟩−i4∣110⟩+34∣111⟩\begin{aligned}\ket{\psi_2} = {}&\tfrac{\sqrt3}{4}\ket{000} + \tfrac{i}{4}\ket{001} \\ + {}&\tfrac{i}{4}\ket{010} + \tfrac{\sqrt3}{4}\ket{011} \\ + {}&\tfrac{\sqrt3}{4}\ket{100} - \tfrac{i}{4}\ket{101} \\ - {}&\tfrac{i}{4}\ket{110} + \tfrac{\sqrt3}{4}\ket{111}\end{aligned}

  2. Regroup

    Collect by the first two labels. The 0000 group is 34∣000⟩+i4∣001⟩=12∣00⟩(32∣0⟩+i2∣1⟩)\tfrac{\sqrt3}{4}\ket{000} + \tfrac{i}{4}\ket{001} = \tfrac12\ket{00}\big(\tfrac{\sqrt3}{2}\ket0 + \tfrac{i}{2}\ket1\big), which is 12∣00⟩∣ψ⟩\tfrac12\ket{00}\ket\psi. The 0101 group is 12∣01⟩(i2∣0⟩+32∣1⟩)=12∣01⟩ X∣ψ⟩\tfrac12\ket{01}\big(\tfrac{i}{2}\ket0 + \tfrac{\sqrt3}{2}\ket1\big) = \tfrac12\ket{01}\,X\ket\psi.

  3. Alice measures

    Alice measures both of her qubits in the 0/1 basis and gets M1M2=11M_1M_2 = 11 (probability 14\tfrac14). Keep only the terms that start with 1111, and multiply by 2 to renormalize.

    Your turn

    Enter ∣ψ3⟩\ket{\psi_3}, the state right after Alice’s result 1111.

    Hints
    1. Which two of the eight terms start with 1111?
    2. They are −i4∣110⟩-\tfrac{i}{4}\ket{110} and 34∣111⟩\tfrac{\sqrt3}{4}\ket{111}. A probability of 14\tfrac14 means you divide by 1/4=12\sqrt{1/4} = \tfrac12.
    Show the answer

    Answer: −12 i∣110⟩+32∣111⟩-\frac{1}{2}\,i|110\rangle + \frac{\sqrt{3}}{2}|111\rangle

    ∣ψ3⟩=−i2∣110⟩+32∣111⟩=∣11⟩(−i2∣0⟩+32∣1⟩)\ket{\psi_3} = -\tfrac{i}{2}\ket{110} + \tfrac{\sqrt3}{2}\ket{111} = \ket{11}\big(-\tfrac{i}{2}\ket0 + \tfrac{\sqrt3}{2}\ket1\big). Bob’s factor is α∣1⟩−β∣0⟩=XZ∣ψ⟩\alpha\ket1 - \beta\ket0 = XZ\ket\psi.

  4. Bob corrects

    The bits 1111 reach Bob, so he applies Z1X1Z^1X^1: first XX, then ZZ.

    Your turn

    Enter ∣ψ4⟩\ket{\psi_4}, the three-qubit state after Bob’s XX and then ZZ.

    Hints
    1. Work on Bob’s factor −i2∣0⟩+32∣1⟩-\tfrac{i}{2}\ket0 + \tfrac{\sqrt3}{2}\ket1; Alice’s ∣11⟩\ket{11} just comes along.
    2. XX swaps the two amplitudes; ZZ then flips the sign of the ∣1⟩\ket1 amplitude.
    Show the answer

    Answer: 32∣110⟩+12 i∣111⟩\frac{\sqrt{3}}{2}|110\rangle + \frac{1}{2}\,i|111\rangle

    XX gives 32∣0⟩−i2∣1⟩\tfrac{\sqrt3}{2}\ket0 - \tfrac{i}{2}\ket1, and ZZ gives 32∣0⟩+i2∣1⟩=∣ψ⟩\tfrac{\sqrt3}{2}\ket0 + \tfrac{i}{2}\ket1 = \ket\psi. So ∣ψ4⟩=∣11⟩∣ψ⟩\ket{\psi_4} = \ket{11}\ket\psi: Alice holds plain ∣1⟩∣1⟩\ket1\ket1, Bob holds ∣ψ⟩\ket\psi. (With ZZ first you would get −∣11⟩∣ψ⟩-\ket{11}\ket\psi: the same physical state, but not the exact vector asked for.)

One more, fully worked, with a different state and result.

Another worked example

Another example: α=3/5\alpha = 3/5, β=4/5\beta = 4/5, result 0101

  1. Regroup

    Take ∣ψ⟩=35∣0⟩+45∣1⟩\ket\psi = \tfrac35\ket0 + \tfrac45\ket1. Every amplitude of ∣ψ2⟩\ket{\psi_2} is α/2=310\alpha/2 = \tfrac{3}{10} or ±β/2=±25\pm\beta/2 = \pm\tfrac25. The 0101 group is 25∣010⟩+310∣011⟩=12∣01⟩(45∣0⟩+35∣1⟩)\tfrac25\ket{010} + \tfrac{3}{10}\ket{011} = \tfrac12\ket{01}\big(\tfrac45\ket0 + \tfrac35\ket1\big).

  2. Alice measures

    Result M1M2=01M_1M_2 = 01 keeps that group and doubles it: ∣ψ3⟩=∣01⟩(45∣0⟩+35∣1⟩)\ket{\psi_3} = \ket{01}\big(\tfrac45\ket0 + \tfrac35\ket1\big). Bob’s factor is ∣ψ⟩\ket\psi with its amplitudes swapped, X∣ψ⟩X\ket\psi.

  3. Bob corrects

    Z0X1=XZ^0X^1 = X. Bob applies XX and gets 35∣0⟩+45∣1⟩=∣ψ⟩\tfrac35\ket0 + \tfrac45\ket1 = \ket\psi, so ∣ψ4⟩=∣01⟩∣ψ⟩\ket{\psi_4} = \ket{01}\ket\psi.

Check yourself

Teleport ∣ψ⟩=12∣0⟩+1+i2∣1⟩\ket\psi = \tfrac{1}{\sqrt2}\ket0 + \tfrac{1+i}{2}\ket1. Alice measures both of her qubits in the 0/1 basis and gets M1M2=01M_1M_2 = 01. Enter Bob’s qubit, as two amplitudes, right after her measurement and before any correction.

Hints
  1. Which Bob factor sits next to ∣01⟩\ket{01} in the regrouped ∣ψ2⟩\ket{\psi_2}?
  2. It is β∣0⟩+α∣1⟩\beta\ket0 + \alpha\ket1. Now substitute α\alpha and β\beta.
Show the answer

Answer: ∣0⟩=1+i2,∣1⟩=12\ket0 = \frac{1+i}{2},\quad \ket1 = \frac{1}{\sqrt{2}}

The 0101 group of the regrouped ∣ψ2⟩\ket{\psi_2} is 12∣01⟩(β∣0⟩+α∣1⟩)\tfrac12\ket{01}(\beta\ket0 + \alpha\ket1). With α=12\alpha = \frac{1}{\sqrt{2}} and β=1+i2\beta = \frac{1+i}{2}, Bob holds 1+i2∣0⟩+12∣1⟩\frac{1+i}{2}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle, and the fix Z0X1Z^{0}X^{1} turns it back into ∣ψ⟩\ket\psi.

12 / 15 · Practice

Practice

Three items, each with new numbers when it comes back. Hints are there if you need them.

Check yourself

Teleport ∣ψ⟩=12∣0⟩+1+i2∣1⟩\ket\psi = \tfrac{1}{\sqrt2}\ket0 + \tfrac{1+i}{2}\ket1. What is the amplitude of ∣110⟩\ket{110} in ∣ψ2⟩\ket{\psi_2}, the state right after Alice’s HH?

Hints
  1. Which of ±α/2\pm\alpha/2 and ±β/2\pm\beta/2 belongs on ∣110⟩\ket{110}? Look at the last two labels first.
  2. The α\alpha branch ends in 0000 or 1111, the β\beta branch in 0101 or 1010. A minus sign appears only on β\beta terms whose first label is 1.
  3. So the amplitude is −β/2-\beta/2. Now substitute β=1+i2\beta = \frac{1+i}{2}.
Show the answer

Answer: −1+i4-\frac{1+i}{4}

∣ψ2⟩=12[α(∣0⟩+∣1⟩)(∣00⟩+∣11⟩)+β(∣0⟩−∣1⟩)(∣10⟩+∣01⟩)]\ket{\psi_2} = \tfrac12\big[\alpha(\ket0+\ket1)(\ket{00}+\ket{11}) + \beta(\ket0-\ket1)(\ket{10}+\ket{01})\big], so ∣110⟩\ket{110} has amplitude −β/2-\beta/2. With α=12\alpha = \frac{1}{\sqrt{2}} and β=1+i2\beta = \frac{1+i}{2} that is −1+i4-\frac{1+i}{4}.

Common slips

  • 1+i4\frac{1+i}{4}: The sign is off. Minus signs come only from H∣1⟩=(∣0⟩−∣1⟩)/2H\ket1 = (\ket0 - \ket1)/\sqrt2, so they sit on the β\beta terms whose first label is 1: ∣101⟩\ket{101} and ∣110⟩\ket{110}.
  • −24−24i-\frac{\sqrt{2}}{4}-\frac{\sqrt{2}}{4}i: That has a factor 1/21/\sqrt2 where 12\tfrac12 belongs: it is the size of a ∣ψ1⟩\ket{\psi_1} amplitude. Alice’s HH brings a second factor 1/21/\sqrt2, so every amplitude of ∣ψ2⟩\ket{\psi_2} is ±α/2\pm\alpha/2 or ±β/2\pm\beta/2.
  • −24-\frac{\sqrt{2}}{4}: You used the other amplitude. Terms whose last two labels are 0000 or 1111 carry α\alpha; terms ending in 0101 or 1010 carry β\beta, because Alice’s CNOT turned the β\beta branch’s pair into ∣10⟩+∣01⟩\ket{10} + \ket{01}.

Check yourself

Alice and Bob run the protocol with a different shared pair, ∣β01⟩=(∣01⟩+∣10⟩)/2\ket{\beta_{01}} = (\ket{01}+\ket{10})/\sqrt2, instead of ∣β00⟩\ket{\beta_{00}}. Everything else is the same. Alice reports M1M2=11M_1M_2 = 11. Which gates does Bob apply, in time order?

  1. ANothing
  2. BXX
  3. CZZ
  4. DXX, then ZZ
Hints
  1. Redo the regrouping with this pair: which Bob state sits next to ∣11⟩\ket{11}?
  2. This pair is ∣β00⟩\ket{\beta_{00}} with an XX on Bob’s half. That gate rides along, so each of Bob’s four states picks it up too.
  3. Bob’s state is α∣0⟩−β∣1⟩\alpha\ket0 - \beta\ket1. Which gates turn it back into α∣0⟩+β∣1⟩\alpha\ket0 + \beta\ket1, up to an overall sign?
Show the answer

Answer: C. ZZ

After result 1111 Bob’s qubit is in α∣0⟩−β∣1⟩\alpha\ket0 - \beta\ket1. Applying ZZ gives α∣0⟩+β∣1⟩\alpha\ket0 + \beta\ket1. With this pair the rule becomes ZM1X1⊕M2Z^{M_1}X^{1\oplus M_2}, XX first.

∣β01⟩\ket{\beta_{01}} is ∣β00⟩\ket{\beta_{00}} with an XX applied to Bob’s half. A gate on Bob’s qubit alone commutes with everything Alice does, so it simply rides along. After result 1111 Bob holds α∣0⟩−β∣1⟩\alpha\ket0 - \beta\ket1; applying ZZ gives α∣0⟩+β∣1⟩\alpha\ket0 + \beta\ket1. In general the fix becomes ZM1X1⊕M2Z^{M_1}X^{1\oplus M_2}, XX first.

Why not the others

  • A. After result 1111 Bob’s qubit is in α∣0⟩−β∣1⟩\alpha\ket0 - \beta\ket1, which is not ∣ψ⟩\ket\psi for general α\alpha and β\beta, so Bob has to act.
  • B. Applying XX turns Bob’s α∣0⟩−β∣1⟩\alpha\ket0 - \beta\ket1 into −β∣0⟩+α∣1⟩-\beta\ket0 + \alpha\ket1, not ∣ψ⟩\ket\psi.
  • D. Applying XX and then ZZ turns Bob’s α∣0⟩−β∣1⟩\alpha\ket0 - \beta\ket1 into −β∣0⟩−α∣1⟩-\beta\ket0 - \alpha\ket1, not ∣ψ⟩\ket\psi. Applying XX and then ZZ would be right for result 1111 with the usual pair ∣β00⟩\ket{\beta_{00}}; with ∣β01⟩\ket{\beta_{01}} Bob’s half starts out with an XX already applied.

Check yourself

Alice teleports ∣ψ⟩=35∣0⟩+45∣1⟩\ket\psi = \tfrac35\ket0 + \tfrac45\ket1. She has measured, and her two bits are on their way. Bob measures his qubit in the 0/1 basis. What is the probability that his result is 11?

Hints
  1. Bob doesn’t know M1M2M_1M_2. What are his four possible states, and how likely is each?
  2. Each of the four has weight 14\tfrac14; average their probabilities for this outcome.
  3. Swapping the amplitudes (XX) and flipping a sign (ZZ) cancel any bias: the four Bloch vectors average to the centre.
Show the answer

Answer: 12\frac{1}{2}

Bob’s qubit, on its own, gives each outcome with probability 12\tfrac12 in every basis at this stage. Once Alice has measured, it is one of ∣ψ⟩\ket\psi, X∣ψ⟩X\ket\psi, Z∣ψ⟩Z\ket\psi, XZ∣ψ⟩XZ\ket\psi, each with probability 14\tfrac14, and these four average to the centre of the Bloch sphere. Before that, it is half of an entangled state, again 50/50. So the answer is 12\tfrac12, not the 1625\frac{16}{25} that ∣ψ⟩\ket\psi itself would give.

Common slips

  • 1625\frac{16}{25}: That is what ∣ψ⟩\ket\psi itself would give. Bob doesn’t hold ∣ψ⟩\ket\psi yet: until Alice’s bits arrive and he corrects, his qubit is 50/50 in every basis, whatever ∣ψ⟩\ket\psi is. If this number could reach him before the bits did, teleportation would signal faster than light.
  • 00: That assumes Bob’s qubit is still the ∣0⟩\ket0 it started as. Since the Bell pair was made, his qubit has had no state of its own; measured alone in the 0/1 basis it gives 0 or 1 with probability 12\tfrac12 each.
  • 14\frac{1}{4}: That is the probability of one of Alice’s four results, not of Bob’s outcome. At this stage Bob’s qubit, on its own, gives each outcome with probability 12\tfrac12, in every basis.
  • 11: That would make Bob’s result certain, one way or the other. At this stage his qubit, on its own, gives each outcome with probability 12\tfrac12 in every basis. Only after the bits arrive and he corrects does it behave like ∣ψ⟩\ket\psi.

13 / 15 · Explain

Explain it

Explain it

Alice’s measurement changes the three-qubit state instantly, however far away Bob is. Why can’t teleportation send Bob information faster than light?

Show a model answer

Before Alice’s bits arrive, Bob’s qubit is one of four versions of ∣ψ⟩\ket\psi, each with probability 14\tfrac14, and averaged over them it is completely random: P(0)=12P(0) = \tfrac12, in every basis, whatever ∣ψ⟩\ket\psi is. So nothing Bob can measure depends on ∣ψ⟩\ket\psi or on what Alice did. Only her two classical bits, which travel no faster than light, tell him which of II, XX, ZZ or ZXZX to apply, and only then does his qubit hold ∣ψ⟩\ket\psi.

A good answer covers

  • Before the bits arrive, Bob’s statistics don’t depend on ∣ψ⟩\ket\psi or on Alice’s actions: his qubit is 50/50
  • Bob needs the two classical bits, which travel no faster than light, to choose the correction
  • Each of Alice’s four results has probability 14\tfrac14
  • No copy is made: Alice’s measurement leaves her qubits in basis states

14 / 15 · Mastery check

Mastery check

Check yourself

Alice and Bob already share a Bell pair. Alice now wants Bob to end up with a qubit in an unknown state ∣ψ⟩\ket\psi that she holds. What must she send him?

  1. AHer qubit itself, through a quantum channel
  2. BTwo classical bits
  3. CNothing more: the shared pair carries ∣ψ⟩\ket\psi over
  4. DThe numbers α\alpha and β\beta, written out
Show the answer

Answer: B. Two classical bits

Alice sends two classical bits, her measurement results M1M_1 and M2M_2. They tell Bob which of four fixed corrections to apply.

Why not the others

  • A. Sending the qubit would work, but the protocol avoids it. Once the pair is shared, only classical bits go from Alice to Bob, and her qubit stays with her, ending in ∣0⟩\ket0 or ∣1⟩\ket1. It seems something quantum must travel; it did, in advance, as Bob’s half of the pair.
  • C. A common picture is that entanglement does the carrying. But nothing Alice does to her half changes what Bob can measure: his qubit stays 50/50 in every basis until a message arrives. He needs her two measurement results to finish, and they travel no faster than light. The name “teleportation” makes the instant picture tempting.
  • D. It feels as if a measurement would let Alice read off α\alpha and β\beta. One measurement of one copy, in any basis, gives one bit and replaces the state, so she can never learn them; and an exact complex number would need unlimited digits anyway. The protocol works without anyone learning α\alpha or β\beta.

Check yourself

Teleport ∣ψ⟩=12∣0⟩+1+i2∣1⟩\ket\psi = \tfrac{1}{\sqrt2}\ket0 + \tfrac{1+i}{2}\ket1. What is the amplitude of ∣101⟩\ket{101} in ∣ψ1⟩\ket{\psi_1}, the state right after Alice’s CNOT? (It may be 0.)

Show the answer

Answer: 24+24i\frac{\sqrt{2}}{4}+\frac{\sqrt{2}}{4}i

∣ψ1⟩=12[α∣0⟩(∣00⟩+∣11⟩)+β∣1⟩(∣10⟩+∣01⟩)]\ket{\psi_1} = \tfrac{1}{\sqrt2}\big[\alpha\ket0(\ket{00}+\ket{11}) + \beta\ket1(\ket{10}+\ket{01})\big]: the terms starting with 1 are ∣110⟩\ket{110} and ∣101⟩\ket{101}, each with β/2\beta/\sqrt2, and ∣100⟩\ket{100}, ∣111⟩\ket{111} are empty. Here β=1+i2\beta = \frac{1+i}{2}, so the amplitude of ∣101⟩\ket{101} is 24+24i\frac{\sqrt{2}}{4}+\frac{\sqrt{2}}{4}i.

Common slips

  • 00: That is the amplitude of ∣101⟩\ket{101} in ∣ψ0⟩\ket{\psi_0}, before the CNOT. CNOT flips the middle label of every term whose first label is 1, so β∣100⟩/2\beta\ket{100}/\sqrt2 moves to ∣110⟩\ket{110} and β∣111⟩/2\beta\ket{111}/\sqrt2 moves to ∣101⟩\ket{101}.
  • −1+i4-\frac{1+i}{4}: That is the amplitude in ∣ψ2⟩\ket{\psi_2}, after Alice’s HH as well. The question stops at ∣ψ1⟩\ket{\psi_1}, right after the CNOT.
  • 12\frac{1}{2}: That uses α\alpha, but the terms starting with 1 carry β\beta: they come from β∣1⟩∣β00⟩\beta\ket1\ket{\beta_{00}}.

Check yourself

Alice reports M1M2=01M_1M_2 = 01. Which gates does Bob apply to his qubit, in time order?

  1. ANothing
  2. BXX
  3. CZZ
  4. DXX, then ZZ
Show the answer

Answer: B. XX

After result 0101 Bob’s qubit is in β∣0⟩+α∣1⟩\beta\ket0 + \alpha\ket1. Applying XX gives α∣0⟩+β∣1⟩\alpha\ket0 + \beta\ket1. That is ZM1XM2Z^{M_1}X^{M_2} with M1=0M_1 = 0 and M2=1M_2 = 1.

Why not the others

  • A. After result 0101 Bob’s qubit is in β∣0⟩+α∣1⟩\beta\ket0 + \alpha\ket1, which is not ∣ψ⟩\ket\psi for general α\alpha and β\beta, so Bob has to act.
  • C. Applying ZZ turns Bob’s β∣0⟩+α∣1⟩\beta\ket0 + \alpha\ket1 into β∣0⟩−α∣1⟩\beta\ket0 - \alpha\ket1, not ∣ψ⟩\ket\psi. It looks as if M1M_1 and M2M_2 were swapped: M2M_2 (from Alice’s half of the pair) controls XX, and M1M_1 (from the ∣ψ⟩\ket\psi qubit) controls ZZ.
  • D. Applying XX and then ZZ turns Bob’s β∣0⟩+α∣1⟩\beta\ket0 + \alpha\ket1 into α∣0⟩−β∣1⟩\alpha\ket0 - \beta\ket1, not ∣ψ⟩\ket\psi.

Check yourself

Alice and Bob share a Bell pair ∣β00⟩\ket{\beta_{00}} and nothing else: no phone, no radio. Can Alice use the pair alone to send Bob one bit instantly, for example by choosing whether or not to measure her half?

  1. ANo: whatever Alice does to her half, Bob’s results have the same statistics
  2. BYes: measuring her half changes Bob’s qubit at once, and Bob can detect the change
  3. CNo, because entanglement fades with distance
  4. DYes, if they agreed in advance which basis Alice would measure in
Show the answer

Answer: A. No: whatever Alice does to her half, Bob’s results have the same statistics

Bob’s half of ∣β00⟩\ket{\beta_{00}} gives 0 or 1 with probability 12\tfrac12 in every basis, and nothing Alice does to her half changes that. A signal would need Bob’s statistics to depend on Alice’s choice, so the pair alone sends nothing.

Why not the others

  • B. It seems Bob should notice the change. Check it: if Alice measures in the 0/1 basis, Bob’s 0/1 measurement gives 0 or 1 with probability 12\tfrac12 each; if she doesn’t measure, it still gives 12\tfrac12 each. Nothing Bob sees on his own depends on what Alice did. The correlation shows up only when they compare notes, which needs a message.
  • C. Distance by itself doesn’t weaken the pair: in the ideal case the qubits stay entangled however far apart they are. The real reason is that Bob’s statistics on his own qubit never depend on what Alice does. Real entanglement is fragile, which makes this reason tempting, but that is noise, not distance.
  • D. Agreeing on a basis doesn’t help. Whatever basis Alice measures in, and whether she measures at all, Bob’s own results are 50/50 in every basis he picks. The correlations are real, but Bob can see them only by comparing his list with Alice’s, and that takes ordinary communication.

Check yourself

Alice and Bob run the protocol with a different shared pair, ∣β11⟩=(∣01⟩−∣10⟩)/2\ket{\beta_{11}} = (\ket{01}-\ket{10})/\sqrt2, instead of ∣β00⟩\ket{\beta_{00}}. Everything else is the same. Alice reports M1M2=01M_1M_2 = 01. Which gates does Bob apply, in time order?

  1. ANothing
  2. BXX
  3. CZZ
  4. DXX, then ZZ
Show the answer

Answer: C. ZZ

After result 0101 Bob’s qubit is in −α∣0⟩+β∣1⟩-\alpha\ket0 + \beta\ket1. Applying ZZ gives −α∣0⟩−β∣1⟩-\alpha\ket0 - \beta\ket1, which is ∣ψ⟩\ket\psi up to the global phase −1-1. With this pair the rule becomes Z1⊕M1X1⊕M2Z^{1\oplus M_1}X^{1\oplus M_2}, XX first.

∣β11⟩\ket{\beta_{11}} is ∣β00⟩\ket{\beta_{00}} with a ZZ and then an XX applied to Bob’s half. A gate on Bob’s qubit alone commutes with everything Alice does, so it simply rides along. After result 0101 Bob holds −α∣0⟩+β∣1⟩-\alpha\ket0 + \beta\ket1; applying ZZ gives −α∣0⟩−β∣1⟩-\alpha\ket0 - \beta\ket1, which is ∣ψ⟩\ket\psi up to a global phase. In general the fix becomes Z1⊕M1X1⊕M2Z^{1\oplus M_1}X^{1\oplus M_2}, XX first.

Why not the others

  • A. After result 0101 Bob’s qubit is in −α∣0⟩+β∣1⟩-\alpha\ket0 + \beta\ket1, which is not ∣ψ⟩\ket\psi for general α\alpha and β\beta, so Bob has to act.
  • B. Applying XX turns Bob’s −α∣0⟩+β∣1⟩-\alpha\ket0 + \beta\ket1 into β∣0⟩−α∣1⟩\beta\ket0 - \alpha\ket1, not ∣ψ⟩\ket\psi. Applying XX would be right for result 0101 with the usual pair ∣β00⟩\ket{\beta_{00}}; with ∣β11⟩\ket{\beta_{11}} Bob’s half starts out with a ZZ and then an XX already applied.
  • D. Applying XX and then ZZ turns Bob’s −α∣0⟩+β∣1⟩-\alpha\ket0 + \beta\ket1 into β∣0⟩+α∣1⟩\beta\ket0 + \alpha\ket1, not ∣ψ⟩\ket\psi.

15 / 15 · Recap

Recap

From memory

Before reading the summary, write down what you remember. Pulling it back from memory is what makes it last.

  1. From memory: what does Alice do, in order, and what does Bob do?

    Show the key points
    • Alice: CNOT from her ∣ψ⟩\ket\psi qubit onto her half of the pair, then HH on the ∣ψ⟩\ket\psi qubit
    • She measures both qubits in the 0/1 basis and sends the two bits M1M2M_1M_2
    • Bob applies XX if M2=1M_2 = 1, then ZZ if M1=1M_1 = 1: the correction ZM1XM2Z^{M_1}X^{M_2}

  2. Write the regrouped ∣ψ2⟩\ket{\psi_2} and say what each group means.

    Show the key points
    • 12[∣00⟩(α∣0⟩+β∣1⟩)+∣01⟩(α∣1⟩+β∣0⟩)+∣10⟩(α∣0⟩−β∣1⟩)+∣11⟩(α∣1⟩−β∣0⟩)]\tfrac12[\ket{00}(\alpha\ket0+\beta\ket1) + \ket{01}(\alpha\ket1+\beta\ket0) + \ket{10}(\alpha\ket0-\beta\ket1) + \ket{11}(\alpha\ket1-\beta\ket0)]
    • Each of Alice’s results has probability 14\tfrac14, whatever ∣ψ⟩\ket\psi is
    • The Bob factor next to ∣M1M2⟩\ket{M_1M_2} is Bob’s state once Alice gets that result

  3. Why does teleportation send nothing faster than light, and why is it not cloning?

    Show the key points
    • Before the bits arrive, Bob’s qubit is 50/50 in every basis, whatever ∣ψ⟩\ket\psi is
    • The two classical bits are needed, and they travel no faster than light
    • Alice ends with plain ∣M1⟩∣M2⟩\ket{M_1}\ket{M_2}: her original is gone, so there is one ∣ψ⟩\ket\psi at the end, not two

Key ideas

  1. Resources: one shared Bell pair plus two classical bits move one qubit’s state. Once ∣ψ⟩\ket\psi is in play, no qubit travels.

  2. Alice’s CNOT and HH turn ∣ψ⟩∣β00⟩\ket\psi\ket{\beta_{00}} into ∣ψ2⟩=12∑M1M2∣M1M2⟩ XM2ZM1∣ψ⟩\ket{\psi_2} = \tfrac12\sum_{M_1M_2}\ket{M_1M_2}\,X^{M_2}Z^{M_1}\ket\psi: the same vector, regrouped by Alice’s results.

  3. Each result has probability 14\tfrac14. Bob’s fix is ZM1XM2Z^{M_1}X^{M_2}, XX first; the other order gives −∣ψ⟩-\ket\psi, the same state up to a global phase.

  4. Until the bits arrive Bob’s qubit is 50/50 in every basis: no faster-than-light signal.

  5. Alice is left with ∣M1⟩∣M2⟩\ket{M_1}\ket{M_2}: the state moved; it wasn’t copied.

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