Self-Assessment Quiz: Quantum Teleportation
Twenty questions on the teleportation protocol, its resource requirements, what it does and does not transmit, and its role as a primitive. Aim for 16+.
Question 1
Teleporting one qubit consumes:
A) One entangled pair and two classical bits B) Two entangled pairs and one classical bit C) One classical bit only D) Nothing — it is free
Question 2
The classical bits are needed because:
A) They carry the quantum state B) Alice's Bell measurement has four equally likely outcomes and Bob must know which C) They synchronize clocks D) They are optional
Question 3
Before the classical bits arrive, Bob's qubit is:
A) Already in the teleported state B) Maximally mixed — it carries no information C) In state $|0\rangle$ D) Entangled with Alice's measured qubit
Question 4
This is required by:
A) The uncertainty principle B) The no-signalling theorem C) The no-cloning theorem D) Conservation of energy
Question 5
After teleportation, the original qubit at Alice's location is:
A) Unchanged B) Destroyed (projected by the Bell measurement) C) Duplicated D) Entangled with Bob's
Question 6
That destruction is required by:
A) No-cloning B) No-signalling C) Unitarity alone D) Thermodynamics
Question 7
Alice's measurement in the protocol is:
A) A computational-basis measurement B) A Bell-basis measurement on her two qubits C) An $X$-basis measurement D) No measurement is needed
Question 8
Bob's correction for the outcome $10$ is typically:
A) $I$ B) $X$ C) $Z$ D) $XZ$
Question 9
Teleportation transmits information faster than light:
A) Yes B) No — the classical channel is limited by light speed C) Only for entangled senders D) Only over fibre
Question 10
Teleportation requires knowing the state being teleported:
A) Yes B) No — it works on unknown states, which is the point C) Only its basis D) Only its phase
Question 11
The entangled pair must be distributed:
A) After the teleportation B) In advance, before the unknown state arrives C) Simultaneously D) It is not needed
Question 12
Entanglement swapping is:
A) Teleporting one half of an entangled pair, entangling two parties who never interacted B) Exchanging two qubits C) A classical protocol D) Impossible
Question 13
Gate teleportation is used in fault tolerance to:
A) Reduce qubit count B) Apply hard gates (e.g. $T$) using prepared resource states C) Eliminate measurement D) Increase gate error
Question 14
Teleportation fidelity above what threshold cannot be achieved by any classical strategy on an unknown qubit?
A) 1/2 B) 2/3 C) 3/4 D) 1
Question 15
The teleportation circuit's classical control can be removed by:
A) The deferred measurement principle B) Removing the entangled pair C) Measuring in the $X$ basis D) It cannot
Question 16
True or false: Teleportation copies a quantum state.
Question 17
True or false: Teleportation could be used to build a faster-than-light telephone if the classical channel were fast enough.
Question 18
True or false: Teleportation is a routing primitive inside quantum computers, not just a communication protocol.
Question 19
Short answer. Explain why teleportation is consistent with the no-cloning theorem.
Question 20
Short answer. A colleague measures teleportation fidelity of 0.71 and claims success. Is that claim defensible?
Answer Key
| Q | Ans | Note |
|---|---|---|
| 1 | A | One ebit + two classical bits per qubit teleported. |
| 2 | B | The Bell measurement yields one of four outcomes uniformly; without knowing which, Bob cannot pick the right correction. Two bits label four outcomes. |
| 3 | B | Bob's reduced state is $I/2$ regardless of the input state, until the classical message arrives. |
| 4 | B | If Bob's local state depended on Alice's input or measurement, that dependence would be a signal, violating no-signalling. |
| 5 | B | The Bell measurement projects Alice's qubits, destroying the original. |
| 6 | A | If the original survived alongside Bob's copy, you would have cloned an unknown state. |
| 7 | B | Implemented as CNOT then $H$, followed by computational-basis measurement. |
| 8 | C | Conventionally the first bit selects $Z$ and the second selects $X$; conventions vary, so verify against your own circuit rather than memorizing. |
| 9 | B | The protocol is useless until two classical bits arrive over an ordinary channel. |
| 10 | B | Teleportation works on arbitrary unknown states — including halves of entangled pairs. If you knew the state you would just describe it classically. |
| 11 | B | Entanglement is a pre-shared resource, consumed by the protocol. |
| 12 | A | The basis of quantum repeaters: extend entanglement beyond the range of direct transmission. |
| 13 | B | $T$ gates are applied by consuming distilled magic states via gate teleportation (Ch. 25) — the standard fault-tolerant route to non-Clifford gates. |
| 14 | B | The classical limit for teleporting an unknown qubit is 2/3; exceeding it certifies genuinely quantum transmission. |
| 15 | A | Replace the classically controlled corrections with quantum-controlled gates and defer measurement to the end (Ch. 4). |
| 16 | False | The original is destroyed. It is a transfer, not a copy — the name is unfortunate. |
| 17 | False | The classical channel is a channel like any other and obeys relativity. Nothing in the protocol removes it. |
| 18 | True | On-chip, teleportation moves quantum information between distant regions without SWAP chains, and it is central to how magic states are consumed in fault-tolerant architectures. |
| 19 | — | Cloning requires two copies to coexist. In teleportation, Alice's Bell measurement destroys her copy at the moment Bob's is created, so there is never a time at which two copies exist. The protocol transfers rather than duplicates, and no-cloning is respected precisely because the original must be destroyed. |
| 20 | — | Partly. 0.71 exceeds the classical bound of 2/3 ≈ 0.667, so the transmission cannot be explained by a classical measure-and-resend strategy — a real result. But it is far from unit fidelity, and the claim needs an error bar: with a modest number of shots, 0.71 may not be statistically distinguishable from 0.667. The defensible statement is "fidelity 0.71 ± σ, exceeding the classical bound by N σ," with the state and shot count specified. |