A Three-Qubit Circuit

Place H, X, Z and CNOT gates on three qubits and step through the state. Then measure a Bell pair and see two random results that always agree.

Quantum computingEntanglementdemo

Objectives

Trace the eight amplitudes of a three-qubit state through a circuit, one gate at a time.
Explain why H, Z, H acts as a bit flip, using the sign of an amplitude.
Read a joint probability table and explain what entanglement does and does not do.

33%

Place gates on three qubits, step through the state, measure it, and find entangled results.

H|0> = (|0> + |1>)/√2 H|1> = (|0> − |1>)/√2 CNOT|c t> = |c, t ⊕ c>

Build a circuit on three qubits

A quantum circuit is a list of gates applied to qubits from left to right. Each qubit starts as |0>. Pick a ready-made circuit or add gates yourself, then move the slider to see the state after each gate.

ready-made circuit

gate

qubit

gates applied so far

2 of 2

After gate 2, CNOT q0 → q1. Only |000> and |110> remain. Each of qubits 0 and 1 alone is 50/50, but the two always agree.

The amplitude and probability of each basis state after 2 gates.
State \|q0 q1 q2>	Amplitude	Probability
\|000>	0.707	50.0%
\|001>	0.000	0.0%
\|010>	0.000	0.0%
\|011>	0.000	0.0%
\|100>	0.000	0.0%
\|101>	0.000	0.0%
\|110>	0.707	50.0%
\|111>	0.000	0.0%

Rows with a rule at the left have a nonzero amplitude. A negative amplitude has the same probability as a positive one of the same size.

Measure the final state

Measuring all three qubits returns one basis state, with the probabilities from the table at the end of the circuit. Run the circuit again on fresh qubits to get another result.

Entanglement: results that always agree

Pick two qubits. The table gives the probability of each pair of results at the end of the circuit. If the qubits were separate, the table would be the product of their own probabilities. Try the Bell pair and the GHZ state.

first qubit

second qubit

Joint probabilities of qubit 0 and qubit 1 at the end of the circuit.
q0 \ q1	q1 = 0	q1 = 1
q0 = 0	50.0%	0.0%
q0 = 1	0.0%	50.0%

P(q0 = 1) = 50.0% P(q1 = 1) = 50.0% P(same result) = 100.0%

what the table shows

Correlated. The table is not the product of the two qubits' own probabilities. Separate (unentangled) qubits could not do this, so these qubits are entangled.

Entanglement does not send a message: each qubit alone still gives a random result. Algorithms use it to tie the input register to the output register. In Shor's algorithm, entanglement links each x to a^x mod N.

step 1/3

CPE 645 / CS 685

A Three-Qubit Circuit

Qubits, Measurement, and Interference

Factoring from a Period