QuantumCore

Hands-on lab · 2 steps

How good is the CNOT copier?

The CNOT gate maps ∣0⟩∣0⟩↦∣00⟩\lvert0\rangle\lvert0\rangle\mapsto\lvert00\rangle and ∣1⟩∣0⟩↦∣11⟩\lvert1\rangle\lvert0\rangle\mapsto\lvert11\rangle, so it copies basis states perfectly. Compute its copy fidelity F=∣⟨ψψ∣ CNOT ∣ψ0⟩∣2F=\lvert\langle\psi\psi\vert\,\mathrm{CNOT}\,\vert\psi0\rangle\rvert^2 for the superpositions ∣+⟩\lvert+\rangle and cos⁡π6∣0⟩+sin⁡π6∣1⟩\cos\tfrac\pi6\lvert0\rangle+\sin\tfrac\pi6\lvert1\rangle.

0/2 steps passedBest score 0/100

Step 1: Copy fidelity for |+⟩

The original is qubit 0 in the state ∣+⟩=12(∣0⟩+∣1⟩)\lvert+\rangle=\tfrac1{\sqrt2}(\lvert0\rangle+\lvert1\rangle) and the blank is qubit 1 in ∣0⟩\lvert0\rangle. Apply CNOT\mathrm{CNOT} (control 0, target 1) and compare the output with the ideal two-copy state ∣++⟩\lvert{+}{+}\rangle:

F=∣⟨++∣ CNOT ∣+0⟩∣2.F=\big\lvert\langle{+}{+}\vert\,\mathrm{CNOT}\,\vert{+}0\rangle\big\rvert^2 .

Enter FF as a real number. Decimals are accepted within 10−310^{-3}; an exact expression such as 1/2 is also accepted.

Enter exact expressions or decimals, e.g. 1/sqrt(2), pi/3. Values must be real.

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Fill every field with a valid expression.

Step 2: Copy fidelity for a real superposition

Now the original is

∣ψ⟩=cos⁡π6 ∣0⟩+sin⁡π6 ∣1⟩=32∣0⟩+12∣1⟩,\lvert\psi\rangle=\cos\tfrac\pi6\,\lvert0\rangle+\sin\tfrac\pi6\,\lvert1\rangle=\tfrac{\sqrt3}{2}\lvert0\rangle+\tfrac12\lvert1\rangle ,

with the blank again in ∣0⟩\lvert0\rangle. Compute F=∣⟨ψψ∣ CNOT ∣ψ0⟩∣2F=\lvert\langle\psi\psi\vert\,\mathrm{CNOT}\,\vert\psi0\rangle\rvert^2 and enter it as a real number. Decimals are accepted within 10−310^{-3}; an exact expression such as ((3*sqrt(3)+1)/8)^2 is also accepted.

Enter exact expressions or decimals, e.g. 1/sqrt(2), pi/3. Values must be real.

enter a value
Fill every field with a valid expression.