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The no-cloning proof assumes a unitary
U
U
U
with
U
∣
ψ
⟩
∣
0
⟩
=
∣
ψ
⟩
∣
ψ
⟩
U|\psi\rangle|0\rangle = |\psi\rangle|\psi\rangle
U
∣
ψ
⟩
∣0
⟩
=
∣
ψ
⟩
∣
ψ
⟩
for every state. Taking the inner product of this relation applied to two states
∣
ψ
⟩
|\psi\rangle
∣
ψ
⟩
and
∣
ϕ
⟩
|\phi\rangle
∣
ϕ
⟩
yields which equation, and what does it imply?
🔬 try it before you answer
⟨
ψ
∣
ϕ
⟩
=
⟨
ψ
∣
ϕ
⟩
2
\langle\psi|\phi\rangle = \langle\psi|\phi\rangle^2
⟨
ψ
∣
ϕ
⟩
=
⟨
ψ
∣
ϕ
⟩
2
, forcing the overlap to be 0 or 1, so no
U
U
U
clones arbitrary states
⟨
ψ
∣
ϕ
⟩
=
2
⟨
ψ
∣
ϕ
⟩
\langle\psi|\phi\rangle = 2\langle\psi|\phi\rangle
⟨
ψ
∣
ϕ
⟩
=
2
⟨
ψ
∣
ϕ
⟩
, which holds for all states, so cloning is always allowed
⟨
ψ
∣
ϕ
⟩
2
=
1
\langle\psi|\phi\rangle^2 = 1
⟨
ψ
∣
ϕ
⟩
2
=
1
, proving all states are identical
⟨
ψ
∣
ϕ
⟩
=
0
\langle\psi|\phi\rangle = 0
⟨
ψ
∣
ϕ
⟩
=
0
for all states, proving every pair of states is orthogonal
Check answer