|q⟩
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Multiple choice
The lesson notes that plain angle encoding,
∣
ϕ
(
x
)
⟩
=
|\phi(x)\rangle =
∣
ϕ
(
x
)⟩
=
tensor of
R
y
(
x
j
)
∣
0
⟩
R_y(x_j)|0\rangle
R
y
(
x
j
)
∣0
⟩
, induces the kernel
K
(
x
,
x
′
)
=
K(x,x') =
K
(
x
,
x
′
)
=
product of
cos
2
(
(
x
j
−
x
j
′
)
/
2
)
\cos^2((x_j - x'_j)/2)
cos
2
((
x
j
−
x
j
′
)
/2
)
. Why does angle encoding alone give no quantum advantage?
It cannot encode real-valued data into qubit states at all
It entangles all qubits, producing a kernel that is provably hard to simulate classically
It requires exponentially deep circuits that no classical computer can match
It is a separable (unentangled) embedding whose kernel is a product of squared cosine similarities — a well-known classical kernel
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