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In time-independent perturbation theory one writes
H
=
H
(
0
)
+
λ
H
′
H = H^{(0)} + \lambda H'
H
=
H
(
0
)
+
λ
H
′
, where
H
(
0
)
H^{(0)}
H
(
0
)
is solvable. The first-order energy shift
E
n
(
1
)
E_n^{(1)}
E
n
(
1
)
turns out to equal which quantity?
The square of the energy gap to the nearest level
The expectation value of the perturbation in the unperturbed state,
⟨
n
(
0
)
∣
H
′
∣
n
(
0
)
⟩
\langle n^{(0)}| H' |n^{(0)}\rangle
⟨
n
(
0
)
∣
H
′
∣
n
(
0
)
⟩
The full unperturbed energy
E
n
(
0
)
E_n^{(0)}
E
n
(
0
)
Zero, because first-order shifts always vanish
Check answer