|q⟩
Bad Qubits
Play
Quest
Questions
Learn
Playground
☕
← Question Bank
Multiple choice
Why is the harmonic potential
V
(
x
)
=
1
2
m
ω
2
x
2
V(x) = \tfrac{1}{2}m\omega^2 x^2
V
(
x
)
=
2
1
m
ω
2
x
2
considered a universal approximation for any smooth potential near a stable equilibrium point
x
0
x_0
x
0
?
🔬 try it before you answer
Every physical potential is exactly parabolic at all distances by definition
The cubic term in the Taylor expansion always dominates near a minimum
A Taylor expansion about
x
0
x_0
x
0
has a vanishing linear term (since
V
′
(
x
0
)
=
0
V'(x_0)=0
V
′
(
x
0
)
=
0
), so the leading correction is the quadratic
1
2
V
′
′
(
x
0
)
(
x
−
x
0
)
2
\tfrac{1}{2}V''(x_0)(x-x_0)^2
2
1
V
′′
(
x
0
)
(
x
−
x
0
)
2
Stable equilibria require
V
′
′
(
x
0
)
<
0
V''(x_0) < 0
V
′′
(
x
0
)
<
0
, which forces a parabolic shape
Check answer