25.6 Examples
25.6.1 1D Harmonic Oscillator
Use
ψ=
(
a^2 −x^2) 2
|x|≤aandψ= 0 otherwise as a trial wave function. Recall the actual wave function ise−mωx
(^2) /2 ̄h 2
. The
energy estimate is
E′=
〈(
a^2 −x^2) 2
|− ̄h2
2 md^2
dx^2 +1
2 mω(^2) x (^2) |(a (^2) −x 2 )^2
〉
〈
(a^2 −x^2 )^2 |(a^2 −x^2 )^2〉.
We need to do some integrals of polynomials to compute
E′=
3
2
̄h^2
ma^2+
1
22
mω^2 a^2.Now we optimize the parameter.
dE′
da^2= 0 =
− 3
2
̄h^2
ma^4+
1
22
mω^2 ⇒a^2 =√
33
̄h^2
mω^2=
√
33
̄h
mωE′=
3
2
̄hω
√
33+
1
22
mω^2√
33
̄h
mω=
(
3
2
√
33
+
√
33
22
)
̄hω=1
2
̄hω(√
33 +
√
33
11
)
=
1
2
̄hω√
4 · 3
√
11
=
1
2
̄hω√
12
11
This is close to the right answer. As always, it is treated as an upper limit on the ground state
energy.
25.6.2 1-D H.O. with exponential wavefunction
As a check of the procedure, take trial functione−ax
(^2) / 2
. This should give us the actual ground state
energy.
E′=
∞∫
−∞ψ∗[
− ̄h2
2 m∂^2 ψ
∂x^2 +1
2 mω(^2) x (^2) ψ
]
dx∫∞
−∞ψ∗ψdx=
{
− ̄h^2
2 m∞∫
−∞e−ax2 [
a^2 x^2 −a]
dx+^12 mω^2∞∫
−∞x^2 e−ax2
dx}
∞∫
−∞e−ax^2 dx