Solutions
247
The momentum matrix element can be expressed in terms of the ladder operators:
χ v
ˆ
p x
χ v
= −i
Mω
2
χ v
ˆ
a − ˆ
a
†
χ v
= 0
Actually there is a more general proof, not based on the properties of the harmonic
oscillator, that the mean value of ˆ
p is zero for real wavefunctions that vanish at
x → ±∞. In fact:
+∞
−∞
ψ(x)
dψ(x)
dx
dx =
ψ
2
(x)
∞
−∞
−
+∞
−∞
dψ(x)
dx
ψ(x) dx
The integral in the LHS is equal to minus itself, so it must vanish.
We now consider the variances:
Δx
2
=
χ v
(x − x e )
2
χ v
=
2Mω
χ v
( ˆ
a + ˆ
a
†
)
2
χ v
=
(2v + 1)
2Mω
Δp
2
=
χ v
p
2
χ v
= −
Mω
2
χ v
( ˆ
a − ˆ
a
†
)
2
χ v
=
Mω(2v + 1)
2
The indetermination product is
Δx Δp =
(2v + 1)
2
which takes the minimum value /2 for v = 0.
The classical amplitude of oscillation is obtained by equating the total energy
with the potential energy:
ω
v +
1
2
=
1
2
Mω
2
Δx
2
cl
We get Δx cl =
(2v + 1)
Mω
1/2
. This result differs from the quantum mechanical
uncertainty Δx by just a factor
√
2, which is quite reasonable since Δx cl is the
maximum elongation whereas Δx is the “root-mean-square” elongation.
4.2 The total energy is made of potential + kinetic energy. With respect to the
minimum of the excited state, ΔE adia , the average potential energy is:
U =
1
2
Mω
2
2
χ 1,0
(R − R 2 )
2
χ 1,0
The eigenfunction χ 1,0 (R) belongs to the initial electronic state 1, so we shall use
the associated ladder operators that allow to write
R = R 1 +
2Mω 1
( ˆ
a + ˆ
a
†
)