Elements of Modern Physics
88
terms of Hermite polynomials H n of order n. The wave functions and their
energies are
φ n (x) =
1/ 2
1/ 4
2 !
n n
−
α
π
exp (– α
2
x
2
/2) H n (αx)
E n =
1
2
n
+
ω
, n = 0, 1, 2, ...
(3.111)
where ω = (k/m)
1/2
. It is observed that again the ground state energy is not zero.
Its value of
1
2
ω
is called the zero-point energy.
The first three Hermite polynomials are
H 0 (αx) = 1
H 1 (αx) = 2αx
(3.112)
H 2 (αx) = 4α
2
x
2
– 2
The harmonic oscillator problem can be solved more elegantly by using
operator algebra. Defining
a =
1/ 2
2
m
x h x
ω
∂
+
ω ∂
†
a =
1/ 2
2
m
x m x
ω
∂
−
ω ∂
(3.113)
it can easily be shown that
†
†
aa
a a
−
= 1
aH – Ha =
a
ω
(3.114)
†
†
a H Ha
−
=
a
+
− ω
with the Hamiltonian H (i.e. energy) being
H =
2
2
2
2
1
2
2
d
kx
m dx
−
+
(3.115)
Therefore if ψ 0 (x) is the ground state with energy E 0 , then using Eq. (3.114)
Haψ 0 (x) =
0
0
(
)
( )
E
a
x
− ω ψ
(3.116)
Ha † ψ 0 (x) =
0
0
(
) † ( )
E
a
x
+ ω
ψ
(3.117)
88
terms of Hermite polynomials H n of order n. The wave functions and their
energies are
φ n (x) =
1/ 2
1/ 4
2 !
n n
−
α
π
exp (– α
2
x
2
/2) H n (αx)
E n =
1
2
n
+
ω
, n = 0, 1, 2, ...
(3.111)
where ω = (k/m)
1/2
. It is observed that again the ground state energy is not zero.
Its value of
1
2
ω
is called the zero-point energy.
The first three Hermite polynomials are
H 0 (αx) = 1
H 1 (αx) = 2αx
(3.112)
H 2 (αx) = 4α
2
x
2
– 2
The harmonic oscillator problem can be solved more elegantly by using
operator algebra. Defining
a =
1/ 2
2
m
x h x
ω
∂
+
ω ∂
†
a =
1/ 2
2
m
x m x
ω
∂
−
ω ∂
(3.113)
it can easily be shown that
†
†
aa
a a
−
= 1
aH – Ha =
a
ω
(3.114)
†
†
a H Ha
−
=
a
+
− ω
with the Hamiltonian H (i.e. energy) being
H =
2
2
2
2
1
2
2
d
kx
m dx
−
+
(3.115)
Therefore if ψ 0 (x) is the ground state with energy E 0 , then using Eq. (3.114)
Haψ 0 (x) =
0
0
(
)
( )
E
a
x
− ω ψ
(3.116)
Ha † ψ 0 (x) =
0
0
(
) † ( )
E
a
x
+ ω
ψ
(3.117)