1549380323-Statistical Mechanics Theory and Molecular Simulation

384 Quantum mechanics

The solution does, indeed, lead to a discrete set of energy eigenvalues given by the
familiar formula

En=

(

n+

1

2

)

̄hω n= 0, 1 , 2 ,.... (9.3.20)

and a set of normalized eigenfunctions

ψn(x) =

(

mω

22 n(n!)^2 π ̄h

) 1 / 4

e−mωx

(^2) /2 ̄h
Hn

(√

mω

̄h

x

)

, (9.3.21)

where{Hn(y)}are the Hermite polynomials

Hn(y) = (−1)ney

(^2) d^2
dyn
e−y
2

. (9.3.22)

The first few of these eigenfunctions are

ψ 0 (x) =

(α

π

) 1 / 4

e−αx

(^2) / 2
ψ 1 (x) =

(

4 α^3

π

) 1 / 4

xe−αx

(^2) / 2
ψ 2 (x) =
(α
4 π
) 1 / (^4) (
2 αx^2 − 1

)

e−αx

(^2) / 2
ψ 3 (x) =

(

α^3

9 π

) 1 / 4

(

2 αx^3 − 3 x

)

e−αx

(^2) / 2
, (9.3.23)
whereα=mω/ ̄h. These are plotted in Fig. 9.1. Note that the number of nodes in
each eigenfunction is equal ton. Doing actual calculations with these eigenfunctions is
mathematically cumbersome. It turns out, however, that there isa simple and conve-
nient framework for the harmonic oscillator in terms of the abstract set of ket vectors
|n〉that define the eigenfunctions through〈x|n〉=ψn(x).
If we exploit the symmetry between ˆpand ˆxin the harmonic-oscillator Hamiltonian,
we can factorize the sum of squares to give

Hˆ=

[

ˆp^2

2 m ̄hω

+

mω

2 ̄h

xˆ^2

]

̄hω

=

[(√

mω

2 ̄h

xˆ−

i

√

2 m ̄hω

pˆ

)(√

mω

2 ̄h

ˆx+

i

√

2 m ̄hω

pˆ

)

+

1

2

]

̄hω. (9.3.24)

The extra 1/2 appearing in eqn. (9.3.24) arises from the nonzero commutator between
xˆand ˆp, [ˆx,pˆ] =i ̄hIˆ. Let us now define two operators