a
a
{
¼
ffiffiffiffiffiffiffi
mω
2ħ
q
i
ffiffiffiffiffiffiffiffiffiffiffiffi
2mħω
p
ffiffiffiffiffiffiffi
mω
2ħ
q
À
i
ffiffiffiffiffiffiffiffiffiffiffiffi
2mħω
p
0
B
B
@
1
C
C
A
q
p
ð2:19Þ
Then we have
a
{ a ¼ 2mħω
ð
Þ
À1 mωq À ip
ð
Þmωq þ ip
ð
Þ
¼ 2mħω
ð
Þ
À1 m
2
ω
2 q
2
þ p
2
þ imω qp À pq
ð
Þ
Â
Ã
¼ ħω
ð Þ
À1 1
2
mω
2 q
2
þ
1
2m
p
2
þ
1
2
iωiħ
h
i
¼ ħω
ð Þ
À1 H À
1
2
ħω
,
ð2:20Þ
where the second last equality comes from (1.140). Rewriting (2.20), we get
H ¼ ħωa
{ a þ
1
2
ħω:
ð2:21Þ
Similarly we get
H ¼ ħωaa
{
À
1
2
ħω:
ð2:22Þ
Subtracting (2.22) from (2.21), we have
0 ¼ ħωa
{ a À ħωaa
{
þ ħω:
ð2:23Þ
That is,
a, a
{
Â
à ¼ 1 or a, a
{
Â
à ¼ E,
ð2:24Þ
where E represents an identity operator.
Furthermore, using (2.21) we have
H, a
{
Â
à ¼ ħω a
{ a þ
1
2
, a
{
h
i
¼ ħω a
{ aa
{
À a
{ a
{ a
À
Á ¼ ħωa
{ a, a
{
Â
à ¼ ħωa
{
: ð2:25Þ
Similarly, we get
H, a
½
Š ¼ Àħωa:
ð2:26Þ
Next, let us calculate an expectation value of H. Using a normalized function jψi,
from (2.21) we have
ψ H
j jψ
h
i¼ ψ ħωa
{ a þ
1
2
ħω
ψ
D
E
¼ ħω ψ a
{ a
ψ
þ
1
2
ħω ψjψ
h
i
¼ ħω aψjaψ
h
iþ
1
2
ħω ¼ ħω aψ
k k
2 þ
1
2
ħω !
1
2
ħω:
ð2:27Þ
2.2 Formulation Based on an Operator Method
35