104 From Classical Mechanics to Quantum Field Theory. A Tutorial
withD(H 0 ):=S(R). Above,x^2 is the multiplicative operator andm, ω >0are
constants.
To go on, we start by defining a triple of operators,a, a+,N:S(R)→L^2 (R,dx)
as
a+:=
√
mω
2
(
x−
mω
d
dx
)
,a:=
√
mω
2
(
x+
mω
d
dx
)
, N:=a+a.
aanda+are called creation and annihilation operators. These operators have
the same domain which is also invariant: a(S(R))⊂S(R),a+(S(R))⊂S(R),
N(S(R))⊂S(R). It is also easy to see, using integration by parts thata+⊂a†
and thatN(callednumber operator) is Hermitian an also symmetric because
S(R)isdenseinL^2 (R,dx). By direct computation, exploiting the given definitions
one immediately sees that
H 0 =
(
a†a+
1
2 I
)
=
(
N+
1
2 I
)
.
Finally, we have the commutation relationsonS(R)
[a, a†]|S(R)=I|S(R). (2.38)
Supposing that there existsψ 0 ∈S(R) such that
||ψ 0 ||=1,aψ 0 = 0 (2.39)
starting form (2.38) and using an inductive procedure on the vectors
ψn:=
√^1
n 1
(a†)nψ 0 (2.40)
it is quite easy to prove that (e.g., see[ 5 ]for details), forn, m=0, 1 , 2 ,...,the
relations hold
aψn=
√
nψn− 1 ,a†ψn=
√
n+1ψn+1, 〈ψn,ψm〉=δnm. (2.41)
Finally, theψnare eigenvectors ofH 0 (andN)since
H 0 ψn=ω
(
a†aψn+
1
2
ψn
)
=ω
(
a†
√
nψn− 1 +
1
2
ψn
)
=ω
(
ψn+1+^1
2
ψn
)
=ω
(
n+^1
2
)
ψn. (2.42)
As a consequence,{ψn}n∈Nis an orthonormal set of vectors. This set is actually
a Hilbertian basis because (1), a solution inS(R) of (2.39) exist (an is unique):
ψ 0 (x)=
1
π^1 /^4
√
s
e−
2 xs^22
,s:=
√
mω