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√2h¯mω −
1
a = (p+imωx). (4.46)
†
√2h¯mω
The reason for giving them these, somewhat esoteric, names will become
clear subsequently. Comparing these definitions with (4.44) suggests taking
z = √2a and z¯ = √2a . However, since there is no reason to choose a par-
†
ticular ordering of the operators, a symmetric ordering will be used. Thus the
Hamiltonian is written
1
H = ¯hω(aa +a a). 4.47
† †
2
Inserting (4.45) and (4.46), and performing some careful algebra, yields
1 1 2
H = ¯hω p imωx p+imωx +
2 (cid:0)√2h¯mω (cid:1) (cid:16)(cid:0) − (cid:1)(cid:0) (cid:1)
p+imωx p imωx =
(cid:0) (cid:1)(cid:0) − (cid:1)(cid:17)
1
(p2+imωpx imωxp+m2ω2x2+p2 imωpx+imωxp+m2ω2x2)=
4m − −
1
(p2+m2ω2x2),
2m
which is the same formula as (4.43) slightly rearranged.
Note that the operator combinations xp and px cancel in the above calcu-
lation. They would not have done that, had not a symmetrical ordering been
chosen.
So far, not very much has been achieved. In order to proceed, some prop-
erties of the creation and annihilation operators must be derived. In quantum
mechanics, the commutators between operators are always important because
muchofthepropertiesofasystemareencodedintothecommutators. Wethere-
forecalculate the commutator[a,a ]using the definitions (4.45) and(4.46)and
†
the basic commutators (4.27)-(4.29)
83
1
[a,a ]= ( i[x,p]+i[p,x])=
†
2h¯ −
1
( i(i¯h)+i( i¯h))=1.
2h¯ − −
This implies that aa =a a+1 and the Hamiltonian can be written as
† †
1
H =h¯ω(a a+ ). (4.48)
†
2
The commutation relations for the creation and annihilation operators can
now be summarized
[a,a†]=1 (4.49)
[a,a]=[a ,a ]=0. (4.50)
† †
Sofarnoreferencehasbeenmadetothestatesoftheharmonicoscillator. It
istimetointroducethemnow. Referringbacktoequation(4.43)weseethatH
is a positive definite operator (the energy is positive classically), and therefore,
on physical grounds, there must be a state with lowest energy. Denote this
ground statewith 0 . Incomputingtheenergyforthisstate,wemustknowthe
| i
effect of the creation and annihilation operators acting on it. We will choose
a0 =0. (4.51)
| i
The intuition behind this choice is that the ground state, being the lowest
energy state, must be annihilated by the annihilation operator, but ultimately
itisjustifiedbythe resultsthatfollow. Theenergyofthe groundstatecannow
be computed
1 ¯hω
H 0 =h¯ω(a a+ )0 = 0 .
†
| i 2 | i 2 | i
If there is a ground state, there ought to be excited states, i.e states with
higher energy. As the terminology suggests, the next excited state above the
ground state is created by the creation operator acting on the ground state.
Denoting this state with 1 , let us define tentatively
| i
1 =a 0 .
†
| i | i
Now,thereisaconsistencyrequirementontheseequations. Since[a,a ]=1
†
it must be the case that
[a,a ]0 =10 = 0 .
†
| i | i | i
But this can now be checked explicitly
0 =[a,a ]0 =aa 0 a a0 =a1 .
† † †
| i | i | i− | i | i
84
Thus it must be the case that
a1 = 0 ,
| i | i
the interpretation of which is that the first excited state is destroyed, or anni-
hilated, by a.
Clearly, it must be possible to generalize this and construct an hierarchy of
excitedstatesbylettingthecreationoperatoractonthegroundstaterepeatedly.