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8.13M
| i | i
belongs inthe interface, being effectively a specificationof a functionality to be
provided by the implementation. The implementation then, could be in terms
ofwavefunctions orin terms ofmatrices,or perhaps interms ofsome other for
the purpose suitable (mathematical) constructs. In the next section on angular
momentum and spin, we will see a concrete example of this.
86
Derivation of the normalization conditions
This section is somewhat technical, and do involve certain concept not yet dis-
cussed. The purpose is to derive the normalization coefficients ξ(n) and η(n).
The reader might want to skip it for now and return after reading chapter 5.
The coefficients ξ and η are subject to some consistency conditions. First,
since [a,a ]=1, we have
[a,a ]n = n
| i | i⇒
ξ(n)η(n+1) ξ(n 1)η(n)=1. (4.59)
− −
Furthermore, the states n are subject to a orthonormalitycondition, anal-
| i
ogous to (??)
nm =δ ,
nm
h | i
and in particular
nn =1.
h | i
Thequestionarises,whatis n? Adetailedexplanationofthiswillbegiven
h |
i chapter 5. Here we can think of n as a form of conjugate to n . As regards
h | | i
the equation (4.52), this conjugation, denoted by a dagger , works as follows
(a n ) =(ξ(n)n+1 )
† † †
| i | i ⇒
na= n+1ξ(n) , (4.60)
h | h |
and as regards equation (4.53)
(an )† =(η(n)n 1 )†
| i | − i ⇒
na = n 1η(n) (4.61)
† ∗
h | h − |
Enforcingthecondition nn =1onthestate n+1 ,thatis n+1n+1 =1
h | i | i h | i
and using (4.52) and (4.53) as well as (4.60) and (4.61) yields
1
n+1n+1 = naa† n =
h | i ξ(n)ξ(n) h | | i
1 1
n[a,a ]+a an = n1+η(n)η(n) n =
† † ∗
ξ(n)ξ(n) h | | i ξ(n)ξ(n) h | | i
∗ ∗
1
1+η(n)η(n)∗ ,
ξ(n)ξ(n)
∗(cid:0) (cid:1)
where the common rewriting trick aa =[a,a ]+a a has been used.
† † †
Concluding, we get
1
1+η(n)η(n) =1,
ξ(n)ξ(n)
∗(cid:0) (cid:1)
87
or more succinctly
ξ(n)ξ(n) η(n)η(n) =1 (4.62)
∗ ∗
Before trying to solve equations (4.59) and (4.62) we will make the simpli-
fying assumption that ξ and η are real. This assumption can be justified after
the fact. So equation (4.62) becomes
ξ(n)2 η(n)2 =1. (4.63)
The solution to (4.59) and (4.63) can now be constructed in an inductive
way. Starting from a0 =0 which implies
| i
η(0)=0
we find that (4.59) implies
ξ(0)=1,
which in its turn using (4.63) implies
η(1)=1.
Goingoninthis way,usingequations(4.59)and(4.63), wegetthe sequence
of equations
ξ(1)=√2
η(2)=√2
ξ(2)=√3
η(3)=√3
.
.
.
η(n)=√n,
or in general
ξ(n)=√n+1