QUANTUM OPERATORS
The proof, which makes repeated use of
[
A, A†]
=, is as follows:Am(A†)m=Am−^1 AA†(A†)m−^1=Am−^1 (A†A+)(A†)m−^1=Am−^1 A†A(A†)m−^1 +Am−^1 (A†)m−^1=Am−^1 A†(A†A+)(A†)m−^2 +Am−^1 (A†)m−^1
=Am−^1 (A†)^2 A(A†)m−^2 +Am−^1 A†(A†)m−^2 +Am−^1 (A†)m−^1=Am−^1 (A†)^2 (A†A+)(A†)m−^3 +2Am−^1 (A†)m−^1
..
.
=Am−^1 (A†)mA+mAm−^1 (A†)m−^1.Now we take the expectation values in the ground state| 0 〉of both sides of
this operator equation and note that the first term on the RHS is zero since it
contains the termA| 0 〉. The non-vanishing terms are
〈 0 |Am(A†)m| 0 〉=m〈 0 |Am−^1 (A†)m−^1 | 0 〉.The LHS is the square of the norm of (A†)m| 0 〉, and, from equation (19.45), it is
equal to
|d 0 |^2 |d 1 |^2 ···|dm− 1 |^2 〈 0 | 0 〉.Similarly, the RHS is equal to
m|d 0 |^2 |d 1 |^2 ···|dm− 2 |^2 〈 0 | 0 〉.It follows that|dm− 1 |^2 =mand, taking all coefficients as real,dm=
√
(m+1).Thus the correctly normalised state of energy (n+^12 ), obtained by repeated
application ofA†to the ground state, is given by
|n〉=(A†)n
(n!n)^1 /^2| 0 〉. (19.48)To evaluate thecn, we note that, from the commutator ofAandA†,
[
A, A†
]
|n〉=AA†|n〉−A†A|n〉|n〉=√
(n+1)A|n+1〉−cnA†|n− 1 〉=√
(n+1)cn+1|n〉−cn√
n|n〉,=√
(n+1)cn+1−cn√
n,which has the obvious solutioncn=
√
n. To summarise:cn=√
n and dn=√
(n+1). (19.49)We end this chapter with another worked example. This one illustrates howthe operator formalism that we have developed can be used to obtain results