Let A, B and C be operators. Then for the commutator [AB, C]

[AB, C] = A[B, C] − [A, C]B

Proof:

[AB, C] = (AB)C − C(AB)
and from the associativity
of operator applications
[AB, C] = A(BC) − (CA)B
and from subtracting and
adding A(CB) on the right
[AB, C] = A(BC) − A(CB) + A(CB) − (CA)B
and hence
[AB, C] = A[B, C] + (AC)B − (CA)B
and finally
[AB, C] = A[B, C] + [A, C]B

Now let A be any operator and A* its adjoint (conjugate operation). Applying the above indentity to [A*A, A] gives

[A*A, A] = A*[A, A] + [A*, A]A
but [A, A]=0, so
[A*A, A] = [A*, A]A
and
since commutator
is antisymmetric
[A*A, A] = −[A, A*]A

Likewise

[A*A, A*] = A*[A, A*] + [A*, A*]A = A*[A, A*]

If A satisfies the canonical quantization condition that [A, A*] is equal to the identity operation then the above two relationships reduce to

[A*A, A] = −A
and
[A*A, A*] = A*

These can also be expressed as

(A*A)A = A(A*A)−A = A(A*A−I)
and
(A*A)A* = A*(A*A)+A* = A*(A*A+I)

where I is the identity operation, which is the same as multiplication by 1.

Let an eigenvalue of A*A be denoted as α. Expressed in the Dirac notation this means

A*A|α> = α|α>

Not only is |α> an eigenfunction of A*A, but A|α> is also an eigenfunction of A*A because

(A*A)A|α> = A(A*A−I)|α> = A{α|α> − |α>} = A(α−1)|α>
thus
(A*A)(A|α>) = (α−1)(A|α>)

So (A|α>) is also an eigenfunction of (A*A) but with an eigenvalue that is 1 less than that of |α>.

A reapplication of the above would show that Aⁿ|α> for n over some range is an eigenfunction of A*A with (α−n) as its eigenvalue. Therefore the eigenfunction of Aⁿ|α> is denoted as |(α−n)>.

Similarly A*|α> is an eigenfunction of A*A with an eigenvalue of (α+1) and therefore (A*)ⁿ|α> is an eigenfunction of A*A with an eigenvalue of (α+n). It is represented as |(α+n)>.

Relationships between Eigenfunctions

Let |α> be an eigenfunction of A*A. Then A|α> and A*|α> are also an eigenfunction of A*A. Thus

A*|α> = |(α+1)> = C|α> A*|(α-1)> = α|(α-1)> A*A(A|α>)

The Integralness of the Eigenvalues of A*A

The eigenfunction of an operator cannot be the zero function. Therefore there must be an integer m such that A^m|α> is an eigenfunction of A*A but A^m+1 |α> is not. This implies that (α−m) is equal to 0 and hence α is equal to that integer Therefore the eigenvalues of A*A are necessarily integers from 0 to some maximum integer m.

Relations that Prevail When the Functions
are Equal to Unity in Squared Magnitude

Letter the inner product of two functions, X and Y, be denoted as (X, Y). Then the squared magnitude of X, |X|², is (X, X). The functions being considered are probability density functions, which means that |X|²=(X, X)=1.

Consider the significance of

|n> = A|(n-1)>

(To be continued.)