San José State University

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 Some General Relationships for Operators on a Function Space that Satisfy the Canonical Quantification Condition

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

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

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

## 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 Am|α> is an eigenfunction of A*A but Am+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.)