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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:
Now let A be any operator and A* its adjoint (conjugate operation). Applying the above indentity to [A*A, A] gives
Likewise
If A satisfies the canonical quantization condition that [A, A*] is equal to the identity operation then the above two relationships reduce to
These can also be expressed as
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
Not only is |α> an eigenfunction of A*A, but A|α> is also an eigenfunction of A*A because
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 A^{n}|α> for n over some range is an eigenfunction of A*A with (α−n) as its eigenvalue. Therefore the eigenfunction of A^{n}|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)>.
Let |α> be an eigenfunction of A*A. Then A|α> and A*|α> are also an eigenfunction of A*A. Thus
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.
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
(To be continued.)
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