Re: Interpretation of canonical/contact transformations

From: Igor Khavkine (igor.kh_at_gmail.com)
Date: 01/10/05 

Date: Mon, 10 Jan 2005 18:23:25 +0000 (UTC)

On Sat, 08 Jan 2005 22:02:37 +0000, Eugene Stefanovich wrote:

> jcooper@ucalgary.ca wrote:
>> It is my understanding that, by means of a contact transformation, the
>> Hamiltonian of a system can be brought into a different form with
>> precisely the same eigenvalues. Because of this, if one expands the
>> Hamiltonian in some basis of operators, one cannot in general expect to
>> determine all of the resulting coefficients experimentally, even in
>> principle, but rather only certain (linear?) combinations of them.
>
> I disagree with your assessment. Two Hamiltonians differing by a unitary
> (contact) transformations are not physically equivalent.

It's already been established in a couple of long threads that your notion
of physical equivalence is not the same as the commonly accepted one.

> They have the
> same eigenvalues, that's true. But their eigenfunctions are different.

There is no need to speak of eigen-*functions* if we are working on an
abstract Hilbert space. Interpretation of Hilbert space vectors as
wave functions requires the position operator. At this point in the
discussion, no observables except that Hamiltonian have been introduced.
No position operator, no wave functions.

A unitary transformation of vectors and operators on a Hilbert space is no
more significant than a change of basis in a finite dimensional vector
space.

Igor

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