Re: operator satisfying relations [H,T] = i hbar

On Dec 13, 3:59 pm, a student <of_1001_nig...@xxxxxxxxxxx> wrote:
Thank you for all the answers. One thing that is still not clear to me
is this:

We can write down the classical (not quantum mechanical) dynamic
equation for the oscillator in the so-called action-angle variables by
the well know canonical transformation found in text books. Suppose we
call P the action and Q the angle, then P and Q are well known
functions of p (the usual momentum) and q (the usual coordinate) found
in textbook. Further more, P and Q are real physical quantities. (it
is also possible to express p and q in terms of P and Q). Here, we are
not talking about quantum mechanics yet.

In classical mechanics, we thus have {P,Q}=1 and H=P in the classical
sense (here, no quantum mechanics is involved). So, {H,Q}=1 in
classical sense. The above Possion brackets are in classical sense.
This is a classical Hamiltonian system. And we can write down the
classical equations in terms of P and Q (not in terms of p and q).

Then, by standard procedure, if we quantize the above classical
Hamiltonian system (written as P and Q), the classical variable will
become quantum mechanical operators, and the Possion bracket become
commutors. This means, for the oscillator, [P,Q]=ihbar and
[H,Q]=ihbar, where P,Q,H are operators and [,] are commutors. Since P
and Q (and H) are real physical quantities in classical limit, I would
argue they are observables in quantum mechanics for the oscillator. We
can just rename our Q to T is we want to.

I would appreciate if someone can point out the mistakes in the
logical steps.

The "mistakes" are all in the para beginning "Then, by the standard
procedure", and in particular in your interpretation of what you are
doing. You are indeed "quantising" an oscillator by your procedure,
but it is not equivalent to the standard quantisation. In general
this is because so-called Dirac quantisation is not actually well-
defined - it is only an ad hoc prescription for trying to find the
"right" quantum Hamiltonian operator for a given system (for which
there are MANY candidates, all having the same classical limit !).
Technically, the inequivalence arises from the fact that if you make a
classical canonical transformation from (q,p) to (Q,P), the two sets
of Dirac operators are not necessarily related by any unitary
transformation.

The case of the oscillator is simple enough to actually make a
connection of sorts between the two inequivalent quantisations.
Briefly, the usual quantisation gives a set of energy eigenstates |0>,
|1>, |2>, .... . Now, compare this to a quantisation of a 1D rotator
(i.e., a particle on a circle), with canonical coordinates (phi, J) -
which are just the prototypical action-angle variables. We know that
the quantisation gives in this case a set of angular momentum
eigenstates ..., |-2>, |-1>, |0>, |1>, |2>, ... . You can see the
inconsistency now already: the spectrum of an "action" variable has
arbitrarily negative values - so the Hilbert space is too large
compared to that of the standard quantisation. In particular,
following your quantisation method will give negative energy
eigenvalues, which are unphysical. For this reason we choose the
first, inequivalent quantisation.

To indicate the formal connection between the two, let's suppose we
follow your quantisation through, so that the energy spectrum has
negative and positive eigenvalues. Now consider the subspace
corresponding to the span of just the zero and positive eigenvalues.
Then, this subspace is clearly unitarily equivalent to the usual
Hilbert space of the oscillator. In this sense, we can interpret the
loss of a Hermitian time operator for the oscillator as being from
having to work on only half the "full" Hilbert space of both positive
and negative eigenvalues. For example, consider the operator
V = sum |n> <n+1|
with summation over positive and negative eigenvalues. This operator
is unitary, and is essentially the same as exp[i phi]. However, if we
truncate to nonnegative eigenvalues, we get the operator
U = |0> <1| + |1> <2| + ....
which is only semi-unitary (one has UU*=1, but not U*U=1). Let me
know if you would like references.- Hide quoted text -

- Show quoted text -

I was always under the impression that canonical trasforms in
classical mechanics corresponds to unitary tranforms in the quantum
mechanics. But as you convienced me, this is not always true. If you
can provide some references, I am interested in knowing under what
condition it is true or not true. There has to be some relationship
even when this simple coorespondence fails.

.

Relevant Pages

  • Re: operator satisfying relations [H,T] = i hbar
    ... equation for the oscillator in the so-called action-angle variables by ... but it is not equivalent to the standard quantisation. ... which are just the prototypical action-angle variables. ... eigenvalues, which are unphysical. ...
    (sci.physics.research)
  • Re: operator satisfying relations [H,T] = i hbar
    ... equation for the oscillator in the so-called action-angle variables by ... no quantum mechanics is involved). ... but it is not equivalent to the standard quantisation. ... which are just the prototypical action-angle variables. ...
    (sci.physics.research)