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Quantization of a free particle...

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Blake Winter...

Posted: Thu Sep 24, 2009 7:32 am

Guest

For simplicity let us take a one dimensional space with a particle in
it described by the standard Hamiltonian H=p^2/2m. The state space for
such a system is generally taken to be L^2(R). The observables q and p
are promoted to operators s.t. if f is a function in L^2(R), then (qf)
(x)=x*f(x), and (pf)(x)=(-i df/dx)(x) (assuming unit hbar).

Now according to standard quantum mechanics, the result of a
measurement of q or p should be an eigenvalue of q or p respectively.
But these operators do not have any eigenvectors and hence do not have
eigenvalues (although as operators on the space of C^1 functions on R
they have continuous spectra). How is this reconciled with the axiom
that measurements should yield eigenvalues?

My guess is that this has something to do with the fact that p and q
are not bounded linear operators?

(I know of course that in practice one assumes that a measurement has
some uncertainty and hence one can still extract meaningful results
from this. But that is not the point; that technique is not what is
postulated for QM).

Bob_for_short...

Posted: Fri Sep 25, 2009 4:41 am

Guest

On 24 sep, 19:32, Blake Winter <valaraukarsb... at (no spam) gmail.com> wrote:
[quote:959a0232d8]
Now according to standard quantum mechanics, the result of a
measurement of q or p should be an eigenvalue of q or p respectively.
But these operators do not have any eigenvectors and hence do not have
eigenvalues (although as operators on the space of C^1 functions on R
they have continuous spectra). How is this reconciled with the axiom
that measurements should yield eigenvalues?

[/quote:959a0232d8]
Any state can be decomposed in eigenstates of coordinate or momentum.
One-time measurement (a single) does not give you a complete
information about the state - about what amplitudes (probabilities) of
eigenstates are in this superposition. One needs a set of experiments
to collect such data.

Arnold Neumaier...

Posted: Fri Sep 25, 2009 4:41 am

Guest

Blake Winter wrote:
[quote:515015ca5a]For simplicity let us take a one dimensional space with a particle in
it described by the standard Hamiltonian H=p^2/2m. The state space for
such a system is generally taken to be L^2(R). The observables q and p
are promoted to operators s.t. if f is a function in L^2(R), then (qf)
(x)=x*f(x), and (pf)(x)=(-i df/dx)(x) (assuming unit hbar).

Now according to standard quantum mechanics, the result of a
measurement of q or p should be an eigenvalue of q or p respectively.
But these operators do not have any eigenvectors and hence do not have
eigenvalues (although as operators on the space of C^1 functions on R
they have continuous spectra). How is this reconciled with the axiom
that measurements should yield eigenvalues?

My guess is that this has something to do with the fact that p and q
are not bounded linear operators?
[/quote:515015ca5a]
No. On the surface, it has something to do with the fact that the
spectrum is continuous. That this invalidates the standard textbook
answer was observed long ago by Wigner.

However, the result of measuring the energy of the electron in a
hydrogen atom does not give an eigenvalue of its Hamiltonian,
although here the eigenvectors exist. It only gives an approximation
to such an eigenvalue. Thus there is a deeper flaw in taking this
``axiom'' for a fundamental truth.

The measurement process as described by von Neumann (and copied from
there to numerous textbooks) is an unrealisitc idealization compared
with most real measurements. The latter are usually much better
described by suitable POVMs than by von Neumann's PVMs.

See Sections 7.3-7.5 of my book
A. Neumaier and D. Westra,
Classical and Quantum Mechanics via Lie algebras,
arXiv:0810.1019
for a realistic account of measurement theory not dependent on this axiom.

Arnold Neumaier

pirillo...

Posted: Fri Sep 25, 2009 8:54 pm

Guest

[quote:7f5d08f407]But these operators do not have any eigenvectors and hence do not have
eigenvalues
[/quote:7f5d08f407]
They have spectra though, even if they don't have point spectrum
(which are the usual eigenvalues)
Now, the spectral theorem will give you a probability measure m_{x,
v} on the spectrum of x for any
state v of unit norm. I guess you match that to experiments.

Arnold Neumaier...

Posted: Sun Sep 27, 2009 1:33 am

Guest

pirillo wrote:
[quote:8f76e1ad89]But these operators do not have any eigenvectors and hence do not have
eigenvalues

They have spectra though, even if they don't have point spectrum
(which are the usual eigenvalues)
Now, the spectral theorem will give you a probability measure m_{x,
v} on the spectrum of x for any
state v of unit norm. I guess you match that to experiments.
[/quote:8f76e1ad89]

The standard textbook measurement theory says that the possible
measurement results in measuring an observable given by a Hermitian
operator A are its possible eigenvalues, with a probability density
depending on the state of the system. This is part of the content of
Born's rule, and counts as one of the cornerstones of the
interpretation of quantum mechanics.

But Born's rule gives only a very idealized account of measurement
theory, and gives no sufficient explanation for what is going on in
many nontrivial measurements.

The spectrum of the Hamiltonian of the electron of a hydrogen atom
has a discrete part, catering for its bound states. According to the
idealized textbook measurement theory, a measurements of the energy
of a bound state should produce an infinitely accurate value agreeing
with one of the values in the (QED-corrected) Balmer (etc.) series.

But this is ridiculous. Repeated preparation and measurement of the
position of the ``same'' spectral line (which provide these energy
measurements, relative to an appropriate zero of the energy) yields
different results, from which the energies themselves can be obtained
only to a certain accuracy.

Thus Born's rule does not account for the interpretation of a
measurement of the energy of an electron. For similar reasons,
measurements of particle masses or resonance energies do not reveal
the exact values (which they should according to Born's rule) but only
approximations whose quality depends a lot on the way the measurement
is done (an aspect that does not figure at all in Born's rule).

Measurements such as that of a particle lifetime or the integral cross
section of a particular reaction do not even have a natural associated
operator of which the measurement result would be an eigenvalue.

The idealized textbook measurement theory based on Born's rule is
appropriate only for the measurement of spin and related variables
that result in recording decisions of finite information content.

Thus the measurement process as described by von Neumann (and copied
from there to numerous textbooks) is an unrealistic idealization
compared with many (and probably most) real measurements.
The latter are usually much better described by suitable POVMs
(positive operator valued measures) rather than by Born's rule,
which corresponds to PVMs (projection-valued measures), a special case
of POVMs in which the positive operators are in fact projections.

See Sections 7.3-7.5 of the book
A. Neumaier and D. Westra,
Classical and Quantum Mechanics via Lie algebras,
arXiv:0810.1019
for a realistic account of measurement theory not dependent on
Born's rule. The latter is derived there as a special case, together
with giving the condition in which it is applicable.

Arnold Neumaier

enders...

Posted: Mon Sep 28, 2009 7:22 am

Guest

On 24 Sep., 19:32, Blake Winter <valaraukarsb... at (no spam) gmail.com> wrote:
[quote:c1020a2521]For simplicity let us take a one dimensional space with a particle in
it described by the standard Hamiltonian H=p^2/2m. The state space for
such a system is generally taken to be L^2(R).
[/quote:c1020a2521]
Yes, but this is too narrow for the case of interactionless particles

[quote:c1020a2521]The observables q and p
are promoted to operators s.t. if f is a function in L^2(R), then (qf)
(x)=x*f(x), and (pf)(x)=(-i df/dx)(x) (assuming unit hbar).

Now according to standard quantum mechanics, the result of a
measurement of q or p should be an eigenvalue of q or p respectively.
But these operators do not have any eigenvectors and hence do not have
eigenvalues (although as operators on the space of C^1 functions on R
they have continuous spectra). How is this reconciled with the axiom
that measurements should yield eigenvalues?
[/quote:c1020a2521]
See the book by Eugene Stefanovich (on arXiv) for f(x)

[quote:c1020a2521]My guess is that this has something to do with the fact that p and q
are not bounded linear operators?
[/quote:c1020a2521]
Yes. More deeply, it roots in that 'quantization as eigenvalue
problem' (Schrödinger 1926) is *not* the appropriate quantization
method, because the maths of eigenvalue problems is maths for
classical physics (strings, resonators), as Schrödinger himself points
out thereself (see my book - Springer 2006 - for a discussion of
Schrödinger's self-criticism). In my book, quantization is treated as
selection problem of states (following Einstein 1907), and all of
Schrödinger's requirements to quantization are fullfilled.

For free particles, quantization as selection problem does not work,
because the classical and quantum energetic spectra are equal.
(Strictly speaking, of a free particle, it cannot be said whether it's
classical or quantum.) For this, it's appropriate to start from
symmetry. |f(x)|^2 and |g(p)|^2 exhibit the symmetry of the
Hamiltonian (see my book, why). Hence, for free particles, up to
normalization,

|f(x)|^2 = const

|g(p)|^2 = delta(p - p_0)

(Dirac's delta function), p_0 being the particle's momentum vector.
This yields Eugene's f(x) mentioned above. (Unfortunately, I have
realized this only after my book had been printed.)

Hope this helps you to overcome the confusion of the treatment in most
textbooks.

Best wishes,
Peter

Blake Winter...

Posted: Mon Sep 28, 2009 7:24 am

Guest

[quote:b6270d9732]Any state can be decomposed in eigenstates of coordinate or momentum.
[/quote:b6270d9732]
I disagree. In L^2(R) there are no eigenfunctions for the momentum
operator, because p is proportional to d/dx, so its eigenfunctions
have to have the form of an exponential. Similarly there are no
eigenstates for q, because qY(x)=xY(x) and so the eigenfunctions would
have to vanish except at a single point. This is precisely what I am
objecting to. There are no eigenstates for p or q in L^2(R).

One can also see that in terms of functions vanishing except at a
single point, a function in L^2(R) will not be a pointwise or measure-
wise limit of finite sums of such functions, so even if they were
there we could not decompose in terms of them.

Furthermore, in general given a bounded linear operator on a Hilbert
space H, its eigenvectors should be a basis. Since L^2(R) is separable
such a basis must be countable. Thus, the fact that q and p are not
bounded seems to have something to do with their spectrum. Or perhaps
technically one should not say they have any spectrum at all as
operators on L^2(R) since they do not have eigenfunctions in that
space!

Arnold Neumaier...

Posted: Wed Sep 30, 2009 1:19 pm

Guest

Blake Winter wrote:

[quote:abbdff3147]Furthermore, in general given a bounded linear operator on a Hilbert
space H, its eigenvectors should be a basis.
[/quote:abbdff3147]
Only if the spectrum is discrete.

1/(1+p^2) is bounded, but there are no eigenvectors.

[quote:abbdff3147]Since L^2(R) is separable
such a basis must be countable. Thus, the fact that q and p are not
bounded seems to have something to do with their spectrum.
[/quote:abbdff3147]
Unbounded self-adjoint operators have an unbounded spectrum.
There is an orthonormal basis of eigenvectors iff this spectrum is discrete,

[quote:abbdff3147]Or perhaps
technically one should not say they have any spectrum at all as
operators on L^2(R) since they do not have eigenfunctions in that
space!
[/quote:abbdff3147]
The spectrum is not defined via eigenvectors but via the set of values
for which A-lambda is not invertible.

Arnold Neumaier

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