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| Jay R. Yablon... |
 Posted: Sat May 09, 2009 8:47 am |
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I seem to recall having this discussion before, but need to refresh.
When apply we the canonical :
[x^i,p_j] = ih delta^i_j, i=1,2,3 (1)
is this really just an approximate relationship in the limiting case
where a mass m-->0?
That is, is (1) really of the form:
[x^i,p_j] = ih delta^i_j + F(m)^i_j (1)
where F(m)^i_j --> 0 as m --> 0?
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jyablon at (no spam) nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm |
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| Hendrik van Hees... |
| Posted: Sun May 10, 2009 9:00 am |
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The socalled "canonical" commutation relations of Cartesian position and
momentum components is exact, supposed you can define self-adjoint
position and momentum operators at all. Its origin is however not some
hand-waving argument called "canonical quantization" (which by the way
doesn't work in almost all other cases than cartesian coordinates  )
but originates from the definition of momentum as the generator of
spatial translations in the sense of representation theory of Lie
algebras and Lie groups on Hilbert space.
In relativistic quantum theory (i.e., relativistic QFT) it's not easy to
define a position operator with the appropriate properties. E.g., it's
far from obvious (and not clear to me) whether one can define a
position operator for massless particles with spin, e.g., for photons.
Fortunately in practice this problem is of academic interest only since
one can define observable quantities like cross sections and the like
without reference to a position operator as is shown in any textbook of
qft. In my opinion the best textbook reference is Weinberg's Quantum
Theory of Fields.
Jay R. Yablon wrote:
[quote:9629c89e4b]I seem to recall having this discussion before, but need to refresh.
When apply we the canonical :
[x^i,p_j] = ih delta^i_j, i=1,2,3 (1)
is this really just an approximate relationship in the limiting case
where a mass m-->0?
That is, is (1) really of the form:
[x^i,p_j] = ih delta^i_j + F(m)^i_j (1)
where F(m)^i_j --> 0 as m --> 0?
Thanks,
Jay
[/quote:9629c89e4b]
--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen
http://theory.gsi.de/~vanhees/faq/ |
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| Arnold Neumaier... |
| Posted: Sun May 10, 2009 9:11 am |
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Jay R. Yablon schrieb:
[quote:b176bb3ed4]
When apply we the canonical :
[x^i,p_j] = ih delta^i_j, i=1,2,3 (1)
is this really just an approximate relationship in the limiting case
where a mass m-->0?
That is, is (1) really of the form:
[x^i,p_j] = ih delta^i_j + F(m)^i_j (1)
where F(m)^i_j --> 0 as m --> 0?
[/quote:b176bb3ed4]
No. it is an exact relationship that is at the basis of quantum
mechanics, valid for total position and momentum of any physical
system with nonzero total mass, even in the relativistic case and
in quantum field theory.
For systems of mass 0 and spin >1/2 , the position operator
is nonexistent, and therefore no associated CCR exists.
See Section S2g ''Particle positions and the position operator''
of my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
Arnold Neumaier |
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| Juan R." González-Álvarez... |
| Posted: Sun May 17, 2009 8:48 am |
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Arnold Neumaier wrote on Thu, 14 May 2009 21:23:51 +0200:
[quote:7e07eb3414]juanREMOVE at (no spam) canonicalscience.com schrieb:
Hendrik van Hees wrote on Sun, 10 May 2009 19:00:59 +0000:
[/quote:7e07eb3414]
(...)
[quote:7e07eb3414]There is not position operator for particles (massless or not) in RQFT.
This is not true. Even though there is a parameter called x and referred
to as 4-dimensional position, there is also an vector defining a
3-dimensional position operator, if a relativistic system is not
massless.
[/quote:7e07eb3414]
First, that four-dimensional parameter x^b = (t,x) has nothing to see with
position or time [1]:
"Every physicist would easily convince himself that all quantum
calculations are made in the energy-momentum space and that the
Minkowski x^b are just dummy variables without physical meaning
(although almost all textbooks insist on the fact that these variables
are not related with position, they use them to express locality of
interactions!)"
[quote:7e07eb3414]For any relativistic theory possesses the Poincare group as a symmetry
group, whose infinitesimal generators satisfy the standard commutation
rules of the Poincare algebra. But given these, the standard
construction by Newton and Wigner gives (in each Lorentz frame) a
3-dimensional position operator with commuting components, and the
associated conjugate momentum operators. These play exactly the same
role as the position and momentum operators in nonrelativistic quantum
mechanics.
See Section S2g ''Particle positions and the position operator'' of my
theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
[/quote:7e07eb3414]
My original claim was about the lack or position operator x for particles
in relativistic quantum field theory (RQFT).
In your FAQ you have a section S2i. "Position operators in relativistic
quantum field theory" where you write:
"In relativistic quantum field theory in its usually given form,
position is promoted to the same status as time, and hence becomes a
parameter in the qwuantum field, while in quantum mechanics it is an
operator vector."
With the typo in the original. This part of your FAQ is right.
Your section S2g ''Particle positions and the position operator'', is
*not* about RQFT but about RQM.
This section is incorrect. In RQM there is an position operator x, but it
cannot be consistently interpreted. The failure to obtain a correct
position operator x was the reason which the founders of RQFT downgraded x
to *parameter*.
You cite several RQM papers about the Newton-Wigner consistent position
operator. Take for instance
L.L. Foldy and S.A. Wouthuysen,
On the Dirac Theory of Spin 1/2 Particles and Its Non-Relativistic
Limit, Phys. Rev. 78 (1950), 29-36.
They start by noticing some difficulties of the position operator x. Then
they introduce an operator X in (23), and refer to it as the new
"position" operator (the "" are in the original and the reason which this
is not the position operator but a "position" operator is given below).
In the same page they do clear that the position operator x can be
splinted as
x = X + (X - x)
They name X the "/mean-position/ operator". Evidently (X - x) measures the
difference with the particle position operator. They also write in a
footnote that X is that Newton and Wigner found.
They also give the "mean velocity operator" associated to X.
Contrary to your claims, X is not the position operator associated to the
particle but only an averaged mean-position operator. The operator
continues being x with all the known deficiencies.
Moreover, the operator X is useless in RQFT; the arguments for quantum
fields continue being the parameters x and t, the scalar interaction in
QED continues being given by an expression (e^2 / |x-y|), the S-matrix
integrations are done over d^3x and dt and so on...
Regards
[1]
The notions of localizability and space: from Eugene Wigner to Alain
Connes 1989: Nucl. Phys. Proc. Suppl. 6, 222. Bacry, H.
--
http://www.canonicalscience.org/
Usenet Guidelines:
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html |
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| Jay R. Yablon... |
| Posted: Mon May 18, 2009 11:57 am |
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"Arnold Neumaier" <Arnold.Neumaier at (no spam) univie.ac.at> wrote in message
news:4A07225B.4030204 at (no spam) univie.ac.at...
[quote:92a0750651]Jay R. Yablon schrieb:
When apply we the canonical :
[x^i,p_j] = ih delta^i_j, i=1,2,3 (1)
is this really just an approximate relationship in the limiting case
where a mass m-->0?
That is, is (1) really of the form:
[x^i,p_j] = ih delta^i_j + F(m)^i_j (1)
where F(m)^i_j --> 0 as m --> 0?
No. it is an exact relationship that is at the basis of quantum
mechanics, valid for total position and momentum of any physical
system with nonzero total mass, even in the relativistic case and
in quantum field theory.
For systems of mass 0 and spin >1/2 , the position operator
is nonexistent, and therefore no associated CCR exists.
See Section S2g ''Particle positions and the position operator''
of my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
Arnold Neumaier
[/quote:92a0750651]
Does this also remain exact for *interacting* fermions, where there is a
potential A^v employed in Dirac's equation and p^v-->p^v+eA^v?
Thanks,
Jay |
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| Arnold Neumaier... |
| Posted: Tue May 19, 2009 6:00 am |
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Jay R. Yablon schrieb:
[quote:17396f6795]
"Arnold Neumaier" <Arnold.Neumaier at (no spam) univie.ac.at> wrote in message
news:4A07225B.4030204 at (no spam) univie.ac.at...
Jay R. Yablon schrieb:
[x^i,p_j] = ih delta^i_j, i=1,2,3 (1)
is an exact relationship that is at the basis of quantum
mechanics, valid for total position and momentum of any physical
system with nonzero total mass, even in the relativistic case and
in quantum field theory.
For systems of mass 0 and spin >1/2 , the position operator
is nonexistent, and therefore no associated CCR exists.
See Section S2g ''Particle positions and the position operator''
of my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
Does this also remain exact for *interacting* fermions, where there is a
potential A^v employed in Dirac's equation and p^v-->p^v+eA^v?
[/quote:17396f6795]
It holds for any Poincare-invariant quantum theory, as I explained
in my mail on this topic that appeared on s.p.r on 2009-05-14.
In this equation, p must be the vector whose components are
the generators of the space translations. In the standard
representation (the instant form in the terminology of Dirac,
see Section S2d of my FAQ), these generators are explicitly
computable for all field theories, as generators of corresponding
symmetries of the action.
For gauge theories, I don't remember whether these are the sum over
the single-particle p^v or the sum over the single-particle p^v+eA^v
(and have currently no time to check).
For particles in a nontrivial external field, however, the field
breaks translation invariance, and the symmetry argument no longer
applies. I can't immediately interpret this situation.
Arnold Neumaier |
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| Posted: Thu May 21, 2009 10:28 am |
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Hi I just saw this post and would like to ask some questions: (esp to
Neumaier) What is the reason why position or momentum operators cannot
be defined for massless particles. Is the reason definitive? Can these
always be defined for massive particles with spin? In other words, I
want to understand the exact obstructions.
For Van Hees, I want to ask exactly what are the observables he claims
that dont depend on position, and WHY it is that only these are
important from a physical point of view, and separately I want to ask
whether we can do away with position altogeteher or do we need it, but
only within a boundary of error-- After all we care that the particles
that we'll do the experiment with ahve a large probability of being
inside the accelerator apparatus, no? |
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| Arnold Neumaier... |
| Posted: Tue May 26, 2009 1:11 pm |
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ultraman2002 at (no spam) hotmail.com schrieb:
[quote:e89dea3cdc]Hi I just saw this post and would like to ask some questions: (esp to
Neumaier) What is the reason why position or momentum operators cannot
be defined for massless particles. Is the reason definitive?
[/quote:e89dea3cdc]
Yes, except for spin 0 or 1/2. See Section S2g (Particle positions
and the position operator) of my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
[quote:e89dea3cdc]Can these always be defined for massive particles with spin?
[/quote:e89dea3cdc]
Yes. See again the above.
[quote:e89dea3cdc]In other words, I want to understand the exact obstructions.
[/quote:e89dea3cdc]
The main reason is that, in the massive case, the algebra of
expressions in p, q and the spin generators is isomorphic to the
algebra of expressions in the Poincare generators and the spin
generators on the corresponding Hilbet space, while in the
massless case, the latter space is too small when only one
irreducible representation is present.
[quote:e89dea3cdc]For Van Hees, I want to ask exactly what are the observables he claims
that dont depend on position,
[/quote:e89dea3cdc]
Scattering cross sections do not depend on position but only on
angles.
[quote:e89dea3cdc]and WHY it is that only these are
important from a physical point of view, and separately I want to ask
whether we can do away with position altogteher or do we need it,
[/quote:e89dea3cdc]
The position operator exists for any massive system separated enough
to be approximated well by an irreducible representation of the
Poincare group. Thus the question whether or not it is needed
is practically irrelevant.
But it is needed to give the concept of ''probability of being
in an arbitrary circular region of space'' a logically coherent
meaning. For a photon, one can attach to this phrase no clear
meaning since the photon is massless.
[quote:e89dea3cdc]but only within a boundary of error-- After all we care that the particles
that we'll do the experiment with have a large probability of being
inside the accelerator apparatus, no?
[/quote:e89dea3cdc]
True, but these particles are all massive, so there is no problem.
Arnold Neumaier. |
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| Hendrik van Hees... |
| Posted: Wed May 27, 2009 3:27 pm |
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Arnold Neumaier wrote:
[quote:0a40b0245b]But it is needed to give the concept of ''probability of being
in an arbitrary circular region of space'' a logically coherent
meaning. For a photon, one can attach to this phrase no clear
meaning since the photon is massless.
[/quote:0a40b0245b]
From a mathematical point of view you may be right, but everyday
experience tells us that we are able to use light for very precise
measurements of position. Every time, you point with a laser pointer
during a talk you are giving, you disprove what you are saying. You
can point pretty precisely where you want to point.
The reason is simply that the wall you point to is massive, and thus
the interaction region of the em. wave with the massive wall, leading
to reflection (and of course also some absorption) of the wave, so
that you can see the spot on the wall. Again, in your eye the light
wave affects some small well-defined region on your retina leading to
the picture of the spot after some signal processing in your brain.
That's why I said previously that we don't need a position operator
for photons to understand its measurable properties: The location of
a photon within the uncertainty given by the detector's resolution at
the moment of measurement is simply defined by the position of the
detector where it is registered.
It's clear that we cannot measure a photon's position without a
detector of any kind. Thus in practice the lack of a position
operator for photons is quite irrelevant.
--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen
http://theory.gsi.de/~vanhees/faq/ |
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| Arnold Neumaier... |
| Posted: Thu May 28, 2009 6:54 am |
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Hendrik van Hees wrote:
[quote:c99c1c0373]Arnold Neumaier wrote:
But it is needed to give the concept of ''probability of being
in an arbitrary circular region of space'' a logically coherent
meaning. For a photon, one can attach to this phrase no clear
meaning since the photon is massless.
From a mathematical point of view you may be right, but everyday
experience tells us that we are able to use light for very precise
measurements of position. Every time, you point with a laser pointer
during a talk you are giving, you disprove what you are saying. You
can point pretty precisely where you want to point.
[/quote:c99c1c0373]
Nobody knowws where a photon in flight is. One at best knows where
a photon had been in the moment it died. A photon in a monochromatic
beam is smeared along the whole beam (or at least a very extended
portion of it) and has no definite position. This is obvious from
the fact that its eigentime is the same everywhere along the beam -
it at all these positions simultaneously.
[quote:c99c1c0373]The reason is simply that the wall you point to is massive, and thus
the interaction region of the em. wave with the massive wall, leading
to reflection (and of course also some absorption) of the wave, so
that you can see the spot on the wall. Again, in your eye the light
wave affects some small well-defined region on your retina leading to
the picture of the spot after some signal processing in your brain.
[/quote:c99c1c0373]
The position where a photon is detected is indeed fixed by where
the beam ends, by interacting with a piece of massive matter that
is absorbing enough.
But it is the position of the absorbing material and not the
position of the photon that is observed.
This is fully consistent with the fact that there is no position
operator for a photon, but that there is one for massive objects.
[quote:c99c1c0373]That's why I said previously that we don't need a position operator
for photons to understand its measurable properties: The location of
a photon within the uncertainty given by the detector's resolution at
the moment of measurement is simply defined by the position of the
detector where it is registered.
It's clear that we cannot measure a photon's position without a
detector of any kind. Thus in practice the lack of a position
operator for photons is quite irrelevant.
[/quote:c99c1c0373]
The lack of a position operator gives just the right insight to
interpret everything consistently.
Arnold Neumaier |
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| Hendrik van Hees... |
| Posted: Thu May 28, 2009 9:00 am |
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Just to make clear that we agree in all points.
Arnold Neumaier wrote:
[quote:75efe3c792]Nobody knowws where a photon in flight is. One at best knows where
a photon had been in the moment it died. A photon in a monochromatic
beam is smeared along the whole beam (or at least a very extended
portion of it) and has no definite position. This is obvious from
the fact that its eigentime is the same everywhere along the beam -
it at all these positions simultaneously.
[/quote:75efe3c792]
Sure, but since a photon is not observable without interaction with a
detector, it doesn't even make sense to ask where it is without
detecting it.
[quote:75efe3c792]But it is the position of the absorbing material and not the
position of the photon that is observed.
[/quote:75efe3c792]
Exactly: Since there's no position operator for a photon, one cannot
define the observable position within QED. Thus within QED the very
notion of a photon's position is meaningless.
[quote:75efe3c792]
This is fully consistent with the fact that there is no position
operator for a photon, but that there is one for massive objects.
[/quote:75efe3c792]
Precisely!
--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen
http://theory.gsi.de/~vanhees/faq/ |
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| Marc Nardmann... |
| Posted: Thu May 28, 2009 9:33 pm |
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ultraman2002 at (no spam) hotmail.com asked:
[quote:c419e08169]What is the reason why position or momentum operators cannot
be defined for massless particles. Is the reason definitive?
[/quote:c419e08169]
Arnold Neumaier replied:
[quote:c419e08169]Yes, except for spin 0 or 1/2. See Section S2g (Particle positions
and the position operator) of my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
[/quote:c419e08169]
A position *observable* for photons can be defined as a positive
operator-valued measure (POVM); this observable has all desired
properties. POVM measures are the today (quite) generally accepted
mathematical formalisation of the notion "observable" in quantum theory
(cf. e.g. "POVM" in Wikipedia). POVM observables are more general than
"operator observables" (which can be described equivalently as
projection-valued measures). Therefore Wightman's no-go theorem does not
apply to them.
This should be mentioned in the FAQ, I guess. I can't give a precise
reference right now, because I don't have access to the literature. If I
remember correctly, the construction of the photon position observable
goes back to Karl Kraus (~1980). It should be mentioned in any decent
book which discusses POVM observables.
-- Marc Nardmann |
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| Hendrik van Hees... |
| Posted: Thu May 28, 2009 11:36 pm |
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That sounds interesting. I cannot find anything with Google Scholar
along the lines you suggested.
I guess the idea is precisely to formalize what I tried to say in the
handwaving physicist's way: The photon position is defined by
interactions with the measurement device (like a photo plate or more
modern versions thereof like CCDs). Isn't this the physics starting
point to introduce POVMs as generalized state-preparation descriptions?
One considers the measurement as coupling of the object (or observable)
to measure with the measurement device, described as another quantum
system (of "macroscopic size") and then to trace out the measurement
device again, leading to a mixed state of the measured object, which in
general is not decomposed into mutually orthogonal projectors but more
generally into non-orthogonal ones.
I think, a nice book to learn about this formalism is
A. Peres, Quantum Mechanics Concepts and Methods, Kluwer
I'm not sure whether the definition (or better formalization) of
photon-position measurements in terms of POVMs can be found in this
book, since I don't have it available at the moment.
Marc Nardmann wrote:
[quote:9caca6d9bc]A position *observable* for photons can be defined as a positive
operator-valued measure (POVM); this observable has all desired
properties. POVM measures are the today (quite) generally accepted
mathematical formalisation of the notion "observable" in quantum
theory (cf. e.g. "POVM" in Wikipedia). POVM observables are more
general than "operator observables" (which can be described
equivalently as projection-valued measures). Therefore Wightman's
no-go theorem does not apply to them.
This should be mentioned in the FAQ, I guess. I can't give a precise
reference right now, because I don't have access to the literature. If
I remember correctly, the construction of the photon position
observable goes back to Karl Kraus (~1980). It should be mentioned in
any decent book which discusses POVM observables.
[/quote:9caca6d9bc]
--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen
http://theory.gsi.de/~vanhees/faq/ |
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| Marc Nardmann... |
| Posted: Fri May 29, 2009 10:53 pm |
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Hendrik van Hees wrote:
[quote:8ded52ed08]That sounds interesting. I cannot find anything with Google Scholar
along the lines you suggested.
[/quote:8ded52ed08]
E.g.: Busch, Grabowski, Lahti: Operational quantum physics. The pages
92-94 (which can be seen on Google Books) contain a short introduction
and references (but one would need a hardcopy of the book to read the
bibliography and find out what their references 3.27 and 3.28 actually are):
http://books.google.com/books?id=anL-mDHBHQcC&printsec=frontcover&dq=busch+grabowski+lahti#PRA1-PA92,M1
The authors give one definition of a photon position observable. They
say that there are many nonequivalent possible definitions.
-- Marc Nardmann |
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| Arnold Neumaier... |
| Posted: Thu Jun 25, 2009 8:41 pm |
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Marc Nardmann schrieb:
[quote:71e5d5dc45]Hendrik van Hees wrote:
That sounds interesting. I cannot find anything with Google Scholar
along the lines you suggested.
E.g.: Busch, Grabowski, Lahti: Operational quantum physics. The pages
92-94 (which can be seen on Google Books) contain a short introduction
and references (but one would need a hardcopy of the book to read the
bibliography and find out what their references 3.27 and 3.28 actually
are):
http://books.google.com/books?id=anL-mDHBHQcC&printsec=frontcover&dq=busch+grabowski+lahti#PRA1-PA92,M1
The authors give one definition of a photon position observable. They
say that there are many nonequivalent possible definitions.
[/quote:71e5d5dc45]
I added the following to Section 2g of my FAQ:
If the concept of an observable is not tied to that of a Hermitian
operator but rather to that of a POVM (positive operator-valued
measure), there is more flexibility, and covariant POVMs for positon
measurements can be meaningfully defined, even for photons. See, e.g.,
A. Peres and D.R. Terno,
Quantum Information and Relativity Theory,
Rev. Mod. Phys. 76 (2004), 93.
[see, in particular, (52)]
K. Kraus, Position observable of the photon, in:
The Uncertainty Principle and Foundations of Quantum Mechanics,
Eds. W. C. Price and S. S. Chissick,
John Wiley & Sons, New York, pp. 293-320, 1976.
M. Toller,
Localization of events in space-time,
Phys. Rev. A 59, 960 (1999).
P. Busch, M. Grabowski, P. J. Lahti,
Operational Quantum Physics,
Springer-Verlag, Berlin Heidelberg 1995, pp.92-94.
Note that a POVM describes the statistics of a measurement process
rather than some underlying reality. This is reflected in the fact
that there are many possible nonequivalent possible definitions of
POVMs, all pertaining to possible different ways to get a measured
position.
Therefore, the POVM does not allow to talk about the position of a
photon - which could exist only if the corresponding operator existed -,
but only about th measured position: The photon is somewhere near the
value obtained by the measurement, without any more definite statement
being possible.
Arnold Neumaier |
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The time now is Sun Mar 14, 2010 6:49 pm
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