On the uncertainty principle for photons. An experimental counter example??by kvblake Tags: counter, experimental, photons, principle, uncertainty 
#1
Oct1807, 05:00 AM

P: n/a

From QM follows that the position of a photon can not be determined
better than L (de Broglie wavelength). Suppose one creates a wave in the meter range  say L=2m. Could the photons in this wave go through a tube of radius 1 cm? If one registers a click of a photon counter placed inside the tube  would this be a counterexample of QM  e.g. the position of the photon determeined better than de Broglie's wavelength??? 


#2
Oct1807, 05:00 AM

P: n/a

kvblake wrote:
> > From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). > Suppose one creates a wave in the meter range  say L=2m. Could the > photons in this wave go through a tube of radius 1 cm? > If one registers a click of a photon counter placed inside the tube  > would this be a counterexample of QM  e.g. the position of the photon > determeined better than de Broglie's wavelength??? "near field" microscopy 107,000 hits Subwavelength apertures are commonplace. Exit intensity rapidly decays with conduit length.  Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 


#3
Oct1907, 05:00 AM

P: n/a

On 17 , 23:48, Uncle Al <Uncle...@hate.spam.net> wrote:
> kvblake wrote: > "near field" microscopy 107,000 hits > > Subwavelength apertures are commonplace. Exit intensity rapidly > decays with conduit length. > >  > Uncle Alhttp://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/lajos.htm#a2 So it's a counterexample which takes common place??? Regards: Kevin 


#4
Oct1907, 05:00 AM

P: n/a

On the uncertainty principle for photons. An experimental counter example??
kvblake <kvblake2003@yahoo.com> wrote:
> From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). If you think so, please show a derivation from first principles. (and please, wavelength will do. No need to involve De Broglie) > Suppose one creates a wave in the meter range  say L=2m. Could the > photons in this wave go through a tube of radius 1 cm? Yes. Try it. > If one registers a click of a photon counter placed inside the tube  > would this be a counterexample of QM  e.g. the position of the photon > determeined better than de Broglie's wavelength??? No, unless you supply the proof required. You can do better than that though. Suppose you throw a photon with sufficient energy at an atom, and detect the resultant photoelectron. If you see an electron you have fixed the position of the photon to an atomic diameter, much smaller than its wavelength. For further amusement you may note that the atom can by a hydrogen atom in a Rydberg state, which allows you to detect even radiofrquency waves. Best, Jan 



#5
Oct2007, 12:36 AM

P: 20

The de Broglie wavelength is not the same as the uncertainty in the particle's position. To quickly convince yourself of this, plug the de Broglie wavelength into the uncertainty principle. You will see that this implies that the uncertainty in momentum has a lower bound, and that this lower bound increases as momentum increases. QM does not limit how precisely we can determine momementum, but rather, how precisely we can simultaneously determine momentum and position.



#6
Oct2007, 05:00 AM

P: n/a

On Oct 17, 12:41 pm, kvblake <kvblake2...@yahoo.com> wrote:
> From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). > Suppose one creates a wave in the meter range  say L=2m. Could the > photons in this wave go through a tube of radius 1 cm? > If one registers a click of a photon counter placed inside the tube  > would this be a counterexample of QM  e.g. the position of the photon > determeined better than de Broglie's wavelength??? I think the misunderstanding is that QM does not say that the energy of a photon cannot be captured in a volume much smaller than its wavelength, but that the exact path that the photon travels between source and detection cannot be measured more precisely than this. As other replys have pointed out, the energy in a radio frequency photon can originate from a volume with dimensions orders of magnitude smaller than the wavelength, and when the photon is detected the location of detection can be determined with much greater precision as well. It is the location of the "photon" in between these two events that is uncertain. Rich L. 


#7
Oct2107, 05:00 AM

P: n/a

kvblake wrote:
> From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). > Suppose one creates a wave in the meter range  say L=2m. Could the > photons in this wave go through a tube of radius 1 cm? > If one registers a click of a photon counter placed inside the tube  > would this be a counterexample of QM  e.g. the position of the photon > determeined better than de Broglie's wavelength??? > de Broglie wave length L = h/p uncertainty (delta L) => h/(4pi(delta p)) There is a difference. Richard 


#8
Oct2207, 05:00 AM

P: n/a

"Rich L." <ralivingston@sbcglobal.net> wrote in message news:1192800026.682129.277410@i38g2000prf.googlegroups.com... > On Oct 17, 12:41 pm, kvblake <kvblake2...@yahoo.com> wrote: >> From QM follows that the position of a photon can not be determined >> better than L (de Broglie wavelength). >> Suppose one creates a wave in the meter range  say L=2m. Could the >> photons in this wave go through a tube of radius 1 cm? >> If one registers a click of a photon counter placed inside the tube  >> would this be a counterexample of QM  e.g. the position of the photon >> determeined better than de Broglie's wavelength??? > > I think the misunderstanding is that QM does not say that the energy > of a photon cannot be captured in a volume much smaller than its > wavelength, but that the exact path that the photon travels between > source and detection cannot be measured more precisely than this. As > other replys have pointed out, the energy in a radio frequency photon > can originate from a volume with dimensions orders of magnitude > smaller than the wavelength, and when the photon is detected the > location of detection can be determined with much greater precision as > well. It is the location of the "photon" in between these two events > that is uncertain. > > Rich L. > Sure, but what about the Delta x times Delta p >= hbar/2 problem? If the position is momentarily at the size of an atom (a few tenths of nm) and much smaller than wavelength, then the momentum uncertainty is more than the wavelength  but it can't be. There must be a good answer, but is it simple? 


#9
Oct2307, 05:01 AM

P: n/a

On Oct 21, 6:58 am, Neil Bates <neil_del...@caloricmail.com> wrote:
> "Rich L." <ralivings...@sbcglobal.net> wrote in message > > news:1192800026.682129.277410@i38g2000prf.googlegroups.com... > > > > > > > On Oct 17, 12:41 pm, kvblake <kvblake2...@yahoo.com> wrote: > >> From QM follows that the position of a photon can not be determined > >> better than L (de Broglie wavelength). > >> Suppose one creates a wave in the meter range  say L=2m. Could the > >> photons in this wave go through a tube of radius 1 cm? > >> If one registers a click of a photon counter placed inside the tube  > >> would this be a counterexample of QM  e.g. the position of the photon > >> determeined better than de Broglie's wavelength??? > > > I think the misunderstanding is that QM does not say that the energy > > of a photon cannot be captured in a volume much smaller than its > > wavelength, but that the exact path that the photon travels between > > source and detection cannot be measured more precisely than this. As > > other replys have pointed out, the energy in a radio frequency photon > > can originate from a volume with dimensions orders of magnitude > > smaller than the wavelength, and when the photon is detected the > > location of detection can be determined with much greater precision as > > well. It is the location of the "photon" in between these two events > > that is uncertain. > > > Rich L. > > Sure, but what about the Delta x times Delta p >= hbar/2 problem? If the > position is momentarily at the size of an atom (a few tenths of nm) and much > smaller than wavelength, then the momentum uncertainty is more than the > wavelength  but it can't be. There must be a good answer, but is it simple? > Hide quoted text  That would be true for the photon, but once it is "detected" it doesn't exist anymore. The energy of the photon is now in an electron confined (possibly) to an atom. The uncertainty principle doesn't say anything about what happens when the wave function "collapses" like this. Rich L. 


#10
Oct2307, 05:01 AM

P: n/a

On Oct 21, 1:58 pm, Neil Bates <neil_del...@caloricmail.com> wrote:
> Sure, but what about the Delta x times Delta p >= hbar/2 problem? If the > position is momentarily at the size of an atom (a few tenths of nm) and much > smaller than wavelength, then the momentum uncertainty is more than the > wavelength  but it can't be. There must be a good answer, but is it simple? The positionmomentum uncertainty as you wrote it applies only to Gaussian wave packets, you can't apply it to a plane wave (unless you think it as a limit for Delta x > infinity). Anyway, if you light the atom with a radio frequency wave, and you get a photoelectron, you know that you've hit it. Now, how do you know where is the atom? The only information you have is that you've hit it, but you don't know where you did it better than the wavelength. If you want to pin down the position of the atom better than that, you can start to use more spatially (or time) compressed wave packets, and you will start to lose the energy information, just as the uncertainty principle says. For example, if you want to know the position within 0.1nm, you'll need a wave packet with an energy dispersion of at least 1keV. Regards, Guillermo 


#11
Oct2307, 05:01 AM

P: n/a

On Oct 21, 6:58 am, Neil Bates <neil_del...@caloricmail.com> wrote:
> "Rich L." <ralivings...@sbcglobal.net> wrote in message > > news:1192800026.682129.277410@i38g2000prf.googlegroups.com... > > > > > > > On Oct 17, 12:41 pm, kvblake <kvblake2...@yahoo.com> wrote: > >> From QM follows that the position of a photon can not be determined > >> better than L (de Broglie wavelength). > >> Suppose one creates a wave in the meter range  say L=2m. Could the > >> photons in this wave go through a tube of radius 1 cm? > >> If one registers a click of a photon counter placed inside the tube  > >> would this be a counterexample of QM  e.g. the position of the photon > >> determeined better than de Broglie's wavelength??? > > > I think the misunderstanding is that QM does not say that the energy > > of a photon cannot be captured in a volume much smaller than its > > wavelength, but that the exact path that the photon travels between > > source and detection cannot be measured more precisely than this. As > > other replys have pointed out, the energy in a radio frequency photon > > can originate from a volume with dimensions orders of magnitude > > smaller than the wavelength, and when the photon is detected the > > location of detection can be determined with much greater precision as > > well. It is the location of the "photon" in between these two events > > that is uncertain. > > > Rich L. > > Sure, but what about the Delta x times Delta p >= hbar/2 problem? If the > position is momentarily at the size of an atom (a few tenths of nm) and much > smaller than wavelength, then the momentum uncertainty is more than the > wavelength  but it can't be. There must be a good answer, but is it simple? > Hide quoted text  That would be true for the photon, but once it is "detected" it doesn't exist anymore. The energy of the photon is now in an electron confined (possibly) to an atom. The uncertainty principle doesn't say anything about what happens when the wave function "collapses" like this. Rich L. 


#12
Oct2307, 05:01 AM

P: n/a

On Oct 21, 1:58 pm, Neil Bates <neil_del...@caloricmail.com> wrote:
> Sure, but what about the Delta x times Delta p >= hbar/2 problem? If the > position is momentarily at the size of an atom (a few tenths of nm) and much > smaller than wavelength, then the momentum uncertainty is more than the > wavelength  but it can't be. There must be a good answer, but is it simple? The positionmomentum uncertainty as you wrote it applies only to Gaussian wave packets, you can't apply it to a plane wave (unless you think it as a limit for Delta x > infinity). Anyway, if you light the atom with a radio frequency wave, and you get a photoelectron, you know that you've hit it. Now, how do you know where is the atom? The only information you have is that you've hit it, but you don't know where you did it better than the wavelength. If you want to pin down the position of the atom better than that, you can start to use more spatially (or time) compressed wave packets, and you will start to lose the energy information, just as the uncertainty principle says. For example, if you want to know the position within 0.1nm, you'll need a wave packet with an energy dispersion of at least 1keV. Regards, Guillermo 


#13
Oct2807, 05:00 AM

P: n/a

"Richard Saam" <rdsaam@att.net> wrote in message
news:gSHRi.232966$ax1.37139@bgtnsc05news.ops.worldnet.att.net... > kvblake wrote: >> From QM follows that the position of a photon can not be determined >> better than L (de Broglie wavelength). >> Suppose one creates a wave in the meter range  say L=2m. Could the >> photons in this wave go through a tube of radius 1 cm? >> If one registers a click of a photon counter placed inside the tube  >> would this be a counterexample of QM  e.g. the position of the photon >> determeined better than de Broglie's wavelength??? >> > de Broglie wave length L = h/p > > uncertainty (delta L) => h/(4pi(delta p)) > > There is a difference. > > Richard > The Heisenberg uncertainty principle requires that when the uncertainty in a photon's position is less than it's wavelength (lambda), the uncertainty in its momentum must be greater than hc/lambda (ignoring factors of 2 and pi). This is possible because the photon momentum is complex, with both a real part, with magnitude hc/lambda, and an imaginary part, which makes the total magnitude of the momentum greather than hc/lambda. The imaginary component of the momentum required for subwavelength resolution corresponds to an attenuation of the photon wave. Using a narrow tube to define the photon position comes at the cost of an exponential loss of intensity. 


#14
Oct3007, 05:00 AM

P: n/a

J.J.Lodder wrote:
kvblake <kvblake2...@yahoo.com> wrote: > From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). If you think so, please show a derivation from first principles. (and please, wavelength will do. No need to involve De Broglie) > Suppose one creates a wave in the meter range  say L=2m. Could the > photons in this wave go through a tube of radius 1 cm? Yes. Try it. > If one registers a click of a photon counter placed inside the tube  > would this be a counterexample of QM  e.g. the position of the photon > determeined better than de Broglie's wavelength??? No, unless you supply the proof required. You can do better than that though. Suppose you throw a photon with sufficient energy at an atom, and detect the resultant photoelectron. If you see an electron you have fixed the position of the photon to an atomic diameter, much smaller than its wavelength. For further amusement you may note that the atom can by a hydrogen atom in a Rydberg state, which allows you to detect even radiofrquency waves. Best, Jan ========================================================= Sorry about the delay. A huge lack of Internet connection... I think that was a trivial fact. Nevertheless the prove I found seems a little strange to me too. It's from QED of Beresteckii, Lifgarbagez (the intro). It's like that: 1. Min uncertainty of x > Dx for a massive particle in its rest frame is Dx=h/mc (under h I mean Dirac's h not Heisenberg's h>h/2PI but I dont know how to write it with this bar up..  that's why ...... was confused with this 4PI   the second 2 is from some preciser formulation of DxDp=h/2) 2. Min uncertainty in the frame where the particle is moving with energy E is then Dx=ch/E 3. For ultrarelativistic particles E almost = cp hence Dx>h/p 4. Photons are always relativistic so Dx>h/p=L (lambda as introduced by de Broglie). In this book also stands the following. The above is valid for experiments from every result of which x can be determined (I would call this experiments of class I) If one uses impacts without probability one for the time of experiment from each deflection of the probe particle one can judge about x. But if deflection is 0 one can not tell anything about x (class II). I think my experiment is from class I and yours is from class II. ================================ To Rich: Rich L. wrote: I think the misunderstanding is that QM does not say that the energy of a photon cannot be captured in a volume much smaller than its wavelength, but that the exact path that the photon travels between source and detection cannot be measured more precisely than this. As other replys have pointed out, the energy in a radio frequency photon can originate from a volume with dimensions orders of magnitude smaller than the wavelength, and when the photon is detected the location of detection can be determined with much greater precision as well. It is the location of the "photon" in between these two events that is uncertain. Rich L. ========================== If there is a source of radiowaves and I put in front of it a tube of lenght 10 m and r=2 cm with a detector at the end and the detector clicks I could be absolute sure of the path and Dp wouldn't also be great provided 'c' is a constant. 


#15
Oct3107, 05:00 AM

P: n/a

On Oct 29, 10:17 am, kvblake <kvblake2...@yahoo.com> wrote:
... cut text... > ================================ > To Rich: > > Rich L. wrote: > > I think the misunderstanding is that QM does not say that the energy > of a photon cannot be captured in a volume much smaller than its > wavelength, but that the exact path that the photon travels between > source and detection cannot be measured more precisely than this. As > other replys have pointed out, the energy in a radio frequency photon > can originate from a volume with dimensions orders of magnitude > smaller than the wavelength, and when the photon is detected the > location of detection can be determined with much greater precision > as > well. It is the location of the "photon" in between these two events > that is uncertain. > > Rich L. > > ========================== > > If there is a source of radiowaves and I put in front of it a tube of > lenght 10 m and r=2 cm with a detector at the end and the detector > clicks I could be absolute sure of the path and Dp wouldn't also be > great provided 'c' is a constant. If you did this experiment, you would find that the probability of detecting the radio frequency photon was extremely small, assuming the walls of the tube are opaque to radio waves and the wavelength is much greater than 2cm. This is no more mysterious than detecting a photon in the evanescent region where the photon theoretically cannot exist at all. The QM wave function exists in such areas, but is attenuating rapidly with distance into it. QM makes experimentally confirmed accurate predictions about such situations. Rich L. 


#16
Nov207, 05:00 AM

P: n/a

On 31 , 07:51, "Rich L." <ralivings...@sbcglobal.net> wrote:
> On Oct 29, 10:17 am, kvblake <kvblake2...@yahoo.com> wrote: > .. cut text... > > > > > > > ================================ > > To Rich: > > > Rich L. wrote: > > > I think the misunderstanding is that QM does not say that the energy > > of a photon cannot be captured in a volume much smaller than its > > wavelength, but that the exact path that the photon travels between > > source and detection cannot be measured more precisely than this. As > > other replys have pointed out, the energy in a radio frequency photon > > can originate from a volume with dimensions orders of magnitude > > smaller than the wavelength, and when the photon is detected the > > location of detection can be determined with much greater precision > > as > > well. It is the location of the "photon" in between these two events > > that is uncertain. > > > Rich L. > > > ========================== > > > If there is a source of radiowaves and I put in front of it a tube of > > lenght 10 m and r=2 cm with a detector at the end and the detector > > clicks I could be absolute sure of the path and Dp wouldn't also be > > great provided 'c' is a constant. > > If you did this experiment, you would find that the probability of > detecting the radio frequency photon was extremely small, assuming the > walls of the tube are opaque to radio waves and the wavelength is much > greater than 2cm. This is no more mysterious than detecting a photon > in the evanescent region where the photon theoretically cannot exist > at all. The QM wave function exists in such areas, but is attenuating > rapidly with distance into it. QM makes experimentally confirmed > accurate predictions about such situations. > > Rich L.  > >   Thanks for consideration: I am sure that probability for detection would be extremely small. But as you say there would be such detections even with opaque walls. What about the path!!! of that photon (even the impulse)  if there is just one exception the Heisenberg principle is threatened??? I can even remove a part of the tube and know the path of that photon in between?! Regards: Kevin 


#17
Nov207, 05:00 AM

P: n/a

On Nov 1, 1:17 pm, kvblake <kvblake2...@yahoo.com> wrote:
> On 31 , 07:51, "Rich L." <ralivings...@sbcglobal.net> wrote: > > > > > > > On Oct 29, 10:17 am, kvblake <kvblake2...@yahoo.com> wrote: > > .. cut text... > > > > ================================ > > > To Rich: > > > > Rich L. wrote: > > > > I think the misunderstanding is that QM does not say that the energy > > > of a photon cannot be captured in a volume much smaller than its > > > wavelength, but that the exact path that the photon travels between > > > source and detection cannot be measured more precisely than this. As > > > other replys have pointed out, the energy in a radio frequency photon > > > can originate from a volume with dimensions orders of magnitude > > > smaller than the wavelength, and when the photon is detected the > > > location of detection can be determined with much greater precision > > > as > > > well. It is the location of the "photon" in between these two events > > > that is uncertain. > > > > Rich L. > > > > ========================== > > > > If there is a source of radiowaves and I put in front of it a tube of > > > lenght 10 m and r=2 cm with a detector at the end and the detector > > > clicks I could be absolute sure of the path and Dp wouldn't also be > > > great provided 'c' is a constant. > > > If you did this experiment, you would find that the probability of > > detecting the radio frequency photon was extremely small, assuming the > > walls of the tube are opaque to radio waves and the wavelength is much > > greater than 2cm. This is no more mysterious than detecting a photon > > in the evanescent region where the photon theoretically cannot exist > > at all. The QM wave function exists in such areas, but is attenuating > > rapidly with distance into it. QM makes experimentally confirmed > > accurate predictions about such situations. > > > Rich L.  > > >   > > Thanks for consideration: > I am sure that probability for detection would be extremely small. But > as you say there would be such detections even with opaque walls. > What about the path!!! of that photon (even the impulse)  if there is > just one exception the Heisenberg principle is threatened??? > > I can even remove a part of the tube and know the path of that photon > in between?! > > Regards: Kevin Hide quoted text  > >  Show quoted text  I don't think so. The Heisenberg principle concerns the product delta_P*delta_X or delta_E*delta_T. In the case of the tube you are confining in the X and Y directions to the diameter of the tube (assuming it extends along the Z direction). Delta X and Delta Y are thus very small. The Uncertainty principle says the uncertainty in the momentum in the X and Y directions will thus be very large, which is is due to the high k (wavenumber) in order to meet the boundary conditions on the walls of the tube (the math is easier if you use a square tube). Thus at the end of the tube you do indeed have a very precise location for the photon, but you have a very poor idea of it's momentum in the X and Y directions. In fact it is because the momentum uncertainty is so large that the amplitude is decaying exponentially down the tube. Does that help? Rich L. 


#18
Nov807, 05:00 AM

P: n/a

On 2 , 07:12, "Rich L." <ralivings...@sbcglobal.net> wrote:
> On Nov 1, 1:17 pm, kvblake <kvblake2...@yahoo.com> wrote: > > On 31 , 07:51, "Rich L." <ralivings...@sbcglobal.net> wrote: > > > > On Oct 29, 10:17 am, kvblake <kvblake2...@yahoo.com> wrote: > > > .. cut text... > > > > > ================================ > > > > To Rich: > > > > > Rich L. wrote: > > > > > I think the misunderstanding is that QM does not say that the energy > > > > of a photon cannot be captured in a volume much smaller than its > > > > wavelength, but that the exact path that the photon travels between > > > > source and detection cannot be measured more precisely than this. As > > > > other replys have pointed out, the energy in a radio frequency photon > > > > can originate from a volume with dimensions orders of magnitude > > > > smaller than the wavelength, and when the photon is detected the > > > > location of detection can be determined with much greater precision > > > > as > > > > well. It is the location of the "photon" in between these two events > > > > that is uncertain. > > > > > Rich L. > > > > > ========================== > > > > > If there is a source of radiowaves and I put in front of it a tube of > > > > lenght 10 m and r=2 cm with a detector at the end and the detector > > > > clicks I could be absolute sure of the path and Dp wouldn't also be > > > > great provided 'c' is a constant. > > > > If you did this experiment, you would find that the probability of > > > detecting the radio frequency photon was extremely small, assuming the > > > walls of the tube are opaque to radio waves and the wavelength is much > > > greater than 2cm. This is no more mysterious than detecting a photon > > > in the evanescent region where the photon theoretically cannot exist > > > at all. The QM wave function exists in such areas, but is attenuating > > > rapidly with distance into it. QM makes experimentally confirmed > > > accurate predictions about such situations. > > > > Rich L.  > > > >   > > > Thanks for consideration: > > I am sure that probability for detection would be extremely small. But > > as you say there would be such detections even with opaque walls. > > What about the path!!! of that photon (even the impulse)  if there is > > just one exception the Heisenberg principle is threatened??? > > > I can even remove a part of the tube and know the path of that photon > > in between?! > > > Regards: Kevin Hide quoted text  > > >  Show quoted text  > > I don't think so. The Heisenberg principle concerns the product > delta_P*delta_X or delta_E*delta_T. In the case of the tube you are > confining in the X and Y directions to the diameter of the tube > (assuming it extends along the Z direction). Delta X and Delta Y are > thus very small. The Uncertainty principle says the uncertainty in > the momentum in the X and Y directions will thus be very large, which > is is due to the high k (wavenumber) in order to meet the boundary > conditions on the walls of the tube (the math is easier if you use a > square tube). Thus at the end of the tube you do indeed have a very > precise location for the photon, but you have a very poor idea of it's > momentum in the X and Y directions. In fact it is because the > momentum uncertainty is so large that the amplitude is decaying > exponentially down the tube. > > Does that help? > > Rich L.  > >   Not much. More questions arise? 1. The uncertianty of impulse means there should be photons of different (even very very high) energies / impulses/ ariving at the end of the tube. Where could this energies come from? [In the tought experiment of Heisenberg this is very clear  an electron gets an impulse/energy from the photon which is intended to locate it.] 2. The wave associated with the photon was originally in the radio wavelength range. If now the energy is high  the wavelength may be already in the visible range. So there should be a change in the wavelength of the particle. I never heard about such phenomenon  change of the wavelength of a particle even if it goes through obstacles. Maybe I just don't know? 3. What about that sectioin where there is no tube. The particle (photon) is not limited in X,Y  the wave diffracts there and many of the photons which had passed part 1 of the tube should spread; but not this one which goes into part 2 of the tube. It is not confined by the walls and nevertheless I can't imagine it has been travelling some other path but straight line into part 2. I can't imagine one of the spread photons can turn and come into part 2 and travel then in straight line. 4. For a moment I thought the Heisenberg principle is for ensembles (not for individual particles) but no it is an absolute prohibition for 1 particle! 5. My original question was not at all concerning impulse  for the massless particles it has been shown (as I cited in this thread before) in Beresteckii, Lifgarbagez that Dx>L (h/p  wavelength of de Broglie; h>h/2.PI)  maybe because Dp couldn't be bigger than p itself (I cited before the whole prove). That's why I thought of that counter example with the tube. Regards: Kevin 


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