I heard the standard interpretation of Heisenberg Uncertainty Principle: Just the measurement affects the position of the body because always you want to see a body (=to measure the position), you need a light - and only the fact that photons that strike the surface of that radiating body have momentum mean that the position of the radiating body isn't certain (even theoretically) only because of your measuring - and therefore the uncertainty in position will always be non-zero.

However, if we knew that a photon that came to us, comes from a perfectly homogeneous radiating body, we also know that the recoil, which gives photon after emitting always corresponds to a photon equally strong on the other side of the radiating body.

Of course, the body can not perfectly homogeneously radiate and moreover there is no way we could get an information that the body perfectly homogeneously radiate, but if it all would possible, where would the position uncertainty disappear?

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    $\begingroup$ I don't understand your question. All measurements have some uncertainty associated with them. For example, if it is a faint source, there is Poisson noise. $\endgroup$ – Virgo Nov 28 '14 at 0:26
  • $\begingroup$ I think I see what you're getting at. A body may emit radiation isotropically (equally in all directions) on average but emission of a photon on one side will not happen at the same time as emission of a photon on the other side. As a result the radiating body will execute a random walk which on average will return it to the starting point, but there will be still be an uncertainty in its position - the average may be zero, but the standard deviation will not be. $\endgroup$ – G. Paily Nov 28 '14 at 21:12
  • $\begingroup$ The position uncertainty of the radiating body would lie in the uncertainty of the photons detected. $\endgroup$ – user56903 Nov 29 '14 at 10:04
  • $\begingroup$ @G.Paily Oh, that (what you´re describing) wouldn´t be perfect isotropic radiation (You says this mean that radiation is equall in all directions - so there for there have to be a photon apart the photon we´ve seen) What happend if we really see perfectly isotropic radiating body? $\endgroup$ – foggy Nov 29 '14 at 13:17
  • $\begingroup$ @DirkBruere Yes, you should be right if there would be some else reason to have uncertainity here, but it seems there is NO reason not to detect the photon perfectly (the basic reason should be affecting of the radiating body, but in this case this cannot be true) $\endgroup$ – foggy Nov 29 '14 at 13:17

This question confuses two frameworks, the classical thermodynamic one with the quantum mechanical one.

A perfectly homogenous radiating body belongs to classical thermodynamics.

Photons belong to quantum mechanics, as well as the uncertainty principle.

Lets look at the problem classically. Electromagnetic radiation is continuous from a classical body and therefore theoretically there is no uncertainty other than the measurement uncertainty given by the method of determining the position of the body, since, as you say, at the same time radiation is leaving equally and oppositely.

The problem quantum mechanically can only be looked at quantum mechanical dimensions, so the body should have nanometer size and smaller . Let us assume a crystal for simplicity, because it has a coherent wave function, and no input energy is keeping it in equilibrium with the environment. The crystal wave function has higher energy levels the higher the temperature, and empty levels below to which it can relax radiating a photon. In these dimensions the quantization of radiation is relevant. There will be a time constant for a photon to be released because the body fell from a higher vibrational level to a lower one, and this will not be compensated at the same delta(t) by an equal and opposite photon in the other direction, because photon emissions are random depending on the wavefunction of the crystal . whose square has the probabilities of emission.

So in the quantum mechanical frame the "opposite" emission of photons cannot hold in a delta(t) so the photon momentum will impose an indeterminacy, assuming the center of the nano crustal to begin at (0,0,0). Using the Heisenberg uncertainty principle, the momentum of the photon is

photon momentum

the uncertainty in momentum is proportional is h*delta(lamda)/lamdda^2 and this will set the uncertainty in position x ,that cannot be known even in theory. This ends up to

delta(lamda)*x> lamda^2 , x>lamda

so the position x cannot be known better than the wavelength of the emitted photon.

This intrinsic quantum mechanical bound of order of nanometers can never be reached in accuracy when measuring the position of a classical body, more so of an astronomical size body. It is irrelevant.

  • $\begingroup$ Thank you very much. But isn't it only our fall that we can't measure at the quantum level the macroscopic systems? $\endgroup$ – foggy Dec 1 '14 at 15:37
  • $\begingroup$ Yes, the Heisenberg uncertainty is always satisfied for macroscopic systems , because h*nu, the energy of a single photon, is so low since h is such a small number for macroscopic systems which are composed by ~ 10^23 , atoms/molecules (avogadro constant) that classical mechanics and thermodynamics describe the systems. $\endgroup$ – anna v Dec 1 '14 at 18:24
  • $\begingroup$ @annav And what causes that crystal's wave function is random (or what makes it random)? I thought that this is proven by the Uncertainity principle. $\endgroup$ – foggy Jul 29 '15 at 10:07
  • $\begingroup$ @Probably The crystal is not random. It is controlled by very precise quantum mechanical functions that give the probability for finding the atoms at specific locations. The Heisenberg uncertainty is an over all tool that reflects the intrinsic structure of the operator functions that are involved in the generation of the state function whose square gives the probabilities. A calculational tool, imo. $\endgroup$ – anna v Jul 29 '15 at 12:29
  • $\begingroup$ @annav Could you please explain me it a little bit deeper (axiom by axiom)? Why is the crystal position given by this wave function? The model I had known before was, as writes foggy, the easy pop-sci interpretation. Thank you very much. $\endgroup$ – foggy Jul 30 '15 at 11:41

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