Naked Science Forum
On the Lighter Side => New Theories => Topic started by: nilak on 04/01/2017 16:00:20
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In the double slit experiments, in terms of Quantum Mechanics (QM) , it is often said that the measurement collapses the wavefunction or more exactly, the measurement destroys the interference. However, we can also think, that is not the measurement that destroys the interference ( except for some experiments). The delayed choice quantum eraser can be used as a proof. Scientists developed the DCQE (Delayed Choice Quantum Eraser by Kim et. al) experiment, where there is no measurement involved at the slits, but they went forward with QM believing that even simply watching the results destroys the interference. If QM is all correct, this is a rational conclusion. However, I think it is not the case. The theory is probably wrong. Apparently, the detectors cannot detect the energy of a photon, if the energy that reaches the detector is by a certain degree less than the electron bind energy . QM says, that there is a significant probability of detection when the wave passes through both slits to have a photoelectron release on either detector A or B. That must be wrong. The probability is almost zero. Unless the other part of the wave hits the orbiting electron that still holds the energy from the previous electromagnetic wave in a very short time, the atom will not release the electron. Let's suppose we have a photon with energy Eϒ. After it passes the slits we get two waves of EϒL and EϒR. Say, the energy required for an atom in ground state to release an electron is Ee- ( this is the workfunction). If a wave with energy EL hits the atom, it will energize it to an intermediate unstable electron orbital, but the atom will not release the electron yet. Since the intermediate state does not correspond to a stable electron orbital, it will quickly loose energy. However, if the wave ER comes quickly enough, there will still be enough energy left in the intermediate unstable orbital and ER+EL>Ee-. Hence the electron will be released.
Notice, that the energy of a photon when it was just released will always exceeds the workfunction (the energy Ee-),otherwise the frequency of the wave is not enough to trigger an photo-electron release. We can, then say that this mechanism should work at any frequency of the photon. More exactly, a wave of any frequency could trigger an electron release, because photons can add up energies to the atom and excite it until releasing the electrons. It is clear that it doesn't since the experiments show that the frequency matters. However, we can then understand that for a photon to interfere with an electron, it needs a minimum frequency. It is possible to explain how this happens. A low frequency wave interferes with an electron attached to an atom as well, just like a high frequency does. However the energised electron , looses energy quicker than the low frequency wave is able to provide the energy for releasing an electron.
Supposing we have a Hydrogen atom.
We can write: EA(t)=EA0+Eϒ- PA*t where, EA0 is the energy of the atom when the electron is a stable configuration, like say first orbital
t is measured in absolute seconds, according to this concept. All symbols for time denote absolute time E(t) is the energy stored in the orbiting electron, which is a function of time. EA is the energy received from an electromagnetic wave. PA is the radiation power of the atom in the unstable state.
Eϒ = Pϒ tA
tA is the emission time of an electromagnetic wave. Eϒ is the energy produced by an electromagnetic wave.
We can also write: Eϒ(t) = Pϒ t
Pϒ- power produced by an electromagnetic wave This power depends on the frequency:
Pϒ=kυ k is a constant and υ is the frequency of the wave. We can now write:
Eϒ(t) = kυ t To produce a photo-electron, the energy EA needs to reach Ee
EA(t)=EA0+ kυ t -PA t
In conclusion, to produce photo-electron, the electromagnetic wave must have a frequency that satisfies the following inequality: υ >= PA /k
These equations need to be expressed in absolute time. Applying relativistic equations to them will generate wrong results in certain situations, because the relativistic effect are generated automatically by the wave behaviour of all particles.
It is important to know that photon emissions are in fixed packets of energy, coresponding
to electron orbital transitions.
The transition times are very small.
A more thorough description implies power densities and wave interference. However these effects seem negligible since it doesn't seem to be possible to create a high amplitude low frequency wave that can extract a photoelectron.
In the DCQE experiment, to get a detection at D3 or D4 the wave must recombine quick enough so that there is no time for the energised electron to revert to initial state. It all has to do with the coherence length.
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Supposing we have a hydrogen atom. There is another option for this mechanism or more precisely the energy release mathematical function (not
the workfunction) can be more accurately described. The energised electron looses energy until it reaches the ground state.
It can do that is a continuous mode, like in the previous message.
However it is more likely that it looses energy by releasing a single pulse, like when wiggling a charge ( this case a single movement).
An energised electron orbiting around the nucleus (as a wave), acquires energy by increasing its frequency (deBroglie). Also the orbiting radius should increase. In my paper, Dynamic Geometry Wave Theory, I've shown that a free electron in terms of shape, is a wave travelling at c in a helical motion. If this motion is preserved within an atom, when the electron gets energised the pitch of the helix increases the deBroglie wavelength reduces and the electron front wave speed is increasing.
If the electron creates a single pulse as it settles to the stable state (which depends on temperature/background radiation) then photons are single wavelength electromagnetic waves.
For the electron to be energised, a short pulse ( small wavelength) can deliver the energy quicker than a long wavelength pulse. A minimum amplitude is also required but since the emissions of photons are at fixed amplitudes the energy of a photon aparently doesn't depend of amplitude. This is not correct, because if you split the wave, the energies left in each half may reduce below the workfunction.
When you increase the intensity of a beam, it doesn't mean the amplitude of the waves increases instead, the number of pulse emissions in a given time increases. However, if there is a constructive interference, of two waves then this pulse can exceed the work function and trigger a photoelectron release. As expected, this has been observed experimentally and it is called TPA (https://en.m.wikipedia.org/wiki/Two-photon_absorption).
This demonstrates photons behave like classical waves from this point of view and not as QM says. The virtual state is real.
It is interesting that in the ground state coresponding to 0 K temperature, the atoms behave differently. For example, if we have few atoms they start to behave like a single one.
In an normal environment the electron seems to be permanently energised by the background radiation and regularly emmits EM waves to try to go to ground state.
It is clear that a world without this background radiation (which in fact dictates the temperature) would be completely different.
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It will take me a little while to understand all that, but for now, you seem to be saying that if a photon having just enkough energy to actuate a detector passes through the left slit only, or the right slit only, a detector being immediately adjacent to each, then one or the other detector may be actuated; but that if it passes through both slits, neither detector will be actuated. Consider the set of all photons that pass through in some manner. Because that set includes the ones that pass through both slits simultaneously, which according to your analysis would be undetectable, the number of detections should be significantly less than the number of total photons. But is that actually what is observed? We could conduct an experiment in which photons are first shot at the slits and detected by one large remote detector that could catch anything that got through, and on the basis measure the overall photon rate. Then we repeat the experiment using the two small local detectors, and according to your analysis, the rate should be less, because those photons that pass through both slits would be compelled to be detected in detectors neither of which received enough of the photon to register. I do not know if that is consistent with current experimental knowledge. I had been under the understanding that in the scenario of small local detectors attached to each slit, every photon that got through would be detected (I here assume ideal apparatus), and would in this situation necessarily appear to have gone through one or the other slit, inasmuch as it would be detected only in one or the other detector. That much I understood as being established, the only question remaining is how can we reconcile this behavior with its known wave properties.
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The large remote detector you are proposing should be able to detect longer wavelength. The problem is at a lower frequency it could catch background more radiation. Also this detector could report double photons when it receives a single higher frequency photon.
There is another option. It is possible to use a BBO crystal to split the wave. If we use uv light it will get converted to red. One beam can go through the double slit and the other to a red light detector that can confirm every photon fired.
In the Delayed Choices Quantum Eraser by Kim et al, on fig.5, the author concludes that there is no interference. However the average red line over-filters the result. The plot shows two peaks similarly to those in D0/D2 joint detection. Following my interpretation, this is because the detectors frequency range is probably able to detect a fraction of lower energy packets. The pattern clearly indicates a small amount of interference.