Dirac said that a 'linear' polarized photon is a superposition of left and right rotating photons. Here is a puzzling aspect of this superposition.
There are dichroic materials which can absorb only left photons. What is their effect on this superposition? They would absorb the left state and let the right. It turns out that there are two photons in a 'linear' photon! The mechanical momentum of the dichroic material must be measurable in principle (like Beth experiment). It would be expected that there is some energy hv absorbed too but it must be in the right photon also. Conservation of energy seems broken? If there is no momentum and no energy absorbed in the dichroic the conservation laws are also severely harmed. At least something happens to that photon (linear to right which is easy to show) but nothing happens to the causer of this event (namely to the dichroic). I see that there must be wavefunction collapse for the photon and in fact the question boils out to:
Is something physically happening to the machinery causing the collapse?
This question is connected to the following question:
Is there any way to split a linear polarised photon into left and right polarised photons? - ResearchGate. Available from: https://www.researchgate.net/post/Is_there_any_way_to_split_a_linear_polarised_photon_into_left_and_right_polarised_photons [accessed Feb 12, 2016].
It's possible to represent the two states of polarization of a photon in many different bases-linear polarization or circular polarization being two possibilities. An eigenstate of any one base is a linear combination of eigenstates in any other base, that's all. These statements, of course, hold for the polarization states of electromagnetic waves as well as for photons. For photons this is standard quantum mechanics, e.g.http://www.feynmanlectures.caltech.edu/III_toc.html
Stam this is trivial. But I don't see any relation with the question.
Stam this is trivial. But I don't see any relation with the question.
Stam this is trivial. But I don't see any relation with the question.
A linearly polarized photon is a superposition of circularly polarized photons. The effect of the medium is to project out one circular polarization, leaving the other. Absorption *means* that the medium gains that amount of energy-so when the remaining circular polarization is projected back in the linear polarization basis, the total energy is, of course, less. It suffices to compute the sum of the squares of the absolute values of the coefficients, that, as standard textbooks in electromagnetism review, represents the energy of the wave in question and the same holds in quantum mechanics, as reviewed in Feynman's lectures. This stuff is taught in undergraduate physics courses.
Saying "crystal" you go to the level of description that forgets about atoms. So, why do you want to use photons? It is light.
A photon is a notion defined and used in quantum electrodynamics. Usually in the scattering problems. It occupies the whole universe and leaves no place to put a crystal in. It does not travel anywhere.
A limited number of quantum problems take into account interaction with photons. There are too many of them to make problems solvable. You cannot describe interaction of even an atom with one photon. All of them have to be taken into account. Lamb shift is one of such problems.
Usually processes of interaction with light use a model in which electromagnetic field is external. The Hamiltonian then is time dependent and no energy conservation law holds. The energy that atomic system takes from the field or passes to the field is not actually counted. You just follow your system.
Physically any light absorption is passing energy to the absorbing body. Heating may be the result. Or luminescence.
Yes, you can split light into two polarizations: split it first and pass through different polarizers. But light.
Stam, despite you started with a statement that the linear photon is a superposition, in the third sentence your point of view changes that the linear polarized photon is in fact two photons - one is absorbed and the other passes thru the dichroic.("Absorption *means* that the medium gains that amount of energy - so when the remaining circular polarization is projected back in the linear polarization basis, the total energy is, of course, less.") Very easy and convenient. But this can not be so. If absorbed (dichroic excluded) the 'linear' photon is always registered with spin -1 (LHP) or +1 (RHP) and with full energy. If it is two photons the registered effective spin would be 0.
By the way I don't understand why you need projection on the linear basis again?
As far as I can see, if a single linearly polarised photon hits a long enough dichroic crystal, it will either be completely absorbed, or go through without absorption. In the latter case, it will come out as a (left) circular polarised photon. The probablility of either event is1/2.
Concerning the claim that a photon ``occupies the whole universe and leaves no place to put a crystal in. It does not travel anywhere.'' That might be true of a photon in a momentum eigenstate. It is certainly not true for a photon in a general state.
Leyvraz your proposal looks very close to the spirit of QM.
I look only at the second probability: the photon goes thru without absorption. It comes out as a left photon. So it is experimentally different from the incoming "'linear" photon (statistically). But as there is no absorption nothing has changed in the dichroic crystal.
So there is an event without any change in the causer (crystal). Some conservation laws are not fulfilled.
``Some conservation laws are not fulfilled.''Conservation laws are tricky in that context. To obtain a correct treatment of this, you should presumably consider the dichroic crystal as part of the quantum mechanical system. The crystal can, of course, absorb enrgy and interchange angular momentum in relevant amounts. It is only the total energy and/or angular momentum that has to be conserved.
In fact this is a projection from undefined spin to certain value of spin. I can't see nonconservation of energy or spin but there is gain of information indeed without any interaction (in the usual sense). Though one must use interraction after the dichroic.
It is somehow reminisent to the dispute Bohr Einstein but should consider it later.
In fact this is a projection from undefined spin to certain value of spin. I can't see nonconservation of energy or spin but there is gain of information indeed without any interaction (in the usual sense). Though one must use interraction after the dichroic.
It is somehow reminisent to the dispute Bohr Einstein but should consider it later.
It is hard to understand why people feel that they understand a linearly polarized photon better when they view it as composed of two circularly polarized photons with opposite angular momentum. One could just as well think of a circularly polarized photon as composed of two linearly polarized photons which are perpendicularly polarized to each other. What's the point of mental acrobatics of this kind? Photons are associated with wave packets of some wave vector and some frequency. These wave packets are solutions of the wave equation of electrodynamics. The most general solutions to this equation are an elliptically polarized wave packets. Linearly polarized wave packets and circularly polarized wave packets are just special cases of them. However, a linearly polarized photon does not possess angular momentum. This has been demonstrated by R. A. Beth, Phys. Rev. 50, 115 (1936). He exposed a quaterwave plate ( a disc suspended at a quartz fiber) to linearly polarized photons. The disc converts these photons into outgoing circularly photons which carry angular momentum. Due to the recoil angular momentum they leave in the disc the latter starts turning (actually oscillating) around the fiber suspension.
At first reading I am very surprised if it is as you state.
The linear polarized photon has energy hv and the circular polarized too hv. There is no energy for rotation of the quartz plate. The Beth experiment is surely with many linear photons. One passes like circular (0% absorption) but other is absorbed (0% passing). Photons can not have 0 spin.
I am going to read Beth paper. I think he has a paper with reflection from a mirror which momentum changes by 2h. This must be another one.
I can only repeat myself: These wave packets are solutions to the wave equation of electrodynamics. That means: A circularly polarized wave for a given wave vector and frequency solves the wave equation just as a linearly polarized wave of the same wave vector and frequency. The mechanism that governs the transfer of the recoil angular momentum of the circularly polarized photon to the quarterwave disc is completely analogous to the transfer of linear recoil momentum to the embedding solid by a gamma quantum of a decaying Fe -atom in the Moessbauer experiment. Because of the large mass of the solid it accepts the recoil momentum without picking up a sizable amount of kinetic energy. But clearly, this amount of kinetic energy is non-vanishing because the mass of the solid is finite. If one would suspend the embedding solid at a quartz fiber and could initiate a succession of gamma quanta emission in the same direction, the piece of solid would swing out and under appropriate conditions start oscillating. Schroedinger's theory of the Doppler-effect with light emitting free atoms is based on exactly this consideration: The excitation energy of the atom goes primarily to the emitted photon. But because of the recoil transferred to the atom the latter changes its kinetic energy. If the light quantum is emitted in the forward direction of the atom's motion, the recoil slows the atom down and this energy drop causes an enhancement of the photon's energy (blue shift). Red shift occurs if the photon is emitted opposite to the motion of the atom. The visibility of the effect is connected to the small mass of the atom compared to the mass of the solid in the Moessbauer experiment. The same applies to the recoil momentum transferred to the quarterwave disc in Beth's experiment. Of course, the energy of the outgoing circularly polarized photon is a tiny amount of energy smaller than that of the incoming linearly polarized photon. Hence, its frequency is minutely, but undetectably smaller than that of the incoming photon. Therefore it takes many linearly polarized photons in succession to cause a motion of the disc.
As for your statement: `"One passes like circular (0% absorption) but other is absorbed (0% passing" I must say: This is far from reality! Even in the simple case of a photon that penetrates into a plate of glass, the incoming wave is completely absorbed within a thin surface layer of some wavelengths thickness. This layer is called the "Oseen extinction- layer". Within this layer the original wave is replaced by a polariton-type wave which constitutes a compound state of an electrodynamic wave and virtual excitations of the atoms in the glass. Because of the coupling to these virtual excitations this polariton-type wave propagates at a different velocity. The probabilty of a photon NOT being absorbed is ZERO!
You are saying: "Photons can not have 0 spin". Who forced that credo on you? Where is the experimental evidence for that? I append a picture of a circularly and linearly polarized wave. Look at the linearly polarized wave: Do you see any feature that could be correlated with an axis of rotation?
Best regards,
L. Fritsche
Lothar
the photons can not have zero spin (or better helicity) because of their zero mass. That's why a linear photon must be superposition of left and right photons.
Dear Ilian,
I meant to send you the attached letter 2 weeks ago. However, I became seriously ill and was completely thrown out of track. I am at a loss to understand your claimed interrelation between masslessness of particles and non-vanishing spin. Heavens! Where does that follow from? All properties of photons that can be derived go back to classical electrodynamics. The rest (that they contain the energy hv and that they are point-like) is just assumption (Planck+Einstein). The wave packets that are associated with them are solutions to the wave equation of classical electrodynamics, and the two fundamental solutions are two plane waves which are orthogonal to each other. Because of the linearity of the wave equation one can combine these two solutions to form elliptically polarized waves, and only them possess angular momentum which follows from Maxwell’s stress tensor. All this has nothing to do with the fact that these wave packets are massless. Read some more in the attached letter.
Best regards,
Lothar
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