Fresnel equations and conservation of energy (phase shifts)

[JerryY]

Quantum mechanically speaking when we split a wave in two the resulting waves must have a 90 degrees phase difference for energy to be conserved. Take the beamsplitter depicted in [1] for example. But the Fresnel equations state that the reflected wave should experience a phase shift of π when reflecting from an optically less dense medium. This results in a π phase difference between the two resulting waves which apparently violates conservation of energy. I tried asking this in other forums but received no answers so I'm guessing I fundamentally misunderstood something.

[1] G. Weihs and A. Zeilinger, “Photon statistics at beam-splitters: an essential tool in quantum information and teleportation,” in Coherence and Statistics of Photons and Atoms, J. Perina, ed. (Wiley, New York, NY, 2001)

 

[Charles Link]

I think I know what you may be trying to do=I think I tried something like that once=if you get all of the energy going in one direction after two beams are incident on a beamsplitter, you might be concluding the wavefronts add in the form $A \cos(\omega t)+B \sin(\omega t)=\sqrt{A^2+B^2} \cos(\omega t-\phi)$ , (so that the energy , which is $A^2+B^2$, is conserved), but this isn't how the calculation is done.

It surprised me a few years back that when I was computing the interference of two beams incident on a single interface from opposite directions in computing the Fabry-Perot effect, that using Fresnel coefficients, you do get interference and conservation of energy. (It normally isn't taught this way in the textbooks=they normally use two interfaces and multiple reflections, but the single interface with its interference is IMO the more fundamental thing that is happening here).
See https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/

This also explains the physics of the beamsplitter case, e.g. with a Michelson interferometer, and it also works for the case of a simple beamsplitter.
See also
https://www.physicsforums.com/threads/interference-puzzle-where-does-the-energy-go.942715/
for a discussion of what may be the more complicated case of a half-silvered interface in another type of beamsplitter. See e.g. posts 5-9.

 

[JerryY]

I think I know what you may be trying to do=I think I tried something like that once=if you get all of the energy going in one direction after two beams are incident on a beamsplitter, you might be concluding the wavefronts add in the form $A \cos(\omega t)+B \sin(\omega t)=\sqrt{A^2+B^2} \cos(\omega t-\phi)$ , (so that the energy , which is $A^2+B^2$, is conserved), but this isn't how the calculation is done.

It surprised me a few years back that when I was computing the interference of two beams incident on a single interface from opposite directions in computing the Fabry-Perot effect, that using Fresnel coefficients, you do get interference and conservation of energy. (It normally isn't taught this way in the textbooks=they normally use two interfaces and multiple reflections, but the single interface with its interference is IMO the more fundamental thing that is happening here).
See https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/

This also explains the physics of the beamsplitter case, e.g. with a Michelson interferometer, and it also works for the case of a simple beamsplitter.
See also
https://www.physicsforums.com/threads/interference-puzzle-where-does-the-energy-go.942715/
for a discussion of what may be the more complicated case of a half-silvered interface in another type of beamsplitter. See e.g. posts 5-9.

Thanks for the detailed reply. The links you've provided look very interesting. However, I'm afraid my original question still hasn't been answered. In the reference I provided (page 2 actually) there is only one beam incident on the beamsplitter, and the resulting waves must be 90 degrees out of phase for energy to be conserved. The Fresnel equations describe the same situation but instead state that the two fields must be 180 degrees out of phase.

 

[Charles Link]

I don't have access to the publication that you referenced, but with a single beam, the phase shouldn't matter=you simply get a 50-50 energy split from the beamsplitter. Interference effects and the phase come into play when you combine beams using a beamsplitter. The emerging energy then gets redistributed in some manner that depends upon the relative phases of the two beams. For the single incident beam, boundary conditions on the fields can, if I'm not mistaken, be used to determine the phases of the emerging beams.

 

[JerryY]

I don't have access to the publication that you referenced, but with a single beam, the phase shouldn't matter=you simply get a 50-50 energy split from the beamsplitter. Interference effects and the phase come into play when you combine beams using a beamsplitter. The emerging energy then gets redistributed in some manner that depends upon the relative phases of the two beams. For the single incident beam, boundary conditions on the fields can, if I'm not mistaken, be used to determine the phases of the emerging beams.

Here's the relevant quote from the publication: 'Consider a 50/50 beam splitter where one particle is incident via beam (a). Obviously, this particle has a 50% chance of ending up in either in output port (c) or in output port (d). Quantum mechanically we may write the operation of the beam splitter as $|a| → 1 √2 (|c \rangle + i|d \rangle)$. Here we have for simplicity assumed that the beam splitter is completely symmetrical . This symmetry implies that a wave experiences a phase shift of π/2 upon reflection relative to transmission, as signified by the phase factor i.' (The figure in question you can substitute with any standard google beamsplitter image, where a is the incident field and c and d are the outgoing fields).

To combine two waves in a manner which conserves energy, the two waves must add under 90 degrees such that no interference occurs. The same goes for when we're splitting a wave. By solving the boundary conditions one arrives at the Fresnel equations, which seemingly contradict with the quantum mechanical description that the resulting waves must be 90 degrees apart.

 

[Charles Link]

I think your question is answered by the second thread that I referenced in post 2. There are indeed two types of beamsplitters:
One type is the asymmetric dielectric beamsplitter, where the splitting occurs off of one surface, and the other surface has an anti-reflection coating. This first one does have a $\pi$ phase change in the reflection from one direction.
The second type is the symmetric one where a thin metallic film is surrounded by glass on both sides. This one is discussed in the second "link" of post 2, (see posts 5-9, etc.). There seems to be some debate over which beam (transmitted or reflected) should get the $\pi/2$ phase change, but in any case, this second type of beamsplitter does have a $\pi/2$ phase change on one of the beams.

 

[Charles Link]

a follow-on: See also the paper that I referenced in post 5 of the second "link" of post 2: https://arxiv.org/pdf/1509.00393.pdf
I think you might find it good reading.

Edit: There is the claim that this $\pi/2$ phase factor is a quantum mechanical result, but it also follows classically when considering two incident beams that are in phase with other. Emerging from each port, (using Fresnel coefficients of $1/\sqrt{2}$ for a 50-50 energy split, where we include some phase factor between reflected and transmitted beams), we have amplitude of $1/\sqrt{2}+e^{i \phi}/\sqrt{2}$. To conserve energy, $\phi$ needs to be $\pi/2$, in order to get an amplitude of unity (with some phase) emerging from each port. (We began with amplitudes of unity feeding into the two input ports, in a completely symmetric problem).

 

[Charles Link]

@JerryY Be sure and see the "Edit" of post 7.

 

[vanhees71]

This has nothing to do with quantum theory or rather the classical and the quantum calculations are identical. The only difference is that in the quantum case the em. field is operator valued ;-).