# FeaturedI Interference puzzle - Where does the energy go?

1. Mar 21, 2018

### Derek P

A laser beam is split into two identical beams which impinge on a half-silvered surface from opposite sides. The reflected wave is out of phase with the transmitted wave. There is destructive interference on both outputs. Where does the energy go?

See my avatar - I can't upload my own picture, not sure if it's possible without a URL. Yes it was triggered by reading about the HOM effect, but as far as I can see it should be true for classical waves - even radio or sound waves.

Normally when someone asks this question of an interference experiment I am the first to say "wherever there's destructive interference somewhere there's constructive interference somewhere else" and waffle on about normalisation and conservation. But I can't see how it would work in this case. If there were random phase jitter it would "all average out" too, but this doesn't occur with a split laser beam.

And I am assuming that the surface is symmetrical with inversion at both reflections. That's why I specified "half-silvered" which in principle could be a thin, self-supporting, film. Or an array of fence posts if you want to use sound. I'd even settle for non-inversion all round and get twice the energy out as I put in, but that probably has a flaw somewhere too.

I dare say the answer is obvious - when it's pointed out. So if you want to me - please do. I will consider it a favour.

Last edited: Mar 21, 2018
2. Mar 21, 2018

### stevendaryl

Staff Emeritus
3. Mar 21, 2018

### ZapperZ

Staff Emeritus
4. Mar 21, 2018

### Derek P

@stevendaryl @ZapperZ
Thanks for the links, but I do understand the principle! The links just tell me that the energy must go somewhere. Which I knew. They don't offer any insight into where it goes in this particular scenario. One reply even mentions two overlapping spots. I don't think that helps in this case.

I can't see how coherence time or the spectrum of the incident beams makes any difference either as the whole thing takes place in a small region. I'm missing something!

I understand why the question is moved to General Physics, but it was the HOM effect that got me thinking. Or rather (Wikipedia alert, ZapperZ!) what the Wikipedia article compares it to. I then realised I don't actually understand the classical case! :(

5. Mar 21, 2018

The half-silvered side is put on a substrate. In general, there will be a $\pi$ phase change that occurs on the external reflection that doesn't occur for the internal one. (This needs to occur if energy is going to be conserved. The alternative is, you could get a 50-50 energy split, but you won't get any interference. With proper choice of the dielectric substrate, no silvering of the beamsplitter is required). I also wrote about this same topic in the Insights article: https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/ $\\$ Editing: I also found this paper of interest, and its claim of some kind of $\pi/2$ phase shift that must occur, in order to conserve energy. https://arxiv.org/pdf/1509.00393.pdf Perhaps there are different types of beamsplitters that can result in this $\pi/2$ phase shift, (and introducing a $\pi/2$ phase shift does conserve energy, but it should occur without any interference [edit: Statement determined to need correction= you can still get interference=as @Derek P points out in post 8], just as the incoherent case will also conserve energy, but without interference). $\\$ In the dielectric type of beamsplitter that I wrote about in the Insights article, everything gets explained classically, without any need for this quantum mechanical $\pi/2$ phase shift. Instead, there is a $\pi$ phase change for the external reflection that doesn't occur for the internal reflection, and the other face of the beamsplitter has an anti-reflection (AR) coating, so that all the reflections take place on just one face.

Last edited: Mar 21, 2018
6. Mar 21, 2018

### Derek P

You get the -DerekP prize! It's the 90 degree bit that rescues everything. People are fond of quoting inversion or non-inversion as if there is no other phase shift. So sanity is restored. I'll plod through the details some time but I'm sure that's what it's all about. Thanks.

On a minor point, a silver layer doesn't have to be on a substrate boundary, it can be embedded in a block. So then the optics remain symmetrical. But with a 90 degree shift that makes all the difference.

So it's back to the HOM effect now :)

7. Mar 21, 2018

### sophiecentaur

Does that really solve the problem? It would be easy to introduce a 90° phase shift into one or more of the optical paths. Would that not just bring us back to the same dilemma?
I was hoping that there would be an explanation involving the basic diffraction limited pattern of the original beam which could cause cancellation at the centre and enhancement off beam. Someone mentioned "two spots" earlier on and why not? There would be a phase slope across the skirt of each spot, I think - I always take this sort of question into the world of RF antennae.

8. Mar 21, 2018

### Derek P

The two spots were mine :)
Yes I was thinking about a geometrical answer but it isn't necessary. The quadrature is why there is no interference.
If you add a shift to one of the inputs, say the green, then it will come more in phase with one of the reds and less with the other because of the inversion. So you can steer the output from one side to the other.

9. Mar 21, 2018

@Derek P I read the paper quickly, (the second "link" of post 5), and I was happy to see the author treats the topic in considerable detail. At this point, I think the half-silvered beamsplitter/quantum mechanical case is different from the dielectric beamsplitter (with one face AR coated) where the classical optics seems to give correct results. I will need to give it further study, but I'm hoping the classical picture remains reasonably valid for the dielectric beamsplitter case.

10. Mar 21, 2018

### Derek P

I reckon this thread has done its job now.

Not sure if the 90 degrees stuff suggested by Charles is in the links @stevendaryl and @ZapperZ gave. If so and I missed the significance, my apologies.

Thanks to y'all for sorting me out.

11. Mar 21, 2018

### Derek P

Likewise!!!

12. Mar 21, 2018

### Cthugha

Yes, it really solves the problem and the somewhat surprising reason is that one cannot really do what you have in mind. You can introduce a 90 degree phase shift in a beam path, but you cannot phase shift a single beam. The required phase shift of 90 degrees occurs between the transmitted and the reflected beam, so introducing another phase shift of x before the beam splitter (in, say, input port 1) just results in the phases behind the splitter being 90°+x and 0°+x instead of 90° and 0°. Introducing the phase shift behind the beam splitter will only shift one of these beams, say the transmitted one from input port 1. But it will also shift the reflected one from input port 2 by the same amount, which leaves the phase difference and therefore also the energy distribution unchanged. What you would need to introduce, is a different phase shift between the transmitted and the reflected beam, but this is impossible without either making the beam splitter lossy or not a 50-50 splitter anymore.

Degiorgio and Zeilinger discussed this already in the 80s in references 5 and 7 of the paper cited a few posts above.

13. Mar 21, 2018

@Derek P It appears having a $+\pi/2$ phase shift on both reflections of a beamsplitter in an interferometer, (and as you pointed out in post 8, and also see the correction that I added in post 5), also works quite well for giving an interference that conserves energy in a completely classical description. The alternative is the dielectric beamsplitter where the phase changes are $\pi$ and zero. $\\$ It does appear these are distinct cases requiring separate treatment, but upon introducing the $+ \pi/2$ phase shift for both reflections off of the half-silvered mirror, (the reasons for this $+\pi/2$ phase shift appear to be quantum mechanical), the classical description is adequate for many purposes. $\\$ See also the addition that I posted to the Insights article discussion: https://www.physicsforums.com/threa...mental-approach-comments.898964/#post-5963690

Last edited: Mar 21, 2018
14. Mar 21, 2018

### stevendaryl

Staff Emeritus
The suggestion made https://physics.stackexchange.com/questions/88422/fully-destructive-interference is that it's not really possible to combine beams in this way to get total destructive interference, because of the way that phases change when light is reflected versus transmitted.

15. Mar 21, 2018

### Derek P

That's neat. With everything else symmetrical, the metallic beam splitter sends half the power in one direction and half in the other, whereas the dielectric sends it all one way. So in neither case is there a hint of a paradox.

I don't think we have to go to a quantum model, much as I would love to blame everything on Schroedinger's cat. I'm pretty sure it can all be modelled with classical waves and different material properties. The conductive reflector responds to the E field by in-phase current flow. The dielectric reflector hosts a displacement current that is in quadrature. Of course, to calculate what that actually means you have to use the continuity conditions. But it's a start.

Last edited: Mar 21, 2018
16. Mar 21, 2018

### sophiecentaur

Best to avoid Schroedinger if possible. It's waves in nearly every case.

17. Mar 21, 2018

### Derek P

Armed with the insights gained from this thread so far, I'd agree. The beam splitter can be one of two types:
1. A metallic, half-silvered mirror. In this case the transmitted and reflected waves are orthogonal so there is no interference and the output beams are equal.
2. A dielectric beam splitter. In this case one reflection must be in the denser medium and one in the rarer. The reflections thus have a $\pi$ phase difference and there is constructive interference in one output and destructive in the other.
What is significantly missing is a beam splitter which changes the component phases the same way for both outputs. That is the source of the OP's confusion in the link and also of mine. And for me it's the second time I've been caught out by the simplistic idea that a beam splitter just splits the beam with no phase shifts.

@ZapperZ I blame Wikipedia.

Thanks again, this thread really can go to sleep now, I think.

Last edited: Mar 21, 2018
18. Mar 21, 2018

### Derek P

Not Schroedinger. Schroedinger's cat. Big difference. Every experiment needs a cat.

19. Mar 21, 2018

@Derek P I am very glad you posted this question. With a combined effort, I think we pretty much solved it. It might be worth pointing out that this (interferometry) is one area where I think the Optics textbooks have been somewhat deficient. My college days were from 1974-1980, and I worked with both Klein, and Jenkins and White in college, and later Hecht and Zajac. I think the interferometry was presented with a couple of key details totally absent. It took me quite a number of years to finally see that the Fresnel coefficients would apply in the dielectric case, and I wrote this up in the Insights article that I authored on interferometry. $\\$ My first instincts were that the half-silvered case might also have $\pi$ and $0$ phase shifts, but the literature suggested otherwise. Surprisingly, the mathematics works equally well with both interference and energy conservation occurring for that case as well. It would seem that it would be beneficial to have an Optics text that teaches these fundamentals (using the Fresnel coefficients) as standard textbook material. In general, the calculations are very straightforward, but there seem to be a select few who have a good handle on this material, because (at least as far as I can tell), it hasn't been part of the standard curriculum. $\\$ Important: One correction to your summary in post 17: The phase change of $+\frac{\pi}{2}$ for both reflections for the half-silvered mirror can produce interference as equally well as that of a dielectric beamsplitter. There just isn't any apparent interference if both incident beams are in phase. Also, the same thing results if the two incident beams are $\pi$ out of phase with each other. Instead, at those phases the energy is equally distributed. Other phase angles results in the same type of interference that is obtained in the dielectric case. In the half-silvered case, the maximum occurs for $+ \frac{\pi}{2}$ phase between the two incident beams, and a minimum for phase of $- \frac{\pi}{2}$ between the two incident beams.

Last edited: Mar 21, 2018
20. Mar 21, 2018

### Derek P

Well it wasn't really a summary. But sure, you can steer the power by phasing the inputs.
I rather agree about standard treatments that gloss over the details!

21. Mar 21, 2018

I don't know that at the time the textbooks were written that the authors knew of these details. The Michelson (in my college days 1974-1980) was treated (and simply only the diffuse source case) as the interference between two point sources from the two virtual images that result as the beam is split and traverses different paths. Meanwhile, the Fabry-Perot (slab) was treated as interference resulting from multiple reflections. $\\$The concept of the interference caused by two beams incident on a single interface from opposite directions apparently had not occurred to them. It appeared to me something was missing, but it wasn't until 2009 that I figured out that the idea of the interference of two beams incident on a single interface from opposite directions, as presented in the Insights article I authored... Around 2011 I did show these calculations to a physics professor at a local university who specialized in optics. I asked him why this material isn't in the textbooks=his response was, if I wanted to learn it, I would need to read his lecture notes for the Physics 522 graduate course in Optics that he taught.

Last edited: Mar 21, 2018
22. Mar 22, 2018

### stevendaryl

Staff Emeritus
I realized that my references talk about the phase changes of $\pi$ versus $0$ for reflection off a surface, depending on what side of the surface the light reflects off of. Your example is using a symmetric beam splitter, so that explanation doesn't apply. I guess $\pi/2$ is splitting the difference.

23. Mar 22, 2018

### sophiecentaur

Isn't the angle (not the phase angle) between the E vectors relevant when they are summed at the final mirror? I have still not come to terms with the idea that it's just that the phase difference along the paths has to be 90°. Any phase delay can be introduced into either paths and so any inherent difference can be achieved.

It may be obvious to some but the explanation of what happens at the final mirror involves how the tangential and normal components of the two incident wave E fields add together. The relative phases will govern how much of the incident energy exits in the two output directions. There is never total cancellation but you can choose which direction gets all the energy by suitable choice of phases. I may have missed this point earlier on but I think it needs to be spelled out.

24. Mar 22, 2018

The Maxwell equations governing the electric field amplitudes are completely linear w.r.t. the electric field(s) $E$ , so that when considering what one of the incident electromagnetic waves does when it encounters the interface (where there is a reflected amplitude and a transmitted amplitude) can be computed with the Fresnel coefficients $\rho$ and $\tau$ completely independent of the other incident wave. $\\$ The incident angle along with the index of refraction, as well as the polarization does affect the Fresnel coefficients in the case of a dielectric beamsplitter. (For normal incidence $\rho=\frac{n_1-n_2}{n_1+n_2}$, and $\tau=\frac{2 n_1}{n_1+n_2}$. For other angles of incidence, the expressions become somewhat more complex, but if $R=1/2$, we still have $\rho=\pm \frac{1}{\sqrt{2}}$, independent of the incident angle that was pre-selected). The calculations can proceed regardless though for the resulting energy distribution. If it is a 50-50 energy split, $R=1/2$ for the energy reflection coefficient when there is a single beam. (Thereby, the index of refraction $n$ is selected to make $R=1/2$ for the particular geometry and polarization that is being implemented). This makes $\rho=\pm \frac{1}{\sqrt{2}}$,(with $\tau=+\frac{1}{\sqrt{2}}$), for the dielectric case, because $|\rho|^2=R$. $\\$ Meanwhile for the half-silvered 50-50 beamsplitter, $\rho=+i \frac{1}{\sqrt{2}}$ where the factor $i$ is the $+ \pi/2$ phase shift factor in the form $i=e^{+i \pi/2}$. The Fresnel transmission coefficient for this case is $\tau=+\frac{1}{\sqrt{2}}$. A sufficiently thick layer of silver is applied, that might depend upon the incident angle, in order to make $R=1/2$. $\\$ The energy reflection coefficients $R$ are not good numbers when two incident beams are present, but the value of the Fresnel coefficients remains unchanged by the presence of a second beam. (The resulting energy distribution that occurs is governed by the relative phases of the two incident beams, rather than by the energy reflection $R$ coefficient of the beamsplitter). There is no requirement for linear behavior for the energy though, because the energy equations are to the second power in the $E$ field amplitude, with intensity $I=n E^2$ where $n$ is the index of refraction. There is however the energy conservation requirement for a lossless system, which these are considered to be. $\\$ The system is completely linear in the electric field amplitudes $E$, so linear principles apply, and what happens to each incident $E$ field amplitude for the two beams can be computed separately, and then the results are summed.

Last edited: Mar 22, 2018
25. Mar 23, 2018

This one may be worth showing in full mathematical detail for the energy for the case of the half-silvered mirror with the $\pi/2$ phase shifts for both reflections. It may also be useful to see it without the complex numbers for the Fresnel reflection coefficients. Here I will proceed to detail the algebra, and will use trigonometric identities to process the calculations: $\\$ Begin with two sources: $E_1=\cos(\omega t)$, and $E_2=\cos(\omega t+\phi)$. $\\$ Receiver 1 will get (electric field as a function of time) $\\$ $R_1=(\frac{1}{\sqrt{2}})(\cos(\omega t)+\cos(\omega t+\phi+\pi/2))$, $\\$ and receiver 2 will get $\\$ $R_2=(\frac{1}{\sqrt{2}})(\cos(\omega t+\pi/2)+\cos(\omega t+\phi))$.$\\$ (The Fresnel reflection and transmission coefficients both have amplitude $\frac{1}{\sqrt{2}}$, but the reflection coefficient comes with a $+\pi/2$ phase change). $\\$ Using the trig identity $\cos{A}+\cos{B}=2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$, we get $\\$ $R_1=\sqrt{2} \cos(\omega t+\pi/4+\phi/2) \cos(\phi/2+\pi/4)$, $\\$ and $\\$ $R_2=\sqrt{2} \cos(\omega t +\pi/4+\phi/2) \cos(\phi/2-\pi/4)$. $\\$ Computing the intensity for these sources, we square the constants, and simply drop the $\cos(\omega t +\pi /4 +\phi /2)$ terms, we get $I_1=2 \cos^2(\phi/2+\pi/4)$, $\\$ and $\\$ $I_2=2 \cos^2(\phi/2-\pi/4)$. $\\$ Doing trigonometric expansions, we get $\\$ $I_1=2 [\frac{1}{\sqrt{2}} (\cos(\phi/2)-\sin(\phi/2))]^2=1-\sin(\phi)$, $\\$ and $\\$ $I_2=2 [\frac{1}{\sqrt{2}} (\cos(\phi/2)+\sin(\phi/2))]^2=1+\sin(\phi)$. $\\$ Thereby, the sum of the intensities will always be equal to $2$, but the intensity at each receiver will vary between $0$ and $2$, depending on the relative phase $\phi$ between the two sources. $\\$ Meanwhile with the complex number mathematics, the two sources would be $E_1=e^{i \omega t}$ and $E_2 =e^{i( \omega t +\phi )}$. The receivers would get $R_1=(\frac{1}{\sqrt{2}})(e^{i \omega t} +i e^{i ( \omega t +\phi)})$ and $R_2=(\frac{1}{\sqrt{2}})(i e^{i \omega t}+e^{i ( \omega t +\phi )} )$. The computation using the trigonometric identities was probably not any more difficult than it would be to process this and compute the intensities.