# Energy conservation in the combination of two coherent beams

I was asked a question today by a student that I couldn't find a satisfactory answer to.

Imagine we have two identical laser beams with average electric field strength $$E$$. Upon combining these beams (assuming perfectly phased to interfere completely constructively), we should find a resultant beam with average electric field strength $$2E$$. Given $$P=IA$$ and $$I\propto |E|^2$$, we see that the resultant beam carries twice as much power as the two incident beams alone. Of course, this is assuming that the beam area remains fixed.

To me this seems a clear violation of conservation of energy. Where does the extra energy in the resultant beam come from?

The answer I gave to the student is that presumably the beam area would be reduced, thereby keeping the total power transfer equal; I am not so sure of the validity of this answer.

Can anyone shed some light on what exactly is going on here?

• See this very nice answer: physics.stackexchange.com/a/196919/128186. They key is that the formulas relating electric field amplitude, electromagnetic field intensity (energy flux), and electromagnetic field energy (within a volume) become more complicated as soon as you are considering anything other than a plane wave. You must be careful and make sure to use the proper definitions for each of these quantities. Oct 21, 2018 at 8:34

Imagine a practical way of doing this - sending the output of a laser into a Michelson interferometer and looking at the on-axis power along the various beam paths.

The "reconstructed beam" contains light from one beam that bounced off the partially silvered beam splitter, bounced off a (perfect) mirror then came straight through the beam splitter. The other beam has gone through the beam splitter, bounced off a mirror and then reflected from the beam splitter.

Each passage through the (perfect) beam splitter halves the power and hence divides the E-field amplitude by $$\sqrt{2}$$.

The combined beam is therefore the in-phase sum of two beams each with amplitude $$E_0/2$$, which gives a resultant wave carrying power proportional to $$E^2$$.

Problem solved? Not quite - you also have to follow what happened to the other recombined beam - the one that goes back towards the laser (note that with an appropriate mirror tilt, it doesn't have to go back exactly towards the laser). The two beams here have (i) gone straight through the beam splitter twice, with one reflection,or been reflected off the beam splitter twice with one reflection. The wrinkle here is that although the physical path difference along each "arm" is the same as for the usual "recombined" beam, there is an additional phase difference caused by the beam splitter.

In the case of the usual recombined beam we have one reflection from a glass-silver interface along one arm and one reflection from a air-silver interface along the other (because the silvering is only on one surface) - this introduced a $$\pi$$ phase shift between the arms which must be removed (by adjusting the arm length) to enable constructive interference.

For the beam going back towards the laser, the light along one arm suffers no reflections in the beam splitter and the other is reflected twice from the glass-silver interface and the net phase change is zero.

Thus if the arm length is adjusted to give constructive interference in the reconstructed beam, then there will be perfectly destructive interference (and hence zero power) in the beam going back towards the laser.

You cannot get constructive interference in one place without destructive interference in another. There is global energy conservation.

• "pathetic difference"? Did you mean "path difference"? Mar 23, 2020 at 9:16
• @ruslan thanks. Autocorrect. Mar 23, 2020 at 9:42

A beamsplitter will combine the two beams along two paths. For a single beam, you would get, say, 50% power along path A and 50% along path B. With two beams, you have the same situation, splitting the field in half into both paths. This means that you’d have along each path $$E_1/2+E_2/2=E$$. The total power from both path A and B is then proportional to $$2E^2$$, which conserves energy.

The key here is that you can’t combine two identical beams 100% into a single path.

• Physically, where are the two paths? Why can two beams not be perfectly combined? Oct 12, 2018 at 3:30
• @RileyJacob for example, think of a beamsplitter cube. There are two inputs and two outputs (= the four sides of the cube). Path A would be E1 transmitted and E2 reflected, and vice versa for path B. Oct 12, 2018 at 3:44
• So effectively there is no possible way to only get one output beam? Oct 12, 2018 at 6:19
• @RileyJacob There is, but not for identical input beams. You need something to distinguish the inputs (like color or polarization) and a carefully designed beamsplitter such that one is 100% reflected and the other is 100% transmitted. But then, they wouldn’t interfere the same way. Oct 12, 2018 at 13:19
• So what then happens if, for example, two laser beams are pointed directly at one another? What is stopping these beams from interfering completely constructively? Oct 12, 2018 at 19:37

Imagine we have two identical laser beams with average electric field strength E. Upon combining these beams (assuming perfectly phased to interfere completely constructively)

is impossible to arrange overall.

You might be able to have the phase of the waves exactly the same in one region but you will also have regions where the waves are not in phase. So what you gain in terms of the rate of energy transfer in one region you lose in another region.

There may well be a region where the rate of energy flow is proportional to $$\left (2E\right)^2= 4E^2$$ but you will also have a region where there is no transfer of energy. The average overall rate of energy transfer will be $$E^2+E^2 = 2E^2$$

This is a valid question. If we are combining two beams coherently in one of the arm of interferometer, the peak intensity increases. Since the energy is conserved, spatial redistribution is the only explanation. The beam size should be narrowed to get high intensity in some parts. But, if we are using perfect plane waves with zero angle between them, how can it be explained? can someone please theoretically explain that?

• This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. To get notified when this question gets new answers, you can follow this question. Once you have enough reputation, you can also add a bounty to draw more attention to this question. - From Review May 26, 2022 at 6:01