Energy conservation in the combination of two coherent beams

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $E^2+E^2 = 2E^2$

Answer 4

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?