Consider a beam with power $P_1$ and electric field amplitude $E_{01}$. It is sent through a 50/50 beam splitter that produces beams with power $P_2=P_3=P_1/2$. What are the electric field amplitudes of the split beams, $E_{02}$ and $E_{03}$?

From what I understand, $P=KE_0^2$ where $K$ is a constant. Therefore,

$E_{02}= \sqrt{P_2/K}=\frac{1}{\sqrt{2}}\sqrt{P_1/K}$,

$E_{03}= \sqrt{P_3/K}= \frac{1}{\sqrt{2}}\sqrt{P_1/K}$

Now say I can recombine the beams in phase and without any losses, then $E_{04}=E_{02}+E_{03}$. The power of the recombined beam is

$P_4=KE_{04}^2 $

$= K[E_{02}^2 + E_{03}^2 + 2E_{02}E_{03}]$

$=\frac{1}{2}P_1+\frac{1}{2}P_1 + P_1$


So $P_4 >P_1$ and I've created power from nowhere! What is wrong with this picture?


2 Answers 2


Your error is in how the electric fields are combined by a 50/50 beam splitter. If you have two entry ports $a$ and $b$ with electric field amplitudes $E_a$ and $E_b$, and exit ports $c$ and $d$ with electric field amplitudes $E_c$ and $E_d$, then the correct way to combine them is $${E_c=\frac1{\sqrt2}(E_a+E_b),\\ E_d=\frac1{\sqrt2}(E_a-E_b).}$$

You know one instance of this already: if port $b$ is shut off, then each of the output ports should get $1/\sqrt2$ of the amplitude. Similarly, if port $a$ is shut off, the same thing should happen, so both $E_a$ and $E_b$ should have equal weights in the expressions for $E_c$ and $E_d$. (Here, of course, I'm invoking the principle of superposition to combine the solutions for different sets of sources.)

The phases are a little trickier. The minus sign along the bottom can be imposed by saying the interferometer is aligned so that no output comes out of that port for equal input intensities. The sign of $E_a$ at the top can also be arbitrarily set, by adding an appropriate phase delay on port $c$.

(A brief note for an intermission. The formulas above can of course be derived rigorously once you know how the beam splitter is implemented. However, it's not necessary to know the details to derive them, since they also follow from the general considerations I'm expounding here.)

By now, your problem has gone away. Even if you had an arbitrary phase in the final coefficient, i.e. $E_c=\frac1{\sqrt2}(E_a+e^{i\theta}E_b)$ for some $\theta$, then you can't create energy:

$$P_c=K|E_c|^2=\frac12K(E_a^2+2\cos\theta E_a E_b+E_b^2)\leq \frac12K(E_a+E_b)^2 = P_1.$$

In fact, in order not to destroy energy, you must have both components coming out in phase, with $\theta=0$ and a plus sign in the equation for $E_c$.

So, what's the bottom line? In order to recombine beams in phase and without any losses, you must put them through a beam splitter and ensure that interference kills the other output port. To do that, though, the amplitude of each beam gets further reduced by $1/\sqrt2$, and this ensures that the total beam energy is conserved.

  • $\begingroup$ Thanks! Makes sense. Do you have a reference for this, like a page out of a book? $\endgroup$
    – cpc333
    Jan 14, 2014 at 19:12
  • $\begingroup$ Just about any optics book must have this, to be honest. The first two links here are good. $\endgroup$ Jan 14, 2014 at 19:15

Obviously, the assumption that you "can recombine the beams in phase and without any losses". Until you show some specifics of how this can be done, the assumption is not obvious.

  • $\begingroup$ A Mach-Zender interferometer can combine beams with a fixed phase relationship, and losses in optical systems can be kept so low as to be negligible for most purposes. $\endgroup$ Jan 15, 2014 at 5:08
  • $\begingroup$ @Chris Mueller: If you believe a Mach-Zender interferometer can increase total energy, could you give specific schematics? $\endgroup$
    – akhmeteli
    Jan 15, 2014 at 5:13
  • $\begingroup$ Lol, somehow we've misunderstood each other. Combining beams with a fixed phase relationship is what an interferometer does. Doing so does not, however, increase the total power. $\endgroup$ Jan 15, 2014 at 5:19

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