# How do you combine two spatial modes of light into one spatial mode in optics?

Is there a lossless way to combine light coming from two different single photon sources into one spatial mode? Either free space of fiber would be fine.

Let's assume the wavelength and polarization are the same in both input spatial modes.

The catch is that the intensity at the single output should be roughly the same as the sum of the intensities at the two inputs. That is why I believe a 1x2 fiber coupler cannot be used because a 1x2 coupler is just a 2x2 coupler with one output chopped off - therefore, the half of light that would normally go into this output is lost.

• there is no reciprocal 3-port matched junction. May 27 '20 at 22:51

If the two inputs are not in phase, the answer is "no". If they are in phase, they can enter in orthogonal linear polarizations, and be combined to form a single mode at 45 degrees polarization. But if they are not in phase, the resultant polarization will vary randomly so the resultant will be mixed mode.

• Do you know what the mathematical relation between inputs and output is? I'm trying to understand your description, but got stuck with an edge case of the input being a in a superposition of polarization (H, V) and input spatial modes (1, 2) with one polarization in phase and the other not: $|H\rangle_1 + |H\rangle_2 + i|V\rangle_1 - i|V\rangle_2$. May 31 '20 at 16:32
• If the polarizations are not in phase, the two cannot be combined into a single mode. With one polarization axis in phase and the other out of phase, the result will be a partially polarized, partially coherent mode. May 31 '20 at 16:51

If you send two optical pulses, in phase, into the two inputs of a 2:1 coupler, they will be combined into a single mode on the other side. If, however the two signals are out of phase, the two signals will not be combined.

The best way to think about it is this. Lets say all outputs/inputs of your 2:1 coupler are identical single-mode waveguides, and let us ignore polarization degree of freedom for now.

Lets say the two waveguides on one side of your waveguide are A and B, and a single waveguide on the other side is S.

The field is the S-waveguide is then: $$E_s=(E_A+E_B)/\sqrt{2}$$, i.e. S takes in the symmetric part of your AB-input, the other anti-symmetric part $$E_A - E_B$$ is the bit that is lost. If your AB-input is purely summetric, i.e. $$E_A$$ and $$E_B$$ are equal in phase and amplitude, then no energy will be lost.

This is all classical optics however. You will need to run more calculations to see what happens, given boson statistics, if $$E_A$$ and $$E_B$$ are single-photon states.

So in answer to your question. I think a 2:1 coupler could still do the job, but you need to tell more about your specific setup. Can you get $$E_A$$ and $$E_B$$ to be in phase and of the same amplitude?

Following the comment.

1. I was to rash to say it is not a beam splitter with one output ignored. Indeed, the output of the 2:1 coupler is one of the standard outputs of a beam splitter. I have corrected it

2. The $$E_A - E_B$$ part will be coupled out of the single-mode '1' waveguide (in the 2:1 coupler), and will therefore be lost to environment, be it housing of the coupler or free-space. I remember seeing papers, published in 1970s or 1980s actually showing detailed derivaions of this, but I cannot find them.

3. In the most general case, I don't think you will be able to merge the two optical modes, at least linearly. Are your signals of definite polarization? If they are, you could mix them via a polarization-sensitive beam-splitter. Of course, if two signals are out of phase, but phase relation-ship is stable, an optical delay line may fix the problem, though I doubt it is that simple. Another option would be to see if non-linear approaches could be harnessed, but then you need a crystal, a pump, etc.

Another option is post-selection. As I understand, in quantum optics it is common to have schemes that do not always work, but work a certain known share of the time, and this is fine as long as one can decide on detection whether the scheme has worked that time. If you do indeed have to single-photon states incident on two different arms of a beam splitted, AFAIK, due to HOM effect those two photons will exit as a two-photon state out of one of the arms of the beam-splitter. So you can have a scheme where there is a conventional 2:2 coupler with inputs A, B and outputs S, X. You send two single-photon states into A and B and make sure their envelopes overlap as well as possible. Then you arrange your optical experiment to be fed by output S, and put a detector on output X.The procedure is then to disregard the optical experiment if detector X clicks, and if it does not click, then you know that, that time, the two photons on the inputs have existed as a two-photon state from output S.

• Thank you, @Cryo! Unfortunately, the two input modes are not in phase. Is there any solution for this case? By the way, isn't your description of input-output relations using $E_S$, $E_A$ and $E_B$ equivalent to the description of a lossless 50:50 beam splitter (with one output lost)? May 28 '20 at 1:36
• And @Cryo, what do you mean by "anti-symmetric part $E_A − E_B$ is the bit that is lost"? Absorbed by the housing of the coupler or lost to destructive interference? I think you meant the former, but I'd rather ask. May 31 '20 at 16:43
• @triclope. I think you have a point in your first comment. See the edits above
– Cryo
Jun 1 '20 at 0:29

No, two single photons coming from two different sources cannot be combined to form a superposition that represents one photon with one spatial mode. In other words, $$\text{two photon state} \neq |a\rangle + |b\rangle .$$ Even though the two photons come from different sources, they still give you a two-photon state. Therefore, what you have is $$\text{two photon state} = |a\rangle|b\rangle .$$ As a result, there is no way to form the superposition of their spatial modes.

You can send then along the same path by changing the polarization of one and combine them with a polarizing beam splitter, but they will each carry their own degrees of freedom, without any superposition.