# Why doesn't light cancel itself?

1. May 12, 2014

### maline

In a beam of unpolarised light, there are thousands of light waves of all polarisations. At a given point along the beam, each of these elements represents an electric and/or magnetic field in some direction. Since all directions are more or less equally represented, shouldn't all the fields sum to zero?

2. May 12, 2014

### ZapperZ

Staff Emeritus
It would take an incredible set of coincidence that two exactly opposite polarization to be at the same location at the same time.

Unpolarised light doesn't mean that all the different polarizations can be found at a particular point in space at the same time, which is what is required for the total cancellation. It just mean that there is random polarization of a beam of light when looked at over time, or if you look at one photon at a time, the polarization comes in randomly over time. Just to get two photons to hit the same spot in space one after the other is not trivial, you need a very intense light source for that.

Zz.

3. May 12, 2014

### maline

Please explain a bit more. Take a circle of radius 1mm in sunlight at noon. about how many photons will be creating electric fields at a given point in time, and about how much will the field total to?

4. May 12, 2014

### ZapperZ

Staff Emeritus
This is not as trivial as you think.

"sunlight" is a combination of not only many different polarizations, but also many different wavelengths! This is not a monochromatic light source. So the "cancellation" here even less probable, because you have light with different frequencies, intensities, etc. Estimating the number of photon density doesn't help in dealing with your question.

But let me emphasize something. The occurrence of multiphoton photoemission using sunlight is negligible. Now, this statement may appear to be a complete tangent to the discussion, but multiphoton photoemission requires that a photon to excite an electron to an excited state, and within a short period of time before it decays back to a lower state, another one happens to hit it at the right time and the right place to cause another excitation. The probability of this occurring depends very much of the photon density, i.e. the higher the number of photons per unit area per unit time, the higher the probability that more than one photon can do the sequential excitation.

Such a process is negligible with sunlight that we receive on earth. In fact, the only time that I've encountered such multiphoton photoemission is using a very high-powered, Class 4 laser with energy of 10 mJ or more, and a spot size of 1 cm in diameter (we published a paper that included this effect).

Now note, this is ONLY tackling the issue of getting more than one photons to be hitting the same location within a very short period of time. This is LESS STRINGENT than having to have two photons, hitting the exact same spot, at the same time, and that just happens to have two completely opposite polarization. What are the odds?

Edit: There is also another issue here that we haven't touched. The typical interference pattern that we see is actually the result of single-photon interference! It is not the interference of two or more photons. The physics of the latter is different than the former.

Zz.

5. May 12, 2014

### Cthugha

They do sum to zero (for a narrow spectral window), but only on average. If you consider that intensity depends on the squared field and thus cannot be negative, this means that one has to look beyond the mean value of the field to get the full picture.

Thermal light sources like light bulbs or the sun can be described in first approximation by a superposition of fields from many emitters with randomly varying phase. Very often all of these fields will cancel. Sometimes they will not. So you will get a probability distribution of possible total fields. This process is equivalent to a random walk and you will get a 2D Gaussian probability distribution for the possible total amplitudes. Starting from there, you can calculate the photon number probability distribution and will find that it is the famous Bose-Einstein distribution. See for example the following demonstration for an example: http://demonstrations.wolfram.com/PhotonNumberDistributions/.

It has the interesting feature that 0 is always the most probable photon number which is a direct consequence of the average field being 0. So loosely speaking you can consider the light from our sun as a consequence of noise and if you look at some distant sun at any instant, it is pretty likely that it is completely dark right now and that it flickers badly if you look for a longer time. However, you will not notice that as our eyes are much too slow to resolve that. The photon number changes on a femtosecond timescale. Integrating over timescales directly accessible to us (and the full spectrum) gives a pretty constant average photon number.

6. May 12, 2014

### sophiecentaur

I think you can ignore the polarisation thing or even the vector nature of the waves - which just complicates the issue. The OP is basically about the sum of a large number of random functions. Imagine an audience, applauding a performance. Each audience member is blindfolded and has ear muffs on (cannot interact with or affect the others) and produces a string of 'almost regular' claps. What is the probability that they will ever all clap at precisely the same time and produce one enormous clap? The most likely thing that you'll hear is a constant level of random, individual claps which will sound like random noise - with no silences and no extreme peaks. When interference can occur (vector addition) the probability of a very high peak is similarly very small and the knife edge condition of a zero is even less likely.

7. May 14, 2014

### maline

Thank you ZapperZ for your long answer. I asked about sunlight to show why I'm so confident that it's not unusual for many wave trains to be moving in parallel in a given small region of spacetime. What you wrote about photons hitting the same location doesn't contradict this, because each wave train can be detected at any point along its wavefront, so its probability of being detected as a photon at the point in question is very low. I think the answer I wanted is Cthugha's-the "random walk" of adding up the field vectors gives a nonzero average magnitude, equal to the square root of the number of vectors. That also explains why the net energy is proportional to the number of waves, rather that to the square of that number, as it is in the case of perfect constructive interference. Thanks!