# Light spreading vs attraction

1. Dec 16, 2009

### rhenretta

First off... woo a physics forum. It's been a while since I have posted in a forum, but I couldn't resist a physics forum :)

I just finished Richard Feynman's QED: The strange theory of light and matter. It brought up many questions, obviously, but the one I am currently trying to figure out is the seemingly paradoxical ideas that:

a) light spreads out and..
b) photons have a tendency to attract to the same point in S/T and same type (the basis of a laser)

a) seems obvious (a laser pointed at a distant target will be bigger than a laser pointed at a close target). b) on the other hand doesn't seem to actually play out. Do I misunderstand something here?

Last edited: Dec 16, 2009
2. Dec 17, 2009

### Rajini

b) it is focusing or coherence nature or light in laser

3. Dec 17, 2009

### rhenretta

hmm, wish I could put a Feynman diagram on here...

In his 3rd lecture in QED, Feynman described the photon's tendency to converge. I'll try to explain what he said sans graphics.

If you have photon a and b (I assume they are of the same frequency) at point 1 and 2, you can figure out the probability of them arriving at points 3 and 4. If 3 and 4 converge to point 5, then P(1 to 5) * P(2 to 5) = P(2 to 5) * P(1 to 5). Thus, as Feynman loved to put it, the arrows line up, and the photons tend to converge.

My only problem with his description is in practice photons of the same type spread out. As I explained earlier, if I take my laser and point it somewhere far away, the dot of light gets bigger, and according to this principle, it should get smaller.

4. Dec 17, 2009

### Cthugha

Although photons have the tendency to bunch, this does not mean that all photons necessarily converge towards being at one single position.

The quantum interpretation of this bunching effect was given by Fano in the sixties. It can be explained by looking at this picture from wikipedia:

Imagine photons are emitted randomly from the points a and b at a constant rate. A and B are detectors. Now one might be interested in the case of detecting photons simultaneously at A and B. Using standard QM formalism (<A| is the probability amplitude connected to detecting a photon at detector A,<B| is the probability amplitude connected to detecting a photon at detector B, |a> is the probability amplitude connected to the emission of a photon at point a and |b> is the probability amplitude connected to the emission of a photon at point b) the expected joint detection rate for random emission will be:

$$R_1=|<A|a><B|b>|^2 + |<B|a><A|b>|^2$$

Assuming the simplifying case that $$|<A|a><B|b>|^2 =|<B|a><A|b>|^2 =N^2$$ this gives a total joint detection rate of $$2 N^2$$

However, for indistinguishable particles one must keep in mind that these two possibilities leading to a joint detection are also in principle indistinguishable and can interfere. Therefore the joint detection rate will read:
$$R_2=|<A|a><B|b> + <B|a><A|b>|^2$$
$$=|<A|a><B|b>|^2 +| <B|a><A|b>|^2 +<A|a><B|b> <B|a><A|b>+<B|a><A|b><A|a><B|b>$$
$$=2 N^2 +<A|a><B|b> <B|a><A|b>+<B|a><A|b><A|a><B|b>$$

To evaluate these additional interference terms one has to keep in mind that photons are bosons. This means the wavefunction is symmetric concerning exchange of particles. Therefore
$$<A|a><B|b> =<B|a><A|b>$$!
So you get
$$R_2=2 N^2 + 2 |<A|a><B|b>|^2=4 N^2$$

The joint detection probability for bosons is therefore twice as high as expected for independent particles. This can also be seen if you shoot exactly 2 indistinguishable photons at a 50/50 beamsplitter. They will always both take the same exit port. However, this does not mean that all photons converge towards one state. In the beam splitter example still 50% of the photons exit through one port and 50% of the photons exit through the other, but the pairs arriving simultaneously will always take the same one. This is the basis of the Hong-Ou-Mandel test, which specifies how indistinguishable two photons are.

Note that the same calculations apply also to fermions. If you do the same experiment with electrons, the math basically stays the same. You now just have to consider that a fermionic wavefunction is antisymmetric in terms of particle exchange. So in this case:
$$<A|a><B|b> =-<B|a><A|b>$$
and accordingly you would get
$$R_2=2 N^2 - 2 |<A|a><B|b>|^2=0$$
for the joint detection rate of fermions, so fermions will never occupy the same state. Sounds familiar?

5. Dec 17, 2009

### rhenretta

Cthugha, thanks for the amazingly well articulated reply. I'm going to like this place :)

If I am understanding this correctly, the additional possibilities would only be indistinguishable in a stream. Could one emit 1 photon simultaneously from each source a and b, thus removing the interference? The interference would be impossible, since it would require 4 particles detected when only 2 were emitted. In this case would photons always pool, and never spread?

6. Dec 17, 2009

### Cthugha

It is unfortunately not that easy. Although since the times of Dirac it is pretty common to talk about interference of particles, Glauber already noted in his Nobel speech that this concept can lead to severe misconceptions. The easiest description is therefore given in terms of interference of probability amplitudes instead of interference of particles.

This means that you - very much in the sense of Feynman - just take all events, which start with the emission of 2 photons at a and b and end with the detection of two photons at A and B and have their wave functions interfere. The small difference is shifting the interference from the particle to the wave function, but this makes the basic conception somewhat less misunderstandable. Therefore you do not need 4 particles, but you need 4 completely indistinguishable events (<A|a>,<B|a>,<A|b>,<B|b>).

So getting back to your question:
If you emit 1 photon simultaneously from a and b and are in principle able to keep track, which is which, those 4 events are distinguishable and the interference vanishes.

If you just have two detections at A and B, but are in principle unable to know, which photon is which, the interference is there. The situation is somewhat similar to the double slit, where the interference also vanishes as soon as you make the two paths via the two slits distinguishable somehow.

7. Dec 17, 2009

### rhenretta

I'm trying to avoid the hypothetical of actually tracking the photon as doing so interferes with the experiment.

I understand the 4 events are indistinguishable (<A|a>,<B|a>,<A|b>,<B|b>). The roadblock for me, is that the event <A|a><B|a><A|b><B|b> is indistinguishable from the event <A|a><B|b>+<A|b><B|a> since the first event would seem to have each detector go off twice, whereas the second has each detector going off once.

8. Dec 17, 2009

### Cthugha

Now we need to have to use clearer notation, I fear. These terms are in principle wavefunctions associated with events, not the probability amplitudes of these events. Therefore you need indeed 4 of these terms. Let me consider

$$\psi_1=<A|a><B|b>$$ and $$\psi_2=<A|b><B|a>$$

It is important to remember, that the physically important thing is the probability density instead of the wave function.

So in the case, which you call $$<A|a><B|b>+<A|b><B|a>$$, this would be just $$\psi_1+\psi_2$$, which is just a sum of two wave functions, but no probability density. The measurable expression would be:
$$|<A|a><B|b>|^2+|<A|b><B|a>|^2 = |\psi_1|^2 + |\psi_2|^2$$ and the event
$$<A|a><B|a><A|b><B|b>$$ is $$\psi_1^* \psi_2$$.

So you get $$|\psi_1|^2 + |\psi_2|^2$$ in the distinguishable case and
$$|\psi_1 + \psi_2|^2 = |\psi_1|^2 + |\psi_2|^2 + Re(\psi_1^* \psi_2) +Re(\psi_2^* \psi_1)$$ in the indistinguishable case. The latter is a typical interference term in QM. Does this make it more understandable or even worse?

9. Dec 17, 2009

### rhenretta

oh, much much worse, but this is a good thing as it opens up the next area of study for me. Obviously, I need more experience in probability theory to really grasp this concept. Are you familiar with any good books or lectures on the subject?

10. Dec 17, 2009

### Cthugha

Oh, this is not really about probability theory. But might I know, how much you know about the basic formalism of QM? From your answers I guessed that you are familiar with it, but if you are not, I will think of a way to explain it differently.

11. Dec 17, 2009

### rhenretta

I have absolutely no formal education on the subject. I have spent, and continue to spend a significant amount of effort in learning quantum theory, but there are plenty of areas that you might consider essential that I have not been exposed to yet. Please, don't feel like you should water anything down, if I am missing important areas, I very much want to remedy this.

12. Dec 17, 2009

### rhenretta

I had assumed <a|A> was equivalent to Feynman's shorthand of P(a to A)

13. Dec 19, 2009

### Cthugha

Ok, so let me explain it a bit differently.

There are two basic quantities: The probability density that some event occurs (like Feynman's P(a to A) I suppose, I have not read his lectures for a while) and the wave function, which contains the physics of the situation, including wave character and stuff. However, the wave function itself can not be measured, but governs, what can be measured (for example the probability density for some event to occur). Mathematically you get the probability density by squaring the wavefunction, so they are closely related.

Now the wave function is basically a description of all possible indistinguishable ways a system can evolve from state A to state B. In the basic double slit experiment (let me call the slits 1 and 2) this means that at a certain position of the screen after the double slit, we have two possibilities: emitter->slit1->screen ($$\psi_1$$) or emitter->slit2->screen ($$\psi_1$$).

Now the basic idea is that we have to sum over all this possible paths beginning from the same event and leading to the same final situation to get the full wave function:
$$\psi=\psi_1+\psi_2$$
To get to the measurable probability density of this event occuring we need to square the wave function:
$$P=\psi^2=|\psi_1|^2+|\psi_2|^2 +2 \psi_1 \psi_2$$

In fact the $$\psi$$ are complex numbers, but that should not bother us in this simple description. However, you now see threre terms in the math: $$|\psi_1|^2$$ corresponding to the photon taking slit 1, $$|\psi_2|^2$$ corresponding to the photon taking slit 2 and a mixed term $$2 \psi_1 \psi_2$$, which is often described as "the photon taking both paths at once" in popular science books. It is important to note that this is the term, which produces interference.

Now if you do the same with balls instead of photons, the situation is different. Although there are still two slits, the possibilities of a ball getting through either of the two slits are distinguishable as you can follow the path of the ball pretty well. Accordingly the probability density that a ball hits a certain spot on the screen will be different because you have two completely different events: You have
$$P=\psi_1^2 + \psi_2^2$$

There is no interference term because the events are distinguishable and different and you cannot add the single wave function and square afterwards, but have to square each of them individually.

Getting back to the photon bunching problem before, the [tex]<A|a><B|b>[Tex] and [tex]<A|b><B|a>[Tex] are the wave functions, but do not actually mean that this exact event happened. It is more like an account of all events, which could have happened to get from the initial to the final situation.

14. Dec 19, 2009

### rhenretta

Very enlightening, thanks Cthugha. At what point does the probability density of the event lose the interference term? If I were to do this with single hydrogen atoms, would there be interference?

15. Dec 21, 2009

### Cthugha

Ah, good question.
That of course depends on how well you can keep your system isolated from interactions with the environment. Of course this works much better for electrons or photons than for large composite particles because the number of possible interactions, which can occur is much larger for composite particles.

As far as I know the largest things, which were tested in a double-slit setup were C60 Buckminsterfullerenes, which are already quite large (60 atoms is quite large on this scale).

The bunching/antibunching-effect is a bit more complicated to show with massive particles. However in 2007, there was a report on the difference between Helium3 (fermionic) and Helium4 (bosonic) in Nature 445, 402-405 (25 January 2007) (or Arxiv: http://arxiv.org/abs/cond-mat/0612278).