# I A few questions about Feynman's QED Lecture

1. Feb 4, 2017

### JohnnyGui

Hello,

Apologies if this has been asked before. After watching Feynman’s QED lectures on the probabilities of different paths that photons can take towards a photomultiplier, a few questions came up in me.

Let’s take the simple scenario in which a photon emitter is faced straight towards a photomultiplier, without any mirrors or glass whatsoever. The photon emitter sends just one photon out. As Feynman stated, that photon can take different paths until it reaches the photomultiplier and you can calculate the probabilities of these paths that the photon would take.
Also as stated, the probability that the photon would take a certain range of more or less straight paths is higher than the probability of it taking a certain range of bent paths.

My questions:
1. Do the probabilities of all possible paths that a photon could take on its way to the photomultiplier together equal 100%?

2. If someone puts a photon detector directed at a certain range of paths, will the probability of the photon taking that range of certain paths be 100% minus the probability of all the possible paths other than that certain range?

3. If the answer to question 2 is yes, does this mean that statistically, if one repeats the experiment from question 2 a 100 times, the detector would go off a number of times according to that probability?

Another question came up when, after watching the lecture, I read further that this one photon actually takes all these possible paths simultaneously. This got me confused because if that’s true, how then can a photon take certain paths based on probabilities while it actually doesn’t have any preference and takes all these paths simultaneously?

4. Does this probability scenario merely arise when one repeats the experiment while using photon detectors directed at a certain range of possible paths?
5. Or is probability merely another term for intensity?

2. Feb 4, 2017

### jfizzix

The probability of taking a particular path is the square of the probability amplitude to do so. This is where the quantumness of quantum physics comes from.
The probability amplitude that a photon emitted at point A will arrive at point B is the sum over the probability amplitudes to take all possible paths beginning at A and ending at B.
Then, the probability (density) that a photon comes from A and ends at B is the square of the sum of these amplitudes.

Since emitters and detectors are not single points, it is important to point out, that this amplitude calculation gives you the conditional probability density of a photon arriving at point B given that it originated at point A. (They're densities, because there are a continuum of possible points to choose from and to sum over to get total probabilities).

Finally, to get the probability that the photon will arrive on a particular area of the detector, given that it originated in some other area on the source, you need to know the probability density of where photons will be emitted from the source, which could be found out experimentally.

As for whether the photon really takes these paths simultaneously, all we really know is that this method of calculation works very well to predict the probabilities that we measure.

Probabilities are not just another term for intensity. In particular, if just one photon hits a detector at a given point, the intensity distribution will just have that one point. Over time, as you gather more and more counts, your intensity distribution becomes smoorther and converges to one particular form, which will be the probability distribution. One can predict the ultimate intensity distribution from the probability distribution, and one can estimate a probability distribution from a given intensity distribution with varying levels of success.

3. Feb 4, 2017

### JohnnyGui

Thanks for your reply.

One other question; imagine a photon emitter shooting just one photon at a glass with such a thickness that it gives a probability of 4% reflection towards a photomultiplier. Does this mean that if you repeat this experiment a 100 times, every time with just one photon, that the photomultiplier would detect a photon 4 times total statistically speaking?

I'm really curious about this, since this does show in some way that a photon doesn't really have a preference based on probability. Which is weird since it is based on probability when you repeat the experiment several times and look at the intensity distribution. I'm talking about shooting just one photon every time.

4. Feb 4, 2017

### Staff: Mentor

You have to stop listening to your classical intuition. The weirdness goes away (or at least is replaced by a different weirdness) if you think in terms of the probability of a photon appearing at a given spot, instead of moving from the source to that spot.

5. Feb 4, 2017

### Mentz114

Yes.

This is one of things that is hard to understand. One could ask how individual photons decide whether to be reflected or transmitted. But this is a simplisitic way of looking at it. The photon is closer to being an excitation of a field (that is everywhere already) than a classical 'corpuscule'.

[beaten to it]

6. Feb 4, 2017

### JohnnyGui

Ah I see now. This makes the whole concept of a photon taking all possible paths at the same time a bit wrong. Since, given your explanation, the photon would be going only one particular way based on the path you've decided to calculate the probability of. Such that, you can only observe a photon taking different paths by repeating the experiment and not just by executing the experiment just once. Am I making sense?

Also, is the probability of a photon taking a particular path calculated by multiplying the probabilities of finding the photon at the different spots on that path?

7. Feb 4, 2017

### Staff: Mentor

No. The photon does not take one path. It does not take all paths. What it is doing has no analogue at all in the classical terminology you are trying to use.

You never observe the photon taking a path. This experiment does not do that at all, no matter how many times you run it.

You apparently did not grasp what Nugatory meant by "you have to stop using your classical intuition". Any talk of the photon "taking a path" is classical intuition. Stop using it.

8. Feb 4, 2017

### Staff: Mentor

It would be better to drop the idea that a photon takes any path. I used the word "appears" quite deliberately in the post you're quoting; the calculation is giving us the probability that a photon will be detected when we put the detector at a given location but says nothing about what's happening or has happened anywhere else. We cannot safely say that the photon at the detector is the same photon ("same" is a tricky concept when applied to quantum particles) as the one that was emitted, nor that it moved on any path through the space between source and detector.

9. Feb 4, 2017

### JohnnyGui

I didn't mean to say that it really does take one or all paths. What I meant is that WE (mistakingly) consider it taking one particular path by calculating the probabilities of different spots on the path that we choose to think the photon is taking.

As for repeating the experiment to see different "paths" let's say we theoretically put photon detectors on all the possible spots between the photon emitter and the photomultiplier. When repeating the experiment several times, won't each try give a different pattern of certain photon detectors going off? Those patterns being based on the probabilities of the different spots where the photon would appear? Since the photon "collapses" when detectors are in place, it won't make 2 or more detectors go off at once. And that would make us THINK that it has taken one particular path based on the probabilities of the spots.
Just like Feynman said that a photon detector sometimes detects a photon that is at an "off-edge" spot and on other times it doesn't.

10. Feb 4, 2017

### JohnnyGui

I have not worded my intention of using paths well. I've rephrased it in my above post. This explantion of yours does give it a whole different view. Which also fits for the double slit experiment when 1 photon is involved I think.

11. Feb 4, 2017

### Staff: Mentor

But we don't do that, at least not in the version of the experiment we've been talking about, where the only photon detection is at the final photomultiplier. We only calculate the amplitude for the photon to reach a photomultiplier at a given location. We don't calculate amplitudes for the photon to reach spots somewhere in space between the source and the photomultiplier.

Then we are running a different experiment, which will be modeled differently in the math. Now we have to compute amplitudes for each detector to fire, and square each one individually to get a probability for that detector to fire. And the probability for the photomultiplier to detect a photon will no longer be the square of a single amplitude; it will be the sum of squares of a lot of amplitudes, one for each in-between detector firing. For example, suppose we have ten detectors in between the source and the photomultiplier, placed so that only one will fire for any given photon. Then the probability for the photon to reach the photomultiplier is the sum of squares of ten amplitudes--the amplitude for the photon to reach the photomultiplier given that detector #1 fires (times the probability that detector #1 fires), the amplitude for the photon to reach the photomultiplier given that detector #2 fires (times the probability that detector #2 fires), etc. And the final prediction for the probability of the photon reaching the photomultiplier will be different than the prediction from the original experiment, with no photon detectors.

In short, this new experiment, while it will give you lots of interesting data, tells you nothing useful about what is or is not happening in the original experiment.

12. Feb 4, 2017

### jfizzix

To answer your first question, the most likely outcome of 100 runs of this experiment is 4 detections (at 19.9 percent probability), but the probabilities of getting 3 detections (19.7 percent), 5 detections (15.6 percent), and others are quite significant. This is part of my favorite version of the law of large numbers. As the number of trials grows large, the probability that your string of outcomes resembles the probability distribution to any fixed tolerance approaches 1. So, for a large enough number of trials, the probability of having less than 3% detections, or more than 5% detections approaches zero (it's below 10 percent for 1000 trials, below $10^{-4}$ percent for 10,000 trials, and below, well, $10^{-9}$ percent for a million trials) .

You can prepare a single photon to be in any quantum state you wish, and you can measure its detection (or lack thereof). Those are what you know, and can control. What happens between creation and detection is not measured and can only be speculated about. What we can predict reliably is what the intensity distribution would look like if we measured the photon between the emitter and the detector, but such a measurement would be another experiment with another detection.

Last edited: Feb 4, 2017
13. Feb 5, 2017

### JohnnyGui

So if I understand this correctly. Suppose we calculate the probability of a certain area of paths of 1 photon between the source and photomultiplier (without any photon detectors in between) according to Feyman, and that probability turns out to be 4%. Now if I put photon detectors only in that certain area of paths between the source and the photomultiplier, filling that area all up, and I repeat the experiment 100 times, there won't be 4 out of those 100 cases in which, cumulatively, all the photon detectors in that area have fired?

This is interesting to read. Isn't this the standard statistical behavior of chances though? That the chance for something would be focused in a narrower range after more and more repeats?

Yes, I was indeed asking about the (intensity) distribution when one repeats the experiment each time with 1 photon. I thought that the probabilities of the different areas in that distribution, calculated in Feynman's way without photon detectors, would agree with the distribution of probabilities of photon detectors firing at those areas.

14. Feb 5, 2017

### Staff: Mentor

Of all the runs in which the photon reaches the photomultiplier, 4% of them will have one of the photon detectors firing. But the fraction of all runs in which the photon reaches the photomultiplier will be different with the detectors present than without them.

15. Feb 6, 2017

### JohnnyGui

This cleared it up for me. Thank you.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted