In this video, it is explained that it is not necessary that photons obey the laws of reflection and each photon can take any possible path to reach the so-called black hole receiver and all the paths have some certain probabilities as shown in the picture.

But if that's the case why do we see a certain region of the room when we stand in front of the mirror as in this picture? Shouldn't we see even the corners of the room standing in the same position? But to see them we have to tilt on one side or move our head.

If mathematics says that probability exists then why don't we observe it in our daily life? Since it never happens shouldn't the probability be zero?

Edit 1: Is it ever possible or are there any phenomenon where we see the effects ( raised in the question)i.e. paths of photon do not interfere to zero and hence have non zero probability ?? How would it look like ??

Edit 2: Is the above mentioned effect more likely to be possible with a curved mirror ( since each part of a spherical mirror can be thought of as a plane mirror )??

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    $\begingroup$ If you roll a die 100 times, is the probability that you get 100 sixes zero? How many dice do you think you'd have to roll to ever see it happen? $\endgroup$ Aug 20, 2020 at 12:39
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    $\begingroup$ We should trust quantum mechanics period. $\endgroup$
    – my2cts
    Aug 20, 2020 at 12:56
  • $\begingroup$ @Peter Shor But do you think we can notice such phenomenon ? $\endgroup$
    – Ankit
    Aug 20, 2020 at 16:03
  • $\begingroup$ Adapt your theory to the facts, not the other way around. The fact says that mirrors do work. Fortunately, the theory is right too. If the probability is less than 1÷the age of the universe, then it's non zero...but "it is zero". $\endgroup$
    – FGSUZ
    Aug 20, 2020 at 21:50

4 Answers 4


Is it ever possible or are there any phenomenon where we see the effects ( raised in the question)i.e. paths of photon do not interfere to zero have non zero probability ?? How would it look like ??

Yes there are experiments to verify this. With good detectors it can be verified that if the edges of the mirrors are covered, the photon count at the detector is affected. This indicates to paths being cutoff.

A more direct way to see if light takes all paths is to change reflectivity of the mirror such that some parts don’t reflect. One such alteration is to make alternative pattern of high and low reflection. Essentially making a reflective grating.

enter image description here

The result of this patterned cutoff of paths is that a diffraction pattern appears in the detector. Indicating that the effect is cumulative of all paths.

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    $\begingroup$ Good answer, +1. The diffraction grating example is exactly what I was thinking of also. It is pretty conclusive $\endgroup$
    – Dale
    Aug 20, 2020 at 22:27
  • $\begingroup$ Can the downvoter kindly share their reason for doing so? $\endgroup$ Aug 20, 2020 at 22:43
  • $\begingroup$ Yeah, I don’t get it either. The diffraction gratings pretty conclusively demonstrate that light does follow all of those paths. Seems solid to me $\endgroup$
    – Dale
    Aug 21, 2020 at 1:01
  • $\begingroup$ @Superfast Jellyfish Do you mean to say that if each of the surface of the mirror between two consecutive black points are painted differently then we will receive all those colours ?? $\endgroup$
    – Ankit
    Aug 21, 2020 at 6:57
  • $\begingroup$ @Superfast Jellyfish And if that's the case then why don't we observe this without affecting the performance of the mirror ?? $\endgroup$
    – Ankit
    Aug 21, 2020 at 7:01

In quantum mechanics we deal with amplitudes, not probabilities directly. Amplitudes are complex numbers, probabilities are the square of these amplitudes. What the first picture says is true in the sense that every path between the light source (call it A) and an observer, maybe you (call it B) have a non-zero amplitude. However the thing about amplitudes is that they interfere. Let's call the amplitude that the photon takes a certain path, $a[x(t)]$, where $x(t)$ is the path it's moving on. Then we can write the amplitude of a photon to leave point A and reach B as, $$K(B,A) = \text{sum of $a[x(t)]$ over all paths between A and B }$$ In this sum, even though $|a[x(t)]|$ might be equal for all paths, when you add them you can get zero. This is in the same way that $|-1| = |1|$ and $1-1 = 0$. This is why you only see certain things in the mirror but not others. Simply for the objects you don't see, the amplitudes from all possible paths sumed up together give zero. When calculating the probability of a photon to leave A and reach B you square $K(B,A)$, and so if $K(B,A)$ is zero, the probability is zero.

  • $\begingroup$ what does it physically mean for amplitudes of the paths taken by each photon to interfere ?? Is it similar to what we study as constructive and destructive interference for waves ?? I am just a high school student 😂 so sorry if I am wrong somewhere. $\endgroup$
    – Ankit
    Aug 20, 2020 at 17:13
  • $\begingroup$ why is the net interference of the amplitudes 0 all the times ?? Does it depend on the set up too ? $\endgroup$
    – Ankit
    Aug 20, 2020 at 17:16
  • $\begingroup$ (1) Yes, it's the same kind of physics as wave interference. What I mean by interference is really what I wrote in the equation for K(B, A). That the amplitudes add up in this way is what I mean by interfere really. (2) No the answer is not always zero. Some other times the amplitude might add up constructively and give you a big probability. I was just trying to point out that the situations you say don't happen, don't happen because the amplitudes interfere in a destructive way. $\endgroup$
    – A. Jahin
    Aug 20, 2020 at 17:25
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    $\begingroup$ If you have a few hours to spare and are interested learning more, try this: youtube.com/watch?v=P9nPMFBhzsI It's really a classic. $\endgroup$
    – A. Jahin
    Aug 20, 2020 at 17:28
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    $\begingroup$ @Ankit From the objects you see in the mirror you experience constructive interference. From the objects you don't see, (even though there is some paths connecting them to you through the mirror) you experience destructive interference. $\endgroup$
    – A. Jahin
    Aug 21, 2020 at 16:54

I don't have a deep enough knowledge of QM to give a complete answer, but I have a pretty good idea of what is confusing you. It's that many physical phenomena can be explained or described at more than one level. What's confusing you is that you are mixing up ideas from different levels. (IMO, the presenter of the video is somewhat to blame because he exploits that confusion in order to make the topic seem more mysterious, and therefore, more entertaining.)

In the classical level of understanding, the picture looks like this (please excuse the crude artwork.)

drawing of classical rays of light reflecting off of a mirror

If the light source is point-like, if the mirror is flat and continuous (e.g., not a diffraction grating as mentioned in some other answers), and if the target is your eye; then all you will see in the mirror is a single point of light. This is an experiment that you can try at home.

Q: So why does my drawing look different from the one at the top of your question?

A: It's because the lines in my drawing represent the "light rays" of clasical optics. But the lines in the drawing from the video represent photons. Photons and light rays do not belong to the same "level" of explanation. Don't mix them up. Photons are not little bursts of light that travel along the classical "rays."

Light is energy. You can't "observe" it in flight. All you can really know is where it came from, and where it ends up. Both the classical explanation and the quantum explanation are models that predict where the light energy leaving any given point can end up.

The video, IMO, leaves a lot out. I'm going to leave out a lot too, but I want to go back to that drawing. It really should show photons "leaving" every point of the mirror in every direction. Something more like this (please excuse the even worse artwork*):

crude drawing, attempting to represent... something... I guess

This is almost where I get off this train, but above I said, "photons are not little bursts of light that follow classical rays." Photons are described by wave equations, and the lines in my second drawing are a crude representation of the normals of the wave fronts of all of the possible waves leaving the light source and, all of the possible waves leaving the mirror.

It turns out (and just trust me as I wave my hands and back away) that if you compute the sum of all of those possible waves, you get a function that you can use to predict the probability that any particular packet of energy (photon) leaving the source will end up depositing that energy at any particular point in space.

And it turns out (backing out of the door and starting to close it), that if you compute those probabilities all the way through the lens of your eye and to the retina, you will find only one tiny spot on the retina where the probabilities add up to significantly more than zero. (i.e., QM predicts that the reflection of the point-like light source that you see in the mirror will look like a single bright point.)

(door closes, jumps off the train, runs.)

* rescued the paper on which I drew the original from the trash bin.

  • $\begingroup$ The first picture is right. $\endgroup$ Aug 20, 2020 at 13:50
  • $\begingroup$ @descheleschilder, OK, I watched the video. And yes, the first picture "right," but only if "right" means "an incomplete and deliberately misleading fragment of a correct explanation." If you fill in the missing parts (which, the video presenter eventually did), then the complete explanation shows at a deep level why the picture that I drew is also right, but at a shallower level. I'll probably delete this "answer" soon. The OP's real question is, "help me understand the video," and I don't really have the time to do that. $\endgroup$ Aug 20, 2020 at 14:26
  • $\begingroup$ Your picture is just a small part of all possible trajectories, for which the classical law of reflection holds (the incoming angle equals the outgoing angle). In QM, trajectories for which both angles are different must be included. $\endgroup$ Aug 20, 2020 at 14:31
  • $\begingroup$ Completely re-wrote my "answer." It's incomplete, but I hope it draws attention to the real source of confusion here, which is the conflation of ideas from two different explanations (classical vs. quantum) of what a mirror does. I can't go deep into the quantum explanation, but I know that it doesn't make sense to try to "import" classical ideas into the quantum explanation. $\endgroup$ Aug 21, 2020 at 15:07
  • $\begingroup$ I think you've got it right this time! +1 $\endgroup$ Aug 21, 2020 at 15:44

Mirrors can be understood by classical electromagnetism. Quantum mechanics just adds to this that the classical image intensity describes the probability distribution of photons.

In classical electromagnetism all paths are possible but only the path directly from the object to your eye survives. The other paths destructively interfere.

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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review. Your answer needs expansion and elucidation. $\endgroup$ Aug 20, 2020 at 19:16
  • $\begingroup$ @CharlesFrancis I could not disagree more. My answer covers the question fully. $\endgroup$
    – my2cts
    Aug 20, 2020 at 19:32
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    $\begingroup$ it contains only the core of the answer, but unless the answer is actually explained, it says nothing of use. $\endgroup$ Aug 20, 2020 at 19:34
  • $\begingroup$ @CharlesFrancis The answer is concise, clear and directly adresses the core of the question. I apologise if this upsets you. $\endgroup$
    – my2cts
    Aug 20, 2020 at 21:41
  • $\begingroup$ In classical electromagnetism all paths are possible This is simply not true. This holds only in the context of quantum mechanics. Please, don't get upset! $\endgroup$ Aug 20, 2020 at 21:52

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