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Taken out f

enter image description here

The First Picture is taken out from the Book The Character of Physical Law By Richard Feynman And the second picture is from his own The Feynman Lectures on Physics.

Both figures correspond to the An experiment with bullets topic which appear on his lecture.

In the first lecture $N_1$ , $N_2$ are taken as the number of bullets that hit on the backstop when hole no. $1$ is open and then again when hole no. $2$ is open.

I'm convinced to some measure of extent what the graph will look like in the individual cases, but not the both holes open case, because it is the sum of the other two(it concluded that $N_{12}=N_1+N_2$?). Please explain me the details of how it came.

And In the second picture, $P_1$ and $P_2$ are taken as the probability of finding the bullets in that place when hole no. $1$ is open, with no. $2$ closed and vice versa.

The sum of the graph is bulged out in the middle in case of two holes open. Also deducing that it is the sum of the probabilities when each hole is open ($P_{12}=P_1+P_2$?). I also don't understand how it adds together to give the probability of both holes open and the bulging out portion of the graph with respect to the slightly curved in portion in the previous graph in the book "The character of physical law".

From what I understood as given in the book The Feynman lectures on physics Chapter 1 "Quantum behaviour": By “probability” we mean the chance that the bullet will arrive at the detector, which we can measure by counting the number which arrive at the detector in a certain time and then taking the ratio of this number to the total number that hit the backstop during that time.

With this definition, I'll try to demonstrate an example where I don't understand how the probability adds in case of two holes open.

Let the source shoots bullet erratically in all directions but with a constant rate say $100$ bullets per hour.

Suppose, Hole $2$ is closed and Hole $1$ is open, Say, $20$ bullets has passed through hole No.$1$ and landed on the backstop (distributing the number of bullets). we are convinced to some level that most of the bullets will land just behind the hole. So, Let $B_1(x)$ be the number of bullets distributed over the backstop. Therefore ,probability of finding a bullet at distance $x$ in an hour is given by $\displaystyle{P_1(x)=\frac{B_1(x)}{20}}$,

Similarly, as the gun is shooting at a constant rate, $\displaystyle{P_2=\frac{B_2(x)}{20}}$ when hole $2$ is open and hole $1$ is closed.

Therefore If I am right, at the same time, when one hole is open and $20$ bullets is passing through the hole in one hour, the other hole which is blocked is blocking $20$ bullets which is inclined to pass through (without interfering the passed bullets) if it were open.

So, when we open the two HOLES, on the average $40$ bullets will pass through the holes, $20$ in Hole No.$1$ and another $20$ in hole no. $2$ under one hour. Therefore lets take the time to count the bullets with respect to $x$ and say the numbers are represented by $B_{12}(x)$, then the probability will be given by
$\displaystyle{P_{12}=\frac{B12(x)}{40}}$ which is clearly,
$P_{12}\neq P_1+P_2 \implies \displaystyle{\frac{B_{12}(x)}{40}\neq \frac{B_1(x)}{20}+\frac{B_2(x)}{20}}$.

Feynman tries to explain why it is bulged out in the middle in his "Lectures on physics" but I didn't quite understand. So, i added an image of his book in his attempt to explain the argument.

enter image description here

Please help me on the parts I find doubt and confusion. I have listed all my doubts in the above section, and help me understand the transition from this classical mechanics problem to the contrast we will see in the quantum behavior and also continue my journey to experience the beauty of quantum mechanics which will only be possible if I understand this experiment. Thanks in advance.

EDIT 1/ 15thNov2016: I still am confused how this probabilities are calculated as Feynman defined it in the book. I hope someone could elaborately explain in the same manner as Feynman defined or maybe some other simpler manner perhaps so that I could understand how the probabilities are calculated in the sense as he defined in the book. Thank you

I have read these for clarification but nothing comes close in explaining the probability calculation as Feynman defined it. http://physics.mq.edu.au/~jcresser/Phys201/LectureNotes/TwoSlitExpt.pdf https://stephenwhitt.wordpress.com/2008/12/17/the-experiment-with-two-holes/ http://physics.mq.edu.au/~jcresser/Phys301/Chapters/Chapter4.pdf

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    $\begingroup$ Yep. If you add two probability distributions whose total probability is one, the total probability is 2, so there's a factor of 1/2. $\endgroup$ – user12029 Sep 18 '16 at 4:00
  • $\begingroup$ so, how should I understand his argument? $\endgroup$ – Jyotishraj Thoudam Sep 18 '16 at 4:02
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    $\begingroup$ In the sense of the first image. $N_1(x)+N_2(x)=N_{12}(x)$. The number of hits adds $\endgroup$ – user12029 Sep 18 '16 at 4:04
  • $\begingroup$ But look at the graphs it doesn't bulge out in the middle in case of $Ns$ $\endgroup$ – Jyotishraj Thoudam Sep 18 '16 at 4:06
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    $\begingroup$ Whether it bulges out in the middle is a function of the distance between the two distributions. Try it out yourself. Plot $\exp (-(x-a)^2)+\exp (-(x+a)^2) $ for various "a". $\endgroup$ – user12029 Sep 18 '16 at 4:35
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First, let me tell you what I think about your graphical question. The shape of the resulting curve depends on two things: first, the width of the slits, wider the slit wider the distribution and second, the distance between the two slits. When the slits are very far away you would see two distinct distributions but as you bring the slits closer they will slowly merge (see the attached animation and my hand drawing). So there is nothing wrong with the graphics. They just depict two different cases.

two Gaussians enter image description here

When it comes to your "mathematical" question, I think, you are right. Feynman did a small mistake there. Obviously, if probability is defined as Feynman did, one cannot have $P_1+P_2=P_{12}$, as you nicely demonstrated. One can simply perform an thought experiment with some numbers and see that it is wrong. I think, the correct way of thinking is to define the probability not as number of hits divided by number of bullets reached to backstop for a given time but rather number of hits divided by number of bullets coming out from the source for a given time. Then, one can write $P_1+P_2=P_{12}$.

Good luck with your journey with learning Quantum Mechanics.

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  • $\begingroup$ Thanks linuxick. I find resort in your elaboration. Would you also provide me the some source of your answer, I mean not all but some parts, like where did you get the graph idea and the slit informations? I will no doubt give you the bounty one way or the other. $\endgroup$ – Jyotishraj Thoudam Nov 18 '16 at 1:31
  • $\begingroup$ I have revised my answer a bit and added a drawing. I hope it is more clear now. $\endgroup$ – physicopath Nov 18 '16 at 9:35
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The experiment can be viewed in different ways. To avoid confusion one needs to be very careful how one defines quantities. For instance if we define $P_1$ as the probability distribution for the bullets hitting the backstop with only hole 1 open, then one needs to say how this probability distribution is calculate. I think the way Feynman defined the probability includes those bullets that did not manage to pass through the hole. Therefore if we integrate the distribution $P_1$ the result would be much smaller than one. A similar small result would be obtained for $P_2$. In that case one can then understand why one would have $P_{12}=P_1 + P_2$. The probability for a bullet to pass through with both holes open would be twice the probability with only one of the two holes open.

Now what about the dip? If we assume the distributions $P_1$ and $P_2$ are Gaussian then we can model the combination by $$ P_{12} = A\exp\left[\frac{-(x-d)^2}{w^2}\right] + A\exp\left[\frac{-(x+d)^2}{w^2}\right] , $$ where $d$ is the shift in the distribution and $w$ is its width. One can determine if the combination contains a dip in the middle or not, by computing the sign of the second derivative at that location. It comes out to be $$ \partial_x^2 P_{12}|_{x=0} = \frac{4A(2d^2-w^2)}{w^4}\exp\left(-\frac{d^2}{w^2}\right) . $$ In other words, for $d<w/\sqrt{2}$ the second derivative at the origin becomes negative, indicating that the dip disappeared.

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  • $\begingroup$ Have you read Feynman's probability definition? He clearly stated that probability is calculated only with sample space as the bullets that hit the backstop. It's in his book Lectures on physics Volume 3, Quantum behaviour! $\endgroup$ – Jyotishraj Thoudam Sep 18 '16 at 15:37
  • $\begingroup$ I see. Well, there is something called the conservation of probability. So if one wants to add up two probabilities to produce another probability they all need to be defined in the same framework. Nevertheless, it is not something you need to break your head about. Could be just a typo. I have that book somewhere. Perhaps I'll dig it out and read through that part to see if there is anything that explains the discrepancy. $\endgroup$ – flippiefanus Sep 19 '16 at 4:20
  • $\begingroup$ you can check this site feynmanlectures.caltech.edu/III_01.html $\endgroup$ – Jyotishraj Thoudam Sep 19 '16 at 10:25
  • $\begingroup$ Any new informations? $\endgroup$ – Jyotishraj Thoudam Sep 24 '16 at 10:41
  • $\begingroup$ Not really. Cannot find a clear definition of what is meant by $P_{12}$, etc. in the link. So I guess one needs to assume it is the propabilities as one would usually define them so that they can be added. This means that the conditions for the different probabilities need to be the same: it is the probability that a bullet is detected given that one was launched toward the holes. $\endgroup$ – flippiefanus Sep 24 '16 at 11:50

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