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I was reading Feynman's lecture in which Feynman invoked his own way of explaining the uncertainty principle using single-slit experiment.

There I found:

To get a rough idea of the spread of momentum, the vertical momentum $p_y$ has a spread which is equal to $p_0\Delta \theta$, where $p_0$ is the horizontal momentum . . .

What is he talking of? What does "spreading" actually mean? And how did he measure it?

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  • $\begingroup$ It is the spread of the possible momenta over an interval. Don't take it too literally, he just means there's an interval in which the possible momenta lie. Nothing is spreading here. $\endgroup$ – ACuriousMind Mar 9 '15 at 17:13
  • $\begingroup$ The edit you did is good, it makes the text clearer. $\endgroup$ – Sofia Mar 9 '15 at 17:42
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"I just want to know, before entering the slit, the particles had no vertical momentum. But after that, they got vertical momentums. Now, how it proved the Uncertainty Principle? Feynman wrote that as the pattern diffracted, the uncertainty of vertical position i.e. vertical momentum increased. Can you tell me how?"

Let's take the things one by one. Look at the picture in my former answer. From the wave inside the slit, Huygens' principle tells us that every point generates a new spherical wave. You can see this more clearly in the animated picture in the site I indicated above. Now, since the slit is very narrow, what gets out from it? A spherical wave. If the slit were very, very wide, the parallel wave coming to the slit would have exited approximately as a parallel wave, but since the slit is so narrow, the spherical waves emitted from the few points of the wave inside the slit, exit the slit in the spherical form. So, here we have a linear momentum component arising, $p_y$.

Thus, the wave comes to the slit with linear momentum $p_0$ along the horizontal axis, and this momentum is simply deflected in all the directions, under all the angles $\theta$. So, its direction spreads out. The horizontal component of the linear momentum decreases a bit, $p_x = p_0 cos\theta$ and there appears a vertical component $p_y = p_0 sin\theta$. As long as $\theta$ is very small, $cos\theta ≈ 1$ and $sin\theta ≈ \theta$, s.t.

$$p_x ≈ p_0, \ \ \ \text {and} \ \ \ p_y ≈ p_0 \theta \tag{i}$$.

Until now it's clear?

Now we go to the uncertainty principle, that says similar things, but in a more mathematical form. For our experiment, the principle says,

$$\Delta y \Delta p_y \ge \hbar/2 \tag{ii}$$

The $\Delta$ from a quantity means standard deviation, but I don't want to complicate you with math. It means in short how big is the undetermination of that quantity. In our case, the undetermination in the height $y$ in the particles' position in the slit is the very slit height, which is very small, as you see in my picture, or in the animations in the site I recommended. So, $\Delta p_y$ is quite big,

$$\Delta p_y \ge \frac {\hbar/2}{\Delta y} \tag{iii}$$

To get some approximate idea of how big is $\Delta p_y$, you can see in my picture the height that I denoted by $\Delta p_y$. Again, I don't want to complicate you with math, but behind the slit we can see, at some distance an interference pattern with maxima and minima of brightness. If the distance from the slit is proportional to $p_0$ then the height of the central maximum is roughly proportional to $\Delta p_y$.

Until now is clear?

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  • $\begingroup$ @user36790 Done! So, please read my new answer, and if further questions, leave me a comment. $\endgroup$ – Sofia Mar 11 '15 at 21:56
  • $\begingroup$ Thanks for replying. So, first parallel waves come with only horizontal momentum. Then they get diffracted & spread in different direction & thus possessing series of momenta in vertical direction also. And so, the particle can have any of the vertical momenta ; so there is a large uncertainty. Right? $\endgroup$ – user36790 Mar 12 '15 at 1:09
  • $\begingroup$ @user36790 completely correct. Just, at a certain time when you'd consider the question answered, please close the question. We have many questions answered, but with no upvotes, also, not closed, and we can't use them later as answers to other similar questions. $\endgroup$ – Sofia Mar 12 '15 at 2:27
  • $\begingroup$ Sorry, I really couldn't understand your last request. $\endgroup$ – user36790 Mar 12 '15 at 5:45
  • $\begingroup$ @user36790 Many times there appears a question that was already posed, but differently worded. We would like to spare our time, and instead of answering the essentially same thing, mark the new question as a repetition (duplicate) of the answered one. However, the rule in this site says that if the former question is neither closed on some answer, nor has upvoted answer(s), it can't be considered solved, i.e. can't be used in the way above. Well, but as you closed your question on my answer, we can use your question it in the future, as I explained. $\endgroup$ – Sofia Mar 12 '15 at 14:26
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As you didn't quote the site with Feynman discourse, I make a guess what was the issue.

The single-slit experiment illustrates diffraction. The wave-front that passes through the slit generates, according to Huygens' principle, a new wave-front, and so on. We obtain in all, beyond the slit, a deviation of the direction of the light. i.e. the linear momentum is no more $\vec p_0$ but takes different directions. Indeed, the uncertainty principle tells us that $\Delta p_y \Delta y > \hbar/2$, where $\Delta y$ is the height of the slit.

If $\Delta \theta$ is the angle beyond the slit within which the intensity of light is appreaciably different from zero, and this angle is sufficiently small, then taking $p_0$ as the horizontal cathete of a triangle in which $p_y$ is the vertical cathete, we can say that the latter is approximately $p_0 \Delta \theta$.

enter image description here

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    $\begingroup$ The question asks about the meaning of $\Delta p_y$. This answer uses $\Delta p_y$ in an expression, but does not define it, so I think it is not really an answer. $\endgroup$ – WillO Mar 9 '15 at 22:38
  • $\begingroup$ @WillO if you have doubts, why don't you ask? What if I'll doubt you that you did smth. wrong, and I'd punish you in base of a doubt? $\endgroup$ – Sofia Mar 10 '15 at 23:47
  • $\begingroup$ I have no doubt that this answer contains no definition of $\Delta p$. $\endgroup$ – WillO Mar 11 '15 at 1:15
  • $\begingroup$ @WillO The user didn't ask for an explanation of the uncertainty principle, s.t. I had no reason to explain him what are $\Delta y$ and $\Delta p_y$, and these are standard deviations, etc. The fact that I showed him the figure and the fact that a $p_y$ appears (because the linear momentum is no more $p_0$) and how it is related to $\theta$ satisfied his requests. If he didn't know what is the uncertainty principle and the quantities involved (standard deviations) he would have asked, the user is not a small child. $\endgroup$ – Sofia Mar 11 '15 at 4:03
  • $\begingroup$ Have you read the question? It asks "What does spreading actually mean?" In other words, he asked. $\endgroup$ – WillO Mar 11 '15 at 14:09
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A spread of momentum is sometimes used literally (as in, the momentum is definitely bigger than $a$ and definitely less than $b$ so the spread of momentum of is $b-a$), and sometimes it is used colloquially just to say that most of the time the momentum is between an $a$ and a $b$ such that $b-a$ equals your spread of momentum.

The reason having a spread of momentum is useful is because it is an overestimate of the standard deviation. If all your momentums are between $a$ and $b$ the largest the standard deviation can be is $(b-a)/2$ (a probability distribution with 50% of the time at $a$ and 50% of the time at $b$ has a standard deviation of $(b-a)/2$, and to keep that same spread you either have to be unbalanced so the mean moves closer to the more common one, thus reducing the average squared deviation or you symmetrically bring some of it in from both sides, which also decreases the averaged squared deviation). So it is easier to compute or understand sometimes and the actual standard deviation is smaller than the spread. So when the spread is small, the standard deviation is even smaller.

For an answer about why the standard deviation is important, consider the answer Uncertainty principle and measurement.

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Taking as given the state of your particle, the result of a momentum measurement is a random variable. $\Delta p$ is the standard deviation of that random variable.

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  • $\begingroup$ The O.P. indicated a formula. How you get it? $\endgroup$ – Sofia Mar 10 '15 at 23:45

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