juanberrio8a

Answered

2022-06-29

What does "spread of momentum" actually mean?

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}\mathrm{\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?

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}\mathrm{\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?

Answer & Explanation

timmeraared

Expert

2022-06-30Added 22 answers

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.

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.

Poftethef9t

Expert

2022-07-01Added 9 answers

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, $cos\theta \approx 1$ and $sin\theta \approx \theta $, s.t.

$\begin{array}{}\text{(i)}& {p}_{x}\approx {p}_{0},\text{}\text{}\text{}\text{and}\text{}\text{}\text{}{p}_{y}\approx {p}_{0}\theta \end{array}$

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,

$\begin{array}{}\text{(ii)}& \mathrm{\Delta}y\mathrm{\Delta}{p}_{y}\ge \hslash /2\end{array}$

The $\mathrm{\Delta}$ 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, $\mathrm{\Delta}{p}_{y}$ is quite big,

$\begin{array}{}\text{(iii)}& \mathrm{\Delta}{p}_{y}\ge \frac{\hslash /2}{\mathrm{\Delta}y}\end{array}$

To get some approximate idea of how big is $\mathrm{\Delta}{p}_{y}$, you can see in my picture the height that I denoted by $\mathrm{\Delta}{p}_{y}$. 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 $\mathrm{\Delta}{p}_{y}$

Until now is clear?

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, $cos\theta \approx 1$ and $sin\theta \approx \theta $, s.t.

$\begin{array}{}\text{(i)}& {p}_{x}\approx {p}_{0},\text{}\text{}\text{}\text{and}\text{}\text{}\text{}{p}_{y}\approx {p}_{0}\theta \end{array}$

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,

$\begin{array}{}\text{(ii)}& \mathrm{\Delta}y\mathrm{\Delta}{p}_{y}\ge \hslash /2\end{array}$

The $\mathrm{\Delta}$ 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, $\mathrm{\Delta}{p}_{y}$ is quite big,

$\begin{array}{}\text{(iii)}& \mathrm{\Delta}{p}_{y}\ge \frac{\hslash /2}{\mathrm{\Delta}y}\end{array}$

To get some approximate idea of how big is $\mathrm{\Delta}{p}_{y}$, you can see in my picture the height that I denoted by $\mathrm{\Delta}{p}_{y}$. 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 $\mathrm{\Delta}{p}_{y}$

Until now is clear?

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