What is $\Delta x$ in Heisenberg principle for a single slit?

Question

In the diffraction from a single slit a particle goes tru the slit of size $D$. Is the $\Delta x$ in HUP $\Delta x~\Delta p>h$ equal to $D$? Please give a QM pdf book where it is said plainly. I tried Feynman and Griffits and some other but there was nothing about.

Answer 1

Yes, the slit width $D$ is the $\Delta x$ in Heisenberg's uncertainty principle $\Delta x~\Delta p>h$.

Searching for "single-slit diffraction uncertainty principle" finds many reputable resources. One of these is in Feynman lectures (Chapter 2 "The Relation of Wave and Particle Viewpoints", Section 2-2 "Measurement of position and momentum"). Here is the essential part of this section which answers your question on the spot:

Fig. 2–2.Diffraction of particles passing through a slit.

Here is one example which shows the relationship between the position and the momentum in a circumstance that is easy to understand. Suppose we have a single slit, and particles are coming from very far away with a certain energy — so that they are all coming essentially horizontally (Fig. 2–2). We are going to concentrate on the vertical components of momentum. All of these particles have a certain horizontal momentum $p_0$, say, in a classical sense. So, in the classical sense, the vertical momentum $p_y$, before the particle goes through the hole, is definitely known. The particle is moving neither up nor down, because it came from a source that is far away — and so the vertical momentum is of course zero. But now let us suppose that it goes through a hole whose width is $B$. Then after it has come out through the hole, we know the position vertically — the $y$-position — with considerable accuracy — namely $\pm B$. That is, the uncertainty in position, $\Delta y$, is of order $B$. Now we might also want to say, since we know the momentum is absolutely horizontal, that $\Delta p_y$ is zero; but that is wrong. We once knew the momentum was horizontal, but we do not know it any more. Before the particles passed through the hole, we did not know their vertical positions. Now that we have found the vertical position by having the particle come through the hole, we have lost our information on the vertical momentum! Why? According to the wave theory, there is a spreading out, or diffraction, of the waves after they go through the slit, just as for light. Therefore there is a certain probability that particles coming out of the slit are not coming exactly straight. The pattern is spread out by the diffraction effect, and the angle of spread, which we can define as the angle of the first minimum, is a measure of the uncertainty in the final angle.

How does the pattern become spread? To say it is spread means that there is some chance for the particle to be moving up or down, that is, to have a component of momentum up or down. We say chance and particle because we can detect this diffraction pattern with a particle counter, and when the counter receives the particle, say at $C$ in Fig. 2–2, it receives the entire particle, so that, in a classical sense, the particle has a vertical momentum, in order to get from the slit up to $C$.

To get a rough idea of the spread of the momentum, the vertical momentum $p_y$ has a spread which is equal to $p_0\Delta\theta$, where $p_0$ is the horizontal momentum. And how big is $\Delta\theta$ in the spread-out pattern? We know that the first minimum occurs at an angle $\Delta\theta$ such that the waves from one edge of the slit have to travel one wavelength farther than the waves from the other side—we worked that out before (Chapter 30 of Vol. I). Therefore $\Delta\theta$ is $λ/B$, and so $\Delta p_y$ in this experiment is $p_0\lambda/B$. Note that if we make $B$ smaller and make a more accurate measurement of the position of the particle, the diffraction pattern gets wider. So the narrower we make the slit, the wider the pattern gets, and the more is the likelihood that we would find that the particle has sidewise momentum. Thus the uncertainty in the vertical momentum is inversely proportional to the uncertainty of $y$. In fact, we see that the product of the two is equal to $p_0\lambda$. But $\lambda$ wavelength and $p_0$ is the momentum, and in accordance with quantum mechanics, the wavelength times the momentum is Planck’s constant $h$. So we obtain the rule that the uncertainties in the vertical momentum and in the vertical position have a product of the order $h$: $$\Delta y \Delta p_y \ge \hbar/2 \tag{2.3}$$