Particles Behaving as Waves Problems and Solutions 14

Q#39.93

A particle with mass m moves in a potential U(x) = A|x|, where A is a positive constant. In a simplified picture, quarks (the constituents of protons, neutrons, and other particles, as will be described in Chapter 44) have a potential energy of interaction of approximately this form, where x represents the separation between a pair of quarks. Because U(x) → ∞ as x → ∞  it’s not possible to separate quarks from each other (a phenomenon called quark confinement). 

(a) Classically, what is the force acting on this particle as a function of x? 

(b) Using the uncertainty principle as in Problem 39.92, determine approximately the zero-point energy of the particle. 

Answer:

(a) U = A|x|, Eq. $F=-\frac{dU}{dx}$ relates force and potential. The slope of the function A x is not continuous at 0 x = so we must consider the regions 0 x > and 0 x < separately. 

For x > 0, |x| = x, so U = Ax and

$F=-\frac{d(Ax)}{dx}=-A$

For x < 0, |x| = -x, so U = -Ax and

$F=-\frac{d(-Ax)}{dx}=+A$

We can write this result as

$F=-\frac{A|x|}{x}$, valid for all x except for x = 0.

(b) Use the uncertainty principle, expressed as , ΔpΔx ≈ h and as in Problem 39.80 estimate Δp by p and Δx by x. Use this to write the energy E of the particle as a function of x. Find the value of x that gives the minimum E and then find the minimum E.

E = K + U = K + A|x|

px  ≈  h, so p ≈ h/x  

Then, E ≈ $\frac{h^2}{2mx^2}$ + A|x|

For, x > 0, E = $\frac{h^2}{2mx^2}$ + Ax

To find the value of x that gives minimum E set $\frac{dE}{dx}=0$

$0=\frac{-2h^2}{2mx^3}+A$

$x^3=\frac{h^2}{mA}$

$x=\left(\frac{h^2}{mA}\right)^{1/3}$

With this x the minimum E is

E = $\frac{h^2}{2m}\left(\frac{mA}{h^2}\right)^{2/3}+A\left(\frac{h^2}{mA}\right)^{1/3}$

E = $\frac{1}{2}h^{2/3}m^{-1/3}A^{2/3}+h^{2/3}m^{-1/3}A^{2/3}$

E = $\frac{2}{3}\left(\frac{h^2A^2}{m}\right)$

The potential well is shaped like a V. The larger A is, the steeper the slope of U and the smaller the region to which the particle is confined and the greater is its energy. Note that for the x that minimizes E, 2K = E.

Q#39.94

Imagine another universe in which the value of Planck’s constant is 0.0663 J.s, but in which the physical laws and all other physical constants are the same as in our universe. In this universe, two physics students are playing catch. They are 12 m apart, and one throws a 0.25-kg ball directly toward the other with a speed of 

(a) What is the uncertainty in the ball’s horizontal momentum, in a direction perpendicular to that in which it is being thrown, if the student throwing the ball knows that it is located within a cube with volume 125 $cm^3$ at the time she throws it? 

(b) By what horizontal distance could the ball miss the second student?

Answer:

(a) Let the y-direction be from the thrower to the catcher, and let the x-direction be horizontal and perpendicular to the y-direction. A cube with volume V = 125 $cm^3$ = 0.125 $\times 10^{-3} \ m^3$ has side length 

l = $V^{1/3}=(0.125 \times 10^{-3})^{1/3}=0.050 \ m$. 

Thus estimate Δx as Δx ≈ 0.050 m. Use the uncertainty principle to estimate Δ$p_x$.

$\Delta p_x \Delta x \geq \frac{\hbar}{2}$ then gives

$\Delta p_x \approx \frac{\hbar}{2\Delta x}$

$\Delta p_x = \frac{0.01055 \ J.s}{2(0.050 \ m)}=0.11 \ kg.m/s$

(The value of $\hbar$ in this other universe has been used.) 

(b) $\Delta x=\Delta v_x t$ is the uncertainty in the x-coordinate of the ball when it reaches the catcher, where t is the time it takes the ball to reach the second student. Obtain $\Delta v_x$ from $\Delta p_x$. The uncertainty in the ball’s horizontal velocity is

$\Delta v_x=\frac{\Delta p_x}{m}=\frac{0.11 \ kg.m/s}{0.25 \ kg}=0.42 \ m/s$

The time it takes the ball to travel to the second student is

$t=\frac{12 \ m}{6.0 \ m/s}=2.0 \ s$

The uncertainty in the x-coordinate of the ball when it reaches the second student that is introduced by $\Delta v_x$ is

$\Delta x=\Delta v_x t = (0.42 \ m/s)(2.0 \ s) = 0.84 \ m$

The ball could miss the second student by about 0.84 m. 

A game of catch would be very different in this universe. We don’t notice the effects of the uncertainty principle in everyday life because h is so small. 

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