Question

Protons, neutrons, and many other particles are made of more fundamental particles called quarks and antiquarks (the antimatter equivalent of quarks). A quark and an antiquark can form a bound state with a variety of different energy levels, each of which corresponds to a different particle observed in the laboratory. As an example, the $\psi$ particle is a low-energy bound state of a so-called charm quark and its antiquark, with a rest energy of 3097 MeV; the $\psi(2 S)$ particle is an excited state of this same quark-antiquark combination, with a rest energy of 3686 $\mathrm{MeV}$ . A simplified representation of the potential energy of interaction between a quark and an antiquark is $U(x)=A|x|$ where $A$ is a positive constant and $x$ represents the distance between the quark and the antiquark. You can use the WKB approximation (see Challenge Problem 40.57$)$ to determine the bound-state energy levels for this potential-energy function In the WKB approximation, the energy levels are the solutions to the equation $$ \int_{a}^{b} \sqrt{2 m[E-U(x)]} d x=\frac{n h}{2} \quad(n=1,2,3, \dots) $$ Here $E$ is the energy, $U(x)$ is the potential-energy function, and $x=a$ and $x=b$ are the classical turning points (the points at which $E$ is equal to the potential energy, so the Newtonian kinetic energy would be zero). (a) Determine the classical turning points for the potential $U(x)=A|x|$ and for an energy $E$ (b) Carry out the above integral and show that the allowed energy levels in the WKB approximation are given by $$ E_{n}=\frac{1}{2 m}\left(\frac{3 m A h}{4}\right)^{2 / 3} n^{2 / 3} \quad(n=1,2,3, \dots) $$ (Hint: The integrand is even, so the integral from $-x$ to $x$ is equal to twice the integral from 0 to $x . )(\mathrm{c})$ Does the difference in energy between successive levels increase, decrease, or remain the same as $n$ increases? How does this compare to the behavior of the energy levels for the harmonic oscillator? For the particle in a box? Can you suggest a simple rule that relates the difference in energy between successive levels to the shape of the potential-energy function?

Name: Problem 59, Chapter 40: University Physics with Modern Physics
Uploaded: 2020-09-13T21:23:26-08:00
Duration: 12 min 42 s
Channel: Guilherme Barros

   Protons, neutrons, and many other particles are made of more fundamental particles called quarks and antiquarks (the antimatter equivalent of quarks). A quark and an antiquark can form a bound state with a variety of different energy levels, each of which corresponds to a different particle observed in the laboratory. As an example, the $\psi$ particle is a low-energy bound state of a so-called charm quark and its antiquark, with a rest energy of 3097 MeV; the $\psi(2 S)$ particle is an excited state of this same quark-antiquark combination, with a rest energy of 3686 $\mathrm{MeV}$ . A simplified representation of the potential energy of interaction between a quark and an antiquark is $U(x)=A|x|$ where $A$ is a positive constant and $x$ represents the distance between the quark and the antiquark. You can use the WKB approximation (see Challenge Problem 40.57$)$ to determine the bound-state energy levels for this potential-energy function In the WKB approximation, the energy
levels are the solutions to the equation
$$
\int_{a}^{b} \sqrt{2 m[E-U(x)]} d x=\frac{n h}{2} \quad(n=1,2,3, \dots)
$$
Here $E$ is the energy, $U(x)$ is the potential-energy function, and $x=a$ and $x=b$ are the classical turning points (the points at which $E$ is equal to the potential energy, so the Newtonian kinetic
energy would be zero). (a) Determine the classical turning points for the potential $U(x)=A|x|$ and for an energy $E$ (b) Carry out the above integral and show that the allowed energy levels in the WKB approximation are given by
$$
E_{n}=\frac{1}{2 m}\left(\frac{3 m A h}{4}\right)^{2 / 3} n^{2 / 3} \quad(n=1,2,3, \dots)
$$
(Hint: The integrand is even, so the integral from $-x$ to $x$ is equal to twice the integral from 0 to $x . )(\mathrm{c})$ Does the difference in energy between successive levels increase, decrease, or remain the same as $n$ increases? How does this compare to the behavior of the energy levels for the harmonic oscillator? For the particle in a box? Can you suggest a simple rule that relates the difference in energy between successive levels to the shape of the potential-energy function?

University Physics with Modern Physics

Hugh D. Young, Roger… 12th Edition

Chapter 40, Problem 59

Instant Answer

Solved by Expert Guilherme Barros

09/13/2020

Step 1

So, we set $U(x) = E$ and solve for $x$. Given $U(x) = A|x|$, we have $A|x| = E$. Therefore, $|x| = \frac{E}{A}$, which gives us the turning points $x = \pm \frac{E}{A}$. Show more…

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Protons, neutrons, and many other particles are made of more fundamental particles called quarks and antiquarks (the antimatter equivalent of quarks). A quark and an antiquark can form a bound state with a variety of different energy levels, each of which corresponds to a different particle observed in the laboratory. As an example, the $\psi$ particle is a low-energy bound state of a so-called charm quark and its antiquark, with a rest energy of 3097 MeV; the $\psi(2 S)$ particle is an excited state of this same quark-antiquark combination, with a rest energy of 3686 $\mathrm{MeV}$ . A simplified representation of the potential energy of interaction between a quark and an antiquark is $U(x)=A|x|$ where $A$ is a positive constant and $x$ represents the distance between the quark and the antiquark. You can use the WKB approximation (see Challenge Problem 40.57$)$ to determine the bound-state energy levels for this potential-energy function In the WKB approximation, the energy levels are the solutions to the equation $$ \int_{a}^{b} \sqrt{2 m[E-U(x)]} d x=\frac{n h}{2} \quad(n=1,2,3, \dots) $$ Here $E$ is the energy, $U(x)$ is the potential-energy function, and $x=a$ and $x=b$ are the classical turning points (the points at which $E$ is equal to the potential energy, so the Newtonian kinetic energy would be zero). (a) Determine the classical turning points for the potential $U(x)=A|x|$ and for an energy $E$ (b) Carry out the above integral and show that the allowed energy levels in the WKB approximation are given by $$ E_{n}=\frac{1}{2 m}\left(\frac{3 m A h}{4}\right)^{2 / 3} n^{2 / 3} \quad(n=1,2,3, \dots) $$ (Hint: The integrand is even, so the integral from $-x$ to $x$ is equal to twice the integral from 0 to $x . )(\mathrm{c})$ Does the difference in energy between successive levels increase, decrease, or remain the same as $n$ increases? How does this compare to the behavior of the energy levels for the harmonic oscillator? For the particle in a box? Can you suggest a simple rule that relates the difference in energy between successive levels to the shape of the potential-energy function?

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Key Concepts

Energy Level Spacing

Energy level spacing refers to the difference in energy between consecutive quantum states in a bound system. This spacing is not universal but depends on the nature of the potential energy function. For example, in some potentials the spacing decreases as the quantum number increases, while in others it might remain constant or even increase. This behavior reveals deeper insights into how the curvature or shape of the potential influences the energy spectrum.

Comparison of Quantum Systems

Comparing different quantum systems, such as the harmonic oscillator and the particle in a box, highlights how various potential energy functions shape their spectra. Each system exhibits a unique pattern of energy level spacing that is directly related to the mathematical form of its potential. Such comparisons can lead to broader rules connecting the specific power-law dependence or features of the potential to the scaling behavior of the quantized energy levels.

Energy Level Quantization

Energy level quantization is the principle that bound quantum systems can only possess specific, discrete energy values rather than a continuous spectrum. The allowed energy states arise from the boundary conditions imposed on the wave functions, and methods like the WKB approximation help derive formulas for these discrete energy levels based on the properties of the potential.

Potential Energy Function

A potential energy function describes how the potential energy of a particle depends on its position. It encodes the forces acting on the particle and determines the behavior of bound states and scattering states. The form of the potential is crucial as it directly influences the quantization conditions and the spacing of energy levels in the system.

Classical Turning Points

In quantum mechanics, classical turning points are the locations where the total energy of a system equals its potential energy, causing the classical kinetic energy to be zero. These points mark the boundaries of the region where a particle can move classically in a given potential, and they serve as integration limits when applying semi-classical methods like the WKB approximation.

WKB Approximation

The WKB (Wentzel-Kramers-Brillouin) approximation is a semi-classical technique used to estimate the solutions of the Schrödinger equation in systems where the potential varies slowly compared to the wavelength of the particle. It provides approximate quantization conditions by relating an integral over the classically allowed region to quantized action values, linking classical mechanics with quantum mechanical energy levels.

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Transcript

00:01 In this exercise, we're given a potential function, u of x that is equal to a positive constant a times the absolute value of x.

00:10 And we're also given the wkb approximation for the allowed energies.

00:15 That's the integral between the two turning points, say, and b, of the square root of 2 times the mass, times the total energy minus the potential, and this is equal to nh divided by 2.

00:29 In question a, we have to find what are the classical turning points for this potential, given an energy e.

00:38 Now, the classical turning points are such that the potential energy u of x is equal to the total energy e.

00:47 Now, u of x is a times the absolute value of x, and this is equal to e.

00:58 So the absolute value of x is equal to e divided by a, meaning that x is the absolute value of x is equal to e divided by a, x is equal to plus or minus e divided by 8.

01:09 These are the classical turning points.

01:14 In question b, we have to use the integral up here in order to find the allowed energies for the wkb approximation.

01:28 So we have to integrate from the first classical turning point, which is minus e over a to the second classical turning point, which is z over a, the following function.

01:43 2m times e minus a times the absolute value of x times the x, and this must be equal to nh divided by 2.

02:00 So let's just do the integral first.

02:05 Notice that this integral can be separated into two parts.

02:09 We have an integral from minus z over a to 0 of the square root, of 2m times e plus a x notice that in the in the range when x is negative the absolute value function is equal to minus x plus 0 the integral from 0 to u over a of 2 times m times e to the minus a x d x okay, so basically what i'm going to do is to take this first integral and do a little change of variables.

02:59 So i'm going to take x prime and call it minus x, such that dx is equal to minus dx prime.

03:12 So the first integral becomes the integral from e over a to 0 of the square root of 2m.

03:26 E minus ax prime times minus dx prime plus the integral from 0 to e over a, sorry, from d over, from e over a to e over a, of the square root of 2m, e minus ax times dx.

03:58 So basically i'm going to, in the first integral, i'm going to change the order of integration.

04:03 And add a minus sign.

04:09 So notice that these two integrals are exactly the same, since it doesn't matter if i'm integrating over x prime or x.

04:16 So this is equal to two times the integral from zero to e over a of the square root of 2m, e minus x, dx.

04:31 Okay.

04:33 So basically, i'm going to do again a change in variables...

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