Protons, neutrons, and many other particles are made of more fundamental particles called quarksand 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 c particle is a low-energy bound

state of a so-called charm quark and its antiquark, with a rest energy of 3097 MeV; the c12S2 particle is an excited state of this same quark-antiquark combination, with a rest energy of 3686 MeV. A simplified representation of the potential energy of interaction between a quark and an antiquark is U1x2 = A0 x0, where Ais a positive constant and xrepresents the distance between the quark and the antiquark. You can use the WKB approximation (see Challenge Problem 40.64) 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

1n= 1, 2, 3, c2

Here Eis the energy, U1x2 is the potential-energy function, and x= aand x= bare the classical turning points (the points at

which Eis equal to the potential energy, so the Newtonian kinetic energy would be zero). (a) Determine the classical turning points for the potential U1x2 = A0 x0 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

(Hint:The integrand is even, so the integral from -xto xis equal to twice the integral from 0 to x.) (c) Does the difference in energy between successive levels increase, decrease, or remain the same as nincreases? 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?

The potential energy of interaction between a quark and an antiquark is given by:

$U\left(x\right)=A\left|x\right|$

We know that, at the turning points, the potential energy equals the total energy; Thus,

$E=U\left(x_{tp}\right)=A\left|x_{tp}\right|$

Solving for$x_{tp}$ We get the turning points,

$$\begin{gathered} \left|x_{tp}\right|=\frac{E}{A} \\ {x}_{tp}=\pm \frac{E}{A} \\ {x}_{tp}=-\frac{E}{A},+\frac{E}{A} \end{gathered}$$

This means that there are two turning points with the same magnitude but on different sides of the equilibrium point.

First, we rewrite the integral in the WKB approximation:

${\int }_a^b \sqrt{2m(E-A(x\left(x\right))}dx=\frac{nh}{2}$ (n = 1,2,:)

To get the energy levels in this approximation, we substitute for $a=-E/A,b=+E/A\mathrm{andU}\left(\mathrm{x}\right)=\mathrm{A}\left|\mathrm{x}\right|$so,

${\int }_{-E/A}^{+E/A} \sqrt{2m(E-A|x\left|\right)}dx=\frac{nh}{2}$

Using the fact that the integral from -r to r is equal to twice the integral from 0 to, we get:

$$\begin{gathered} \frac{nh}{2}={\int }_{-E/A}^{+E/A} \sqrt{2m(E-A|x\left|\right)}dx \\ =2{\int }_0^{+E/A} \sqrt{2m(E-A|x\left|\right)}dx \\ =2\sqrt{2m}{\int }_0^{+E/A} (E-A|x|{)}^{1/2}dx \\ =2\sqrt{2mA}{\int }_0^{+E/A} {\left(\frac{E}{A}-x\right)}^{1/2}dx \\ =2(2mA{)}^{1/2}\frac{-2}{3}{\left[{\left(\frac{E}{A}-x\right)}^{3/2}\right]}_0^{E/A} \\ =\frac{-4}{3}(2mA{)}^{1/2}\left[-{\left(\frac{E}{A}\right)}^{3/2}\right] \end{gathered}$$

Now, solving for E, we get,

$$\begin{gathered} \frac{2^2}{3}{2}^{1/2}{m}^{1/2}{A}^{1/2}\frac{E^{3/2}}{A^{3/2}}=\frac{nh}{2} \\ {E}^{3/2}=2^{-7/2}3m^{-1/2}Anh \\ E=\left(2{)}^{-7/3}\right(3{)}^{2/3}\left(m{)}^{-1/3}\right(A{)}^{2/3}\left(n{)}^{2/3}\right(h{)}^{2/3} \\ \therefore E=\frac{1}{2m}{\left(\frac{3mAh}{4}\right)}^{2/3}{n}^{2/3} \end{gathered}$$

…….(1)

From equation (1), the ground-level energy corresponds to n= 1 and it is given by:

$\mathrm{E}=\frac{1}{2m}{\left(\frac{3mAh}{4}\right)}^{2/3}(1{)}^{2/3}=\frac{1}{2m}{\left(\frac{3mAh}{4}\right)}^{2/3}$

So, equation (1) can be written in terms of the ground-level energy as

$E_n=n^{2/3}{E}_1$ (2)

Using equation (2), for two successive levels T and n +l, the energy separation is:

$\begin{array}{c}\mathrm{\Delta }E=E_{n+1}-E_n=E=(n+1{)}^{2/3}{E}_1-n^{2/3}{E}_1\\ \mathrm{\Delta }E=\left.⌈(n+1{)}^{2/3}-n^{2/3}\right]E_1\end{array}$

$E_1$is constant, so$△E$ depend only on$\left[(n+1{)}^{2/3}-n^{2/3}\right]$

But since the expansion$(n+1{)}^{2/3}$ is complicated, it is better to see the change$∆E$as n increases by trial.

For n= 1, we get:

$△E=[2^{2/3}-1^{2/3}]E_1=0.5847E_1$

For n=2, we get

$△E=[3^{2/3}-2^{2/3}]E_1=0.4927E_1$

For n=3, we get:

$△E=[4^{2/3}-3^{2/3}]E_1=0.4398E_1$

As we can see the energy separation between two adjacent levels decreases with increase.

The difference in energy between successive levels decreases with increasing when (color(red)$U\left(x\right)\propto x$ as in this problem.

The difference in energy is constant when$U\left(x\right)\propto x$ as with the harmonic oscillator.

And the difference in energy increases with increasing n when U(x) changes from 0 to$\infty$in a very sharp step.

Putting it in a simple rule,

If the curvature (the rate at which a curve is changing direction at a given point) of the potential energy is less than that of a parabola$U\left(x\right)\propto x$, the level differences will decrease; For example,$U\left(x\right)\propto x$.

If the curvature of the potential energy is greater than that of a parabola, the level differences will increase; For example, the potential energy of a potential box.