2. EL POLO DEMOCRÁTICO
2.2 La izquierda en Colombia
In addition to the Coulomb potential, there are a variety of other central potentials that are of interest, although some of them may only be solved
numerically. Examples of such potentials are molecular potentials, which exhibit relatively weak attraction far away but strong repulsion when very close, and nuclear potentials, where neutrons are essentially particles in a spherical box and quarks appear to have a constant attractive force which leads to a linear potential, at least in the limit where the separation becomes large.
In treating these more general spherical potential problems it is useful to first note the identity
1 r2
d dr
r2dR
dr
≡1 r
d2 dr2(rR) .
With this identity, we may write the general radial equation as
−~2 2µ
d2
dr2(rR) +
V (r) + ~2 2µ
`(` + 1) r2
(rR) = E(rR) , (3.89) where we have multiplied by r. Letting u = rR, this result may be written more conveniently as
u00+ 2µ
~2(E − V ) −`(` + 1) r2
u = 0 . (3.90)
3.4.1 Nuclear Potentials
The simplest model of neutrons is that they are particles in a box with a finite radius and finite but constant potential in the box. Another model is to consider the nucleus to be a water droplet, where there are no forces between particles except for the surface tension when they try to escape. These two models are similar, in that the potential inside is constant, but the former considers a rigid spherical box, while the water droplet model constrains the volume to remain fixed, but the surface can be deformed from a spherical shape. The latter case is outside the scope of this book, but the former, with
` = 0, leads simply to
u00+2µ
~2
[E − V (r)]u = 0 , with V (r) =
∞, r < 0,
−V0, 0 ≤ r ≤ a,
0, r ≥ a.
The nuclear potential sketched in Figure 3.3 for neutrons is nearly the same as the particle in a finite well as sketched in Figure 2.3, except that in this case the region r < 0 is excluded by an infinite potential barrier at the origin (to get to the other side, one goes around in angle, not by passing through the origin to −r), and outside the well the potential is taken to be zero so that outside the neutron is a free particle. Once a proton or alpha particle leaves the nucleus, however, it experiences a Coulomb repulsion, so its well is effectively deeper, but there is a finite chance of tunneling through the barrier if E > 0.
The Schr¨odinger Equation in Three Dimensions 97
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r V (r)
0
−V0
neutrons protons
FIGURE 3.3
A nuclear potential well, showing the additional Coulomb potential for protons or alphas.
Problem 3.21 The neutron potential problem is one we have solved already, with a few observations. If you were to put an impenetrable barrier in the center of the rectangular well problem of Figure 2.3, relate the energies of this neutron well problem of Figure 3.3 to those of that problem. (Hint: The symmetry of the wave function, i.e., even or odd parity, and the boundary condition at the origin are important in making the connection between the problems.)
Problem 3.22 Neutron potential well depth. A neutron is in the ground state of a well as depicted in Figure 3.3. The lowest possible energy of a gamma ray that can remove the neutron from the well is 8.0 MeV. The radius of the well is 3.0 · 10−15 m. Obtain the depth V0 in MeV.
Problem 3.23 Deuteron potential well depth. The deuteron is the ground state of a bound proton and neutron in a potential well as depicted in Figure 3.3 (solid line since the neutron is neutral). Assuming that the neutron is just barely bound (E = V0− , ∼ 0), estimate the depth of the potential well if the width is r0 ∼ 10−15 m and the proton and neutron masses are each m = 1.67 × 10−27 kg. Express the depth in MeV (1 MeV = 106 eV or 1.602 × 10−13 J).
(a) Solve for the radial function u(r) = rR(r) for 0 ≤ r ≤ r0 and for r > r0. Match boundary conditions at r0 so that both u(r) and u0(r) are continuous and assume V = ∞ for r ≤ 0.
(b) From these boundary conditions, show that we find a relationship like Equation (2.21). Since there is only one bound state and since it is just barely bound, estimate the value of k0 with E ∼ V0 and then estimate V0 (in MeV).
Problem 3.24 Spherical harmonic oscillator. A particle is bound in a three dimensional symmetric harmonic oscillator well with V = 12Kr2. If it is in the the ground state with ` = 0, find the normalized eigenfunction R(r) and the energy eigenvalue.
3.4.2 Quarks and Linear Potentials
There are a number of physical problems where the force is constant, giving rise to linear potentials. Examples include uniform electric fields or gravitation near the earth’s surface, but one of particular interest is the force between quarks. At least at large separations (but still less than the size of a nucleus), the force between two quarks appears to be constant and attractive. This force is due to “color charge” and entirely independent of the usual electrostatic forces and electric charge. We can then model the force between two color charges by F = −∇V = −ger and the corresponding potential is V (r) = gr − V0where V0gives the potential at the bottom of the potential well. With this potential, the radial equation of Equation (3.90) becomes
u00+2µ
~2
(E + V0− gr)u = 0 .
If we change to dimensionless variables so that ξ = α[r − (E + V0)/g], and set α3= 2µg/~2, then we obtain
u00− ξu = 0 . Airy equation (3.91) Equation (3.91) is called the Airy equation, and although deceptively simple in appearance, its solutions may not be represented in terms of elementary functions. The solutions can be related to Bessel functions of one-third in-tegral order, but this is of little direct use. It has general properties such that the solutions are either exponentially growing or decaying for ξ > 0, while the solutions oscillate for ξ < 0. The two linearly independent solutions are usually designated Ai(ξ) and Bi(ξ), where Ai(ξ) is exponentially decaying for positive ξ and Bi(ξ) is exponentially growing for positive ξ, and is there-fore inadmissible for our purposes, since we require bounded solutions. Many properties of the Ai(ξ) solution are evident in Figure 3.4.
While the study of Airy functions is generally very difficult, the solution can be written as an integral, such that
Ai(ξ) ≡ 1 π lim
→0
Z ∞ 0
e−ucos u3 3 + ξu
du. (3.92)
Direct substitution into the Airy equation, interchanging the order of integra-tion and differentiaintegra-tion, leads to
d2
dξ2πAi(ξ) − ξπAi(ξ)
The Schr¨odinger Equation in Three Dimensions 99
The Airy function, Ai(ξ), showing the first three zeros.
= lim
The only purpose of the e−u factor was to guarantee convergence as u → ∞, and the factor is frequently omitted with the understanding that the contri-bution from the upper limit vanishes. From the form of the argument of the oscillatory term, it is clear that the oscillation frequency increases in the large argument limit so that the integral is summing rapidly oscillating contribu-tions of substantially equal amplitudes, so it is evident that the integral will converge even without the exponential convergence factor.
For large arguments, i.e., as |ξ| → ∞, the asymptotic forms are given by
Ai(ξ) ∼
These expressions show the rapid convergence for positive ξ, as well as the oscillatory character with a slowly decreasing amplitude for negative ξ. While this is not an elementary function, it is a tabulated function [1].
Returning to our quark potential problem, we note that at r = 0 we require u = 0, but in terms of the variable ξ, we must have u(ξn) = 0 where
ξn =
0 − En+ V0 g
2µg
~2
1/3 ,
and ξn is one of the zeros of the Airy function. Hence for the energy eigen-values, we have
En= |ξn| g2~2 2µ
1/3
− V0. (3.94)
If our model of the quark force and potential are reasonable approximations to the real forces and potentials, then we should be able to find systems of two quarks (a quark and an antiquark form a meson) and our candidate is the J/ψ meson made of a pair of charmed quarks. This meson has more massive quarks than those found in protons and neutrons and hence the relativistic corrections are expected to be smaller and provide a better test of the model than a pair of up or down quarks that form π mesons. When the mass of the two quarks is equal, the reduced mass is just half the mass of either quark, and the system is analogous to positronium, which is a bound system of a positron and an electron. The only thing we can measure about the meson is its mass in its lowest and some of its first few excited states. If we call the total rest energy of the meson Mnc2, then this energy is comprised of
Mnc2= 2M c2+ |ξn| g2~2 M
1/3
− V0, (3.95)
where the total energy is made up of the rest energy of the two charmed quarks with mass M plus the potential energy. We do not know M , g or V0, but there are really only two unknown constants, 2M c2− V0 and g2~2/M . By doing a least squares fit to the first four energy levels of charmonium (the ground state of charmonium is the J/ψ), which are 3.097 GeV, 3.686 GeV, 4.1 GeV, and 4.414 GeV, with
Mnc2= A + B|ξn| ,
we find A = 2.42 GeV and B = 0.300 GeV. The fit with just these two parameters leads to values of Mnc2that are within about 1% or better of the measured values. That one can fit all four levels within a percent with only two constants indicates that the model has some validity, especially when one considers that relativity, spin and other effects are neglected.
Problem 3.25 Charmonium energy levels.
(a) Find the percent error for the Mnc2 for each of the first four states between the calculated and measured values.
The Schr¨odinger Equation in Three Dimensions 101 (b) With M c2 = 1.84 GeV, evaluate V0 in GeV and g in GeV/f , where f
(≡ 10−15m) is a Fermi.
(c) If one considers that the classical turning point occurs when the kinetic energy vanishes, estimate the classical radius of the turning point in Fermis.
(d) Using half this classical radius value (the average is approximately half the maximum), estimate the energy of charmonium when ` = 1 by adding the additional potential term in Equation (3.89) to the ground state energy. Compare this value with the average of the measured values of the first three charmonium values with ` = 1, which are 3.413 GeV, 3.508 GeV, and 3.554 GeV.
Problem 3.26 Charmonium levels — Bohr model. Solve the Charmonium problem using the Bohr method, where L = n~ for the angular momentum.
The answer may be written in the form
E(B)n + V0= aE0n2/3, (3.96) where E0≡ (g2~2/2µ)1/3so that the solution from Equation (3.94) is written
En+ V0= E0|ξn| . (a) Find a in Equation (3.96).
(b) Find the percent error for each of the two lowest states (percentage En(B)
is below En).
3.4.3 Potentials for Diatomic Molecules∗
When two neutral atoms approach one another, the weak interactions between the electrons cause slight rearrangements of the charge distribution that leads to an attractive force which varies as r−6. On the other hand, when atoms get very close so that there is a lot of overlapping, there is a strong repulsion from the nuclei of the atomic cores, and there is therefore some equilibrium distance where there is a potential minimum. There is a variety of models for this molecular potential, but the first one we note is of the form
V (r) = A r12− B
r6. (3.97)
The r−12 term represents the repulsive core while the r−6 term dominates at large distance and gives the attraction that draws the atoms into a molecular configuration. This potential is difficult to solve directly, but if one only needs to know the energy levels of the first few states, we can expand about the minimum, and represent the potential as
V (r) = 12k(r − r0)2− V0+ O(r3) . (3.98)
If we restrict our attention to the lowest lying states where ` = 0, then this is just the harmonic oscillator potential, where
u00+2µ
~2
[E + V0−12k(r − r0)2]u = 0 . The energy eigenvalues of this problem are then given simply by
En = (n +12)~ω − V0, (3.99) where ω2= k/µ.
Another model potential for a diatomic molecule is called the Morse poten-tial, and is given by
V (r) = V0[1 − e−a(r−r0)]2− V0. (3.100) This potential has the property of having a minimum of −V0 at r = r0, and goes to zero exponentially as r → ∞. The potential at the origin is finite, whereas it should be infinite, but this is generally insignificant since the potential is quite large at the origin. With this potential, we will keep all the terms in Equation (3.90), obtaining
u00+ 2µ
~2[E + 2V0e−a(r−r0)− V0e−2a(r−r0)] −`(` + 1) r2
u = 0 . (3.101) At this point, we change variables to
y = e−a(r−r0), We then approximate the r02/r2 term by expanding about y = 1 (which is equivalent to expanding r about r0) and find that
r20
so that Equation (3.102) becomes d2u
The Schr¨odinger Equation in Three Dimensions 103 We could factor out the asymptotic form and examine the resulting power series, but if we simply assume a solution of the form (which is the same form as the radial solution for the one-electron atom)
u = e−αzzmF (z) , with z = βy, then Equation (3.103) can be written as
F00+ 2m + 1 z − 2α
F0+
"
m2+ 2µ
~2a2(E − c0)
z2 +
2µ(2V0−c1)
~2a2β − α(2m + 1) z
+α2−2µ(V0+ c2) β2~2a2
F = 0 . (3.104) Then if we choose
α = 12, β2= 8µ
~2a2(V0+ c2) , m2= 2µ
~2a2(c0− E) , Equation (3.104) can be written simply as
zF00+ (2m + 1 − z)F0+ nF = 0 ,
which is the associated Laguerre equation [which is not surprising since the form of u(z) was the same as R(ρ)] if we let
n = 2µ
~2a2β(2V0− c1) − m −12, (3.105) so that with 2m = k and n an integer, F = Lkn+k(z). Rearranging Equation (3.105) and eliminating m, we find
n +12− 2µ
~2a2β(2V0− c1)
2
= m2= 2µ
~2a2(c0− E) , and solving for the energy, we find
En = c0−~2a2
2µ (n +12)2+ ~a(2V0− c1)
p2µ(V0+ c2)(n +12) −(2V0− c1)2
4(V0+ c2) . (3.106) Expanding in the smallness of c1/V0and c2/V0, with ω ≡ ap2V0/µ, the result may be expressed as
En= ~ω(n+12)
1 − 3~2C 2IV0
`(` + 1) − ~ω 4V0
(n +12)
−V0+~2
2I`(` + 1) , (3.107)
where I = µr20 is the rotational moment of inertia and C = 1/ar0− 1/a2r02. From a comparison of Equation (3.107) to Equation (3.99), it is apparent that with ` = 0, the solutions are nearly identical except for a term of the order of ~ω/V0 and the inclusion of angular momentum terms. We will later be able to show that including cubic and quartic terms in the expansion of Equation (3.98) will bring the two results even closer.
Problem 3.27 For the potential of Equation (3.97),
(a) Find r0, k, and V0in Equation (3.98) in terms of A and B.
(b) Graph the potentials of Equations (3.97) and (3.98) on the same graph, matching the values of V0, r0, and k for the two cases.
(c) If, for a particular molecule, V0 = 2 eV and r0 = 1.5 · 10−10 m, find A, B, and 12~ω (in eV) for a hydrogen molecule.
Problem 3.28 Fill in the missing steps starting with Equation (3.90) with the potential from Equation (3.100) and ending with Equation (3.107).