PHY6095/PHZ6166: homework assignment # 3
due Friday, March 22
2
Problems 1 and 3 deal with the Thomas-Fermi model. You may want to supplement the discussion of this model in
class by reading a corresponding chapter of your favorite QM textbook.
1. Problem 1 Charge density distribution in the Thomas-Fermi model.
Using the Thomas-Fermi scaling form for the electrostatic potential inside a heavy atom
Z
φ(r) = f (bZ 1/3 r), b =
r
√ !2/3
8 2
≈ 0.885
3π
(1)
(all quantities are in the atomic units, in which energy is measured in Rydbergs Ry = me4 /h̄2 , distance in Bohr
radii aB = h̄2 /me2 , etc.), find the scaling form for the local electron density n(r). Estimate the local density at
distances of the order of the Thomas-Fermi radius aT F ∼ Z −1/3 . Determine the asymptotic behavior of n(r) at
distances Z −1 r Z −1/3 .
Laplace’s equation (negative charge for electrons)
n = ∇2r φ/4π
where
φ(r) =
Z
χ(ar), a ≡ bZ 1/3 .
r
Rescaling r = ρ/a and noticing that ∇2r = a2 ∇2ρ , we obtain one gets
1
χ
1
3 2
3 1 ∂
2 ∂ χ
n=
Za ∇ρ
=
Za 2
ρ
4π
ρ
4π
ρ ∂ρ
∂ρ ρ
0
1
1 ∂
χ
χ
1
1 ∂
χ00
=
Za3 2
ρ2 − 2 +
=
Za3 2
(−χ + ρχ0 ) = Za3 .
4π
ρ ∂ρ
ρ
ρ
4π
ρ ∂ρ
ρ
where g(x) ≡ f 00 (x)/x.
According to the Thomas-Fermi equation,
√
χ00 = χ3/2 / ρ,
and thus
3/2
3/2
χ
χ
2 3
n = Za
=Z b
.
ρ
ρ
3
At r ∼ Z −1/3 , ρ ∼ 1, χ(∼ 1) ∼ 1, and n ' Z 2 .
At distances (Z −1 r Z −1/3 ), χ ≈ 1, thus χ/ρ ≈ ρ−3/2 and
3/2 Zb 3/2
n(r) ≈ Z 2 b3 / bZ 1/3 r
=
.
r
2. Problem 2 Two-dimensional hydrogen atom.
For more details on this interesting problem see, e.g., K. Eveker et al. Am. J. Phys. 58, 1183 (1990); T. Garon
et al.,ibid. 81, 92 (2013).
Suppose that a “nucleus” is an infinitely thin line of positive charge λ per unit length. A two-dimensional
hydrogen atom is formed by a “nucleus” and an electron which is free to move along the line but is confined
in the transverse direction. (This situation is common for, e.g., charged polymer molecules or dislocations in
solids.)
(a) Determine the appropriate set of “atomic units” (Ry 0 , a0B , . . .), describing a bound state in such an atom.
The wavefunction of an electron bound to a line of charge is
Ψ(r⊥ , z) = Aeikz z Φ(r⊥ ),
3
where A is the normalization factor and kz is the wavevector of 1D motion along the line. By Gauss’
theorem, the potential energy of electron at distance r⊥ from the line is
U (r⊥ ) = 2eλ ln
r⊥
,
a
where e > 0 is the magnitude of the electron charge and a is some distance defining the reference point. It
can be chosen to be, e.g., the Bohr radius to be determined later. The potential energy increases toward
the ∞, which means that the electron is bounded. The Schroedinger equation takes the form
−∇2⊥ Φ +
2m
4meλ r⊥
ln 0 Φ = 2
aB
h̄2
h̄
h̄2 kz2
2m 2
E−
Φ = 2 k⊥
Φ.
2m
h̄
The choice of
1
2
(a0B )
=4
meλ
,
h̄2
or
s
a0B =
h̄2
2meλ
makes this equation dimensionless (for an arbitrary normalization of Ψ). The corresponding Rydberg is
Ry 0 =
h̄2
2
m (a0B )
= 2eλ.
Note that Ry 0 does not contain h̄.
A blind application of the rules of dimensional analysis to this problem faces with some difficulties. Indeed,
there are four basic quantities which may enter the answer:
h̄, m, e, λ
so one might require that
L = [h̄]a [m]b [e]c [λ]d .
However, [e] and [λ] are related to each other: [e] = [λ]L. Thus the right-hand-side of the equation can
be re-written in terms of three basic units: m, L, and T (time), which means that we will have three
equations for four exponents–a, b, c, d, and the problem does not have a unique solution. The way to select
an appropriate solution is to notice that e and λ can enter only as a product (as is evident from the
Schroedinger equation). Thus d = c, and now we have three equations for three unknowns.
(b) Using the Bohr-Sommerfeld quantization condition for the radial momentum, find the spectrum of the
bound states. Bohr-Sommerfeld condition reads
Z rt
α
p(r)dr = 2πnh̄,
(2)
0
where rt is the classical turning point, p(r) is the radial momentum, and α is the number of times the
classicalptrajectory traverses the interval (0, rt ). Consider motion with zero angular momentum, when
p(r) = 2m(E − U )/h̄ and α = 2. Writing E as
E = 2eλ ln(rt /a0B )
makes sure that rt is the turning point, when E = U . Then
r
1√
rt
p(r) =
4emλ ln .
h̄
r
4
Integrating in (2) over r with the help of
R1
0
dx ln x = −1, we obtain
πnh̄2
=
rt = √
2 meλ
√
2πn 0
aB .
2
Thus
√
E = 2eλ ln( 2πn).
Note that the constant under the log is not well-defined: it owes its origin to an arbitrary choice of the
reference point at a = a0B . This means that the ground state energy cannot be determined uniquely.
However, the energy difference between any two states is determined uniquely
Em − En = 2eλ ln
m
.
n
3. Problem 3 Two-dimensional Thomas-Fermi atom.
Suppose that more electrons are added to the atom of Problem 2 in such a way that the density of electrons,
n(r⊥ ), satisfy the electroneutrality condition
Z
(3)
e d2 r⊥ n(r⊥ ) = λ.
[n(r⊥ ) is the number of electrons per unit volume which depends only on the coordinate transverse to the line,
r⊥ .]
(a) Construct the Thomas-Fermi model for such an atom.
(b) Find the scaling forms for the electrostatic potential, φ(r⊥ ), and local density, n(r⊥ ).
(c) Determine the Thomas-Fermi radius, a0T F .
Laplace’s equation (m = e = h̄ = 1)
∇2r⊥ φ = 4πn(r⊥ ).
Fermi statistics
3
2
4π (kF (r⊥ ))
= n(r⊥ )
3
3
(2π)
1/3
kF (r) = 3π 2 n(r⊥ )
Total energy
−φ0 =
kF2 (r⊥ )
− φ(r⊥ )
2
Boundary conditions: r → ∞, n(r⊥ ), φ(r⊥ ) → 0 ⇒ φ0 = 0.
kF2 (r⊥ ) = 2φ(r⊥ )
2/3
3π 2 n
= 2φ
23/2 3/2
φ
2
3π
√
8 2
n(r⊥ )
∇2⊥ φ =
3π
n=
The same equation as in 3D, except for the Laplacian is a 2D one and the asymptotic behavior at r → 0 is
different. Choose the scaling form for φ as
φ = λζ(br/aT F )
5
([λ] = [Q] / [L] = [φ]). Substituting this ansatz into the Laplace’s equation, one gets
λb2
a2T F
" √ #3/2
ζ0
8 2
00
ζ +
=
λ3/2 ζ 3/2 ⇒
x
3π
aT F = λ−1/4
" √ #3/4
8 2
b=
3π
ζ 00 +
ζ0
= ζ 3/2
x
For x → 0, equation for ζ reduces to
ζ 00 +
ζ0
=0⇒
x
ζ = ln x
which is the correct result for the potential of a charged line.
Restoring the fundamental units,
h̄6
=
m3 e5 λ
aT F
1/4
.
4. Problem 4 Bound states in 2D
R∞
Find an approximate ground-state energy in a weak, attractive potential U (r) in 2D. Assume that 0 drrU (r) ≤
∞, and that the maximum value of |U | and typical length-scale a of U (r) are such that |Umax | h̄2 /ma2 .
Assume that |E| Ū , where Ū some typical value of U . (This assumption will be justified later.) Within the
radius of the potential well, one can then neglect the term with E in the Schroedinger equation:
dψ<
2m
1 d
r
+ 2 U (r)ψ< = 0.
−
r dr
dr
h̄
Since U is itself weak, one can neglect the variation of ψ within the potential well, and replace ψ< in the U ψ<
term by its value at the origin, which can be chosen to be equal to unity. Then,
1 d
dψ<
2m
−
r
+ 2 U (r) = 0.
r dr
dr
h̄
p
Integrate this equation from 0 to some distance r0 such that a r0 1/k, where k = 2m|E|/h̄. As U falls
off rapidly, the upper limit in the RHS can be replaced by ∞:
Z ∞
dψ< 2m
drrU (r).
(4)
= 2
dr r0
h̄ r0 0
Far away from the potential, the motion is free, hence
1 d
dψ>
−
r
= −k 2 ψ> .
r dr
dr
A solution of this equation, corresponding to the bound state, must fall off at r → ∞. Such a solution is the
modified Bessel function ψ> = K0 (kr). At distances kr 1, the solution reduces to ln kr. Now we can match
0
0
the log derivatives of ψ< (ψ<
/ψ< = ψ<
and ψ> (= 1/r ln(kr)) at r = a. Using Eq. (4), we obtain
1
2m
= 2
ln ka
h̄
Z
∞
drrU (r).
0
6
From here, we find
" Z
−1 #
∞
h̄2
h̄
.
exp
drrU (r)
|E| =
ma2
m
0
Since U < 0, |E| is exponentially small, which justifies our initial assumption.
The
actual small parameter of
R
∞
2
the problem is the combination occurring in the exponential (h̄ /m) 0 drrU (r).