PHYS 771, Quantum Mechanics, Homework 6, Fall 2011
Instructor: Dr. A. G. Petukhov
Merzbacher, Chapter 6, Solutions
Exercise 6.14
The first boundary condition (continuity of ψ) at x = 0 reads:
A+B =C +D
(1)
The second boundary condition is Eq. (6.56) from the book. It can be expressed as:
ik(C − D − A + B) = (2λ/a)(C + D),
(2)
where λ = m|g|a/h̄2 . Solving Eqs. (1) and (2) for A and B we obtain:
A
1 − λ/ika −(λ/ika)
C
=
B
(λ/ika)
1 + λ/ika
D
(3)
Imposing boundary condition D = 0 we obtain the transmission coefficient:
2
C E
1
=
(4)
T = =
2
2
2
A
1 + λ /k a
E + mg 2 /2h̄2
Problem 6.1
Following the same steps as in the book we obtain:
e−2ik1 a
F
=
,
A
cos(2k2 a) − (i/2)[(k12 + k22 )/k1 k2 ] sin(2k2 a)
p
√
where k1 = 2mE/h̄ and k2 = 2m(E − V )/h̄, and
T =
4E(E − V )
4E(E − V ) + V 2 sin2 (2k2 a)
(5)
(6)
The transmission coefficient is shown in Fig. 1.
Problem 6.3
The nature of the boundary conditions (6.56) implies that we have to use
three regions: (1) −∞ < x < −a; (2) −a < x < a; (3) a < x < ∞, and
consider only negative energies. The solutions in all three regions must be
exponentially decaying (evanescent) because all these regions are classically
1
Transmission coefficient
1
0.8
0.6
0.4
0.2
0
1
2
E/V
1.5
2.5
3
Figure 1: Transmission coefficient versus E/V for (2ma2 V )1/2 = (3π/2)h̄.
forbidden for the negative energies. In (1) and (3) the solutions are single
exponents. In region (2) the solution is a superposition of the two exponents.
The wavefunction reads:
ψ1 (x) = Aeκx ,
ψ2 (x) = Beκx + Ce−κx ,
ψ3 (x) = De−κx ,
−∞ < x < −a
−a < x < a
a<x<∞
√
where κ = −2mE/h̄.
As we demonstrated in class, using boundary conditions (6.56) and continuity of the wavefunctions we obtain:
2mg
Be−κa + Ceκa ,
2
h̄
2mg
= 2 Beκa + Ce−κa ,
h̄
2κCeκa =
2κBeκa
x = −a
x=a
This is a homogeneous system of linear equations, which has non-trivial
solution only if its determinant is zero. The two roots of this determinant
correspond to two solutions with even (B = C) and odd (B = −C) parity.
2
This leads to the transcendental equations for the energy eigenvalues:
λ=
ξ
1 ± exp(−2ξ)
(7)
where ”+” and ”-” signs correspond to even and odd solutions respectively.
Here λ and ξ are dimensionless variables:
λ = mga/h̄2 = a/a0
√
ξ = −2mEa/h̄,
(8)
(9)
where a0 = h̄2 /mg is the parameter defining length scale of the problem. We
can plot the solutions of the transcendental Eq. (7), ξ± (λ), by, first, plotting
λ(ξ), then rotating the plot 90◦ counterclockwise and, finally, flipping x-axis,
x → −x.
Using Eqs. (8) and (9)we obtain the energy levels
2
2
h̄2
ξ± (λ)
ξ± (λ)
E± = −
= −E0
,
2ma20
λ
λ
(10)
where ξ± (λ) are the solutions of Eq. (7) and E0 = h̄2 /2ma20 = mg 2 /2h̄2 . The
energy levels versus λ = a/a0 are plotted in Fig. 2. Notably, the odd bound
state exists only for λ > 1/2. In the limit of λ → 0, E+ → −4E0 . Naturally,
this solution corresponds to V (x) = 2gδ(x) when two wells merge. Finally,
in the limit of large λ → ∞ we have two degenerate levels E± = −E0 , which
is the case of the two isolated wells that are very far apart.
In order to establish asymptotic behavior of the energy levels at large λ
we rewrite Eq. (7) as:
ξ = λ(1 ± exp(−2ξ)) ' λ(1 ± exp(−2λ)),
(11)
where we used the fact that for large λ, ξ ' λ and replaced ξ with λ in
the exponent, which is a first iterative correction to ξ = λ in this limit.
Substituting Eq. (11) in Eq. (10) we obtain:
E± ' −E0 (1 ± 2 exp(−2λ))
(12)
The difference between the energy levels reads:
E− − E+ ' 4E0 exp(−2λ) = 4E0 exp(−2a/a0 )
3
(13)
0
E-
E/E0
-1
E+
-2
-3
-4
0.5
1
1.5
a/a0
2
2.5
3
Figure 2: Energy levels versus distance between the centers (solid lines).
Approximate solutions (see Eq. (12)) are shown with dashed lines.
Problem 6.4
This problem is based on the adiabatic (aka Born-Oppenheimer) approximation. Since the nuclei are moving much slower than the electron we can
treat them as static when considering the electron Schcrödinger equation. If
the distance between the nuclei is changed infinitesimally slow it will lead to
the change of the energies E± (a) and will exert the force F = −∂E± (a)/∂a
acting upon the nuclei. Thus E± (a) can be treated as a potential energy of
the nucleus-nucleus attraction caused by the nucleus-electron interaction. In
other words the negatively charged electron works as a ”glue” that keeps two
positively charged nuclei together. To calculate the total potential energy
of the inter-nuclei interaction we have to add the energy of their Coulomb
repulsion UN N ∼ 1/2a. As we have seen in class the addition of this energy
can induce a minimum of E+ (a) for sufficiently small UN N . However E− (a)
will never have a minimum, i.e. this state is always unstable. The total
potential energies of the nuclei in the states with energies E± are plotted in
Fig. 2 with UN N = 0.2a0 /a.
4
0
-0.2
E/E0
-0.4
E-+UNN
-0.6
-0.8
-1
E++UNN
-1.2
-1.4
0.5
1
1.5
a/a0
2
2.5
3
Figure 3: Energy levels versus distance between the centers (solid lines).
Approximate solutions (see Eq. (12)) are shown with dashed lines.
Problem 6.5
The three regions are defined as previously. The wavefunction reads:
ψ1 (x) = Aeikx + Be−ikx
ψ2 (x) = Ceikx + De−ikx
ψ2 (x) = F eikx + Ge−ikx
The first boundary condition (continuity of ψ) at x = −a reads:
Ae−ika + Beika = Ce−ika + Deika
(14)
The second boundary condition is Eq. (6.56) from the book. It can be expressed as:
ik(Ce−ika − Deika − Ae−ika + Beika ) = (2λ/a)(Ce−ika + Deika ),
(15)
where λ = m|g|a/h̄2 . Solving Eqs. (14) and (15) for A and B we obtain:
A
1 − λ/ika
−(λ/ika)e2ika
C
=
(16)
B
(λ/ika)e−2ika
1 + λ/ika
D
5
A similar equation can be obtained for x = a:
C
1 + λ/ika
(λ/ika)e−2ika
F
=
D
−(λ/ika)e2ika
1 − λ/ika
G
Substitution of Eq. (17) in Eq. (18) yields:
A
M11 M12
F
=
,
B
M21 M22
G
(17)
(18)
where matrix M is the product of matrices of Eqs. (18) and (17). We now
impose the boundary condition G = 0, which means that A = M11 F . Thus
for our purposes we need to calculate M11 only:
M11 = 1 +
λ2
λ2 4ika
−
e
k 2 a2 k 2 a2
(19)
The transmission coefficient reads:
T =
1
E2
p
=
,
|M11 |2
E 2 + 4E1 (E + E1 ) sin2 ( E/E0 )
(20)
where E0 = h̄2 /8ma2 and E1 = 4λ2 E0 . The plot of T versus E/E0 is shown
in Fig. 4.
6
Transmission coefficient
1
0.8
0.6
0.4
E1=10E0
E1=20E0
E1=40E0
0.2
0
50
100 150 200
Energy (E/E0)
250
300
Figure 4: Transmission coefficient (Eq. (20)) for different values of E1 .
7