Solid State Physics: Essential Concepts (2nd Edition, 2020) – David W. Snoke – Solutions Manual – PDF Download

ALL 11 CHAPTERS COVERED




SOLUTIONS MANUAL

,1 Solutions to Chapter 1 Exercises

1.1

Exercise 1.1.1 According to the prescription of the text, we assume that the solutions in the two
regions take the form


Ψ1 (x) = A1 sin(Kx) + B1 cos(Kx) , (1.1.1)

and
Ψ2 (x) = A2 eκx + B2 e−κx . (1.1.2)

There is no need to write down the solution for Ψ3 (x), since the complete solution is either
symmetric, in which case the derivative of Ψ2 (x) vanishes at x = 0, or antisymmetric, therefore,
Ψ2 (x) itself vanishes at x = 0.
The boundary conditions of the problem are
       
b b b
Ψ1 x = −a − = 0 −→ A1 sin K −a − + B1 cos K −a − = 0 . (1.1.3)
2 2 2

         
b b b b b b
Ψ1 x=− = Ψ2 x=− −→ A1 sin K − +B1 cos K − = A2 e−κ 2 +B2 eκ 2
2 2 2 2
(1.1.4)

     
Ψ1 Ψ2 b b b b
= −→ A1 K cos K − −B1 K sin K − = A2 κeκ 2 −B2 κe−κ 2 ,
dx x=− 2b dx x=− 2b 2 2
(1.1.5)
and finally,
Ψ2 (x = 0) = 0 −→ A2 + B2 = 0 (1.1.6)

i.e., A2 = −B2 for the antisymmetric case, or
Ψ2
= 0 −→ A2 κ − B2 κ = 0 , (1.1.7)
dx x=0

i.e., A2 = B2 for the symmetric case. Regarding the amplitudes A1 , B1 , A2 and B2 as the
unknowns, we get a homogeneous set of four linear equations, which means that in order to
1

,2 Solutions to Chapter 1 Exercises


have a solution, the determinant of the corresponding matrix must vanish. From this condition,
we arrive at the equation
κb κb
κ sin aK cosh + K cos aK sinh =0 (1.1.8)
2 2
for the antisymmetric case, and
κb κb
κ sin aK sinh + K cos aK cosh =0 (1.1.9)
2 2
bκ bκ
for the symmetric one. Dividing by cos aK and cosh , or sinh , respectively, we get
2 2
bκ
tan aK tanh
=− 2 (1.1.10)
K κ

bκ
tan aK coth
=− 2 (1.1.11)
K κ
for the antisymmetric and symmetric case, respectively.
We have an additional equation, which links K and κ. Namely,
h̄2 K 2
=E (1.1.12)
2m
2 2
h̄ κ
= U0 − E , (1.1.13)
2m
i.e.,
r
2mU0
κ= − K2 . (1.1.14)
h̄2
Then the two equations above, Eqs. (1.1.10-1.1.11), lead to two equations for K, where a, b and
2mU0 /h̄2 play the role of parameters. Let us note that in Eq. (1.1.14), κ becomes imaginary,
when h̄2 K 2 /2m > U0 . This means that in that case, we have real sine and cosine solutions in
the barrier, which is a simple consequence of the fact that the particle’s energy is larger than
the “confining” potential, i.e., the particle is not bound in that region.
The attached Mathematica code contains the derivation and the graphical solutions of the
two equations above. A typical case is shown in Fig. 1.1, where a = 1, 2mU0 /h̄2 = 100, and
b = 0.1, or b = 0.02. We plotted only K > 0, since the equations are invariant under the
transformation K ↔ −K. By trying various values for b, we notice that as we increase b, the
energy of the symmetric solution drops rapidly, while that of the antisymmetric is more or less
constant. In particular, when b → 0, the right hand side of Eq. (1.1.11) tends to −∞, which
means that the first solution will be at K = π/2. At the same time, the right hand side of
Eq. (1.1.10) tends to 0, i.e., all solutions of that equation will be at integer multiples of π.
This immediately answers the second question of the problem, because the wave number of
the symmetric solution is exactly half of that of the asymmetric solution. Also, if we keep the
thickness of the barrier constant, and increase the potential, the two solutions separate more
and more.

, 3



1 1



0 0



-1 -1



-2 -2
0 2 4 6 8 10 0 2 4 6 8 10
k [a.u.] k [a.u.]
t
Fig. 1.1 The functions in Eqs. (1.1.10)-(1.1.11) of Exercise 1.1.1 for a = 1, b = 0.1, and
2mU0 /h̄2 = 100 on the left hand side, and for a = 1, b = 0.02, and 2mU0 /h̄2 = 100
on the right hand side. The common left hand side of the equations is shown in solid
red; the long-dashed green is the right hand side of Eq. (1.1.10) for the asymmetric
solution; while the short-dashed blue line is the right hand side of Eq. (1.1.11) for the
symmetric solution.


Next, we calculate the first roots of Eqs.(1.1.10)-(1.1.11) as a function of the barrier width,
b. We see from Fig. 1.1 that both of these roots are in the interval [π/2, π], and no other roots
are to be found there. This means that we can use this interval for bracketing the solutions.
The results are shown in Fig. 1.2. We can notice that for b → 0, we indeed have a factor of 2 in
the values of the wave number, while for b → ∞, the two energies will virtually be the same.
This behavior can be understood, if we notice that as b → ∞, the overlap of the wavefunctions
in the central region goes to zero, so the solutions become decoupled.
Once we have the value of K, we can solve for A1 , B1 , A2 and B2 , which give the wavefunc-
tions. Two typical solutions are shown in Fig. 1.3, for a = 1, 2mU0 /h̄2 = 100, and b = 0.1, or
b = 0.3.
Mathematica code:

(* ========= Matrix describing the antisymmetric configuration ======== *)
MatAnti := {{Sin[k*(-a - b/2)], Cos[k*(-a - b/2)], 0, 0}, {Sin[-k*b/2],
Cos[-k*b/2], -Exp[-kappa*b/2], -Exp[kappa*b/2]}, {k*Cos[-k*b/2], -k*
Sin[-k*b/2], -kappa*Exp[kappa*b/2], kappa*Exp[-kappa*b/2]}, {0, 0, 1,
1}}
MatrixForm[MatAnti]
DetMat = Det[MatAnti]
DetMat2 = FullSimplify[DetMat]

(* =========== Matrix describing the symmetric configuration ========== *)
MatSim := {{Sin[k*(-a - b/2)], Cos[k*(-a - b/2)], 0, 0}, {Sin[-k*b/2],
Cos[-k*b/2], -Exp[-kappa*b/2], -Exp[kappa*b/2]}, {k*Cos[-k*b/2], -k*
Sin[-k*b/2], kappa*Exp[kappa*b/2], -kappa*Exp[-kappa*b/2]}, {0, 0,
kappa, -kappa}}
MatrixForm[MatSim]