The Kronig-Penney one-dimensional model
Purpose: to demonstrate that in solids, where many atoms stay
closely, the interference between atoms will create allowed and
forbidden bands of energy for electrons.
To simplify the analysis, we only consider a one-dimensional
system where atoms are aligned and equally spaced. This
constructs a one-dimensional potential function:
V(x)
V0
II I II I II I II
x
-(b+a) -b 0 a (b+a)
Where V0 is the value of potential barrier; a and b are lattice
constant, represent distance between atoms.
For an electron traveling in the x-direction in free-space, the
general solution of the wave equation is,
! ( x) = exp( jkx )
Now, within this periodic potential structure, the solution should
be modified,
! ( x) = u ( x) exp( jkx )
Bring this assumed solution back to the Schrodinger equation,
d 2" ( x) 2m
2
+ 2 [E ! V ( x)]" ( x) = 0
dx !
In region I, where V(x) = 0, we have,
d 2u1 ( x) du1 ( x)
2
+ 2 jk − ( k 2
− α 2
)u1 ( x) = 0 (1)
dx dx
where α 2 = 2mE / 2
In region II, where V(x) = V0, we have,
d 2u 2 ( x ) du2 ( x)
2
+ 2 jk − (k 2 − β 2 )u2 ( x) = 0 (2)
dx dx
2m 2mV0
where β 2
= (E − V 0 ) = α 2
−
2 2
Equations (1) and (2) are two new equations for envelop u1(x) and
u2(x) in regions I and II, respectively.
The general solutions for (1) and (2) is,
u1 ( x) = Ae j (α − k ) x + Be − j (α + k ) x For region I (0 < x < a)
u2 ( x) = Ce j ( β − k ) x + De− j ( β + k ) x For region II (-b < x <
0)
Boundary conditions:
Field continuity u1 (0) = u2 (0)
du1 du 2
=
dx x =0 dx x =0
Periodic structure u1 (a) = u2 (−b)
du1 du 2
=
dx x=a dx x = −b
This results in 4 equations for coefficients A, B, C, and D,
A+ B −C − D = 0
(α − k ) A − (α + k ) B − ( β − k )C + ( β − k ) D = 0
Ae j (α − k ) a + Be − j (α + k ) a − Ce − j ( β − k )b − De j ( β + k )b = 0
(α − k ) Ae j (α −k ) a − (α + k ) Be − j (α +k ) a − ( β − k )Ce − j ( β −k )b + ( β + k ) De j ( β +k )b = 0
In order to have nontrivial solutions for A, B, C, and D, the
determinant must be zero. That is
1 1 −1 −1
(α − k ) − (α + k ) − (β − k ) (β + k )
=0
e j (α − k ) a e − j (α + k ) a e − j ( β − k )b e j ( β + k )b
(α − k )e j (α − k ) a − (α + k )e − j (α + k ) a − ( β − k )e − j ( β − k ) b ( β + k )e j ( β + k ) b
This is equivalent to,
(α 2 + β 2 )
− sin(αa) sin( βb) + cos(αa) cos( βb) = cos k (a + b)
2αβ
We are mostly interested in the case of V0 > E (electrons are
bounded inside the crystal structure). In this case,
2m 2m
β2 = 2
(E − V0 ) < 0 and β = j 2
(V0 − E ) = jγ
where γ is real and,
(γ 2 − α 2 )
sin(αa) sinh(γb) + cos(αa) cosh(γb) = cos k (a + b)
2αγ
To further simplify the analysis, we assume δ-type potential
barriers with V0 = ∞ , b = 0 and V0b = u, which is a constant,
2m 2m u
Then, γb = b 2
V0 = b 2
→0
V0
cosh(γb) → 1 and sinh(γb) → γb
γ 2 −α 2 γ 1 2m mu
sinh(γb) → γb = V0 b =
2αγ 2α 2α 2 α 2
Therefore, we have,
mV0ba sin(αa)
+ cos(αa) = cos( ka)
2
αa
sin(αa)
or, M + cos(αa) = cos( ka) (3)
αa
mV0ba
where M ≡
2
On the right-hand-side of equation (3),
− 1 < cos(ka) < 1
While on the left-hand-side of equation (3), the value of
sin(αa )
M + cos(αa ) is not bounded within ±1.
αa
Therefore, in order to have non-trivial solution of equation (3),
the parameter α = 2mE / or ultimately the electron energy E
2
only has certain allowed values, while other values are forbidden.
This gives an explanation of allowed and forbidden energy bands:
sin(αa )
Allowed energy band: M + cos(αa ) ≤ 1
αa
sin(αa )
Forbidden energy band: M + cos(αa ) > 1
αa
sin(αa)
Although M + cos(αa) = cos( ka) can be solved
αa
numerically, we only look at two extreme cases:
(1) No periodic potential barrier V0 = 0 or V0b = 0 and M=0,
Equation (3) becomes, cos(αa) = cos(ka)
Therefore α =k
E
2 2
2mE k
that is, = k or, E =
2m
Obviously any E-value is allowed, no k
restriction.
(2) Very high periodic potential barrier V0b >> 1 and therefore, M
>> 1.
sin(αa )
Equation (3) becomes, M = cos( ka)
αa
Since M >>1, the solutions can only be found around sin(αa) = 0 ,
or,
2mE nπ
α≡ =± with n = 1, 2, 3….
a
That is,
2
⎛ nπ ⎞
2
E =⎜ ⎟ with n = 1, 2, 3….
⎝ a ⎠ 2m
Obviously, E has only discrete values. E
