7.6 The Schrodinger equation of electron in a periodic potential

7.6.1. central equation

The Schrodinger equation:

(7.67)-Ñ

2

2 m

¶x2

ΨHxL + UHxL ΨHxL = Ε ΨHxLFor a periodic potential UHxL, we can expand it as a Fourier series

(7.68)UHxL = âG

UG ãä G x

where G = 2 Π n � a is the reciprocal lattice vector and

(7.69)UG =1

aà

0

a

â x UHxL ã-ä G x

Because UHxL is real, it is easy to show that UG* = U-G.

(7.70)UG*

=1

aà

0

a

â x UHxL ã-ä G x

*

=1

aà

0

a

â x UHxL ãä G x

=1

aà

0

a

â x UHxL ã-ä H-GL x

= U-G

We can use these plan waves as our basis, and therefore we can write any wavefunction in the following form

(7.71)ΨHxL = âk

CHkL ãä k x

(7.72)CHkL =1

aà

0

a

â x ΨHxL ã-ä k x

Consider a system with size L and periodic boundary conditions, ΨHx + LL = ΨHxL. With such a boundary condition, we know that the wavevector

for plan waves can only take discrete values

(7.73)k =2 Π

L

n with n = … - 3, -2, -1, 0, 1, 2, 3 …

Thus, the Schordinger equation we wrote above turns into

(7.74)-Ñ

2

2 m

¶x2

ΨHxL + UHxL ΨHxL = Ε ΨHxL

(7.75)

Ñ2

k2

2 m

âk

CHkL ãä k x

+ âG

UG ãä G x â

kCHkL ã

ä k x= Ε â

kCHkL ã

ä k x

(7.76)

Ñ2

k2

2 m

âk

CHkL ãä k x

+ âG

âk

UG CHkL ãä HG+kL x

= Ε âk

CHkL ãä k x

The second term on the l.h.s. has two sums ÚG Úk . Define k ' = G + k and change the sum over k into the sum of k '. As a result, this term becomes

(7.77)âG

âk

UG CHkL ãä HG+kL x

= âG

âk '

UG CHk ' - GL ãä k ' x

= âk

âG

UG CHk - GL ãä k x

So

(7.78)âk

B Ñ2

k2

2 m

CHkL + âk

âG

UG CHk - GL F ãä k x

= âk

Ε CHkL ãä k x

(7.79)âk

B Ñ2

k2

2 m

- Ε CHkL + âk

âG

UG CHk - GL F ãä k x

= 0

For plane waves, if I have a equation

(7.80)âkaHkL ã

ä k x= 0

the only solution to this equation is aHkL = 0 for every k. This is because plane waves with different wave-vectors are linear independent

Xk k '\ = 0.Therefore, if the sum over planes with different k is zero, every term in the sum must be zero. For the Schrodinger equation we

considered above, this means that

Phys463.nb 61

the only solution to this equation is aHkL = 0 for every k. This is because plane waves with different wave-vectors are linear independent

Xk k '\ = 0.Therefore, if the sum over planes with different k is zero, every term in the sum must be zero. For the Schrodinger equation we

considered above, this means that

(7.81)

Ñ2

k2

2 m

- Ε CHkL + âG

UG CHk - GL = 0

In our text book, it is defined Λk = Ñ2

k2 � 2 m, so we get

(7.82)HΛk - ΕL CHkL + âG

UG CHk - GL = 0

This equation is known as the central equation. It is just the Schrodinger equation rewritten in the plane wave basis.

The central equation implies that CHkL is coupled with CHk + GL, which includes CHk ± Π � aL, CHk ± 2 Π � aL, CHk ± 3 Π � aL, …In the same time, it is easy to notice that if k - k ' ¹ G, CHkL and CHk 'L decouple from each other. CHk 'L never appears in any equation of CHkL and

vice versa. In other words, the value of CHkL doesn’t care about the value of CHk 'L. The bottom line: Unless k and k ' differs by a reciprocal lattice vector, CHkL doesn't care about the value of CHk 'L.

7.6.2. Solutions of the central equation: Bloch waves

Here, let’s focus on the momentum k. The central equation tells us that to compute CHkL, we will also need to consider CHk ± Π � aL, CHk ± 2 Π � aL,CHk ± 3 Π � aL, … , because they all couples together in the central equation.

In the same time, we set CHk 'L = 0 for any k ' ¹ k + G. Since the value of CHk 'L has no influence on the value of CHkL, we can set CHk 'L = 0 without

loss of generality.

(7.83)ΨHxL = âk

CHkL ãä k x

= âG

CHk - GL ãä Hk-GL x

= IâG

CHk - GL ã-ä G xM ã

ä k x

Define

(7.84)ukHxL = âG

CHk - GL ã-ä G x

So we get

(7.85)ΨHxL = ukHxL ãä k x

It is easy to prove that ukHxL is a periodic function ukHx + TL = ukHxL where T is a lattice vector

(7.86)ukHx + TL = âG

CHk - GL ã-ä GHx+TL

= âG

CHk - GL ã-ä G x

ã-ä G T

= âG

CHk - GL ã-ä G x

Here we used the fact that G T = 2 Π n, so ã-ä G x = 1.

If we recall the definition of Bloch waves

(7.87)Ψn,kHxL = un,kHxL ãä k x

we find immediately that the solution we show here is exactly the Bloch waves.

62 Phys463.nb

7.6.3. Solutions of the central equation: perturbation theory

(7.88)Λk CHkL + âG

UG CHk - GL = Ε CHkLWe can write it in a matrix form

(7.89)

B

… … … … … … …

… Λk-4 Π�a 0 0 0 0 …

… 0 Λk-2 Π�a 0 0 0 …

… 0 0 Λk 0 0 …

… 0 0 0 Λk+2 Π�a 0 …

… 0 0 0 0 Λk+4 Π�a …

… … … … … … …

+

… … … … … … …

… 0 U-2 Π�a U-4 Π�a U-6 Π�a U-8 Π�a …

… U2 Π�a 0 U-2 Π�a U-4 Π�a U-6 Π�a …

… U4 Π�a U2 Π�a 0 U-2 Π�a U-4 Π�a …

… U6 Π�a U4 Π�a U2 Π�a 0 U-2 Π�a …

… U8 Π�a U6 Π�a U4 Π�a U2 Π�a 0 …

… … … … … … …

F

…

CHk - 4 Π � aLCHk - 2 Π � aL

CHkLCHk + 2 Π � aLCHk + 4 Π � aL

…

=

Ε

…

CHk - 4 Π � aLCHk - 2 Π � aL

CHkLCHk + 2 Π � aLCHk + 4 Π � aL

…

Here we set that UG=0 =1

aÙ

0

aâ x UHxL = 0.

This is an eigen-equation

(7.90)HH0 + H1L

…

CHk - 4 Π � aLCHk - 2 Π � aL

CHkLCHk + 2 Π � aLCHk + 4 Π � aL

…

= Ε

…

CHk - 4 Π � aLCHk - 2 Π � aL

CHkLCHk + 2 Π � aLCHk + 4 Π � aL

…

where H0 is a diagonal matrix and H1 is an off-diagonal matrix. Here H0 comes from the kinetic energy Λk = k2 � 2 m and H1 comes from the

potential energy UG.

If we consider a very weak potential HH1 << H0L, we can treat H0 as an “unperturbed Hamiltonian” and H1 as a small perturbation.

To the zeroth order, the eigen-energies (Ε) are just the eigenvalues of the H0 matrix. Because H0 is diagonal, its eigenvalues are just the diagonal

component Εn,k = Ñ2Hk - 2 Π n � aL2 �2 m. This is just the free fermion p

2 � 2 m dispersion folded into the first BZ.

Phys463.nb 63

Fig. 6. Reduced zone for a free particle. (textbook p 177 Fig. 8). This is also Εk to the zeroth order in the perturbation theory.

7.6.4. non-degenerate perturbation

Here, we assume that all eigenvalues of H0 are different from one another (non-degenerate). This condition is valid for most k points away from

the zone boundary. For this situation, one needs to use non-degenerate perturbation theory.

To the first order, the correction to the eigen energy is the same diagonal component of the H1 matrix. But we know that all diagonal terms of H1

are 0. So the first order correction is always 0.

To the second order, the correction is in the order of OHU2L.So we can write Εk = Λk + OHUG

2L

7.6.5. degenerate perturbation: near zone boundary

The momentum at zone boundary is k = G � 2, where G is a reciprocal lattice vector. For k = G � 2, it is easy to notice that

(7.91)ΛHkL = ΛHk - GLHere, the l.h.s. is ΕHk = G � 2L, while the r.h.s. is ΕHk - G = -G � 2L. Because ΛHkL = Ñ

2k

2 � 2 m is an even function of k, the value of Λ at G � 2 and

-G � 2 is the same.

This means that for the unperturbed Hamiltonian HH0L there are two degenerate eigen-states. For these two states, we should us degenerate

perturbation theory, instead of the non-degenerate perturbation theory. To the leading order, the eigen equation is

(7.92)ΛG�2 U-G

UG Λ-G�2 K CHG � 2LCH-G � 2L O = ΕK CHG � 2L

CH-G � 2L OFor this equation, it is easy to solve. The eigen-energy is

(7.93)Ε± = ΛG�2 ± UG

At zone boundary, there are two possible energies. In other words, the band opens a gap and the gap is D = Ε+ - Ε- = 2 UG

64 Phys463.nb