Transcript FREE ELECTRON THEORY - West Virginia University
For review/introduction of Schrodinger equation: http://web.monroecc.edu/manila/webfiles/spiral/6schrodingereqn.pdf
Chapter 8 and 9 lectures run together a little Bloch’s Theorem and Krönig-Penney Model
Learning Objectives for Today
After today’s class you should be able to: Calculate free electron energy bands and the number of k states in the bands Apply Bloch’s theorem to the Kronig- Penney model or any other periodic potential Explain the meaning and origin of bands and “forbidden band gaps” Another source on today’s topics, see Ch. 7 of Kittel or search Bloch theorem in others
Revisiting the Quantum Metal
H
0 14 18 17 16 15 5 4 3 2 1 0 (
k
) 13 12 11 10 9 Inside U=0 7 U 6 Fill lowest L
h
2 2
k
2 (3,2,1)
m
Energy of “free” electrons
Will the band fill up?
The# of allowed k states (dots) is equal (2,2,1) to the number of primitive cells in the crystal (1,1,1) Consider examples Cube V=L 3 – /a E /a k
Why have I drawn this?
Looking at the energy distribution of multiple electrons (
k
)
h
2 (
k
2 E 2
m
/
a
) 2
nk
More generically:
h
2 (
k
2
n
/
a
) 2 2
m
(
k
) – /a
h
2
k
2 2
m
/a
nk
nk
(
r
)
n
,
k
K
n
,
k
K
(
r
) k
Free Electron Brillouin Zone
Common Methods of Representing the Dispersion Relation E(k) 3
k
2 1
k k
Repeated Zone Scheme contains a lot of redundant info First Brillouin Zone
Why does this dispersion relation look different?
fcc in real space # of nearest neighbors Nearest-neighbor distance # of second neighbors Second neighbor distance 8 ½ a 3 6 a 12 a/ 2 6 a
“Realistic” Potential in Solids
Multi-electron atomic potentials are complex Even for hydrogen atom with a “simple” Coulomb potential solutions are quite complex U 0 U=0 L
U
(
x
)
Bringing Atoms Close Together in a Periodic Fashion
For one dimensional case where atoms (ions) are separated by distance a, we can write the condition of periodicity as
U
(
x
)
U
(
x
a
n
)
a a
Bloch electrons
are electrons that obey the 1D Schrodinger equation with a periodic potential.
They reduce to free electrons if you take U(x) = 0 = U(x+an)
Bloch’s Theorem
This theorem gives the electron wavefunction in the presence of a periodic potential energy. We will prove 1-D version, AKA Floquet’s theorem.
(3D proof in the book) When using this theorem, we still use the time-indep. Schrodinger equation for an electron in a periodic potential 2 2
m
2
U
(
r
)
E
where the potential energy is invariant under a lattice translation of a In 3D (vector):
R
U u
(
a x
)
v b
U
(
x w
c
a
n
)
T R U
(
r
)
U
(
r
R
)
Bloch Wavefunctions
a
Bloch’s Theorem states that for a particle moving in the periodic potential, the wavefunctions
ψ nk
(x) are of the form
nk
(
x
)
u nk
(
x
)
e ikx
,
where u nk u nk
(
x
)
u nk
(
x
)
is a periodic function
(
x
a
)
u nk
(x) has the periodicity of the atomic potential The exact form of u(x) depends on the potential associated with atoms (ions) that form the solid
Main points in the proof of Bloch’s Theorem in 1-D
nk
(
x
)
u nk
(
x
)
u nk
(
x
)
e ikx
,
u nk
(
x
a
) 1. First notice that Bloch’s theorem implies:
n k
(
r
R
)
u n k
(
r
R
)
e i k
r
e i k
R
u n k
(
r
)
e i k
r
e i k
R
n k
(
r
)
e i k
R
Or just:
n k
(
r
R
)
n k
(
r
)
e i k
R
Can show that this formally implies Bloch’s theorem, so if we can prove it we will have proven Bloch’s theorem.
2. To prove the statement shown above in 1-D: Consider N identical lattice points around a circular ring, each separated by a distance a. Our task is to prove: 3 2 1 N (
x
a
) (
x
)
e ika
Built into the ring model is the periodic boundary condition: (
x
Na
) (
x
)
Proof of Bloch’s Theorem in 1-D: Conclusion
3 The symmetry of the ring implies that we can find a solution to the wave equation (QM reason too): If we apply this translation N times we will return to the initial atom position: (
x
Na
) (
x
a
)
c
(
x
)
c N
(
x
) (
x
) 2 This requires 1 N
c N
1 And has the most general solution:
c N
e
2
ni n
0 , 1 , 2 ,...
Or:
c
e
2
ni
/
N
e ika
Where we define the Bloch wavevector:
k
2
n Na
Now that we know C we can rewrite (
x
a
)
c k
(
x
)
e ika
(
x
)
Q
.
E
.
D
.
Consequence of Bloch’s Theorem Probability of finding the electron
P
(
x
)
P
(
x
a
)
Each electron in a crystalline solid “belongs” to each and every atom forming the solid
Very accurate for metals where electrons are free to move around the crystal!
Understanding the notation of the second proof of Bloch’s theorem
Another proof unnecessary, but the notation present will come up again (
r
N i a
i
) (
r
) Known as the Born-von Karman boundary condition As in Phys 314, you can always expand a wave packet in terms of plane waves You can expand any function obeying this condition as the set of plane waves that satisfy the B.C.s
Ψ 𝑟 = 𝑞 𝑐 𝑞 𝑒 𝑖 𝑞 𝑟 The potential U obeys the Born-von Karman condition
Understanding the notation of the second proof of Bloch’s theorem
The potential U(r)=U(r+Na) obeys the Born-von Karman condition U 𝑟 = 𝐾 𝑈 𝐾 𝑒 𝑖 𝐾 𝑟 Ψ 𝑟 = 𝑞 𝑐 𝑞 𝑒 𝑖 𝑞 𝑟 U K U K = U -K are the Fourier coefficients of U if the crystal has inverse symmetry Plugging U into the Schrodinger’s Equations gives
H
(
h
2 2
m
2
U
)
E
h
2 2
m
2 (
q c q e iq
r
)
Understanding the notation of the second proof of Bloch’s theorem
U 𝑟 = 𝐾 𝑈 𝐾 𝑒 𝑖 𝐾 𝑟 Ψ 𝑥, 𝑡 = 𝑞 𝑐 𝑞 𝑒 𝑖 𝑞𝑥−𝜔𝑡
U
(
U K e iK
r
)(
c q e iq K q
Combining sums and then defining q=q’-K
r
Kq U K c q e i
(
K
q
)
r
Kq
'
U K c q
'
K e iq
'
r
) Some more manipulation gives an alternate form of S.E.
h
2 ( 2
m
(
k
K
) 2 )
c k
K
'
K U K
'
K c k
K
' 0
It looks more complicated that it is.
Schrodinger’s equation in momentum space simplified by condition of periodicity 2 ( 2
h m
(
k
K
) 2 )
c k
K
K
'
U K
'
K c k
K
' 0 In the free electron case, all U K are 0, so this simplifies:
h
2 ( 2
m
(
k
K
) 2 )
c k
K
0 (
k
0
K
)
c k
K
We will look at more complicated examples in the next chapter.
Propagation on a crystal lattice To understand some generic features of electron conduction on a crystal lattice, let’s model a 1D crystal, i.e. a lattice with a periodic potential.
V
(
x
) Ion core
a x
The exact shape of the periodic potential will not matter.
.
Propagation on a crystal lattice
But the exact shape doesn’t matter, so let’s try something easier!
V
( (
x
) )
x d
Before we do a whole crystal, let’s remind ourselves how to deal with step
𝑈(𝑥) 𝐸 𝑈 0 How to determine transmission probability?
𝑥 I 0 II Solve time-independent Schrodinger equation to find 𝜓 𝑥 .
ℏ 2 − 2𝑚 𝑑 2 𝜓 𝑥 𝑑𝑥 2 𝑑 2 𝜓 𝑥 𝑑𝑥 2 = − + 𝑈 𝑥 𝜓 𝑥 = 𝐸𝜓 𝑥 2𝑚 ℏ 2 𝐸 − 𝑈 𝑥 𝜓 𝑥 𝑑 2 𝜓 𝑥 𝑑𝑥 2 Region I : = − 2𝑚𝐸 ℏ 2 𝜓 𝑥 Where Ψ 𝑥, 𝑡 = 𝜓 𝑥 𝑒 −𝑖 𝑑 2 𝜓 𝑥 𝑑𝑥 2 Region II : = − 2𝑚 ℏ 2 𝐸 − 𝑈 0 𝜓 𝑥
Transmission Probability
𝑈(𝑥) How to determine transmission probability?
𝐸 𝑈 0 Solve time-independent Schrodinger equation to find 𝜓 𝑥 𝑥 0 I II 𝑑 2 𝜓 𝑥 𝑑𝑥 2 Region I : = − 2𝑚𝐸 ℏ 2 𝜓 𝑥 𝑑 2 𝜓 𝑥 𝑑𝑥 2 Region II : = − 2𝑚 ℏ 2 𝐸 − 𝑈 0 𝜓 𝑥 𝑘 1 2 Ψ 1 𝜓 1 𝑥 = 𝐴𝑒 𝑖𝑘 1 𝑥 + 𝐵𝑒 𝑥, 𝑡 = 𝐴𝑒 𝑖 𝑘 1 𝑥−𝜔𝑡 −𝑖𝑘 1 + 𝐵𝑒 𝑥 −𝑖 𝑘 1 𝑥+𝜔𝑡 𝑘 2 2 Ψ 2 𝜓 2 𝑥 = 𝐶𝑒 𝑖𝑘 2 𝑥 + 𝐷𝑒 𝑥, 𝑡 = 𝐶𝑒 𝑖 𝑘 2 𝑥−𝜔𝑡 −𝑖𝑘 2 + 𝐷𝑒 𝑥 −𝑖 𝑘 2 𝑥+𝜔𝑡
Region I
The general solution is: Ψ 1 𝑥, 𝑡 = 𝐴𝑒 𝑖 𝑘 1 𝑥−𝜔𝑡 + 𝐵𝑒 −𝑖 𝑘 1 𝑥+𝜔𝑡 Concept Test: What do the terms starting with 𝐴 A.
B.
C.
D.
𝐴 𝐴 𝐴 𝐴 is kinetic energy, 𝐵 is the potential energy.
is a wave traveling to the right, 𝐵 is a wave traveling to the left, is a standing wave peak, 𝐵 and 𝐵 𝐵 represent, physically?
is a wave traveling to the left.
is a wave traveling to the right.
is a standing wave trough.
E. These terms have no physical meaning.
𝑈(𝑥) 𝐸 𝑈 0 𝑥 I 0 II
Coefficients
𝑈(𝑥) 𝐴 𝐸 𝑈 0 𝐵 𝑒 − 𝑥 I 0 II 𝑑 2 𝜓 𝑥 𝑑𝑥 2 Region II : = − 2𝑚 ℏ 2 𝐸 − 𝑈 0 𝜓 𝑥 𝐶 𝐷 Ψ 1 𝑑 2 𝜓 𝑥 𝑑𝑥 2 Region I : = − 2𝑚𝐸 ℏ 2 𝜓 𝑥 𝑘 1 2 𝑥, 𝑡 = 𝐴𝑒 𝑖 𝑘 1 𝑥−𝜔𝑡 + 𝐵𝑒 −𝑖 𝑘 1 𝑥+𝜔𝑡 Ψ 2 𝑘 2 2 𝑥, 𝑡 = 𝐶𝑒 𝑖 𝑘 2 𝑥−𝜔𝑡 + 𝐷𝑒 −𝑖 𝑘 2 𝑥+𝜔𝑡 Right-going Left-going Right-going What do each of these waves represent?
A.
𝐴 = 𝐵 = 𝐶 = left reflected transmitted incoming from 𝐷 = right incoming from B.
𝐴 = 𝐵 = 𝐶 = left transmitted reflected incoming from 𝐷 = right incoming from C.
𝐴 = left incoming from 𝐵 = 𝐶 = reflected transmitted 𝐷 = right incoming from Left-going D.
𝐴 = 𝐵 = 𝐶 = 𝐷 = right incoming from left transmitted reflected incoming from
Coefficients
𝑑 2 𝜓 𝑥 𝑑𝑥 2 Region I : = − 2𝑚𝐸 ℏ 2 𝜓 𝑥 Ψ 1 𝑘 1 2 𝑥, 𝑡 = 𝐴𝑒 𝑖 𝑘 1 𝑥−𝜔𝑡 + 𝐵𝑒 −𝑖 𝑘 1 𝑥+𝜔𝑡 𝑈(𝑥) 𝐴 𝐸 𝑈 0 𝐵 𝑥 I 0 II 𝑑 2 𝜓 𝑥 𝑑𝑥 2 Region II : = − 2𝑚 ℏ 2 𝐸 − 𝑈 0 𝜓 𝑥 𝐶 𝐷 Ψ 2 𝑘 2 2 𝑥, 𝑡 = 𝐶𝑒 𝑖 𝑘 2 𝑥−𝜔𝑡 0 + 𝐷𝑒 −𝑖 𝑘 2 𝑥+𝜔𝑡 Incoming from left Reflected Transmitted Incoming from right Boundary and initial conditions determine values of coefficients.
Initial conditions: electron incoming from left Boundary conditions: Continuity 𝜓 1 0 = 𝜓 2 0 𝐷 = 0 Smoothness 𝑑𝜓 2 𝑑𝑥 𝑥=0 𝑑𝜓 1 𝑑𝑥 𝑥=0 = 𝐴 + 𝐵 = 𝐶 𝑖𝑘 1 (𝐴 − 𝐵) = 𝑖𝑘 2 𝐶
Transmission
Region I : Ψ 1 𝑥, 𝑡 = 𝐴𝑒 𝑖 𝑘 1 𝑥−𝜔𝑡 + 𝐵𝑒 −𝑖 𝑘 1 𝑥+𝜔𝑡 𝐴 𝐵 I 𝑈(𝑥) 0 Region II : II Ψ 2 𝑥, 𝑡 = 𝐶𝑒 𝑖 𝑘 2 𝑥−𝜔𝑡 𝐸 𝑈 0 𝑥 𝐶 Incoming from left Reflected Transmitted Concept Test: If an electron comes in with an A.
Reflected 𝐵 amplitude 𝐴 , what is the probability that it is reflected or transmitted?
Transmitted 1 − 𝐵 𝑃 reflect. = reflected prob. density incident prob. density B.
C.
D.
E.
𝐵/𝐴 𝐵 𝐵 𝐵 2 2 /𝐴 2 2 𝐴 2 1 − 𝐵/𝐴 1 − 𝐵 1 − 𝐵 1 − 𝐵 2 2 /𝐴 2 2 𝐴 2 𝑃 reflect. = 𝑃 reflect. = 𝐵 𝐴 Ψ Ψ ∗ ∗ ∗ ∗ 𝐵 𝐴 Ψ refl.
Ψ inc.
= 𝐵 𝐴 2 2 = 𝑅
Trans./Refl. Probabilities
Use boundary conditions to find 𝑅 and 𝑇 .
Continuity 𝜓 1 0 = 𝜓 2 0 𝐴 + 𝐵 = 𝐶 Smoothness 𝑑𝜓 1 𝑑𝑥 𝑥=0 = 𝑖𝑘 1 (𝐴 − 𝐵) = 𝑖𝑘 2 𝐶 𝑑𝜓 2 2 equation, 3 unknowns!?
𝐵 𝐴 and 𝐶 𝐴 . 𝐵 = 𝐶 − 𝐴 𝑖𝑘 1 (𝐴 − (𝐶 − 𝐴)) = 𝑖𝑘 2 𝐶 𝐵 = 𝑘 1 2𝑘 1 + 𝑘 2 𝐴 − 𝐴 2𝑖𝑘 1 𝐴 − 𝑖𝑘 1 𝐶 = 𝑖𝑘 2 𝐶 2𝑘 1 𝐶 = 𝑘 1 + 𝑘 2 𝐴 𝑘 1 − 𝑘 2 𝐵 = 𝐴 𝑘 1 + 𝑘 2 Transmission and Reflection coefficients: 𝐵 𝐴 2 2 = 𝑇 = 1 − 𝑘 1 𝑘 1 − 𝑘 2 + 𝑘 2 𝐵 2 𝐴 2 = 𝑘 1 4𝑘 1 2 2 𝑘 + 𝑘 2 2 2
Concept Test
An electron approaches the end of a long wire 𝑒 − 𝑈(𝑥) 𝑈 0 𝐸 𝑥 0 Concept Test: If the total energy, 𝐸 , of the electron is
less
than the work function of the metal, 𝑈 0 , when the electron reaches the end of the wire, it will… A. …stop.
B. …be reflected.
C. …exit the wire and keep moving to the right.
D. …either be reflected or exit the wire with some probability.
Wavefunction
𝑈(𝑥) I 0 𝑑 2 𝜓 𝑥 𝑑𝑥 2 𝑈 0 𝐸 𝑥 II Region I : Solve time-independent Schrodinger equation to find 𝜓 𝑥 Region II : = − 2𝑚𝐸 ℏ 2 𝜓 𝑥 𝑑 2 𝜓 𝑥 𝑑𝑥 2 = − 2𝑚 ℏ 2 𝐸 − 𝑈 0 𝜓 𝑥 𝑘 1 2 𝛼 2 > 0 Ψ 1 𝜓 1 𝑥 = 𝐴𝑒 𝑖𝑘 1 𝑥 + 𝐵𝑒 𝑥, 𝑡 = 𝐴𝑒 𝑖 𝑘 1 𝑥−𝜔𝑡 −𝑖𝑘 1 + 𝐵𝑒 𝑥 −𝑖 𝑘 1 𝑥+𝜔𝑡 Ψ 2 𝜓 2 𝑥 = 𝐶𝑒 −𝛼𝑥 + 𝐷𝑒 𝑥, 𝑡 = 𝐶𝑒 −𝛼𝑥−𝑖𝜔𝑡 𝛼𝑥 + 𝐷𝑒 𝛼𝑥−𝑖𝜔𝑡 Note, 𝛼 2 is like −𝑘 except this time 𝑘 2 2 2 from before, is purely imaginary, while 𝛼 is purely real
Coefficients
𝑈(𝑥) 𝐴 𝑑 2 𝜓 𝑥 𝑑𝑥 2 𝐵 𝑈 0 𝐸 𝑥 I 0 Region II : II = − 2𝑚 ℏ 2 𝐸 − 𝑈 0 𝜓 𝑥 𝐶 𝐷 𝑑 2 𝜓 𝑥 𝑑𝑥 2 Region I : = − 2𝑚𝐸 ℏ 2 𝜓 𝑥 𝑘 1 2 Ψ 1 𝑥, 𝑡 = 𝐴𝑒 𝑖 𝑘 1 𝑥−𝜔𝑡 + 𝐵𝑒 −𝑖 𝑘 1 𝑥+𝜔𝑡 𝛼 2 Ψ 2 𝑥, 𝑡 = 𝐶𝑒 −𝛼𝑥−𝑖𝜔𝑡 + 𝐷𝑒 𝛼𝑥−𝑖𝜔𝑡 0 Incoming from left Reflected Note: no transmitted wave in the solution.
Boundary conditions: Continuity 𝜓 1 Normalizable lim 𝑥→∞ 𝜓 2 0 = 𝜓 2 0 Exponential decay Exponential growth (Non-physical) 𝑥 = 0 𝐷 = 0 𝐴 + 𝐵 = 𝐶 Smoothness 𝑑𝜓 2 𝑑𝑥 𝑥=0 𝑑𝜓 1 𝑑𝑥 𝑥=0 = 𝑖𝑘 1 𝐴 − 𝐵 = −𝛼𝐶
Using Bloch’s Theorem: The Krönig-Penney Model
Bloch’s theorem allows us to calculate the energy bands of electrons in a crystal if we know the potential energy function. First done for a chain of finite square well potentials model by Krönig and Penney in 1931 with E -b 0 a a+b 2a+b 2(a+b) x We can solve the SE in each region of space: 0 < x < a -b < x < 0 I II ( x ) ( x ) Ae iKx Be iKx Ce x De x E 2 K 2 2 m V 0 E 2 2 2 m 2 2 m d 2 dx 2 V ( x ) E I ( x ) Ae iKx Be iKx Boundary Conditions and Bloch’s Theorem The solutions of the SE require that the wavefunction and its derivative be Thus, at the two boundaries (which are infinitely repeated): continuous across the potential boundaries. x = 0 A B C D (1) iK ( A II ( x ) B ) Ce ( C x De D ) (2) x What else can we do? T R is a translation operator that shifts the position by vector R T R ( r ) ( r R ) T R Since the Hamiltonian/energy is periodic, it doesn’t matter if we translate the wavefunction before or after the hamiltonian H H ( r R ) ( r R ) H ( r ) ( r R ) HT R Since T and H commute, T must have the same eigenstates as H T and H Commute