NCN www.nanohub.org
EE‐606: Solid State Electronics Lecture 3: Elements of Quantum Mechanics Muhammad Ashraful Alam
[email protected]
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Outline
1) Why do we need quantum physics 2) Quantum concepts 3) Formulation of quantum mechanics 4) Conclusions
Reference: Vol. 6, Ch. 1 (pages 23‐32)
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Do I really need Quantum Mechanics ? Original Problem
Periodic Structure
Electrons in periodic potential: Problem we want to solve
If it were large objects, like a skier skiing past a set of obstacles, Newton’s mechanics would work fine, but in a micro-world …… Alam ECE‐606 S09
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Carrier Density
Carrier number = Number of states x filling factor Chapters 2‐3
Chapter 4
Total number of occupants = Number of apartments X The fraction occupied
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Outline
1) Why do we need quantum physics 2) Quantum concepts 3) Formulation of quantum mechanics 4) Conclusions
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Four Quantum Concepts ..
• Blackbody Radiation
• Photoelectric Effect • Bohr Atom • Wave Particle Duality
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(1) black‐body radiation 2000 K
1000 K
300 K
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Rayleigh-Jeans Formula
2000 K
(u)
u ∝ k BT / λ 4 log(u ) = −4 log(λ ) + log(T ) Wein’s Formula
u∝
e
− β / λT
λ
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Plank’s fitting formula
1 1 u ∝ 5 β / λT − 1 λ e Alam ECE‐606 S08
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Interpretation of Plank’s Formula dλ 1 1 dλ u ( f , T ) = u (λ , T ) ~ 5 β / λT df λ e − 1 df 1 2 ~ f × hf × hf / kT −1 e nos. of modes
c λ= f
Occupation Probability
Energy of mode
EM emission occurs in discrete quanta of
E = hf
n=1,2, …….. N
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Recent Example: COBE Data
J.C. Mather, Astrophysics J., 1990.
Show that the cosmic background temperature is approximately 3K. Can you “see” this radiation? Alam ECE‐606 S09
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(2) Photoelectric Effect
E = (hf‐W) Electrons
Light (hf)
υ
hf
mυ 2 = hf − W
cathode
VR
VR
W
( 2)
VR ≈ 1
W
hf
Absorption occurs in quanta as well, consistent with photons having E=hf Alam ECE‐606 S09
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Origin of Quantization
frequency
1 1 Em ,n = const × 2 − 2 n m Alam ECE‐606 S09
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(3) Bohr Atom .. Assume that angular momentum is quantized:
υ = n! / m0 rn
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(3) Bohr Atom (continued) …
1 1 Em ,n = const × 2 − 2 n m Alam ECE‐606 S09
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(4) Wave‐Particle Duality Photons act both as wave and particle, what about electrons ?
E = m0 c + p c 2 4
hf = pc
2 2
m0=0 (photon rest mass)
p = hf / c = h / λ (because c = λf ) =!k (because k = 2π / λ ) Alam ECE‐606 S09
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Outline
1) Why do we need quantum physics 2) Quantum concepts 3) Formulation of Schrodinger Equation 4) Conclusions
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Schrodinger Equation 2 2 2 4 E = m0 c + p c ≈ m0 c 1 + p c / 2m0 c + ... 2 4
E − m0c
2 2
2
2
= V + (p / 2m0 ) 2
hf = !ω = V + (! k / 2m0 ) 2 2
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Schrodinger Equation (continued)
!ω = V + (! k / 2m0 ) 2 2
Assume,
Ψ ( x, t ) = A exp(−i (ωt − kx))
d Ψ / dt = −iωΨ and d Ψ / dx = −k Ψ 2
2
2
! d Ψ dΨ − + V Ψ = i! 2 2m0 dx dt 2
2
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Conclusions 1. Given chemical composition and atomic arrangements, we are ready to compute electron density by using quantum mechanics. 2. We discussed the origin of quantum mechanics – experiments p were inconsistent with the classical theory. y 3. We saw how Schrodinger equation could arise as a consequence of quantization and relativity, but this is not a derivation. 4. We will solve some toy problems in the next class to get a feeling of how to use quantum mechanics. Alam ECE‐606 S09
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