MTLE-6120: Advanced Electronic Properties of Materials Quantum - PowerPoint PPT Presentation

1 MTLE-6120: Advanced Electronic Properties of Materials Quantum kinetics Reading: ◮ Kasap: 3.7 ◮ Griffiths QM: 9.1 - 9.2

2 Operators and expectation values r ) | 2 probability distribution of � ◮ | ψ ( � r ◮ Average value of � r : � r ) | 2 � � � r � ≡ d � r | ψ ( � r Expectation value of operator � r in state with wavefunction ψ � � ψ | � r | ψ � ≡ d � r ψ ∗ ( � r ) � r ψ ( � r ) ◮ Expectation value of r 2 : � r ) r 2 ψ ( � � r 2 � ≡ � ψ | r 2 | ψ � ≡ d � r ψ ∗ ( � r ) ◮ Uncertainty in x , ∆ x ≡ � � x 2 � − � x � 2

3 Momentum operator √ k , ψ = e i� p = � � ◮ For a free particle with momentum � k · � x / L ◮ Consider expectation value of gradient � r ψ ∗ ( � � ψ |∇| ψ � ≡ d � r ) ∇ ψ ( � r ) = 1 � r e − i� x ∇ e i� k · � k · � x d � L = 1 � r e − i� x i� ke i� k · � k · � x d � L = i� k ◮ Momentum operator ˆ p � = � ψ | ˆ � p defined by � � � p | ψ � ◮ So the momentum operator is ˆ � p = − i � ∇

4 Hamiltonian operator ◮ If we define the Hamiltonian operator as ˆ H ≡ � 2 ∇ 2 p 2 � ˆ 2 m + V ( � r ) = 2 m + V ( � r ) the Schrodinger equation becomes ˆ Hψ = Eψ ◮ Expectation value of the Hamiltonian is � ψ | ˆ H | ψ � = E ◮ What about time dependence? Remember ψ ( � r ) e − iEt/ � r, t ) = ψ ( � � � ψ | ˆ r, t ) ˆ rψ ∗ ( � H | ψ � ≡ d � Hψ ( � r, t ) � r ) e iEt/ � ˆ r ) e − iEt/ � rψ ∗ ( � = d � Hψ ( � � r ) ˆ rψ ∗ ( � = d � Hψ ( � r ) � = d � rψ ∗ ( � r ) Eψ ( � r ) = E

5 Time dependence due to perturbations ◮ Let Hamiltonian ˆ H have two eigenstates Hψ 1 = E 1 ψ 1 and Hψ 2 = E 2 ψ 2 ◮ Eigenstates are orthogonal � ψ ∗ 1 ψ 2 = 0 and complete: any ψ = c 1 ψ 1 + c 2 ψ 2 H ′ e − iωt with ◮ Say apply electric field � E e − iωt , changes Hamiltonian to ˆ H + ˆ H ′ = e� ˆ E · � r ◮ Time-dependent Schrodinger equation ( ˆ H + ˆ H ′ e − iωt ) ψ = i � ∂ψ ∂t ◮ Substitute expansion ψ ( t ) = c 1 ( t ) e − iE 1 t/ � ψ 1 + c 2 ( t ) e − iE 2 t/ � ψ 2 ( ˆ H + ˆ H ′ e − iωt )( c 1 e − iE 1 t/ � ψ 1 + c 2 e − iE 2 t/ � ψ 2 ) c 1 + E 1 c 1 ) e − iE 1 t/ � ψ 1 + ( i � ˙ c 2 + E 2 c 2 ) e − iE 2 t/ � ψ 2 = ( i � ˙ ◮ Rewrite using eigenvalues of ˆ H c 1 e − i ( E 1 + � ω ) t/ � ˆ H ′ ψ 1 + c 2 e − i ( E 2 + � ω ) t/ � ˆ H ′ ψ 2 c 1 e − iE 1 t/ � ψ 1 + i � ˙ c 2 e − iE 1 2 / � ψ 2 = i � ˙

6 Time dependence due to perturbations (contd.) ◮ Equation in terms of expansion ψ ( t ) = c 1 ( t ) e − iE 1 t/ � ψ 1 + c 2 ( t ) e − iE 2 t/ � ψ 2 c 1 e − i ( E 1 + � ω ) t/ � ˆ H ′ ψ 1 + c 2 e − i ( E 2 + � ω ) t/ � ˆ H ′ ψ 2 c 1 e − iE 1 t/ � ψ 1 + i � ˙ c 2 e − iE 1 2 / � ψ 2 = i � ˙ r, t ) ∗ to get ◮ Now integrate equation � ψ 2 ( � H ′ | ψ 1 � e i ( E 2 − E 1 − � ω ) t/ � + c 2 � ψ 2 | ˆ H ′ | ψ 2 � e − iωt = i � ˙ c 1 � ψ 2 | ˆ c 2 ◮ If we start at t = 0 in state ψ 1 i.e. c 1 (0) = 1 , c 2 (0) = 0 , then at t = 0 c 2 = � ψ 2 | ˆ H ′ | ψ 1 � e i ( E 2 − E 1 − � ω ) t/ � i � ˙ which is the rate at which state ψ 2 starts appearing ◮ c 2 ( t ) oscillates in time with zero average value as long as E 2 � = E 1 + � ω (Energy conservation) ◮ If E 2 = E 1 + � ω , then c 2 ( t ) grows in time

7 Fermi’s Golden rule ◮ Upon applying a perturbation Hamiltonian H ′ e iωt , Γ 1 → 2 = 2 π � |� ψ 2 | ˆ H ′ | ψ 1 �| 2 δ ( E 2 − ( E 1 + � ω )) is the rate of transitioning from ψ 1 to ψ 2 ◮ More generally, Γ i = 2 π � |� ψ f | ˆ H ′ | ψ i �| 2 δ ( E f − ( E i + � ω )) � f is the rate of transitioning out of initial state ψ i ◮ Fundamental equation of ‘quantum kinetics’ (say analogous to Arrhenius equation)

8 Example: particle in a box absorption spectrum ◮ States with discrete energies and (normalized) wavefunctions: E n = n 2 � 2 π 2 � L sin nπx 2 2 mL 2 , ψ n ( x ) = L ◮ Start at n = 1 , apply EM potential e E xe − iωt 0 L ◮ Absorb photons and go to higher n ◮ Will excitations occur to all n with equal probability? ◮ Matrix element for transition: � L 4 e E Ln π 2 ( n 2 − 1) 2 [1 − ( − 1) n − 1 ] � ψ n | e E x |� ψ 1 � ≡ e E d xψ ∗ n ( x ) xψ 1 ( x ) = 0 ◮ Transition (absorption) rate: 2 Γ = 2 π � 8 e E Ln � � � � δ ( E n − E 1 − � ω ) π 2 ( n 2 − 1) 2 � � � � � even n ◮ Selection rule: transitions from n = 1 only to even n

9 Orbital angular momentum ◮ Classical picture: electrons revolving around nuclei with � L = � r × � p ◮ In quantum picture, ˆ � r × ˆ L = � p = − i � � � r × ∇ ◮ In particular ˆ L z = − i � ( x∂ y − y∂ x ) = − i � ∂ φ ◮ In atomic orbitals, angular dependence Y lm l ( θ, φ ) = P m l (cos θ ) e im l φ l ◮ Azimuthal angular momentum � ˆ L z � = m l � ◮ Account for all directions, magnitude of angular momentum � ˆ L 2 � = l ( l + 1) � 2 ◮ Number of projections quantized to 2 l + 1 (number of allowed m l )

10 Spin angular momentum ◮ Electrons have spin s = 1 / 2 ◮ Corresponding m s = ± 1 / 2 (2 = 2 s + 1 values) ◮ Projected angular momentum S z = m s � ◮ Angular momentum magnitude S 2 = s ( s + 1) � 2 ◮ Both orbital and spin angular momentum for electron z ◮ Total angular momentum � J = � L + � S = + L S ◮ Also quantized, with quantum numbers j, m j S ◮ j = | l − s | to l + s in increments of 1 L ◮ m j = − j, − j + 1 , . . . , + j y ◮ Projected angular momentum J z = m j � ◮ Angular momentum magnitude J 2 = j ( j + 1) � 2 x

11 Angular momentum consequence: magnetic moments ◮ Consider particle with charge q and mass m moving with speed v in circle of radius r ◮ Angular momentum L = mvr qv ◮ Current I = 2 πr r × d � ◮ Magnetic moment µ = 1 lI = 1 2 r (2 πr ) qv � � 2 πr = qvr/ 2 2 q ◮ Classical particle µ = 2 m L ◮ Exactly true for orbital angular momentum µ z = − e 2 mm l � = − m l µ B where µ B ≡ e � 2 m is the Bohr magneton ◮ What about spin? µ z = − g e m s µ B e 2 where g e ≈ 2 . 0023 = 2 + 4 πǫ 0 hc + · · · is called the gyromagnetic ratio (Relativity ⇒ g e = 2 , rest quantum correction)

12 Angular momentum conservation: selection rules ◮ Light absorption: photon excites electron from lower state nlm l to higher state n ′ l ′ m ′ l ◮ Dominant electron-photon interaction through electric field ⇒ involves only L of electron (not S ) ◮ Initial angular momentum s = 1 in photon and l of electron ⇒ j = l − 1 , . . . , l + 1 ◮ Projection m j = m s + m l = m l − 1 , . . . , m l + 1 ◮ Angular momentum conservation ( l ′ , m ′ l ) must equal ( j, m j ) ◮ Process allowed only if ∆ l = 0 , ± 1 and ∆ m l = 0 , ± 1 ◮ More careful analysis ∆ l = 0 disallowed (because � ψ 2 | ˆ H ′ | ψ 1 � = 0 ), ⇒ ∆ l = ± 1 and ∆ m l = 0 , ± 1