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

1 MTLE-6120: Advanced Electronic Properties of Materials Band theory of solids Contents: ◮ Classical analogue: coupled oscillators ◮ 1D δ -crystal band structure ◮ Bloch theorem and wavefunctions ◮ Tight-binding and nearly-free limits ◮ Band strctures of real materials (in 3D) ◮ Effective mass, transport and density of states Reading: ◮ Kasap: 4.1 - 4.5

2 Crystals ◮ Periodic arrangement of atoms (ions: nuclei + core electrons) ◮ Previously: classical motion of electrons in crystal ◮ What was the role of the periodicity? None! ◮ Next: quantum motion of electrons in crystal ◮ Bravais lattice: regular grid of points + + + + + + + + + + + + + + + ◮ Lattice vectors � a 1 ,� a 2 , � a 3 + + + + + ◮ Any point on grid n 1 � a 1 + n 2 � a 2 + n 3 � a 3 + + + + + + + + + + + + + + + ◮ Unit cell: repeat at grid points to fill space + + + + + ◮ Multiple atoms possible per unit cell + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

3 Classical example: one oscillator ◮ Mass m attached to a spring with constant k ◮ Equation of motion in displacement x : m ¨ x = − kx � ◮ In terms of ω 0 = k/m , x = − ω 2 ¨ 0 x ◮ Oscillatory solutions e ± iω 0 t with angular frequency ω 0

4 Classical example: two oscillators ◮ Two masses m each attached to a spring with constant k ◮ Equation of motion in displacements x 1 and x 2 : m ¨ x 1 = − kx 1 m ¨ x 2 = − kx 2 � ◮ Solutions: both x 1 and x 2 oscillate with ω 0 = k/m

5 Classical example: two coupled oscillators ◮ Two masses m each attached to a spring with constant k ◮ Attached to each other with another spring with constant K ◮ Equation of motion in displacements x 1 and x 2 : m ¨ x 1 = − kx 1 − K ( x 1 − x 2 ) m ¨ x 2 = − kx 2 − K ( x 2 − x 1 ) ◮ Coupled equations, x 1 and x 2 are not each an oscillator ◮ Add and subtract m (¨ x 1 + ¨ x 2 ) = − k ( x 1 + x 2 ) m (¨ x 1 − ¨ x 2 ) = − ( k + 2 K )( x 1 − x 2 ) ◮ Oscillator differential equations in ( x 1 ± x 2 ) � ◮ ( x 1 + x 2 ) oscillates with frequency k/m � ◮ ( x 1 − x 2 ) oscillates with frequency ( k + 2 K ) /m

6 Coupled oscillators: eigenvalue problem ◮ Equation of motion in matrix form: � � � � � � m d 2 x 1 k + K − K x 1 = − · d t 2 x 2 − K k + K x 2 � �� � A ◮ Coupled equations because of off-diagonal terms in A ◮ Diagonalize A = V · D · V − 1 using eigenvalue diagonal matrix D and eigenvector diagonal matrix V � x 1 � x 1 � � m d 2 d t 2 V − 1 · = − D · V − 1 · x 2 x 2 ◮ V − 1 combines x into normal modes x 1 + x 2 and x 1 − x 2 ◮ Corresponding eigenvalues are mω 2 = ( k + K ) ∓ K i.e. k and k + 2 K

7 Three coupled oscillators ◮ For three oscillators   k + K − K 0 A = − K k + 2 K − K   0 − K k + K ◮ Eigenvalues mω 2 :   k k + K   k + 3 K

8 N coupled oscillators ◮ For N oscillators, get N × N matrix   k + K − K − K k + 2 K − K     ...   − K k + 2 K   A =   ... ...   − K     − K k + 2 K − K   − K k + K ◮ Consider eigenvector of displacements X = ( x 1 , x 2 , . . . , x N ) with eigenvalue mω 2 = λ (say) ◮ n th row ( n � = 1 , N ) of eigenvalue equation AX = λX is Cx n − 1 + Dx n + Cx n +1 = λx n where C = − K is the coupling (off-diagonal term) and D = k + 2 K is the diagonal term

9 Tri-diagonal matrix eigenvalue problem ◮ n th row ( n � = 1 , N ) of eigenvalue equation AX = λX is Cx n − 1 + Dx n + Cx n +1 = λx n ◮ Try solution of form x n = e ik ( na ) where a is spacing between oscillators (just so that k has usual dimensions of wavevector) Ce ik ( n − 1) a + De ikna + Ce ik ( n +1) a = λe ikna Ce − ika + D + Ce ika = λ D + 2 Ccos ( ka ) = λ (Each spring has phase e ika relative to previous one) ◮ All rows satisfy this condition except first and last e ika ( D − K ) + Ce 2 ika = λe ika ( n = 1) Ce i ( N − 1) ka + ( D − K ) e Nika = λe Nika ( n = N ) ◮ Connect last spring to first

10 Periodic boundary conditions (Born-von Karman) ◮ Edge equations after connecting last spring to first + e ika D + Ce 2 ika = λe ika Ce Nika ( n = 1) � �� � from n = N Ce i ( N − 1) ka + De Nika + Ce ika = λe Nika ( n = N ) � �� � from n = 1 ◮ Canceling common factors: ( e Nika ) Ce − ika + D + Ce ika = λ ( n = 1) Ce − ika + D + Ce ika ( e − Nika ) = λ ( n = N ) ◮ These are the same as the other equations when e Nika = 1 i.e. k = 2 πj/ ( Na ) ◮ Here, j can be any integer, but for j → j + N , k → k + 2 π/a and e ika → e i ( ka +2 π ) = e ika , so nothing changes

11 N → ∞ coupled oscillators ◮ Eigenvalues mω 2 = λ = D + 2 C cos ka = ( k + 2 K ) − 2 K cos ka ◮ k = 2 πj/ ( Na ) for j = 0 , 1 , 2 , . . . , N − 1 becomes continuous k ∈ [0 , 2 π/a ) ◮ Spacing in k = 2 π/ ( Na ) = 2 π/L , where L = Na is length of system m ω 2 k+4K k+2K k -3 π /a -2 π /a -1 π /a 0 π /a 1 π /a 2 π /a 3 π /a k ◮ Eigenvalues mω 2 form a ‘band’ with range [ k, k + 4 K ] ◮ Band center set by diagonal element k + 2 K ◮ Band width set by off-diagonal element K (coupling strength) ◮ Any interval in k of length 2 π/a equivalent; conventionally [ − π/a, π/a ]

12 1D crystal of δ -atoms � ◮ An array of δ -potentials at na i.e. V ( x ) = − V 0 n δ ( x − na ) √ ◮ Consider E < 0 case first, with κ = − 2 mE/ � ◮ Wavefunctions e ± κx in each segment, say ψ ( x ) = A n e − κ ( x − na ) + B n e κ ( x − ( n +1) a ) , na < x < ( n + 1) a

13 δ -crystal: matching conditions Continuity at x = na and ( n + 1) a : A n + B n e − κa = A n − 1 e − κa + B n − 1 A n e − κa + B n = A n +1 + B n +1 e − κa Derivative matching conditions at x = na and ( n + 1) a : − 2 mV 0 � A n + B n e − κa � � − A n + B n e − κa � � − A n − 1 e − κa + B n − 1 � = κ − κ � 2 − 2 mV 0 � A n e − κa + B n � � − A n +1 + B n +1 e − κa � � − A n e − κa + B n � = κ − κ � 2 Simplify using q ≡ 2 mV 0 / � 2 and t ≡ e − κa : A n + B n t = A n − 1 t + B n − 1 A n t + B n = A n +1 + B n +1 t − q κ ( A n + B n t ) = − A n + B n t + A n − 1 t − B n − 1 − q κ ( A n t + B n ) = − A n +1 + B n +1 t + A n t − B n

14 δ -crystal: eliminate B ’s ◮ Add first and third equations, subtract fourth from secondequation: (1 − q/κ )( A n + B n t ) = 2 A n − 1 t − A n + B n t (1 + q/κ )( A n t + B n ) = 2 A n +1 − A n t + B n ◮ Collect all A terms to left side and B to right side: (2 − q/κ ) A n − 2 A n − 1 t = ( q/κ ) B n t (2 + q/κ ) A n t − 2 A n +1 = − ( q/κ ) B n ◮ Eliminate B n : � 1 + t 2 + ( t 2 − 1) q � − A n − 1 t + A n − A n +1 t = 0 2 κ

15 δ -crystal: eigenvalue condition 1 + t 2 + ( t 2 − 1) q � � − A n − 1 t + A n − A n +1 t = 0 2 κ ◮ Substitute A n = e ikna : 1 + t 2 + ( t 2 − 1) q � � − te ik ( n − 1) a + e ikna − te ik ( n +1) a = 0 2 κ 1 + t 2 + ( t 2 − 1) q � � − te − ika + − te ika = 0 2 κ 1 + t 2 + ( t 2 − 1) q = 2 t cos( ka ) 2 κ 1 /t + t − q 2 κ · 1 /t − t = cos( ka ) 2 2 ◮ Since t ≡ e − κa cosh( κa ) − q 2 κ sinh( κa ) = cos( ka )

16 δ -crystal: eigenvalues ◮ Energy eigenvalue is E = � ω = − � 2 κ 2 2 m ◮ For the δ -atom, we got a unique κ = mV 0 / � 2 ≡ q/ 2 ◮ But now, an entire family of solutions for k ∈ ( − π/a, π/a ) : cosh( κa ) − q 2 κ sinh( κa ) = cos( ka ) ◮ First consider limit of deep potential i.e. large q ; rearrange: q sinh( κa ) κ = 2(cosh( κa ) − cos( ka )) ◮ Large q ⇒ large κ ⇒ sinh , cosh extremely large ◮ For q → ∞ , κ → q/ 2 as in the δ -atom ◮ For large q , substitute κ = q/ 2 in RHS to get q sinh qa 2 − cos( ka )) ≈ q � � 1 + 2 e − qa/ 2 cos( ka ) 2 κ ≈ 2(cosh qa 2

17 δ -crystal: tight-binding limit ◮ For large q , κ ≈ q � � 1 + 2 e − qa/ 2 cos( ka ) 2 ◮ Correspondingly, the energy E = − � 2 κ 2 2 m ≈ − � 2 q 2 − � 2 q 2 e − qa/ 2 cos( ka ) 8 m 2 m � �� � E 0 ◮ Exactly like the coupled springs! ◮ First term = band center = energy E 0 of isolated δ -atom ◮ Second term ∝ band width and coupling strength ∝ e − qa/ 2 ◮ Coupling proportional to overlap between wavefunctions at adjacent atoms

18 δ -crystal: free states ◮ For bound states E = − � 2 κ 2 < 0 , we derived eigenvalue condition 2 m cosh( κa ) − q 2 κ sinh( κa ) = cos( ka ) ◮ For free states E = � 2 K 2 > 0 (using K since k is taken), substitute 2 m κ = iK above: q cosh( iKa ) − 2 iK sinh( iKa ) = cos( ka ) which simplifies to q cos( Ka ) − 2 K sin( Ka ) = cos( ka )

19 δ -crystal: energy conditions cosh( κa ) − q cos( Ka ) − q 2 κ sinh( κa ) = cos( ka ) or 2 K sin( Ka ) = cos( ka ) ◮ Plot LHS as a function of E = � 2 K 2 or − � 2 κ 2 2 m ; note | RHS | ≤ 1 2 m cos( ka ) For qa = 5 1 E - 2 /( ma 2 ) h 0 -10 -5 0 5 10 15 20 25 30 35 40 -1 - 2 κ 2 /(2 m ) < 0 - 2 K 2 /(2 m ) > 0 E = - h E = h ◮ When LHS magnitude > 1 , those κ or K not valid solutions ◮ Correspondingly, not all E valid: bands!