Christophe De Beule. Graphical interpretation of the Schrödinger equation

Transcription

1 Christophe De Beule Graphical interpretation of the Schrödinger equation

2 Outline Schrödinger equation Relation between the kinetic energy and wave function curvature. Classically allowed and forbidden regions. Examples Linear potential. Finite well. Hybridization. Electron in a 1D lattice. 2/26

3 Schrödinger equation d dx ( ) dψ dx = 2m (E V (x)) ψ(x) 2 What is the meaning of the Schrödinger equation? curvature of ψ at x kinetic energy at x ψ at x 3/26

4 Allowed region In classically allowed regions, the kinetic energy is positive. sign curvature of ψ at x = sign kinetic energy at x ψ at x = sign ψ at x The curvature and ψ have opposite sign, so that ψ is concave towards the axis, therefore oscillating around zero. ψ(x) > 0 ψ(x) < 0 : ψ(x) is concave : ψ(x) is convex 4/26

5 Allowed region ψ(x) > 0 ψ(x) < 0 : ψ(x) is concave : ψ(x) is convex ψ(x) x 5/26

6 Forbidden region In classically forbidden regions, the kinetic energy is negative. sign curvature of ψ at x = sign kinetic energy at x ψ at x = sign ψ at x The curvature and ψ have the same sign sign, so that ψ is convex towards the axis, curving away from zero. 6/26

7 Summary In regions of small (large) positive kinetic energy, the wave function oscillates slowly (rapidly) with large (small) amplitude. ψ(x) x In regions of negative kinetic energy, the wave function curves away from zero. ψ(x) x 7/26

8 Example 1: Linear potential Consider a free particle (E > 0) in a 1D linear potential, V (x) = ax, a > 0, with kinetic energy T (x) = E V (x) = E ax. T (x) E E/a x 8/26

9 Linear potential In the classically allowed region x < E/a, the kinetic energy, T (x) = E ax > 0. ψ oscillates around zero, faster as x becomes more negative. ψ(x) E/a x 9/26

10 Linear potential In the classically forbidden region x > E/a, the kinetic energy, T (x) = E ax < 0. A physical ψ has to curve away from zero exponentially slow. ψ(x) E/a x 10/26

11 Example 2: Finite well Consider the bound states (E < 0) of a finite potential well, with V (x) = V 0 θ( a < x < a), ( ) d dψ = 2m ( E + V ) ψ(x). dx dx 2 V 0 x = a x = a 11/26

12 Finite well When x < a, the kinetic energy ( E ) is negative, and ψ increases exponentially. ψ(x) V 0 x = a x = a 12/26

13 Finite well At x = a, the kinetic energy (V 0 E ) becomes positive, and since ψ is positive, the curvature becomes negative. ψ curves back towards zero, at a rate dependent on V 0 E. ψ(x) V 0 x = a x = a 13/26

14 Finite well If V 0 E is large enough, the slope will be negative at x = a, and ψ begins curving upwards again. ψ diverges unless E is such, that the slope at x = a is just right, so that it increases at an exponentially slow rate. ψ(x) V 0 x = a x = a 14/26

15 Example 3: Hybridization Consider a two state system, where A and B are two s like states that hybridize, e.g. the hydrogen molecule. 1 2 ( A B ) A E 2 A Ĥ B B 1 2 ( A + B ) 15/26

16 Hybridization Symmetric combination The wave function is always positive and convex in the center, Classically forbidden region with negative kinetic energy. sign curvature of ψ at x = sign kinetic energy at x ψ at x 16/26

17 Hybridization Antisymmetric combination The wave function has a single node in the center, Classically allowed region with positive kinetic energy. 17/26

18 Hybridization The kinetic energy is given by, T (x) = e2 x + b + e2 x b E. T (x) T S = 0 T A = 0 b b x 18/26

19 Hybridization The total kinetic energy is given by, ˆp 2 ψ 2m ψ = dx ψ(x) 2 T (x). The ground state is given by the combination with the fewest nodes, which is the symmetric combination for s like states. 19/26

20 Example 4: Electron in a 1D lattice The wave function of the electron can be written in a basis of localized (atom-like) states n, ψ = n n n ψ. n 1 n n /26

21 Stationary states The stationary states are given by, ψ k = n n e ikna. All solutions are contained in the Brillouin zone k ( π/a, π/a]. The wave function is a superposition of localized orbitals modulated by the phase factor e ikx. a x n a x n x n + a 21/26

22 s like states: wave function Consider the case where the basis states n are s like. φ k=0 = n n, s φk=π/a = ( 1) n n, s. n /26

23 s like states: energy band For s like basis states, the wave function has, no nodes at k = 0, one node for each unit cell at k = π/a. The symmetric combination (k = 0) has the fewest nodes. E π/a π/a k 23/26

24 p like states: wave function Consider the case where the basis states n are p like. φ k=0 = n n, p φ k=π/a = ( 1) n n, p. n /26

25 p like states: energy band For p like basis states, the wave function has, one node for each unit cell at k = 0, no nodes at k = π/a. The antisymmetric combination (k = π/a) has the fewest nodes. E π/a π/a k 25/26

26 Question: dimer lattice Now we have a lattice with a two atom basis. Take one s or p like basis state for each atom. What does the band structure look like, qualitatively? Hint: first look at a single unit cell. a (n 1)B na nb (n + 1)A (n + 1)B 26/26