Solving the Schrodinger Equation

Transcription

1 Time-dependent Schrödinger Equation: i!!!2 " (x,t) =!t 2m! 2 " (x,t) + U(x)" (x,t) 2!x Stationary Solutions:! (x,t) = "(x)(t)!(t) = e "it, = E! Time-independent Schrödinger equation:!!2 2m d 2 "(x) + U(x)"(x) = E"(x) 2 dx This equation is not always easy to solve analytically, but can be solved numerically on a computer. However we can analytically solve some special cases.

2 Consider a particle inside a box of length L with impenetrable walls: Outside the box the particle wave function is 0. Inside the box, U(x) = 0 Inserting into the TISE,!!2 2m d 2 "(x) + U(x)"(x) = E"(x) 2 dx!!2 2m d 2 "(x) = E"(x) 2 dx Solutions inside the box:!(x) = Acos(kx) + Bsin(kx), k = 2mE!

3 The wave function is continuous so the wave function inside the box must match the wave function outside the box at x=0 and x=l. Outside the box, the wavefunction is always 0. Solutions inside the box at x=0 and L:!(0) = Acos(k0) + Bsin(k0) = A = 0!(L) = Acos(kL) + Bsin(kL) = 0 A = 0, kl = n!, n = 1,2... Acceptable solutions are thus (x) = Bsin n" x & (, n = 1,2,3.. (For n=0, ψ(x)=0, particle not found in the box, hence not acceptable solution)

4 Solutions : (x) = Bsin n" x & (, n = 1,2,3.., 0 ) x ) L!(x) = 0, x > L, x < 0 Energy is quantized: E n = 1 2 mv2 = p2 2m =!2 k 2 2m =!2 n 2! 2 2mL 2 Non-zero minimum energy E 1 (zero-point energy) This satisfies the uncertainty principle. If the minimum energy were 0, then the momentum would be precisely 0, and then the location of the particle would be unknown - it would not be confined to the box.

5 Solutions : (x) = Bsin n" x & (, n = 1,2,3.., 0 ) x ) L!(x) = 0, x > L, x < 0 Normalization: L (x) 2 L L n x " dx = B 2 sin 2 ' 1 " & L ( ) dx = B 2 2n x " 1 * cos 2 & & L 0 0 L = B2 x 2 0 * L 2n x ' sin 2n x & L ( ) 0 L 0 = B2 L 2 = 1 ' ( ) ' ( ) dx B = 2 L

6 Solutions : (x) = 2 L!(x) = 0, x > L, x < 0 n" x & sin (, n = 1,2, 3.., 0 ) x ) L E n =!2 n 2! 2 2mL 2, n = 1,2, 3...

7 A particle is in a box of length L in the ground state (lowest energy state). Where is the particle most likely to be found? What is the probability of finding the particle in the middle half of the box?! 1 (x) = 2 L sin " x & (, n = 1 The particle is most likely to be at x=l/2 where the probability density curve is a maximum.

8 The middle half of the box is the region from x=l/4 to x=3l/4. Probability of finding the particle in the middle half of the box:! 1 (x) = 2 L sin " x & (, n = 1 P(L / 4! x! 3L / 4) = 3L /4 L /4 3L /4 L /4 " 1 (x) 2 dx 2 x ( L sin2 & ' L ) * dx = x L + 1 sin x ( & ' L ) * cos x ( & ' L ) * 3L /4 L /4, P(x) = = = 81.8

9 Solutions : (x) = 2 L!(x) = 0, x > L, x < 0 Orthonormality: Notice that n" x & sin (, n = 1,2, 3.., 0 ) x ) L " (x)! m (x)dx = mn i.e. the above integral is zero except when m=n. For the case of m=n, the integral is 1 (normalization). This property holds for stationary solutions of other potentials as well.

10 Solutions : (x) = 2 L!(x) = 0, x > L, x < 0 n" x & sin (, n = 1,2, 3.., 0 ) x ) L Completeness: Any function f(x) can be written as a linear combination of ϕ n (x) f (x) = " n=1 c n (x) This is nothing but the use of Fourier series or Dirichlet s theorem

11 Solutions : (x) = 2 L!(x) = 0, x > L, x < 0 n" x & sin (, n = 1,2, 3.., 0 ) x ) L Completeness: Any function f(x) can be written as a linear combination of ϕ n (x) f (x) = " n=1 c n (x) How to calculate the coefficients c n? Well, " c n = c m! mn = c m * n (x) m (x)dx = * n (x) f (x)dx m=1 " m=1

12 Correspondence principle: Recall that the correspondence principle as formulated by Bohr states that generally when the quantum number of a system becomes large it should behave classically. For a particle moving with constant speed v inside a box, the classical probability density is constant: P classical (x) = 1 L Average classical probability density matches the quantum value for large quantum numbers n