Introduction to Schrödinger Equation: Harmonic Potential

Transcription

1 Introduction to Schrödinger Equation: Harmonic Potential Chia-Chun Chou May 2, 2006 Introduction to Schrödinger Equation: Harmonic Potential

2 Time-Dependent Schrödinger Equation For a nonrelativistic particle with mass m moving along the x axis in a potential V (x, t), the time-dependent Schrödinger equation is given by Ψ(x, t) i h t = h2 2 Ψ(x, t) 2m x 2 + V (x, t)ψ(x, t). (1) If the potential V is independent of t, the Schrödinger equation can be solved by the method of separation of variables: Substituting Eq. (2) into Eq. (1) gives Ψ(x, t) = ψ(x)f(t). (2) df dt = iē h f (3) Introduction to Schrödinger Equation: Harmonic Potential 1

3 h2 d 2 ψ + V (x)ψ = Eψ. (4) 2m dx2 Eq.(3) is easy to solve; the solution is f(t) = e iet h. (5) Eq.(4) is called the time-independent Schrödinger equation and it can be expressed by Ĥψ = Eψ (6) d 2 where Ĥ is called Hamiltonian operator. Ĥ = h2 + V (x) (7) 2mdx2 Introduction to Schrödinger Equation: Harmonic Potential 2

4 These separable solutions are called stationary states The eigenvalue E is the energy of the state ψ(x). Ψ(x, t) = ψ(x)e iet h. (8) Born s statistical interpretation says that Ψ(x, t) 2 gives the probability of finding the particle at point x at time t. Therefore, Ψ(x, t) 2 dx = 1. (9) Physically realizable states correspond to the square-integrable solutions to the Schrödinger equation. Boundary condition is that Ψ(x, t) must go to zero as x goes to ±. For stationary states, although the wave function depends on t, the Introduction to Schrödinger Equation: Harmonic Potential 3

5 probability density does not: Ψ(x, t) 2 = ψe iet h ψ e iet h = ψ(x) 2. (10) Principle of Superposition: The general solution is a linear combination of separable solutions. Ψ 1 (x, t) = ψ 1 (x)e ie 1 t h (11) Ψ 2 (x, t) = ψ 2 (x)e ie 2 t h (12). Once we have found the separable solutions, we can construct a much Introduction to Schrödinger Equation: Harmonic Potential 4

6 more general solution Ψ(x, t) = n=1 c n ψ n (x)e ie nt h. (13) Given the starting wave function Ψ(x, 0), the coefficients in the expansion can be determined by Ψ(x, 0) = c n = c n ψ n (x) (14) n=1 ψ n (x) Ψ(x, 0)dx. (15) Introduction to Schrödinger Equation: Harmonic Potential 5

7 Harmonic Potential The harmonic potential is given by V (x) = 1 2 mω2 x 2 (16) where m is the mass of the particle and ω is the angular frequency of the oscillation. We want to solve the time-independent Schrödinger equation If we introduce the dimensionless variable h2 d 2 ψ 2m dx mω2 x 2 ψ = Eψ. (17) ξ = mω h x, Introduction to Schrödinger Equation: Harmonic Potential 6

8 the Schrödinger equation becomes d 2 ψ dξ = 2 (ξ2 K)ψ and K = 2E hω (18) where K is the energy. Our problem is to solve Eq. (18). At very large ξ, ξ 2 completely dominates over the constant K, so in this region The approximate solution is d 2 ψ dξ 2 ξ2 ψ. (19) ψ(ξ) Ae ξ2 2 + Be ξ 2 2. (20) Introduction to Schrödinger Equation: Harmonic Potential 7

9 According to the boundary condition, ψ(ξ) 0 as ξ ±, so B = 0. Therefore, the physically acceptable solutions have the asymptotic form This suggests that ψ(ξ) Ae ξ2 2, at large ξ. (21) Then, we substitute Eq. (22) into Eq. (18): ψ(ξ) = h(ξ)e ξ2 2. (22) dψ dξ = ( ) dh dξ ξh e ξ2 2 (23) Introduction to Schrödinger Equation: Harmonic Potential 8

10 d 2 ψ dξ 2 = so the Schrödinger equation, Eq. (18), becomes ( d 2 ) h dξ 2ξdh 2 dξ + (ξ2 1)h e ξ2 2, (24) d 2 h dξ 2 2ξdh dξ + (K 1)h = 0. (25) We use the power expansion method to solve Eq. (25) to look for solutions in the form of power series in ξ: h(ξ) = a j ξ j. (26) j=0 Introduction to Schrödinger Equation: Harmonic Potential 9

11 Differentiating the series term by term, dh dξ = j=0 ja j ξ j 1 (27) d 2 h dξ 2 = (j + 1)(j + 2)a j+2 ξ j. (28) j=0 Putting these into Eq. (18), we find [(j + 1)(j + 2)a j+2 2ja j + (K 1)a j ] ξ j = 0. (29) j=0 It follows that the coefficient of each power of ξ must vanish, (j + 1)(j + 2)a j+2 2ja j + (K 1)a j = 0, Introduction to Schrödinger Equation: Harmonic Potential 10

12 and hence that a j+2 = (2j + 1 K) (j + 1)(j + 2) a j. (30) This recursion formula is entirely equivalent to the Schrödinger equation. Starting with a 0, it generates all the even-numbered coefficients and starting with a 1, it generates all the odd coefficients. We write the complete solution as h(ξ) = h even (ξ) + h odd (ξ). However, not all the solutions so obtained are normalizabled. At very large j, the recursion formula becomes (approximately) a j+2 2 j a j. Introduction to Schrödinger Equation: Harmonic Potential 11

13 The approximate solution is a j C (j/2)! for some constant C, and this yields (at large ξ where the higher powers dominate) h(ξ) C (j/2)! ξj C 1 j! ξ2j Ce ξ 2. j j Therefore, if h goes like exp(ξ 2 ), the ψ goes like exp(ξ 2 /2): h(ξ) e ξ2 ψ(ξ) = h(ξ)e ξ2 2 e ξ 2 2 as ξ ±. Therefore, for normalizable solutions, the power series must terminate. Introduction to Schrödinger Equation: Harmonic Potential 12

14 There must occur some highest j (call it n), such that a n+2 = 0. a n+2 = (2n + 1 K) (n + 1)(n + 2) a n = 0 K = 2n + 1. For physically acceptable solutions, K = 2n + 1 = 2E/ hω. The energy level is ( E n = n + 1 ) hω, for n = 0, 1, 2, 3,. (31) 2 For the allowed values of K, the recursion formula reads a j+2 = (2j + 1 (2n + 1)) a j = (j + 1)(j + 2) 2(n j) (j + 1)(j + 2) a j. (32) Introduction to Schrödinger Equation: Harmonic Potential 13

15 If n = 0 (a 1 = 0), there is only one term in the series: h 0 (ξ) = a 0 (33) ψ 0 (ξ) = a 0 e ξ2 2. (34) For n = 1 we take a 0 = 0 and h 1 (ξ) = a 1 ξ (35) ψ 1 (ξ) = a 1 ξe ξ2 2. (36) Introduction to Schrödinger Equation: Harmonic Potential 14

16 For n = 2 (a 1 = 0), a 2 = 2a 0, a 4 = 0, (37) h 2 (ξ) = a 0 + a 2 ξ 2 = a 0 (1 2ξ 2 ) (38) ψ 2 (ξ) = a 0 (1 2ξ 2 )e ξ2 2. (39) In general, h n (ξ) will be a polynomial of degree n in ξ. Apart from the overall factor (a 0 or a 1 ), they are the Hermite polynomials, H n (ξ). By tradition, the arbitrary multiplicative factor is chosen so that the coefficient of the highest power of ξ is 2 n. Introduction to Schrödinger Equation: Harmonic Potential 15

17 Hermite polynomials H 0 = 1 H 1 = 2ξ H 2 = 4ξ 2 2 H 3 = 8ξ 3 12ξ H 4 = 16ξ 4 48ξ H 5 = 32ξ 5 160ξ ξ Introduction to Schrödinger Equation: Harmonic Potential 16

18 The normalized stationary states for the harmonic potential are ψ n (x) = ( mω ) 1/4 1 π h 2n n! H n(ξ)e ξ2 2. (40) The orthonormality condition of the eigenfunctions ψ n (x) and ψ m (x) is ψ n(x)ψ m (x)dx = 1 π 1/2 2 n n! H n (ξ)h m (ξ)e ξ2 dξ = δ nm. (41) Introduction to Schrödinger Equation: Harmonic Potential 17

19 First few eignefunctions Introduction to Schrödinger Equation: Harmonic Potential 18

20 Energy Estimate The time-dependent Schrödinger equation is given by Ψ(x, t) i h t = h2 2 Ψ(x, t) 2m x 2 + V (x, t)ψ(x, t) (42) and its complex conjugate is given by i h Ψ (x, t) t = h2 2 Ψ (x, t) + V (x, t)ψ (x, t). (43) 2m x 2 We multiply Eq. (42) by Ψ (x, t) and multiply Eq. (43) by Ψ(x, t). Then, taking the difference of the two resulting equations gives i h ( ) Ψ t Ψ + Ψ t Ψ = h2 2m ( 2 Ψ x 2 Ψ 2 Ψ ) x Ψ. (44) 2 Introduction to Schrödinger Equation: Harmonic Potential 19

21 t (Ψ Ψ) = i h 2m t Ψ 2 = x ( Ψ 2 Ψ [ i h 2m We integrate the equation over all space: d dt Ψ 2 dx = t Ψ 2 dx = i h 2m ) x Ψ 2 x 2 Ψ 2 ( Ψ Ψ x Ψ x Ψ )] ( ) Ψ Ψ x Ψ x Ψ (45) (46) = 0 (47) It follows that Ψ(x, t) 2 dx = constant = Ψ(x, 0) 2 dx. (48) If Ψ is normalized at t = 0, it stays normalized for all future time. Introduction to Schrödinger Equation: Harmonic Potential 20

22 Reference David J. Griffiths, Introduction to Quantum Mechanics, Prentice Hall, Englewood Cliffs, N.J., Introduction to Schrödinger Equation: Harmonic Potential 21