Notes on Quantum Mechanics

Transcription

1 Notes on Quantum Mechanics Kevin S. Huang Contents 1 The Wave Function The Schrodinger Equation Probability Normalization Momentum The Uncertainty Principle The Time-Independent Schrodinger Equation 4.1 Stationary States The Infinite Square Well The Harmonic Oscillator Algebraic Method Analytic Method The Free Particle The Delta-Function Potential Bound States: E < Scattering States: E > Finite Square Well Bound States: E < Scattering States: E > The Scattering Matrix The Wave Function 1.1 The Schrodinger Equation Goal of Quantum Mechanics: Determine the wave function Ψ(x, t) of the particle. Schrodinger Equation: i Ψ t = Ψ m x + V Ψ Given the initial conditions: Ψ(x, 0); the Schrodinger equation determines Ψ(x, t) for all future time. 1

2 1. Probability Ψ(x, t) dx represents the probability of finding the particle between x and x + dx. N(j) - number of people with j: P (j) - probability of getting j: Expectation value (average): N = N(j) j=0 P (j) = N(j) N P (j) = 1 j=0 j = jn(j) N = jp (j) j=0 Variance: Standard deviation - σ f(j) = jn(j) = f(j)p (j) N j=0 j = j j σ = ( j) σ = j j ρ(x) dx - probability that individual is between x and x + dx ρ(x) - probability density Continuous distributions: P ab = b a ρ(x) dx ρ(x) dx = 1 x = f(x) = xρ(x) dx f(x)ρ(x) dx σ = x x

3 1.3 Normalization Particle must be somewhere: Ψ(x, t) dx = 1 Non-normalizable solutions cannot represent particles. Wave function remains normalized as time goes on. 1.4 Momentum Average of measurements performed on particles all in the state Ψ: x = x Ψ(x, t) dx Through integration by parts and discarding boundary terms: d x dt = i m Ψ Ψ x dx Expectation value of velocity: Expectation value of momentum: Notation with operators: p = m d x dt Expectation value of dynamical variables: 1.5 The Uncertainty Principle v = d x dt = i x = Ψ (x)ψ dx ( Ψ Ψ ) dx x ) ( p = Ψ Ψ dx i x ( Q(x, p) = Ψ Q x, ) Ψ dx i x de Broglie formula: Heisenberg uncertainty principle: p = h λ = π λ σ x σ p 3

4 The Time-Independent Schrodinger Equation.1 Stationary States Separation of Variables: Using the Schrodinger equation: i 1 f Ψ(x, t) = ψ(x)f(t) df dt = m 1 ψ ψ x + V Crucial argument: Left side is a function of only t, right side is a function of only x, therefore both sides are constant denoted by E. df dt = iē h f f(t) = e iet/ Time-independent Schrodinger equation: Separable Solutions d ψ m dx + V ψ = Eψ Ψ(x, t) = ψ(x)e iet/ 1. They are stationary states - the probability density does not depend on time.. They are states of definite total energy. Hamiltonian: H(x, p) = p m + V (x) H = E σ H = 0 3. The general solution is a linear combination of separable solutions. There is a different wave function for each allowed energy. Ψ 1 (x, t) = ψ 1 (x)e ie 1t/ Ψ (x, t) = ψ (x)e ie t/ Time-dependent Schrodinger equation has the property that any linear combination of solutions is itself a solution. General solution: Ψ(x, t) = c n ψ n (x)e ient/ n=1 4

5 . The Infinite Square Well V = 0 if 0 x a, and V = otherwise. Outside the well: ψ(x) = 0 Inside the well: V = 0 Simple harmonic oscillator: General solution: Boundary Conditions: Hence: Distinct solutions: d ψ dx = k ψ me k = ψ(x) = A sin(kx) + B cos(kx) ψ(0) = ψ(a) = 0 ψ(x) = A sin(kx) ka = 0, ±π,, ±π, ±3π,... k n = nπ a Normalize ψ: E n = kn m = n π ma A = a Solutions: ( ) nπ ψ n (x) = a sin a x ψ 1 - ground state, other waves are excited states Important properties of ψ n (x) 1. They are alternately even and odd (true when potential is an even function).. Each successive state has one more node (universal). 3. They are mutually orthogonal (quite general). ψ m (x) ψ n (x) dx = 0, m n δ mn - Kronecker delta ψ m (x) ψ n (x) dx = δ mn 5

6 4. They are complete, in the sense that any other function, f(x), can be expressed as a linear combination of them ( ) nπ f(x) = c n ψ n (x) = c n sin n=1 a n=1 a x Expansion coefficients can be evaluated by Fourier s trick: ψ m (x) f(x)d = c n ψ m (x) ψ n (x) dx = n=1 c m = ψ m (x) f(x) dx c n δ mn = c m n=1 Stationary States for Infinite Square Well: Ψ n (x, t) = a sin ( nπ a x ) e i(n π /ma )t General Solution: Ψ(x, t) = n=1 c n a sin ( nπ a x ) e i(n π /ma )t c n = a ( ) nπ sin a 0 a x Ψ(x, 0) dx 6

7 .3 The Harmonic Oscillator Practically any potential is approximately parabolic. Virtually any oscillatory motion is approximately simple harmonic. V (x) = 1 V (x 0 )(x x 0 ) V (x) = 1 mω x Solve: d ψ m dx + 1 mω x ψ = Eψ.3.1 Algebraic Method Ladder Operators: Schrodinger equation: 1 m ( i a a + = 1 m a + a = 1 m d dx a ± = 1 m ( i ( i ( i ) + (mωx) d dx d dx ψ = Eψ ) d dx ± imωx ) + (mωx) ) + (mωx) a a + a + a = ω 1 (a a + = Eψ ω)ψ 1 (a + a + = Eψ ω)ψ + 1 ω 1 ω Important: If ψ satisfies the Schrodinger equation with energy E, then a + ψ satisfies the Schrodinger equation with energy (E + ω). a ψ is a solution with energy (E ω). There must exist a lowest rung : a ψ 0 = 0 ) 1 ( dψ 0 m i dx imωxψ 0 ψ 0 = A 0 e mω x E 0 = 1 ω 7 = 0

8 Excited states: ψ n (x) = A n (a + ) n e mω x E n = (n + 1 )ω.3. Analytic Method Substitution: Schrodinger equation: Large ξ: Asymptotic form: d ψ dξ ξ = mω x = (ξ K)ψ K = E ω d ψ dξ ξ ψ ψ(ξ) Ae ξ / + Be +ξ / ψ(ξ) ()e ξ / ψ(ξ) = h(ξ)e ξ / d h dh ξ + (K = 1)h = 0 dξ dξ Look for a solution in the form of a power series: h(ξ) = a 0 + a 1 ξ + a ξ +... = a j ξ j j=0 Recursion formula: a j+ = (j + 1 K) (j + 1)(j + ) a j The power series must terminate for the solution to be normalizable. One series must truncate, the other must be zero from the start. a j+ = K = n + 1 E n = (n + 1 )ω (n j) (j + 1)(j + ) a j 8

9 Stationary states: ψ n (x) = ( ) mω 1/4 1 π n n! H n(ξ)e ξ /.4 The Free Particle V (x) = 0 everywhere. d ψ dx = k ψ me k = k ik(x Ψ(x, t) = Ae m t) k ik(x+ + Be m t) Special combination: (x ± vt) represents a wave of fixed profile, traveling in the x direction, at speed v. k > 0: Wave traveling to the right k < 0: Wave traveling to the left k i(kx Ψ k (x, t) = Ae m t) v quantum = k E m = m A free particle cannot exist in a stationary state; there is no such thing as a free particle with a definite energy. 9

10 Wave packet: Plancherel s theorem: Ψ(x, t) = 1 π ω = k m φ(k) = 1 π v group = dω dk φ(k)e i(kx ωt) dk Ψ(x, 0)e ikx dx v phase = ω k.5 The Delta-Function Potential Bound State: If V(x) rises higher than the particle s total energy E on either side, then the particle is stuck in the potential well. Scattering State: If E exceeds V(x) on one side or both, the the particle comes in from infinity and returns to infinity. Tunneling allows the particle to leak through any finite potential barrier. E < V () and V (+ ) - Bound State E > V () or V (+ ) - Scattering State Real life: Most potentials go to zero at infinity. E < 0: Bound State E > 0: Scattering State Infinite square well and harmonic oscillator admit bound states only. Free particle only allows scattering states. Dirac delta function δ(x) = 0, x 0 δ(x) =, x = 0 δ(x) dx = 1 Potential: f(x)δ(x a) = f(a)δ(x a) f(x)δ(x a) dx = f(a) V (x) = αδ(x) δ(x a) dx = f(a) 10

11 d ψ αδ(x)ψ = Eψ m dx.5.1 Bound States: E < 0 General Solution: Boundary Conditions: x < 0, V (x) = 0 d ψ dx = κ ψ me κ = ψ(x) = Be κx, x < 0 ψ(x) = F e κx, x > 0 1. ψ is always continuous.. dψ dx is continuous except at points where the potential is infinite. F = B Integrate the Schrodinger equation from ɛ to ɛ and take the limit as ɛ : One bound state: ( ) dψ = m dx ɛ lim ɛ 0 ɛ ψ(x) = V (x)ψ(x) dx = mα ψ(0) = Bκ κ = mα mα B = mα E = mα e mα x 11

12 .5. Scattering States: E > 0 d ψ dx = k ψ me k = ψ(x) = Ae ikx + Be ikx, x < 0 ψ(x) = F e ikx + Ge ikx, x > 0 In a typical scattering experiment particles are fired in from one direction. In that case the amplitude of the wave coming in from the right will be zero. A - amplitude of the incident wave B - amplitude of the transmitted wave Reflection Coefficient Transmission Coefficient R = B A = T =.6 Finite Square Well V (x) = V 0 for a < x < a V (x) = 0 for x > a G = 0 β = mα k B = iβ 1 iβ A F = 1 1 iβ A β 1 + β = ( E/mα ) F A = β = (mα / E) R + T = 1 1

13 .6.1 Bound States: E < 0 x < a, V (x) = 0: a < x < a, V (x) = V 0 : x > a, V (x) = 0: 1. Wide, deep well d ψ dx = κ ψ me κ = ψ(x) = Be κx d ψ dx = l ψ m(e + V 0 l = ψ(x) = C sin(lx) + D cos(lx) ψ(x) = F e κx E n + V 0 = n π m(a). Shallow, narrow well: There is always one bound state, no matter how weak the well becomes..6. Scattering States: E > 0 x < a: ψ(x) = Ae ikx + Be ikx me k = a < x < a: ψ(x) = C sin(lx) + D cos(lx) m(e + V 0 l = x > a (assuming there is no incoming wave in this region): ψ(x) = F e ikx F = B = i sin(la) (l k )F kl e ika A cos(la) i sin(la) kl (k + l ) 13

14 T 1 = 1 + Energies for perfect transmission:.7 The Scattering Matrix Arbitrary Localized Potentials v 0 4E(E + V 0 sin E n + V 0 = n π m(a) ( ) a m(e + V 0 Region 1: Region : Region 3: ψ(x) = F e ikx + Ge ikx ψ(x) = Cf(x) + Dg(x) ψ(x) = F e ikx + Ge ikx Scattering Matrix Scattering from the left: Scattering from the right: B = S 11 A + S 1 G F = S 1 A + S G S = ( B F ( S11 S 1 ) S 1 S ) ( ) A = S G R l = S 11 T l = S 1 R r = S T r = S 1 If you want to locate the bound states, put in k iκ. 14