() 5.73 Lecture #4 4 - 1 Lecture #4: Stationary Phase and Gaussian Wavepackets Last time: tdSE → motion, motion requires non-sharp E phase velocity began Gaussian Wavepacket goal: 〈x〉, ∆x, 〈p〉 = h〈k〉, ∆p = h∆k by construction or inspection Ψ(x,t) is a complex function of real variables. Difficult to visualize. What are we trying to do here? techniques for solving series of increasingly complex problems illustrate philosophical points along the way to solving problems. free particle  So far: infinite well  very artificial δ - function  * nothing particle-like * nothing molecule-like * no spectra Minimum Uncertainty (Gaussian) Wavepacket -- QM version of particle. We are going to construct a Ψ(x,t) for which | Ψ(x,t)|2 is a Gaussian in x and the FT of Ψ(x,t), gives Φ(k,t), for which |Φ(k,t)|2 is a Gaussian in k. center of wavepacket follows Newton’s Laws extra stuff: spreading interference tunneling Today: (improved repeat of material in pages 3–4 through 3–1 infer ∆k by comparing g(k) to std. G(x; x0, ∆x) gk α ()gk =| ( ) | eik for k near k0 dα ≡−x0 STATIONARY PHASE dk =kk0 2x tΨ(, ) moving, spreading wavepacket how is it possible that the center of the wavepacket vG ≠ vφ moves at a different velocity than its center k- component revised 9/4/02 2:33PM 45.73 Lecture #4 4 - 2 Here is a normalized Gaussian (see Gaussian Handout) Gx; x0, ∆x)= (2π)−1/2 ∆ 1x e−(x −x0 )2 [2(∆x)2 ]( ∞normalized ∫−∞ G(x; x0 , ∆x)dx = 1     center x = x0 by construction  std. dev. ∆x ≡[ x2 − x 2 ]1/2   Now compare this special form against a1/2 ∞ −(a2 4)(k −k0 )2 eikx F.T. of aΨ(x,0) = (2π)3/4 ∫−∞ e12g(k) free particle a Gaussian in k, but what width and 〈k〉? 444 3{ dk Gaussian in k by analogy ( a  a2 1 Gk;k0, ∆k)= (2π)−1/2  21/2  g(k) 4 = 2(∆k)2 by analogy with G(x;x0,∆x) 123 21/2 ∴∆k = a 1 ∆k So casual inspection of this form of Ψ(x,0) gives us 〈k〉 and ∆k. Not quite so easy to get 〈x〉 and ∆x. If we actually carry out the F.T. specified in the definition of Ψ(x,0) above (see bottom of page 3–4), we get 14 Ψ(,0) =  π 2a2  / eik0xe−xa2 2 2(∆ 1x)2 = a12x x = x0 = 0 ∆x = a21/2 / x/∆= 2−12a, previously k = k0, ∆k = 212; a revised 9/4/02 2:33PM∆∆ gk     5.73 Lecture #4 4 - 3 But the square of a Gaussian is a Gaussian and its ∆x or ∆k is a factor of 2-1/2 smaller than the original value. a x x∆x for Ψ(,0) is 2-1/ 2 a, ∆x for Ψ(,0)2 is . 2 1/2 k k∆k for Φ(,0) is 2 , ∆k for Φ(,0)2 is 1. a a a 1 1xk = = See CTDL, p. 231 [∆x,∆k are defined 2 a 2 rigorously in contrast to treatment on p. 23.] This is a very special Gaussian wavepacket * minimum uncertainty * x0 = 0 What about more general Gaussian wavepackets.? g(k) is a complex function of k sharply peaked near k = k0 g(k) =|g (k)|eiα(k) amplitude, argument form If |g(k)| is sharply peaked near k = k0, then the only relevant part of α(k) is the part for k near k0 − higher termsExpand α(k) = α(k0 )+(kk0 ) dα+ neglected123 dkkk0 = α0 12a/ kx2π / ∞ |( ) | eiα()eikxdkΨ(,0) = ()34 ∫−∞ 1442443   ik − k0 ) dα+ kx  g(k)eiα 0e ( dk k=k0  We want to “cook” Ψ(x,0) so that it is localized near x = x0. In order for this to happen, −the factor (kk0 ) dα+ kx , must be indpendent of k near k = k0. Stationary Phase! =dkkk0 revised 9/4/02 2:33PM 5.73 Lecture #4 4 - 4 How does integral of a wiggly function accumulate? k ′ e.g., I(k) = ∫−∞ eikxdk′ I(k) but if phase factor stops wiggling near k = k0 k0 F(k0)δk where δk is range of k over which the phase factor changes by π. So, arrange for phase factor to become stationary near k = k0 0 = d (k − k0 ) dα+ kx dk dk dα0 = + x satisfied ifdk ddk x kkα =≡− 0 0 ! k revised 9/4/02 2:33PM() () 5.73 Lecture #4 4 - 5 Thus − dα dk k = k 0 ↓ a1/2 ∞ −(a2/4)(k − k0 )2 (Ψ(x,0) = (2π)3/4 eiα 0 ∫−∞ e144244 e−ik −k0 )x0eikx 3144244 dk3 g(k) eik(x − x0 )eik0x0 (stops wiggling only when ( insertion of e±ik − k0 )x 0 phase factor ↓ x ≈ x0)  to center w.p. at x0.  δ(x − x0 ) shifts Ψ to any desired x0 Now put in time-dependence by adding e−iωkt factor ωk = Ek = h2k2 1 h 2m  h hk2 ωk =2m xt 12 ∞ gk e −ik − k0 )x0 ikxa/ (Ψ(,) = ()34 ∫−∞ 1442443e{e −iω ktdk2π / gk eigenstate of H revised 9/4/02 2:33PM  35.73 Lecture #4 4 - 6 This FT is evaluated and simplified in CTDL, page 64   4m2a4 exp− mΨ(x,t) 2 =(π2 a2)1/ 2 1+ 4h2t2 2a2(x − hk0 t)2  1243  a4+4h2t2 time dependent  42444normalization 144 m2 Gaussian with time dependent width and center position Maximum of Gaussian occurs when numerator of exp –[ ] is 0. MOTION: 0= x − hk0 t x0 (t) = hk0 t m m = dx0 (t) = hk0= p0= vclassicalvGdt m m This is 2× larger than vφ. Classically expect free particle to move at constant v = p m WIDTH: compare coefficient of (x - x0(t))2 in exp – 12/ ∆= a4 + 4h2t2 /m2  ≈ a + ht x 2 ma 4a2  {{ minimum width increases width at linearly in t at long t = 0 time (quadratically at early time). [] to standard G(x;x0 , ∆x) in handout 1 2242 2 4 22 2 ∆()= + x a a t m h 〈x〉 and ∆x are time dependent, but what about 〈k〉 and ∆k? recall original definition of Ψ(x,0) (page 4-2), where Ψ(x,0) is written as the FT of a Gaussian in k g(k,t) = e−iωktg(k, 0) revised 9/4/02 2:33PM(, ∆∆ 5.73 Lecture #4 4 - 7 k = k0 ∴Φ kt)2 has 1 time independent∆= k a  We know free particle must have time independent k0 and ∆k (no forces — divide w.p. into ∆k slices) 1  4h22 1/2 txk =1 + 24  minimum uncertainty at t = 0 (and linearly increasing at long t).2  ma  For free particle, build w.p. with any desired x0, k0, ∆k starting from ∞ xtΨ(,) = ∫−∞ g(k)eikxe −iω ktdk ω k = hk2 2m dαfind x0 from – dkkk0 = xt0() = x0 + vGt vG = hk0 m / a  4h22 12 t x∆= 2 1 + 24  ma if we want a value of ∆x other than a/2 at t = 0, replace x by x′ = …