8.04 Quantum Physics Lecture XIV Figure I: A wavepacket ψ(x) in position space or φ(k) in momentum space whose wavefunction for large x (or k) falls off faster than x 1/2 (k 1/2 ) can be directly normalized. 2. Periodic boundary conditions Assume box of finite length L, require periodic boundary conditions ψ(0) = ψ(l) (14-7) For plane waves e ipx/h this implies that e ipl/h = 1 or pl = kl = n2π, n integer, h 2π i.e. momentum is quantized, p n = n hk o with k 0 =. The corresponding L momentum states are normalizable in the interval [0, L], L dx Ce ipx/ h 2 = L C 2 = 1 (14-8) 0 1 u pn (x) = L e ipnx/ h (14-9) normalized momentum eigenstates in box of size L with p n = n hk o L dxu pn (x)u pm (x) = δ nm orthonormality condition in box (14-10) 0 We perform all calculations for fixed size box, then take the limit L (i.e. 0, momentum spectrum becomes continuous). All physically sensible k 0 Massachusetts Institute of Technology XIV-2

8.04 Quantum Physics Lecture XIV Figure II: Wavefunction in box of length L with periodic boundary conditions. results will be independent of the initially chosen box size L as long as L is large compared to distances of interest. Time evolution of free-particle wavepackets In free space we often work with normalized Gaussian wavepackets 1 x 2 4w 2 Ψ(x, t = 0) = 0 (2π) 1/4 w 1/2 e (14-11) 0 Written in this form we have 2 x Ψ(x, 0) 2 1 = (2π) 1 /2 w 0 e 2w 2 0 dx Ψ(x, 0) 2 = 1 x = 0 2 2 x = w 0 (δx) 2 = x 2 x 2 = w 0 2 Δx = w 0 is the uncertainty or rms width (root-mean-square width) of the wavepacket. Why do we prefer this Gaussian form of wavepacket? 1. Particularly simple and symmetric, the Fourier transform is also a Gaussian wavepacket: 1 4k 2 φ(k) = k 2 0 1/2 e (14-12) (2π) 1/4 k 0 with k 0 = 2w 1 0. (Δk)2 = k 2 k = k 0 2 2. This is a wavepacket with the minimum uncertainty ΔxΔk = 2 1 (ΔxΔp = h 2 ) allowed by QM 3. Physical system after give rise to Gaussian broadening in momentum or position, e.g., thermal distribution of atomic momenta in a gas is a Gaussian distribution. Massachusetts Institute of Technology XIV-3

8.04 Quantum Physics Lecture XIV How do we make a wavepacket move at velocity v 1? We displace the distribution in momentum space from p = hk = 0 to p = hk = hk 1 = mv 1 (see Fig. III). φ(k) = 1 (2π) 1/4 k 0 (k k 1 ) 2 4k 2 0 1/2 e. (14-13) The inverse Fourier transform, i.e., the spatial wavefunction 1 Ψ(x, t = 0) = (2π) 1/ x 2 4wo 4 w 1/2 e 2 e ik 1x 0 (14-14) is still a Gaussian, but now with a phase variation e ik 1x, rather than a constant phase over the wavepacket (compare Eq. (14-11). This phase variation e ik 1x in position Figure III: Moving Gaussian wavepacket with average velocity v 1 = hk 1 /m and spatial 1 wavefunction ψ 1 (x) = (2π) 1/ 2 e x 4w2 eik 1 x 0. 4 w 1/2 0 hk 1 space encodes the motion of the wavepacket at velocity v 1 = m : The dominant de Broglie wavelength in the wavepacket corresponds to a wavevector k 1, or a momentum hk 1. Massachusetts Institute of Technology XIV-4

8.04 Quantum Physics Lecture XIV How does a free-space Gaussian wave packet evolve in time? In general, we expand a wavefunction Ψ(x, 0) into energy eigenfunctions u E (x), and then evolve the energy eigenfunctions as e iet/ h. In free space, there is only KE. Then the momentum eigenstates u p (x) are simultaneous eigenstates of energy: or ˆ pˆ2 Hu p (x) = u p (x) (14-15) 2m ( )2 1 1 = hi e ipx/ h (14-16) 2m x 2πh p 2 = u p (x) (14-17) 2m 2 Hu ˆ (x) = p p u p (x) 2m (14-18) = E p u p (x) (14-19) in free space. The energy eigenstates are said to be doubly degenerate: For each eigenvalue of energy E > 0 there are two different momentum states (namely u ±p (x) with p = 2mh ) that have the same energy. It follows that a momentum eigenstate with eigenvalue p evolves in time as e iept/ h, so that the wavefunction in momentum space evolves in time as 2 Φ(p, t) = Φ(p, 0)e i p t/ h 2m (14-20) time evolution of momentum eigenfunctions in free space. The wavefunction in real space is given by the inverse Fourier transform Ψ(x, t), or equivalently, as the superposition of energy eigenfunctions with their corresponding phase evolution factors e iept/ h : 1 ipx/h Ψ(x, t) = dpφ(p, t)e (14-21) 2πh ( = dpφ(p, t)u p (x)) (14-22) 1 p = dpφ(p, 0)e ipx/ h e 2 2m t/h (14-23) 2πh = dpφ(p, 0)U p (x, t) (14-24) = dpφ(p, t)u p (x) (14-25) Massachusetts Institute of Technology XIV-5

8.04 Quantum Physics Lecture XIV 2 eipx/ h 2 e i p π h where U p (x, t) = u p (x)e p t/ h 2m = 1 t/ h 2m 2 are the time-dependent momentum eigenfunctions in free space. The above equation shows that the phases of different Fourier components u p (x) = 1 eipx/ h 2 evolve in time at different speeds, the π h running out of phase of different Fourier components leads to a spreading of the wavepacket in position space. In the problem sets you will show that the rms width Δx(t) = w(t) of the wavepacket grows in time as h 2 k w(t) = w 2 0 1 + (14-26) m 2 w 2 0 Since a wavepacket contains different momentum components, it changes in time in free space even though there are non external forces acting. For long times t t 0 = mwo 2 h h the wavepacket spreads as w(t) t, i.e. at a speed v 0 = that h mw 0 mw 0 is inversely proportional to its initial size. That speed is negligible for macroscopic wavepacket size, but can be appreciable for initially well-localized microscopic objects. The spreading of a wavepacket in free space was early evidence that the wavepacket size cannot be identified with the particle size. The spreading is due to the quadratic (i.e. not linear) dependence of the energy, and hence the phase evolution rate, on momentum. Note that the wavepacket of a massless particle, e.g. a photon, with E = pc would not spread. (The SE is non-relativistic and does not apply to photons.) Motion of wave packets, group velocity, and stationary phase Why is it that a wavefunction 1 Ψ(x, 0) = (2π) 1/ x 2 4 w 1/2 e 4w0 2 e ik 1x 0 (14-27) hk 1 represents a particle moving at velocity v 1 = m? Since a crest of a single momentum 1 hk 2 2π 2m component u k1 (x, t) = 2π e k1x e i t moves forward a distance λ = k1 in a time hk T = 2 π (remember that ω 1 ω1 1 = 2 and e ie 1t/ h = e iω1t ), the velocity of the crest is 2m λ 2π ω ω 1 hk v 1 ph = T = = k1 2π k 1 = 2m ω 1 hk 1 p 1 v ph = = = (14-28) k 1 2m 2m This is the phase velocity of a momentum component. The particle does not move at the phase velocity v ph = ω 1 k 1 at which the plane wave associated with a single momentum moves forward. At what velocity then? Massachusetts Institute of Technology XIV-6

8.04 Quantum Physics Lecture XIV Look at exponent and write: hk E 1 1 = E 1 (k 1 ) = hω 1 = hω(k 1 ) = 2m e ( ) x 2 4w0 2 +ik 1x iω(k 1 )t 2 Remember Fermat s principle of stationary phase: path is defined by region of space x 2 where phasors point mostly in one direction, i.e. where the phase φ(k) = 4wo 2 + ikx + iω(k 1 )t does not vary between different momentum components k to lowest order ( ) ( ( )) φ ω ω 0 = = ix i t = i x t, (14-29) k k x or ( ) Fermat s principle leads us to the concept of group velocity ω x(t) = t. (14-30) k ω hk 1 p 1 v gr = (k 1 ) = = (14-31) k m m Group velocity of the wavepacket at which the wavepacket, i.e. the region of constructive interference, propagates. The difference between group and phase velocity is due to the fact the ω = ω, or E = ( hω) = E, i.e. the quadratic dependence of KE k k p ( hk) p on momentum in free space. This is in contrast to photons with a linear dispersion relation ω = ω = c in vacuum, where group and phase velocity are the same. k k Massachusetts Institute of Technology XIV-7