Gaussian Wave Packet

Free-Particle Schrödinger Equation: Wave Packets

3.2 Schrödinger’s Equation

3.2.1 Wave Packets

3.2.1.1 Gaussian Wave Packet

AttD0, I take the wave function for a particle having massmto be .x; 0/D 1

. ²/¹⁼⁴e^{x2=2 2}e^ik0x; (3.24) such that

j .x; 0/j²D 1

. ²/¹⁼²e^{x2= 2} (3.25)

is a Gaussian centered at the origin with a full-width at half maximum (FWHM) equal to 1.67 . The normalization has been chosen such that

Z 1 1

dxj .x; 0/j²D1: (3.26) I want to find .x;t/:To do so, I must first findˆ.k/and then use Eq. (3.14) to get .x;t/:The factor ofe^ik0xin Eq. (3.24) leads to an average velocity for the packet equal to„k₀=m;as you shall see.

It follows from Eq. (3.14) that .x; 0/D 1

p2 Z 1

1

dkˆ.k/e^ikx: (3.27) I take the inverse Fourier transform of this equation to obtain

ˆ.k/D 1 p2

Z 1 1

dx .x; 0/e^ikx

D 1 p2

1 . ²/¹⁼⁴

Z 1 1

dx e^{x2=2 2}e^ik0xe^ikx: (3.28) The integral is tabulated or can be evaluated using contour integration. In either case, one finds

ˆ.k/D 2

¹⁼⁴

e^{.kk0/2 2=2}I (3.29)

jˆ.k/j²D 2

¹⁼²

e^{.kk0/2 2}: (3.30)

The k-space distribution is alsoa Gaussian, centered at k D k0; having FWHM equal to 1.67/ :

The variance ofxattD0is Œx.tD0/²D˝

x².tD0/˛ D

Z 1 1

dx x²j .x; 0/j²D

2

2 (3.31)

and the variances ofkandpare .k/²DD

.kk0/²E D

Z 1 1

dk.kk0/²jˆ.k/j²D 1

2 ²I (3.32a) .p/²D „²

2 ²; (3.32b)

such that

x.tD0/kD 1

2I (3.33a)

x.tD0/pD „

2: (3.33b)

As you shall see, this corresponds to what is called aminimum-uncertainty wave packet, having the minimum value ofxpallowed for solutions of Schrödinger’s equation. The momentum distribution andpdo not change in time since no forces act on the particle.

Owing to the spread of momenta in the wave packet, however,x doeschange as a function of time. The wave packet is no longer a minimum uncertainty packet fort> 0. Using Eqs. (3.14) and (3.29), I calculate

.x;t/D 1 p2

2

¹⁼⁴Z 1 1

dk e^{.kk0/2 2=2}eⁱ.^kx„k2t=2m/

D e^ik0.^xv0₂^t/ p2

2

¹⁼⁴Z 1 1

dk⁰e^k02. ^2Ci„t=m/⁼²e^ik0.xv0t/;

(3.34) wherek⁰Dkk0and

v0D „k0=mDp0=m: (3.35)

The integral is tabulated or can be evaluated using contour integration and the result is

.x;t/D 2

¹⁼⁴

e^ik0.^xv0₂^t/ h 2C ^i„t_mi1=2exp

0

@.xv0t/² 2h

2C^i„t_mi 1

A: (3.36)

60 3 Free-Particle Schrödinger Equation: Wave Packets As a consequence,

j .x;t/j²D 1

.t/² 1=2

e^{.xv0t/2= .t/2}; (3.37) with

.t/²D ²C „t

m 2

: (3.38)

The FWHM at any time is 1.67 .t/andx.t/D .t/=p

2;such that

x.t/pD „ .t/

2 D „ 2

"

1C „t

m ² 2#1=2

„

2: (3.39)

The packet remains Gaussian but spreads owing to the spread of momenta in the original packet. To see howx.t/depends onp, I use the relationshipsx.0/D =p

2andvDp=mD „=p 2m

to rewritex.t/²as

x.t/² D .t/²

2 D

2

2 C 1 2

„t m

2

Dx.0/²C.v/²t²: (3.40)

Although I have chosen a Gaussian wave packet, Eq. (3.40) turns out to beexactfor any square-integrable initial wave function of the form .x; 0/Df.x/e^ik0x, for real f.x/(see Problem 5.14–15in Chap.5). The variance of the wave packet is itsinitial variance plus a contribution attributable to the variance of the velocity components contained in the packet. For sufficiently large times,x.t/vt.

I am now in a position to see when the wave packet can correspond to a classical particle. Free particles have never heard about wave packets; wave packets are a construct of physicists. For the wave packet to correspond to a particle, however,the uncertainties in position and momentum must satisfy

x.t/x0; pp0 (3.41)

subject to the restriction

x.0/p „

2: (3.42)

The quantities x0 and p0 are determined by the problem. You can think of them as the smallest possible resolution in position and momentum that can be detected in a given experiment. In bound state problems they could correspond

to some typical bound state radius and magnitude of bound state momentum for the bound particle. For the free particle, let’s takex0 D 10⁸ m andv0 D 10⁷ m/s, which locates the particle to better than an optical wavelength and fixes its velocity to about three meters per year. We might be able to accomplish this by using spatial filters (e.g., slits) to select both the position and range of velocities.

For a one gram mass, suppose we take x.0/ D 10¹¹ m which implies that v D p=m „=Œmx.0/ 10²⁰ m/s [admittedly, it would be difficult to create such a small wave packet]. This is a “classical” particle attD0according to my definition since it obeys conditions (3.41). At what timetwould the spreading be sufficient to render the particle “unclassical” ? Arbitrarily, let’s say the particle is no longer classical ifx.t/Dx0=100D10¹⁰m, which occurs fortD10¹⁰s, 300 years! The bottom line is that spreading is unimportant as long as the de Broglie wavelength is much smaller than any characteristic length in the problem, such as the width of the initial wave packet. On the other hand, if you confine a free particle wave packet to a distance equal to its de Broglie wavelength, the spread in momentum in the wave packet is of order of the average momentum in the packet;

as such spreading is important and the particle can no longer be viewed as a classical particle.

A simple example that illustrates the necessity of using a quantum description of matter can be found in an experiment related to atom optics. Suppose a well- collimated, pulsed atomic beam having velocityv0in thezdirection is incident on a circular aperture having diameterdthat is located in thexyplane. Moreover, assume that the de Broglie wavelength of the atoms,dBd. The atoms are treated as point particles, so another implicit assumption is that the atomic size is much smaller thand as well. After traversing the aperture att D 0, you can think of the initial wave packet as a short pulse in thezdirection having a cross-sectional area equal tod²=4. The transverse uncertainty in the momentum of this beam is of order p? „=d. Matter wave effects become important when the transverse spreading is of orderd;that is, for timestgreater than some critical timetF defined by

p?tF=m „tF=.dm/d; (3.43) wheremis the mass of an atom in the beam. Sincetz=v0wherezis the distance from the screen containing the aperture, the distancezF corresponding to the time tFis

zFDv0tF mv₀d²

„ d² dB

: (3.44)

Forz zF, the scattering of the particles by the slit is in theshadow regionand the atomic motion can be treated classically. However forz& zF, diffraction plays an important role and a wave theory is needed. The situation is analogous to the scattering of optical radiation having wavelengthby an aperture having diameter d. For distanceszzF D d²=from the diffracting screen, a geometrical picture of light rays can be used, but once z & zF, diffraction effects become important and a wave theory of light is needed. In the optical case the region withz zF

corresponds toFresnel diffraction.

62 3 Free-Particle Schrödinger Equation: Wave Packets Perhaps the best way to create a well-defined wave packet is to trap and cool an atom in the potential well of anoptical lattice. An optical lattice is formed by using pairs of counter-propagating laser beams. These pairs of fields form standing wave patterns that can be used to trap neutral atoms owing to a spatially varying potential that is experienced by the atoms in the fields. Moreover the atoms can be cooled to the point where they are in the ground state of the potential. As such, if you trap one atom in one well, you have a pretty good idea of its wave function. If you suddenly remove the potential by turning off the fields, you have an initial condition in which the atom is in its ground state and has a center-of-mass wave function given by the ground state of the potential. You could let this wave packet propagate for some time and then restore the lattice and determine how far the packet has moved by seeing which well it is in. Although this experiment has yet to be carried out, the technology is now at the point where it is feasible.