130_notes.dvi

(Frankie) #1

To cover the general case, lets expandω(k) around the center of the wave packet in k-space.


ω(k) =ω(k 0 ) +


dk





k 0

(k−k 0 ) +

1

2

d^2 ω
dk^2





k 0

(k−k 0 )^2

We anticipate the outcome a bit and name the coefficients.


ω(k) =ω 0 +vg(k−k 0 ) +β(k−k 0 )^2

For the photon,vg=candβ= 0. For the NR electron,vg= ̄hkm^0 andβ= 2 ̄hm.


Performing the Fourier Transform (See section 5.6.5), we get


ψ(x,t) =


π
α+iβt

ei(k^0 x−ω^0 t)e

−(x−vgt)^2
4(α+iβt)

|ψ(x,t)|^2 =

π

α^2 +β^2 t^2

e

−α(x−vgt)^2
2(α^2 +β^2 t^2 ).

We see that the photon will move with the velocity of light and that thewave packet will not
disperse, becauseβ= 0.


For the NR electron, the wave packet moves with the correctgroup velocity,vg=mp, but the


wave packetspreads with time. The RMS width isσ=



α^2 +

( ̄ht
2 m

) 2

A wave packet naturally spreads because it contains waves of different momenta and hence different
velocities. Wave packets that are very localized in space spread rapidly.


5.6 Derivations and Computations


5.6.1 Fourier Series*.


Fourier series allow us toexpand any periodic functionon the range (−L,L), in terms of sines
and cosines also periodic on that interval.


f(x) =

∑∞

n=0

Ancos

(nπx
L

)

+

∑∞

n=1

Bnsin

(nπx
L

)

Since the sines and cosines can be made from the complex exponentials, we can equally well use
them for our basis for expansion. This has the nice simplification of having only one term in the
sum, using negativento get the other term.


f(x) =

∑∞

n=−∞

ane

inπxL

The exponentials are orthogonal and normalized over the interval(as were the sines and cosines)


1

2 L

∫L

−L

e

inπxL
e

−imπxL
dx=δnm
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