130_notes.dvi

We write explicitly thatwdepends onk. For our free particle, this just means that the energy
depends on the momentum. To cover the general case, lets expandω(k) around the center of the
wave packet in k-space.

ω(k) =ω(k 0 ) +

dω

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

We still need to do the integral as before. Make the substitutionk′=k−k 0 givingA(k′) =e−αk
′ 2

Factor out the constant exponential that has nok′dependence.

ψ(x,t) = ei(k^0 x−w^0 t)

∫∞

−∞

A(k′)ei(k

′x−vgt)

e−ik

′ (^2) βt
dk′
ψ(x,t) = ei(k^0 x−w^0 t)

∫∞

−∞

e−αk

′ 2

ei(k

′x−vgt)

e−ik

′ (^2) βt
dk′
ψ(x,t) = ei(k^0 x−w^0 t)

∫∞

−∞

e−[α−iβt]k

′ 2

ei(k

′x−vgt)

dk′

We nowcompare this integral to the one we did earlier(so we can avoid the work of completing
the square again). Dropping the constants, we had

f(x) =eik^0 x

∫∞

−∞

e−αk

′ 2

eik

′x

dk′=eik^0 xe−

x 4 α^2

Our new integral is the same with thesubstitutionsk 0 x→k 0 x−ω 0 t,k′x→k′(x−vgt), and
α→(α+iβt). We can then write down the answer

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+iβt

√

π

α−iβt

e

−(x−vgt)^2

4(α+iβt)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 )

5.6.6 Numbers

The convenient unit of energy (mass and momentum too) is the electron volt.

1 eV= 1. 602 × 10 −^12 erg = 1. 602 × 10 −^19 Joule