130_notes.dvi

Similarly the phase goes through 2πin one periodτin time.

ωτ= 2π

ωis the angular frequency. It changes by 2πevery cycle. The frequencyνincreases by 1 every cycle
so
ω= 2πν.

There is no reason to memorize these equations. They should be obvious.

Lets see how fast one of the peaks of the wave moves. This is called the phase velocity. At time
t= 0, there is a peak atx= 0. This is the peak for which the argument of cosine is 0. At time
t= 1, the argument is zero whenkx=ωtor atx=ωk. If we compute the phase velocity by taking
∆x
∆t, we get
vphase=

ω

k

That is, one of the peaks of this wave travels with a velocity ofωk.

v=

ω

k

=

2 πν

2 π

λ

=νλ

In non-relativistic QM, we have ̄hk=p,E= ̄hω, andE=p

2

2 m, so

ω(k) =

E

̄h

=

̄h^2 k^2

2 m ̄h

=

̄hk^2

2 m

You may remember that a pulse will move at the group velocity which is given by

vg=

(

dω

dk

)

=

2 ̄hk

2 m

=

̄hk

m

=

p

m

(The phase velocity for the non-relativistic case isvp= 2 pm.)

4.2 Sample Test Problems

1. Write down the two (unnormalized) free particle wave functions for a particle of kinetic energy
   E. Include the proper time dependence and expressions for otherconstants in terms of E.

2.