130_notes.dvi

Its already easy to see that the phase in a wave like

ei(~p·~x−Et)/h ̄

is Lorentz invariant.

To read off the wavelength, lets pick off the part of the expression that corresponds toe^2 πi(x/λ−νt).
We see that^2 λπ=p ̄hand therefore the DeBroglie wavelength is.

λ=

2 π ̄h

p

=

h

p

DeBroglie derived this result in 1923.

We can also read off the frequencyν.

ν=

E

2 π ̄h

=E/h

This was in some sense the input to our calculation.

The deBroglie wavelength will be ourprimary physics inputfor the development of Quantum
Mechanics. Its not that this work was the most significant, but, this wavelength summarizes most
of what happened before 1923.

3.4.1 Computing DeBroglie Wavelengths

We usually quote the energy of a particle in terms of itskinetic energyin electron Volts, eV (or
Million electron Volts, MeV). The reason for this is that particles are usually accelerated to some
energy by an electric field. If I let an electron (or proton...) be accelerated through a 100 Volt
potential difference, it will have a kinetic energy of 100eV.

The whole problem of computing a deBroglie wavelength is to convert from kinetic energy to mo-
mentum. If you always want to be correct without any need for thinking, use therelativistically
correct formulafor the kinetic energy

T=

√

(mc^2 )^2 +p^2 c^2 −mc^2

and solve it forpc,
pc=

√

(T+mc^2 )^2 −(mc^2 )^2

then use this handy formula to get the answer.

λ=

h

p

=

2 π ̄hc

pc

I remember that ̄hc= 1973 eV ̊A allowing me to keep the whole calculation in eV. I also know the
masses of the particles.
mec^2 = 0.51 MeV
mpc^2 = 938.3 MeV