An interesting take on the De Broglie wavelength. This is
adapted from Ch1 Sect 5 of Quantum Mechanics of Particles and Fields by Arthur
March, available from Dover Publications.
Consider a particle as an object of great or even infinite
length along the x axis. Coordinates for this particles frame are x’,y’,z’,t’.
Assume it oscillates say in the y direction with period tau’. Assume it is now
moving in the x direction with respect to a set of coordinates K ( x,y,z, t) at
velocity v. For simplicity we set t=0 when t’=0 at x=x’=0. Because the times in
the particle frame and K differ, different positions, x, will see different
phases of the oscillation. In particular let’s find a position, x*, in K where
at time t=0 the phase is the same at x =0. This means that the time in the
particle frame at this location must be tau’, i.e. a full period, ahead or
behind t’=0. Let’s assume a period head i.e. t’ at x=x* and t=0 =tau’
Using the Lorentz transformation we have
t=0=gamma(tau’+vx’/c^2). [gamma =1/sqrt(1-v^2/c^2)
].
This means that x’ corresponding to t=o and x=x*
is given by vx’/c^2 = -tau’ or x’ = -tau’ c^2/v.
From this we can, using the Lorentz transformation for
x, find x*
x* = gamma (x’+vt’) = gamma (-c^2tau’/v +v tau’)
=gamma v tau’(-c^2/v^2 +1) = gamma v tau’
v (1-c^2/v^2)
taking a factor of c^2/v^2 out of the parentheses we have
= gamma tau’
c^2/v(v^2/c^2- 1) note that the term in () is –1/gamma^2
we have x* = (- c^2/v) tau’/gamma.
Since the phase at x* is the same as at x=0 the distance
between the two can be viewed as an apparent wavelength, lambda. So lambda apparent = (c^2/v) tau’/gamma. Since frequency nu = 1/tau
we have lambda= (c^2/v) /nu or
lambda nu = (c^2/v)
so that (c^2/v) can be thought
of as the phase velocity.
Since E = gamma mc^2 = h nu
and momentum, p
= gamma mv= mc^2 (v/c^2) = h nu (v/c^2)
p = h nu / (c^2 /v)= h nu/ lambda nu = h/ lambda or p = h/
lambda, the De Broglie relation.
Thus when a particle exhibits wave nature it is because the
particle is intrinsically oscillating in some sense, which determines tau’, and it has motion relative to
an observer the velocity of which determines, with tau’, the observed wavelength.
Note that this also makes the De Broglie wavelength something of an artifact of
special relativity, i.e. the fact that the clocks in the two frames are not
synchronized.
