On De Broglie Wavelength and Uncertainty

DeBroglie Waves and Heisenberg's Uncertainty

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 ( plus 2pi) 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.

E =h nu or E tau = h tau =1/nu which is more intuitive. Tau is simply the characteristic time [or period] it takes for a particle of energy E to accumulate one h worth of increase in total action. 

Similarly, De Borglie's relation, p lambda = h indicates that a particle of momentum p must travel a distance lambda to accumulate one h of increase in total action.

If total action is truly quantized, then there is no measurable change until total action changes by h.

Heisenberg’s uncertainty principal can be explained as follows: Suppose one measures the change in total action over a time interval, t,  less than tau. Say sometime during that interval, t,  the total action changes by one h.  The only things you know are that sometime during t total action changed by h. From this you conclude that E could be as high as h/t. On the other hand if you think about it a bit you realize that your interval t could have started after a large fraction of tau had passed since the last increase in total action by h, so that most of the accumulation of energy time leading to an addition of h in total action occurred prior to your beginning your measurement. Therefore the energy could be much lower( near zero?). Thus you are uncertain about E by h/t.

Now suppose your measurement takes place over a period, t’,  several times tau, lets say  t’ = 7 tau for example. Depending on exactly when you begin your measurement vis a vis  when a period begins [ i.e. the time when the last increase in total action by h occurred prior to the measurement]  you will measure either 6 h or 7h as the change in total action. So you know the energy is between 6h/t’ and 7h/t’  but since t’ is 7 x t , you have reduced your uncertainty by a factor of 7. 

Similarly, for momentum if you measure the particle’s total action over a length , r, shorter than lambda , you may detect a change in total action of h. Then you can conclude that the momentum may be as high as h/r. However, most of the accumulation of momentum times distance since the last change in total action may have occurred in the particle path before your measurement, so p could be much lower (near zero?) so your uncertainty in p is h/r. If you measure over r’ = 7 lambda  you will measure between 6h and 7h change in total action, and the uncertainty becomes h/r’ or 1/7 of h/r.