# Dispersion relation and group velocity

The group velocity for example of electron waves is given by the derivative of the dispersion relation: $\frac{dE}{dp}=v$ (this is for free electrons) $^{1}$. Now the Heisenberg's uncertainty principle has two forms, one for position and momentum and the other for energy and time, namely:

$\Delta x\Delta p \gtrsim h$ [1] $^{2}$

$\Delta E\Delta t \gtrsim h$ [2] $^{2}$

(note these are approximate relations)

Dividing [2] nd expression from the [1] st gives:

$\frac{\Delta E}{\Delta p}\frac{\Delta t}{\Delta x} \gtrsim 1$

By letting the limits of the time and position changes tend to infinitesimally small values i.e. $t\rightarrow 0$ and $x\rightarrow 0$ we get the differential form:

$\frac{dE}{dp}\frac{dt}{dx} \gtrsim 1$

but $\frac{dx}{dt} = v$

Hence

$\frac{dE}{dp} \gtrsim v$ which is the dispersion relation for a free particle.

I realise this may not be the most rigorous derivation, and it may be just a crackpottery as a result of my daydreaming. However, if there is any significance to this, I would like to hear an explanation to it. I could not find anything remotely close to this on the quantum book that I have which is by Alastair I. M. Rae.

Any input would be greatly appreciated,

Kind regards,

Cygni.

References:

(1) Experimental Physics, Modern Methods by R.A. Dunlap page 15.
(2) http://en.wikipedia.org/wiki/Uncertainty_principle

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You can't simply divide those two relations, the result can also be a false statement, for example:

2>1

and

1.5>1.

If you divide the second by the first, you end up at

0.75>1,

which is obviously false.

I understand the reasoning behind the example you have showed, which is fine, it proves easily that I'm wrong. But could you elaborate more on why one can't simply divide those two relations?