What is the relationship between wavepacket uncertainty and probability density?

[T-7]

Hi,

I'm puzzled by a couple of formulae in the answer sheet to a problem set I'm working on.

To calc. the new uncertainty in the position of a group of electrons, initially localised to \pm1\mum, after time t, it uses the factor:

\left(1+\frac{\hbar^{2}(\Delta k^{4}t^{2})}{4m^{2}}\right)^{1/2}

The equation I would use for the probability density of this wave is

<br /> \left|\Psi(z,t)\right|^{2} = \frac{\pi.(\Delta k^{2})}{\left(1+\frac{\hbar^{2}(\Delta k^{4}t^{2})}{4m^{2}}\right)^{1/2}}.exp\left(\frac{(\Delta k^{2})(z-vt)^{2}}{2\left(1+\frac{\hbar^{2}(\Delta k^{4}t^{2})}{4m^{2}}\right)}\right)<br />

The denominator of the exp. component describes the increasing width of the wave packet (whereas \left(1+\frac{\hbar^{2}(\Delta k^{4}t^{2})}{4m^{2}}\right)^{1/2} describes the decreasing peak height). I am puzzled by the square root sign they've introduced. Shouldn't they have simply applied it without?

Clearly, one needs to calc. \Delta k to determine the new uncertainty in position, but I'm also puzzled by the claim that \Delta p\Delta z \approx \sqrt{2} (nowhere justified). How do they figure that one?

Cheers!

 

[genneth]

The uncertainty of a gaussian is usually taken to be its std. dev. The denominator in the exponent is usually 2\sigma^2 where sigma is the S.D.

 

[T-7]

Thank you.