# Uncertainty in the wavepacket

1. Aug 30, 2007

### 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

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

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!

Last edited: Aug 30, 2007
2. Aug 30, 2007

### 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.

3. Aug 30, 2007

Thank you.