What does it mean that a wave packet has an uncertainty

[ehrenfest]

What exactly does it mean that a wave packet has an uncertainty deltaX = delta/2^(1/2)? I thought deltaX represented the standard deviation of the observable X for that particle but what does delta mean by itselt on the right-hand side?

 

[meopemuk]

What exactly does it mean that a wave packet has an uncertainty deltaX = delta/2^(1/2)? I thought deltaX represented the standard deviation of the observable X for that particle but what does delta mean by itselt on the right-hand side?

It would be nice if you could explain the context in which you saw this formula. My guess is that the shape of the wave packet was expressed through some function (e.g., a gaussian) and delta was a parameter of this function. Is it true?
Eugene.

 

[ehrenfest]

I'm sorry. You're right. The equation of the wave packet is

$$\psi(x',0) = e^{i p_0 x'/\hbar}\frac{e^{-x'^2/2\Delta^2}}{(\pi\Delta)^{1/4}}$$.

So delta just represents some constant, characteristic of a given wave packet?

 

[meopemuk]

I'm sorry. You're right. The equation of the wave packet is

$$\psi(x',0) = e^{i p_0 x'/\hbar}\frac{e^{-x'^2/2\Delta^2}}{(\pi\Delta)^{1/4}}$$.

So delta just represents some constant, characteristic of a given wave packet?

That's right. Parameter delta is proportional to the width of the Gaussian wave packet. Thus delta is a measure of the position uncertainty in this particular case.

Eugene.

 

[olgranpappy]

I thought deltaX represented the standard deviation of the observable X for that particle

yeah, it does. and you can calculate that quantity in the case of the state you have given. i.e., calculate
$$\sqrt{<\psi|(x-<\psi|x|\psi>)^2|\psi>}$$

for the $$\psi(x)$$ you have given and my guess is that you will arrive at the result you want.

 

[olgranpappy]

...indeed, it's quite easy for your case since <x>=0. you just need to find $$\sqrt{<x^2>}$$