# Rate of the increase of width of a Gaussian wavepacket

So, I'm following the MIT OCW lectures on 8.04 quantum mechanics by Prof. Allan Adams. I have the expression for the probability distribution of a gaussian wavepacket for a free particle situation. No initial momentum is imparted. This is a non-relativistic treatment.

$$\mathbb{P}(x,t) = a(t) e^{\frac{-x^{2}}{2(a^{2}+(\frac{\hbar}{2ma})^2t^{2})}}$$

$a(t)$ decreases with time. Allan says that the velocity with which the width of this gaussian increases is $\frac{\hbar}{2ma}$, pretty much from dimensional analysis.

However, the width of this gaussian at time = $t$ is $w = \sqrt{a^{2}+(\frac{\hbar}{2ma})^2t^{2}}$. I tried to differentiate $w$ with respect to time but did not get the proposed velocity. Can someone help me here.

Differentiating $$\sigma=\sqrt{a^2+\left({\hbar t \over 2 m a}\right)^2}$$ you get - $$\frac{\partial \sigma}{\partial t} = {\hbar^2t \over 4 a^2 m^2\sqrt{a^2+\left({\hbar t \over 2 m a}\right)^2}}$$
In the limit of $t \rightarrow\infty$ you get $$\left(\frac{\partial \sigma}{\partial t}\right) _{t\rightarrow \infty} = {\hbar \over 2 m a}$$
• I don't sure what you mean by "velocity limit". The expectation value of the gaussian don't change in time (it is a static particle), but the uncertainty in position grows with $t$. – Alexander May 13 '16 at 15:34
• By velocity limit, I meant the time rate of change of the 'width' of the gaussian in the limit that $t \rightarrow\infty$, i.e., $\left(\frac{\partial \sigma}{\partial t}\right) _{t\rightarrow \infty}$ – IanDsouza May 16 '16 at 6:27