This Quantum World/Implications and applications/How fuzzy positions get fuzzier

How fuzzy positions get fuzzier

We will calculate the rate at which the fuzziness of a position probability distribution increases, in consequence of the fuzziness of the corresponding momentum, when there is no counterbalancing attraction (like that between the nucleus and the electron in atomic hydrogen).

Because it is easy to handle, we choose a Gaussian function

$\psi (0,x)=Ne^{-x^{2}/2\sigma ^{2}},$

which has a bell-shaped graph. It defines a position probability distribution

$|\psi (0,x)|^{2}=N^{2}e^{-x^{2}/\sigma ^{2}}.$

If we normalize this distribution so that $\int dx\,|\psi (0,x)|^{2}=1,$ then $N^{2}=1/\sigma {\sqrt {\pi }},$ and

$|\psi (0,x)|^{2}=e^{-x^{2}/\sigma ^{2}}/\sigma {\sqrt {\pi }}.$

We also have that

• $\Delta x(0)=\sigma /{\sqrt {2}},$
• the Fourier transform of $\psi (0,x)$ is ${\overline {\psi }}(0,k)={\sqrt {\sigma /{\sqrt {\pi }}}}e^{-\sigma ^{2}k^{2}/2},$
• this defines the momentum probability distribution $|{\overline {\psi }}(0,k)|^{2}=\sigma e^{-\sigma ^{2}k^{2}}/{\sqrt {\pi }},$
• and $\Delta k(0)=1/\sigma {\sqrt {2}}.$

The fuzziness of the position and of the momentum of a particle associated with $\psi (0,x)$ is therefore the minimum allowed by the "uncertainty" relation: $\Delta x(0)\,\Delta k(0)=1/2.$

Now recall that

${\overline {\psi }}(t,k)=\phi (0,k)e^{-i\omega t},$

where $\omega =\hbar k^{2}/2m.$ This has the Fourier transform

$\psi (t,x)={\sqrt {\sigma \over {\sqrt {\pi }}}}{1 \over {\sqrt {\sigma ^{2}+i\,(\hbar /m)\,t}}}\,e^{-x^{2}/2[\sigma ^{2}+i\,(\hbar /m)\,t]},$

and this defines the position probability distribution

$|\psi (t,x)|^{2}={1 \over {\sqrt {\pi }}{\sqrt {\sigma ^{2}+(\hbar ^{2}/m^{2}\sigma ^{2})\,t^{2}}}}\,e^{-x^{2}/[\sigma ^{2}+(\hbar ^{2}/m^{2}\sigma ^{2})\,t^{2}]}.$

Comparison with $|\psi (0,x)|^{2}$ reveals that $\sigma (t)={\sqrt {\sigma ^{2}+(\hbar ^{2}/m^{2}\sigma ^{2})\,t^{2}}}.$ Therefore,

$\Delta x(t)={\sigma (t) \over {\sqrt {2}}}={\sqrt {{\sigma ^{2} \over 2}+{\hbar ^{2}t^{2} \over 2m^{2}\sigma ^{2}}}}={\sqrt {[\Delta x(0)]^{2}+{\hbar ^{2}t^{2} \over 4m^{2}[\Delta x(0)]^{2}}}}.$

The graphs below illustrate how rapidly the fuzziness of a particle the mass of an electron grows, when compared to an object the mass of a $C_{60}$ molecule or a peanut. Here we see one reason, though by no means the only one, why for all intents and purposes "once sharp, always sharp" is true of the positions of macroscopic objects.

Above: an electron with $\Delta x(0)=1$ nanometer. In a second, $\Delta x(t)$ grows to nearly 60 km.

Below: an electron with $\Delta x(0)=1$ centimeter. $\Delta x(t)$ grows only 16% in a second.

Next, a $C_{60}$ molecule with $\Delta x(0)=1$ nanometer. In a second, $\Delta x(t)$ grows to 4.4 centimeters.

Finally, a peanut (2.8 g) with $\Delta x(0)=1$ nanometer. $\Delta x(t)$ takes the present age of the universe to grow to 7.5 micrometers.