Question 1.10.16: A neutron is confined in space to 10^-14 m. Calculate the ti...

Since the rest mass energy of a neutron is equal to m_nc^2=939.6MeV,  we can infer the time constant for the neutron from (1.227) \tau=\frac{2mc^2(\Delta x_0)^2}{\hbar c^2}\simeq \frac{2\times 0.5  MeV\times 10^{-12}m^2}{197\times 10^{-15}MeV  m\times 3\times 10^8 m  s^{-1}} =1.7\times 10^{-8}s, :

\tau =\frac{2m_nc^2(\Delta x_0)^2}{\hbar c^2}\simeq \frac{2\times 939.6  MeV\times (10^{-14}m)^2}{197\times 10^{-15}MeV  m\times 3\times 10^{8}m  s^{-1}}=3.2\times 10^{-21}s.              (1.231)

Inserting this value in (1.226) we obtain the time it takes for the neutron’s packet to grow from an initial width \Delta x_0  to a final size \Delta x(t):

t=\tau \sqrt{\left(\frac{\Delta x(t)}{\Delta x_0} \right)^2-1 }=3.2\times 10^{-21}s \sqrt{\left(\frac{\Delta x(t)}{\Delta x_0} \right)^2-1.}                (1.232)

The calculation of t reduces to simple substitutions.
(a) Substituting \Delta x(t)=4\Delta x_0  into (1.232), we obtain the time needed for the neutron’s packet to expand to four times its original size:

t=3.2\times 10^{-21}s\sqrt{16-1}=1.2\times 10^{-20}s.                (1.233)

(b) The neutron’s packet will expand from an initial size of 10^{-14}m  to  12.7 \times 10^{6}m   (the diameter of the Earth) in a time of

t=3.2\times 10^{-21}s\sqrt{\left(\frac{12.7\times 10^{6}m}{10^{-14}m} \right)^2-1 }=4.1  s.                  (1.234)

(c) The time needed for the neutron’s packet to spread from 10^{-14}m  to  3.84 \times 10^{8}m   (the distance between the Earth and the Moon) is

t=3.2\times 10^{-21}s\sqrt{\left(\frac{3.84\times 10^{8}m}{10^{-14}m} \right)^2-1 }=12.3  s.              (1.235)

The calculations carried out in this problem show that the spread of the packets of microscopic particles is significant and occurs very fast: the size of the packet for an earthly neutron can expand to reach the Moon in a mere 12.3 s! Such an immense expansion in such a short time is indeed hard to visualize classically. One should not confuse the packet’s expansion with a growth in the size of the system. As mentioned above, the spread of the wave packet does not mean that the material particle becomes bloated. It simply implies a redistribution of the probability density. In spite of the significant spread of the wave packet, the packet’s norm is always conserved; as shown in (1.149) \int_{-\infty }^{+\infty }{|\psi (x,t)|^2}dx=\sqrt{\frac{2}{\pi a^2} }\frac{1}{\gamma }\int_{-\infty }^{+\infty }exp\left\{-\frac{2(x-\hbar k_0t/m)^2}{(a\gamma )^2} \right\} dx=\sqrt{\frac{2}{\pi a^2} }\frac{1}{\gamma }\sqrt{\frac{\pi a^2\gamma ^2}{2} }=1,   it is equal to 1.