(a) If the initial width of the wave packet of the particle is \Delta x_0, the width at time t is given by
\Delta x(t)=\Delta x_0\sqrt{1+\left(\frac{\delta x}{\Delta x_0} \right)^2 }, (1.216)
where the dispersion factor is given by
\frac{\delta x}{\Delta x_0}=\frac{2 \hbar t}{ma^2}=\frac{\hbar t}{2m(a/2)^2}=\frac{\hbar t}{2m(\Delta x_0)^2}. (1.217)
(i) For the 25 eV electron, which is clearly not relativistic, the time to travel the L = 100 m distance is given by t=L/\upsilon =L\sqrt{mc^2/2E}/c, since E=\frac{1}{2}m\upsilon ^2=\frac{1}{2} mc^2(\upsilon ^2/c^2) or \upsilon = c\sqrt{2E/(mc^2)}. We can therefore write the dispersion factor as
\frac{\delta x}{\Delta x_0}=\frac{\hbar }{2m\Delta x^2_0}t=\frac{\hbar}{2m\Delta x^2_0}\frac{L}{c}\sqrt{\frac{mc^2}{2E} }=\frac{\hbar cL}{2mc^2\Delta x^2_0}\sqrt{\frac{mc^2}{2E} }. (1.218)
The numerics of this expression can be made easy by using the following quantities: \hbar c\simeq 197\times 10^{-15}MeV m, the rest mass energy of an electron is mc^2=0.5 MeV,\Delta x_0=10^{-6} m,E=25 eV=25\times 10^{-6} MeV, and L=100m. Inserting these quantities into (1.218), we obtain
\frac{\delta x}{\Delta x_0}\simeq \frac{197\times 10^{-15 } MeV m\times 100 m}{2\times 0.5 MeV\times 10^{-12}m^2}\sqrt{\frac{0.5 MeV}{2\times 25\times 10^{-6} MeV} }\simeq 2 \times 10^{3}; (1.219)
the time it takes the electron to travel the 100 m distance is given, as shown above, by
t=\frac{L}{c}\sqrt{\frac{mc^2}{2E} }=\frac{100 m}{3\times 10^{8}ms^{-1}}\sqrt{\frac{0.5 MeV}{2\times 25\times 10^{-6} MeV} }=3.3\times 10^{-5} s. (1.220)
Using t=3.3\times 10^{-5} s and substituting (1.219) into (1.216), we obtain
\Delta x(t=3.3\times 10^{-5}s)=10^{-6}m\times \sqrt{1+4\times 10^6}\simeq 2\times 10^{-3}m=2mm. (1.221)
The width of the wave packet representing the electron has increased from an initial value of 10^{-6}m to 2\times 10^{-3}m,i.e., by a factor of about 10^{3}. The spread of the electron’s wave packet is thus quite large.
(ii) The calculation needed here is identical to that of part (i), except the value of \Delta x_0 is now 10^{-8}m instead of 10^{-6}m. This leads to \delta x/\Delta x_0\simeq 2\times 10^7 and hence the width is \Delta x(t)=20 cm; the width has therefore increased by a factor of about 10^{7}. This calculation is intended to show that the narrower the initial wave packet, the larger the final spread. In fact, starting in part (i) with an initial width of 10^{-6}m, the final width has increased to 2\times 10^{-3}m by a factor of about 10^{3}; but in part (ii) we started with an initial width of 10^{-8}m, and the final width has increased to 20 cm by a factor of about 10^{7}.
(iii) The motion of a 100 MeV electron is relativistic; hence to good approximation, its speed is equal to the speed of light, \upsilon \simeq c. Therefore the time it takes the electron to travel a distance of L = 100 m is t \simeq L/c=3.3\times 10^{-7}s. The dispersion factor for this electron can be obtained from (1.217) where \Delta x_0=10^{-3}m:
\frac{\delta x}{\Delta x_0}=\frac{\hbar L}{2mc\Delta x_0^2}=\frac{\hbar cL}{2mc^2\Delta x^2_0} \simeq \frac{197\times 10^{-15}MeVm\times 100m}{2\times 0.5 MeV\times 10^{-6}m^2}\simeq 2\times 10^{-5}. (1.222)
The increase in the width of the wave packet is relatively small:
\Delta x(t=3.3\times 10^{-7}s)=10^{-3}m\times \sqrt{1+4\times 10^{-10}}\simeq 10^{-3}m =\Delta x_0. (1.223)
So the width did not increase appreciably. We can conclude from this calculation that, when the motion of a microscopic particle is relativistic, the width of the corresponding wave packet increases by a relatively small amount.
(iv) In the case of a macroscopic object of mass m = 0.1 kg, the time to travel the distance L = 100 m is t=L/\upsilon =100m/50 ms^{-1}=2 s. Since the size of the system is about \Delta x_0=1 cm=0.01 m and \hbar =1.05\times 10^{-34}J s, the dispersion factor for the object can be obtained from (1.217):
\frac{\delta x}{\Delta x_0}=\frac{\hbar t}{2m\Delta x_0^2} \simeq \frac{1.05\times 10^{-34}J s\times 2 s}{2\times 0.1 kg\times 10^{-4}m^2}\simeq 10^{-29}. (1.224)
Since \delta x/\Delta x_0=10^{-29}\ll 1, the increase in the width of the wave packet is utterly undetectable:
\Delta x(2s)=10^{-2}m\times \sqrt{1+10^{-58}}\simeq 10^{-2}m=\Delta x_0 . (1.225)
(b) Using (1.216) and (1.217) we obtain the expression for the time t in which the wave packet spreads to \Delta x(t):
t=\tau \sqrt{\left(\frac{\Delta x(t)}{\Delta x_0} \right)^2-1 }, (1.226)
where \tau represents a time constant \tau=2m(\Delta x_0)^2/\hbar (see (1.148)) \Delta x(t)=\Delta x_0\sqrt{1+\left(\frac{t}{\tau } \right)^2 }, . The time constant for the electron of part (i) is given by
\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, (1.227)
and the time constant for the object of part (iv) is given by
\tau=\frac{2m(\Delta x_0)^2}{\hbar}\simeq \frac{2\times 0.1 kg\times 10^{-4}m^2}{1.05\times 10^{-34}J s}=1.9\times 10^{29} s . (1.228)
Note that the time constant, while very small for a microscopic particle, is exceedingly large for macroscopic objects.
On the one hand, a substitution of the time constant (1.227) into (1.226) yields the time required for the electron’s packet to spread to 10 mm:
t =1.7\times 10^{-8}s\sqrt{\left(\frac{10^{-2}}{10^{-6}} \right)^2 -1}\simeq 1.7\times 10^{-4}s. (1.229)
On the other hand, a substitution of (1.228) into (1.226) gives the time required for the object to spread to 10 cm:
t =1.9\times 10^{29}s\sqrt{\left(\frac{10^{-1}}{10^{-2}} \right)^2 -1}\simeq 1.9\times 10^{30}s. (1.230)
The result (1.229) shows that the size of the electron’s wave packet growsin a matter of 1.7\times 10^{-4}s from 10^{-6}m to 10^{-2}m, a very large spread in a very short time. As for (1.230), it shows that the object has to be constantly in motion for about 1.9\times 10^{30}s for its wave packet to grow from 1 cm to 10 cm, a small spread for such an absurdly large time; this time is absurd because it is much larger than the age of the Universe, which is about 4.7\times 10^{17}s. We see that
the spread of macroscopic objects becomes appreciable only if the motion lasts for a long, long time. However, the spread of microscopic objects is fast and large.
We can summarize these ideas in three points:
• The width of the wave packet of a nonrelativistic, microscopic particle increases substantially and quickly. The narrower the wave packet at the start, the further and the quicker it will spread.
• When the particle is microscopic and relativistic, the width corresponding to its wave packet does not increase appreciably.
• For a nonrelativistic, macroscopic particle, the width of its corresponding wave packet remains practically constant. The spread becomes appreciable only after absurdly long times, times that are larger than the lifetime of the Universe itself!