## Flying electron

An electron moves in one direction with the velocity of (1000 ± 1) km · s−1. With which highest uncertainty can we theoretically determine its location? In the end, discuss if it is necessary to use relativistic effects.

• #### Hint 1

We know that, using the principle of uncertainty, with the specification of the electron’s motion comes the price of lost accuracy in measuring another quantity. What is the quantity (or quantities) in question?

• #### Hint 2

What is the relative uncertainty in the measurement? How does the relative uncertainty propagate from one variable to another? What if the one variable is the product of two other which carry some uncertainty themselves?

• #### Notation

 me = 9.11·10−31 kg Rest mass of the electron v = 1000 km·s−1 Electron’s velocity Δv = 1 km·s−1 Absolute uncertainty of velocity Δx = ? (m) Uncertainty of location
• #### Solution

We need to know the uncertainty Δp to estimate Δx from the principle of uncertainty $\Delta x \,\cdot\, \Delta p \ge \frac{\hbar}{2}$ . As uncertainty of the momentum, we consider the absolute uncertainty of determination of the momentum. Since the ball’s momentum is the product of its mass and velocity p = me·v. If we consider the mass of the electron as a constant the uncertainty of the momentum is given only by uncertainty of the velocity. Therefore we can use the formula η(p) = η(v) to calculate the uncertainty of these variables. We have also the formula between absolute and relative uncertainty Δp = η(pp. With the use of the principle of uncertainty we can calculate Δx $\Delta x \,\cdot\, \Delta p \ge \frac{\hbar}{2}$ and we obtain

$\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\Delta p} ,$ $\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\eta(p)\cdot p} ,$ $\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\eta(v)\cdot mv} ,$ $\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\frac{\Delta v}{v}\cdot mv} ,$ $\Delta x \ge \frac{1.055\,\cdot\,10^{-34}}{2\,\cdot\,\frac{1}{1000}\,\cdot\, 9.11\,\cdot\, 10^{-25}}\ \mbox{m} ,$ $\Delta x \ge 5.8\,\cdot\,10^{-8}\ \mbox{m}\,\dot{=}\,60\ \mbox{nm} .$

Out of principle we cannot determine the location of the mentioned electron with the accuracy higher than tens of nanometres, which is two orders of magnitude higher than the characteristic dimensions of an atom. From the electron’s perspective, this uncertainty is quite important.

• #### Discussion

Lorentz factor

$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-\left( \frac{1}{300}\right)^2}}=\sqrt{\frac{90\,000}{89\,999}}\,\dot{=}\, 1.000 006 .$

If the momentum of our electron is in the order of 10−24 kg·m·s−1 we could see the relativist effects in order of 10−30 kg·m·s−1 so we can neglect it.