18.8 Uncertainty Principle

18.8 Uncertainty Principle

Position and momentum of a particle cannot be measured simultaneously with perfect accuracy. There is always a fundamental uncertainty associated with any measurement. This uncertainty is not associated with the measuring instrument. It is a consequence of the wave-particle duality of matter and radiation. This was first proposed by Werner Heisenberg in 1927 and hence, is known as Heisenberg uncertainty principle. This fundamental uncertainty is completely negligible for measurements of position and momentum of macroscopic objects in daily life but is a predominant fact of life in the atomic domain. For example, a stream of light photons striking a flying tennis ball hardly affects its path, but one photon striking an electron drastically alters its motion. Since light has also wave properties, we would expect to be able to determine the position of the electron only to within one wavelength of the light being used. Hence, in order to observe the position of an electron with less uncertainty, we must use light of short wavelength. But it will alter the motion drastically making momentum measurement less precise. If light of wavelength $\lambda$ is used to locate a micro-particle moving along $x$-axis, the uncertainty in its position measurement is: $\Delta x \approx \lambda$. At most, the photon of light can transfer all its momentum $h/\lambda$ to the micro-particle whose own momentum will then be uncertain by an amount: $\Delta p = \dfrac{h}{\lambda}$ Multiplying these two uncertainties gives:

Equation 18.14 is the mathematical form of position-momentum uncertainty principle. It states that:

The product of the uncertainty $\Delta x$ in the position of a particle at some instant and the uncertainty $\Delta p$ in the $x$-component of its momentum at the same instant approximately equals Planck's constant $h$. Using uncertainty principle, we can prove that electrons are not present inside atomic nucleus. As the diameter of a nucleus is of the order of $10^{-14}\,\mathrm{m}$ for an electron to be in the nucleus, the uncertainty in its position can be $10^{-14}\,\mathrm{m}$. Using uncertainty principle, the uncertainty in the momentum will be: $\Delta p = \dfrac{h}{\Delta x} = \dfrac{6.63 \times 10^{-34}\,\mathrm{J\,s}}{1 \times 10^{-15}} = 6.63 \times 10^{-19}\,\mathrm{J\,s}$ The corresponding energy uncertainty should be of the order of $\mathrm{GeV}$. No such election in the atom has been found. Thus electrons do not exist inside the nucleus. There is another form of uncertainty principle which relates the energy of a particle and the time at which it had that energy. i.e.,

$\dfrac{x}{v}.\Delta pv \ge h$ $\Delta t.\Delta E \ge h$ or $\Delta E.\Delta t \ge h$ where $\hbar = \dfrac{h}{2\pi} = 1.05 \times 10^{-34}\,\mathrm{J\,s}$

Thus, more accurately we determined the energy of a particle, the more uncertain we will be of the time during which it has that energy. According to Heisenberg's more careful calculations, he found that at the very best: $\Delta x.\Delta p \ge \hbar$ where $\hbar = \dfrac{h}{2\pi} = 1.05 \times 10^{-34}\,\mathrm{J\,s}$ and $\Delta E.\Delta t \ge \hbar$ Werner Heisenberg was awarded Nobel Prize in 1932 for his contribution towards quantum physics.

Example 18.4

What can be the velocity of an electron enclosed in a box about the size of an atom of the order of $1.0 \times 10^{-10}\,\mathrm{m}$?

Solution

Size of atom $\Delta x = 1.0 \times 10^{-10}\,\mathrm{m}$ Mass of electron $m = 9.1 \times 10^{-31}\,\mathrm{kg}$ Velocity of electron $\Delta v = ?$ Using the formula $\Delta p.\Delta x \approx h$ As $\Delta p = m\Delta v$ Hence $m\Delta v.\Delta x = h$ $\Delta v = \dfrac{h}{m \times \Delta x}$ Putting the values $\Delta v = \dfrac{6.63 \times 10^{-34}\,\mathrm{J\,s}}{9.1 \times 10^{-31}\,\mathrm{kg} \times 1.0 \times 10^{-10}\,\mathrm{m}}$ $\Delta v = 7.3 \times 10^{6}\,\mathrm{m\,s^{-1}}$

Example 18.5

Find the uncertainty in the energy of a photon which is emitted from an atom radiating for about $10^{-9}$ second.

Solution

$\Delta t = 10^{-9}\,\mathrm{s}, \quad h = 6.63 \times 10^{-34}\,\mathrm{J\,s}$ and $\Delta E = ?$ Using uncertainty principle: $\Delta E.\Delta t \approx h$ or $\Delta E \approx \dfrac{h}{\Delta t}$ $\Delta E \approx \dfrac{6.63 \times 10^{-34}\,\mathrm{J\,s}}{10^{-9}\,\mathrm{s}} \approx 6.63 \times 10^{-25}\,\mathrm{J}$ As $1.6 \times 10^{-19}\,\mathrm{J} = 1\,\mathrm{eV}$, hence $\Delta E \approx \dfrac{6.63 \times 10^{-25}}{1.6 \times 10^{-19}}\,\mathrm{eV}$ $\Delta E \approx 4 \times 10^{-7}\,\mathrm{eV}$