The Uncertainty Principle and Its Applications

By the end you should be able to: State Heisenberg's uncertainty principle for the conjugate pairs position-momentum and energy-time, and apply it to show that an electron cannot exist inside the nucleus and to estimate natural line broadening.

The syllabus asks for the uncertainty principle as a statement, not a derivation, and for its applications. So the statement is given, and the effort goes where the marks are: using it.

Conjugate observables

Certain pairs of quantities cannot both be known precisely at the same time. Such pairs are called conjugate observables. The two that matter here are position with momentum, and energy with time.

This is not a limitation of instruments. It is a property of the objects themselves. A particle with a perfectly definite momentum does not possess a definite position that we are failing to measure — it does not have one. Any wave-like entity of a single, definite wavelength must extend over all space, which is exactly what "no definite position" means.

The statement

\boxed{\Delta x \, \Delta p_x \geq \frac{\hbar}{2}} \qquad \boxed{\Delta E \, \Delta t \geq \frac{\hbar}{2}}

where $\hbar = h/2\pi = 1.055\times10^{-34}\ \text{J s}$ .

Textbooks vary: you will also see $\Delta x\,\Delta p \geq \hbar$ and $\Delta x\,\Delta p \geq h/4\pi$ (which is the same as $\hbar/2$ ). These differ by factors of order one and none of the physical conclusions change. Use $\hbar/2$ and state which form you used — the marks are for the reasoning and the order of magnitude, not for a factor of 2.

Application 1 — no electrons inside the nucleus

Before the neutron was discovered, a nucleus of mass number $A$ and charge $Z$ was thought to hold $A$ protons and $A-Z$ electrons. Beta decay, in which a nucleus emits an electron, seemed to support it: how could it emit an electron it did not already contain?

The uncertainty principle rules it out, and the argument is worth following because it is the cleanest example of the principle doing real physical work.

Confine an electron to a nucleus, diameter about $10^{-14}\ \text{m}$ . Then $\Delta x \approx 10^{-14}\ \text{m}$ , so

\Delta p \geq \frac{\hbar}{2\Delta x} = \frac{1.055\times10^{-34}}{2\times10^{-14}} = 5.3\times10^{-21}\ \text{kg m s}^{-1}

The momentum is uncertain by at least this much, so the electron must have momentum of at least roughly this size. Now find the corresponding energy. Check first whether the electron is relativistic: compare $pc$ with the rest energy $m_ec^2 = 0.511\ \text{MeV}$ .

pc = (5.3\times10^{-21})(3\times10^{8}) = 1.58\times10^{-12}\ \text{J} = 9.9\ \text{MeV}

That is about twenty times the rest energy, so the electron is ultra-relativistic and $E \approx pc \approx 10\ \text{MeV}$ .

The conclusion. An electron inside a nucleus would need a kinetic energy of about 10 MeV. But the electrons actually emitted in beta decay are observed with energies of a few MeV at most, and the Coulomb attraction binding an electron to a nucleus is only of order a few MeV. Nothing available can hold a 10 MeV electron in. Therefore electrons do not exist inside the nucleus; the beta-decay electron is created at the moment of decay, when a neutron converts into a proton, an electron and an antineutrino.

Note the shape of the argument, which is the transferable part: confine a particle tightly, and the uncertainty principle forces a large momentum, hence a large energy. Confinement costs energy. That idea returns in the particle-in-a-box problem, where it produces the zero point energy.

Application 2 — natural line broadening

An atom in an excited state does not stay there. It survives for a characteristic lifetime $\Delta t$ , typically about $10^{-8}\ \text{s}$ for an allowed optical transition.

Because the state exists for only a finite time, the energy-time relation says its energy cannot be perfectly sharp:

\Delta E \geq \frac{\hbar}{2\Delta t}

The emitted photon's energy therefore has a spread, so the spectral line has a finite width. This is natural line broadening, and it is irreducible: cool the gas to remove Doppler broadening, thin it to remove collision broadening, and this width remains.

Only a state of infinite lifetime — the ground state — has perfectly sharp energy. The shorter the lifetime, the broader the line, which is why forbidden transitions with long lifetimes give very narrow lines and short-lived states give broad ones.