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Heisenberg's $\Delta x\,\Delta p \ge \hbar/2$ and $\Delta E\,\Delta t \ge \hbar/2$, confinement energies of electrons in atoms and nuclei, natural line widths, zero-point energy, and the resolution of atomic clocks.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to use the uncertainty relations to estimate momentum spreads, confinement energies and line widths.
From the last two lessons you know that particles have wavelengths $h/p$ and that passing through a narrow slit spreads a beam sideways. From waves you know that a short pulse contains a spread of frequencies. This lesson turns those facts into Heisenberg's uncertainty relations and uses them to estimate the energies of confined particles.
| Term | What it means |
|---|---|
| Uncertainty | The spread, $\Delta x$ or $\Delta p$, of a quantity's possible values in a quantum state. |
| Heisenberg's relation | $\Delta x\,\Delta p \ge \hbar/2$. |
| Reduced Planck constant | $\hbar = h/2\pi = 1.055 \times 10^{-34}$ J·s $= 6.582 \times 10^{-16}$ eV·s. |
| $\hbar c$ | $197.3$ eV·nm $= 197.3$ MeV·fm, handy for estimates. |
| Energy–time relation | $\Delta E\,\Delta t \ge \hbar/2$, linking a state's lifetime to its energy spread. |
| Natural line width | The energy spread of a spectral line set by the lifetime of the excited state. |
| Zero-point energy | The least energy a confined particle can have, never zero. |
A wave with a single, exact wavelength goes on forever. To make a wave confined to a small region, you must add many wavelengths, and the narrower the region, the wider the spread. Since $p = h/\lambda$, a spread of wavelengths means a spread of momenta. Werner Heisenberg made this exact in 1927:
$$\Delta x\,\Delta p \ge \frac{\hbar}{2}, \qquad \hbar = \frac{h}{2\pi}.$$
The same reasoning for time and frequency gives
$$\Delta E\,\Delta t \ge \frac{\hbar}{2}.$$
These are not limits of instruments but properties of waves. A particle squeezed into a small space has a large momentum spread, and so a kinetic energy of at least about $(\hbar/2\Delta x)^2/2m$. That minimum, the zero-point energy, is why atoms have a size and electrons do not fall into nuclei.
Another way: picture
Picture a musical note. A long, sustained note has a clear pitch; a sharp clap lasting a millisecond has no definite pitch at all, only a spread. Squeeze a sound into a short time and its frequencies spread out. A particle's wave works the same way in space: squeeze it into a short distance and its momenta spread out.
Another way: steps
A wave packet of width $\Delta x$ is built by adding waves with wavenumbers $k = 2\pi/\lambda$ spread over $\Delta k$. For the waves to cancel outside the packet, their phases must drift apart across its width, which needs $\Delta k\,\Delta x$ of at least about 1. With $p = \hbar k$, this becomes $\Delta x\,\Delta p \gtrsim \hbar$, and a careful calculation with the best-shaped packet, a Gaussian, gives exactly $\hbar/2$.
Nothing about measurement entered the argument. The relation describes the state itself: a particle with a well-defined position has a poorly defined momentum, not just a poorly measured one.
Heisenberg first explained the relation with a thought experiment. To see an electron's position precisely, you must use light of short wavelength, since a microscope cannot resolve details much smaller than $\lambda$. But short-wavelength photons carry large momentum, $h/\lambda$, and kick the electron unpredictably as they scatter.
Sharpening the position blurs the momentum, and the product comes out about $h$. The story is useful but misleading if taken as the whole truth, because it suggests the electron had an exact position and momentum that the measurement disturbed. Quantum mechanics says it had no exact values of both to begin with.
The relation gives quick, surprisingly good estimates. An electron in a hydrogen atom is confined to about $0.1$ nm. Then $\Delta p\,c \ge 197.3/0.2 \approx 1000$ eV, and its kinetic energy is about $(1000)^2/(2 \times 511{,}000) \approx 1$ eV, the right order of magnitude for atomic energies of a few eV.
Make the region smaller and the kinetic energy grows as $1/\Delta x^2$, while the electric potential energy falls only as $1/\Delta x$. The balance fixes the atom's size. Minimizing the total energy with this estimate gives a radius close to the true Bohr radius, $0.053$ nm, from nothing but the uncertainty relation and Coulomb's law.
Before the neutron was discovered in 1932, some physicists thought nuclei contained electrons, since beta decay emits them. The uncertainty relation rules this out. A nucleus is about $5$ fm across, so an electron inside would have $\Delta p\,c \approx 197/10 \approx 20$ MeV, and so kinetic energy of about $20$ MeV.
Nuclei bind their particles by only about $8$ MeV each, and beta-decay electrons emerge with at most a few MeV. An electron with $20$ MeV would escape at once. The electrons of beta decay must therefore be created in the decay, not stored beforehand, as Enrico Fermi's theory of 1934 proposed.
An excited atom that lives a time $\Delta t$ before emitting a photon has an energy spread $\Delta E \approx \hbar/2\Delta t$. For a lifetime of $10$ ns, $\Delta E$ is about $3 \times 10^{-8}$ eV. That sets the natural width of the spectral line: sharp, compared with photon energies of a few eV, but not perfectly sharp.
Very short-lived particles have large energy spreads. The Z boson, which lives about $3 \times 10^{-25}$ s, has a mass spread of about $2.5$ GeV, measured as the width of its peak in collider data. Particle physicists read lifetimes directly from such widths, far too short to time any other way.
Checking an answer. Smaller confinement must give larger momentum and energy. Estimates are order-of-magnitude, so factors of 2 do not matter, but powers of ten do. Units: $\hbar c$ in eV·nm with lengths in nm gives eV. And results for electrons in atoms must come out near eV, for nucleons in nuclei near MeV.
Treating the minimum spread as a typical momentum is an estimate, not a calculation. A particle confined to $\Delta x$ has momenta spread around zero, and its average kinetic energy is about $(\Delta p)^2/2m$, so the estimate captures the scale. The exact answer, from solving the Schrödinger equation, differs by a factor of order one.
The energy–time relation is subtler than the position–momentum one, because time is not a quantity a particle has. It relates how quickly a state changes to how sharp its energy is. Using a lifetime for $\Delta t$ is the standard, well-tested application.
A particle can never be completely at rest in a confined space, since zero momentum would need infinite spread in position. Its lowest possible energy, the zero-point energy, is above the bottom of the well. Atoms in a crystal jiggle even at absolute zero.
Zero-point motion explains why helium stays liquid at absolute zero at ordinary pressure: its atoms are light and weakly attracted, and their zero-point motion is enough to prevent them from locking into a solid. The same effect makes hydrogen bonds and chemical reaction rates depend on whether ordinary hydrogen or heavier deuterium is involved.
The uncertainty relations set limits that engineers now design against. The LIGO detectors in Hanford, Washington, and Livingston, Louisiana, measure mirror motions of a thousandth of a proton's width. At that level, the quantum uncertainty of the laser light itself, in photon number and phase, is a leading source of noise.
LIGO beats part of it by injecting squeezed light, a quantum state that reduces the uncertainty in one quantity at the cost of increasing it in another, exactly as Heisenberg allows. The trade lets the detectors hear more colliding black holes and neutron stars each year.
The uncertainty relation applies to baseballs too, but $\hbar$ is so small that it never shows. Locate a $0.15$ kg ball to a micrometer and the minimum spread in its velocity is $\hbar/(2m\Delta x) = 3.5 \times 10^{-28}$ m/s: in a billion years the ball would drift a distance smaller than an atomic nucleus. Every classical trajectory is, in quantum terms, a wave packet so narrow in both position and momentum that the spreads are far below anything measurable.
The relation bites only when the product of a system's size and momentum is comparable to $\hbar$. For electrons in atoms, with sizes of $10^{-10}$ m and momenta of $10^{-24}$ kg·m/s, the product is about $10^{-34}$ J·s, right at the scale of $\hbar$. For protons in nuclei, with sizes of $10^{-15}$ m and momenta of $10^{-19}$ kg·m/s, it is again about $\hbar$. That is the quick test of whether quantum mechanics is needed: compare the natural action of the system, size times momentum, with $\hbar$.
The cesium fountain clocks at NIST in Boulder, Colorado, define American civil time. They toss a ball of laser-cooled cesium atoms upward through a microwave cavity; the atoms rise and fall back through it about a second later. The long interval between the two passes lets the clock pin the atoms' resonance at $9{,}192{,}631{,}770$ Hz very sharply.
The energy–time relation sets the limit: probing for $1$ s resolves the frequency to about $0.08$ Hz, and averaging many tosses narrows the uncertainty further. That is why fountain clocks, with their slow, cold atoms, beat older beam clocks, whose atoms flew past in milliseconds. NIST-F2 keeps time to about a second in 300 million years.
Two protons in the Sun's core repel each other electrically, and at the core's temperature of $15$ million kelvin they rarely have enough energy to get close enough to fuse. Classically, the Sun could not shine. Quantum mechanics lets them tunnel through the electrical barrier, a consequence of the wave nature of the protons.
The uncertainty relation gives the scale: a proton's wave, spread over a region comparable to its wavelength, reaches into the classically forbidden zone. Hans Bethe at Cornell worked out the resulting fusion rates in 1939, explaining why the Sun burns slowly enough to last ten billion years. He received the 1967 Nobel Prize.
It is natural to read the uncertainty relation as a statement about imperfect instruments: measure more carefully and you could know both position and momentum. That is wrong. The relation is a property of the particle's wave, which cannot be both narrow and made of a single wavelength. It holds whether or not anyone is measuring.
A related error is to think the uncertainty means the particle is jittering around a hidden exact position. Quantum mechanics says there is no such hidden exact value; the spread is all there is until a measurement produces a result.
An electron is confined to $\Delta x = 0.10$ nm. Find the minimum $\Delta p\,c$.
$\Delta p\,c = \dfrac{197.3}{2 \times 0.10} = 987\ \text{eV}$
$\hbar c/2\Delta x$.
Check that Newton's formula is fine.
$987\ \text{eV} \ll 511{,}000\ \text{eV}$
Nonrelativistic.
Estimate its kinetic energy.
$K \approx \dfrac{987^2}{2 \times 511{,}000} = 0.95\ \text{eV}$
$(\Delta p\,c)^2/2mc^2$.
Compare with hydrogen's binding energy.
$13.6\ \text{eV}$
The same order: this is atomic physics.
Halve the region and re-estimate.
$K \approx 3.8\ \text{eV}$
Four times as much.
A proton is confined to a nucleus $\Delta x = 5.0$ fm wide. Find the minimum $\Delta p\,c$.
$\Delta p\,c = \dfrac{197.3}{10} = 19.7\ \text{MeV}$
$\hbar c = 197.3$ MeV·fm.
Compare with the proton's rest energy.
$19.7 \ll 938\ \text{MeV}$
Nonrelativistic.
Estimate its kinetic energy.
$K \approx \dfrac{19.7^2}{2 \times 938} = 0.21\ \text{MeV}$
Newton's formula.
Compare with nuclear binding energies.
$\text{about } 8\ \text{MeV per nucleon}$
Comfortably bound.
Repeat for an electron in the same nucleus.
$K \approx 19.7 - 0.511 \approx 19\ \text{MeV}$
Relativistic: $E \approx pc$.
Draw the conclusion.
$\text{electrons cannot be bound in nuclei}$
Their confinement energy is too high.
Write the energy of an electron confined to radius $r$.
$E(r) \approx \dfrac{\hbar^2}{2m_er^2} - \dfrac{ke^2}{r}$
Taking $p \approx \hbar/r$.
Put the constants in eV and nm.
$E(r) \approx \dfrac{0.0381}{r^2} - \dfrac{1.44}{r}$
$\hbar^2/2m_e = 0.0381$ eV·nm², $ke^2 = 1.44$ eV·nm.
Differentiate and set to zero.
$-\dfrac{0.0762}{r^3} + \dfrac{1.44}{r^2} = 0$
Minimum energy.
Solve for the radius.
$r = \dfrac{0.0762}{1.44} = 0.053\ \text{nm}$
The Bohr radius.
Find the minimum energy.
$E = \dfrac{0.0381}{0.053^2} - \dfrac{1.44}{0.053} = 13.6 - 27.2 = -13.6\ \text{eV}$
Hydrogen's ground-state energy.
Note the agreement.
$\text{exact, by a happy choice of } p \approx \hbar/r$
Estimates usually land within a factor of two.
State why atoms do not collapse.
$\text{confinement energy rises faster than the attraction falls}$
$1/r^2$ beats $1/r$ at small $r$.
Write the energy–time relation.
$\Delta E \ge \dfrac{\hbar}{2\Delta t}$
Heisenberg.
Substitute the values.
$\Delta E = \dfrac{6.582 \times 10^{-16}}{4.0 \times 10^{-9}}$
In eV.
Evaluate the spread.
An electron confined to a region has a minimum momentum spread of $2 \times 10^{-25}$ kg·m/s. The region is made half as wide. What is the new minimum momentum spread?
Complete the worked solution: suppose an electron were confined inside a nucleus, a region $10$ fm wide. With $\hbar c = 197.3$ MeV·fm and $m_ec^2 = 0.511$ MeV, find its minimum $\Delta p\,c$, its total energy for that momentum, and its kinetic energy, all in MeV.
Find the momentum spread.
$\Delta p\,c = \dfrac{\hbar c}{2\Delta x} =$ p
Heisenberg.
Find the total energy.
$E = \sqrt{(pc)^2 + (m_ec^2)^2} =$ e
Relativistic: $pc \gg m_ec^2$.
Subtract the rest energy.
$K = E - m_ec^2 =$ k
Kinetic energy.
Compare with nuclear binding.
$\text{nuclei bind particles by only a few MeV}$
So no electron can live inside a nucleus.
Match each statement to its uncertainty relation or consequence.
| $\Delta x\Delta p \ge \hbar/2$ | $\Delta E\Delta t \ge \hbar/2$ | $h/2\pi$ | raises kinetic energy | |
|---|---|---|---|---|
| position and momentum | ||||
| energy and time | ||||
| reduced Planck constant | ||||
| confinement |
An electron is confined to a region $\Delta x = 0.2$ nm wide. With $\hbar c = 197.3$ eV·nm and $m_ec^2 = 511{,}000$ eV, fill in $2\Delta x$ in nm, the minimum momentum spread times $c$ in eV, and the estimated kinetic energy $(\Delta p\,c)^2/2mc^2$ in eV.
| value | |
|---|---|
| 2Δx (nm) | |
| Δp·c (eV) | |
| kinetic energy estimate (eV) |
Estimate the kinetic energy of a proton confined to a region $x$ fm wide, taking its momentum to be the minimum spread $\hbar/2x$. Here $(\hbar c)^2/8mc^2 = 5.187$ MeV·fm². Write the energy in MeV as a formula in $x$.
Answer:
An excited atomic state lives $1.6$ ns on average before emitting a photon. With $\hbar = 6.582 \times 10^{-16}$ eV·s, what is the minimum spread in the photon's energy, in neV?
Answer: neV
In a cesium fountain clock at NIST in Boulder, Colorado, atoms are probed for $2$ s as they rise and fall. By the energy–time uncertainty relation, what is the smallest spread in the frequency they can resolve, in Hz?
Answer: Hz
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Estimate the kinetic energy of an electron confined to a region $x$ nm wide, taking its momentum to be the minimum spread $\hbar/2x$. Here $(\hbar c)^2/8mc^2 = 0.009525$ eV·nm². Write the energy in eV as a formula in $x$.
Answer:
You can reason with the uncertainty relations. Explain to someone why an electron cannot live inside a nucleus.
21. Your turn: an excited state lives $2.0$ ns. Find the minimum energy spread of the photon it emits., step 3
$\Delta E = 1.6 \times 10^{-7}\ \text{eV}$
About $160$ neV.