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Heisenberg's relation $\sigma_x\sigma_p \geq \hbar/2$ checked against real wave functions, the Gaussian as the minimum, the energy-time relation, and estimates of atoms, nuclei and spectral line widths.
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 state and apply the uncertainty principle, check it for a given wave function, use it to estimate the size and energy of bound systems, and use the energy-time relation for spectral line widths.
You can compute $\sigma_x$ and $\sigma_p$ from a wave function, and you know that the momentum operator is $-i\hbar\,d/dx$. From modern physics you have met the uncertainty principle as a slogan. This lesson states it precisely, checks it against real wave functions, and turns it into a tool for estimating the sizes and energies of quantum systems.
| Term | What it means |
|---|---|
| Uncertainty principle | $\sigma_x\sigma_p \geq \hbar/2$ for any state. |
| Reduced Planck constant | $\hbar = h/2\pi = 1.0546 \times 10^{-34}$ J s $= 6.582 \times 10^{-16}$ eV s. |
| Minimum-uncertainty state | A Gaussian wave packet, for which $\sigma_x\sigma_p = \hbar/2$ exactly. |
| Commutator | $[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}$; for position and momentum it is $i\hbar$. |
| Energy-time relation | $\sigma_E\,\tau \geq \hbar/2$, where $\tau$ is the time over which the state changes appreciably. |
| Natural linewidth | The spread in photon energy from an excited state of finite lifetime. |
| Zero-point energy | The least energy a confined particle can have, forced by the uncertainty principle. |
| $\hbar c$ | $197.33$ MeV fm $= 197.33$ eV nm, a convenient constant for estimates. |
A wave with a well-defined wavelength, and so by de Broglie a well-defined momentum, stretches out indefinitely. A wave packet confined to a small region is built from many wavelengths, and so from many momenta. The quantitative statement, proved from the wave function alone, is Heisenberg's uncertainty principle:
$$\sigma_x\sigma_p \geq \frac{\hbar}{2}.$$
It is not a statement about measurement disturbing a particle. It is a property of every possible wave function: no state exists in which the spread of position measurements and the spread of momentum measurements are both small. Its origin is that the position and momentum operators do not commute, $\hat{x}\hat{p} - \hat{p}\hat{x} = i\hbar$.
The states met so far all obey it with room to spare. The parabolic well state has $\sigma_x = a/\sqrt{28}$ and $\sigma_p = \sqrt{10}\,\hbar/a$, a product of $\sqrt{10/28}\,\hbar = 0.598\hbar$. The exponential state $e^{-|x|/a}/\sqrt{a}$ has $\sigma_x = a/\sqrt{2}$ and $\sigma_p = \hbar/a$, a product of $0.707\hbar$. Only the Gaussian, $\psi \propto e^{-x^2/4\sigma^2}$, reaches exactly $\hbar/2 = 0.5\hbar$.
A companion relation links energy and time, $\sigma_E\,\tau \geq \hbar/2$: a state that lasts only a time $\tau$ cannot have a sharply defined energy. It has a different status, since time is a parameter rather than an operator, but it is just as reliable in practice.
Another way: picture
Picture trying to describe a sound. A single pure note has a precise pitch but lasts forever; a sharp click happens at a precise instant but contains every pitch at once. Any real sound trades one against the other, and a musician cannot play a note both perfectly brief and perfectly in tune. The quantum wave function obeys the same mathematics of waves, with position in place of time and momentum in place of pitch.
Another way: steps
The uncertainty principle is most often used for estimates, and the method is a short chain.
Checks. The product $\sigma_x\sigma_p$ computed from any real wave function must be at least $\hbar/2$; a smaller result means an error in an integral. Estimates are good to factors of a few, not to decimals, and they should reproduce known scales: electron energies in atoms of order electronvolts, nucleon energies in nuclei of order MeV. And tighter confinement must always mean higher kinetic energy.
Why does the electron in hydrogen not fall into the proton, as classical physics predicts? Confining it within a radius $r$ gives it a momentum of at least about $\hbar/r$ and a kinetic energy of about $\hbar^2/(2mr^2)$, while the potential energy is $-ke^2/r$. The total,
$$E(r) \approx \frac{\hbar^2}{2mr^2} - \frac{ke^2}{r},$$
rises steeply as $r$ shrinks, because the kinetic term goes as $1/r^2$ and the potential only as $1/r$. The minimum, from $dE/dr = 0$, is at
$$r = \frac{\hbar^2}{mke^2} = 0.0529\ \text{nm},$$
the Bohr radius, with $E = -13.6$ eV — exactly the hydrogen ground state, which the estimate hits by the luck of its factors. The stability of every atom, and so of all ordinary matter, rests on this balance: the uncertainty principle is what keeps electrons from spiraling into nuclei.
Before the neutron was discovered in 1932, physicists supposed a nucleus might contain electrons, since some nuclei emit them in beta decay. The uncertainty principle rules it out. An electron confined to a nucleus $5$ fm across would have a momentum spread of at least $\hbar c/(2 \times 5\ \text{fm}) \approx 20$ MeV/$c$. That is forty times its rest energy of $0.511$ MeV, so the electron would be relativistic with kinetic energy near $20$ MeV — far more than any known nuclear force could hold, and far more than the few MeV that beta-decay electrons actually carry. The electrons of beta decay are created at the moment of the decay, not stored beforehand.
Nucleons, $1836$ times heavier, fare differently. The same $20$ MeV/$c$ of momentum gives a proton a kinetic energy of only $(20)^2/(2 \times 938) \approx 0.2$ MeV, easily held by nuclear forces; the full calculation with realistic confinement gives tens of MeV, matching the depths of nuclear potentials.
Heisenberg's first explanation imagined a microscope: to see an electron you must hit it with light, and the photon's kick disturbs its momentum. That story is memorable and misleading. The relation $\sigma_x\sigma_p \geq \hbar/2$ is derived without any mention of a measuring device; it follows from the wave function itself and would hold for the best conceivable instruments.
The experimental test confirms it. Prepare many electrons in the same state; measure position on half of them and momentum on the other half, each once, so that no electron is disturbed twice. The spreads of the two sets of results still obey the bound. What the relation limits is not our knowledge of a particle with secret sharp values, but the sharpness the particle's state can have in the first place.
The state that meets the bound exactly is the Gaussian, $\psi(x) = (2\pi\sigma^2)^{-1/4}e^{-x^2/4\sigma^2}$. Its position spread is $\sigma_x = \sigma$ by construction. Its Fourier transform — its distribution of momenta — is another Gaussian, with spread $\sigma_p = \hbar/(2\sigma)$, and the product is exactly $\hbar/2$. Every other shape has some extra structure, sharp edges or long tails, that costs additional spread in one variable or the other.
That is why lasers, which produce nearly Gaussian beams, reach the diffraction limit, and why the ground state of the harmonic oscillator, a Gaussian, has the lowest energy any state in a quadratic well can have. It is also why a Gaussian packet is the natural model of a free particle: it is as close to a classical point with a definite velocity as quantum mechanics allows.
The energy-time relation sets the sharpness of spectral lines, and that is what limits atomic clocks. The cesium clocks that define the second, including NIST-F2 in Boulder, Colorado, lock a microwave oscillator to a transition in cesium atoms at $9\,192\,631\,770$ Hz. The precision depends on how long each atom can be observed: in a fountain clock, atoms are tossed upward and fall back through the microwave cavity over about a second.
An observation time $\tau \approx 1$ s means an energy spread of at least $\hbar/(2\tau) \approx 3.3 \times 10^{-16}$ eV, or a frequency spread of about $0.1$ Hz, against a frequency of nine billion hertz: a line sharp to about one part in $10^{11}$. Averaging over many tosses and days then pins the center of the line to about one part in $10^{16}$.
The newer optical clocks, such as the strontium lattice clocks at JILA, use transitions whose excited states live for over a hundred seconds and whose frequencies are a hundred thousand times higher. Both effects shrink the relative line width, which is why optical clocks now reach one part in $10^{18}$, precise enough to detect the change in the rate of time between two heights a centimeter apart.
The uncertainty principle is ultimately why solid matter is solid. Squeeze an atom and you confine its electrons more tightly; their momentum spread, and so their kinetic energy, rises as $1/r^2$. The electrostatic energy gained by squeezing grows only as $1/r$, so there is a size below which squeezing costs more than it gains. Combined with the exclusion principle, which forbids electrons from sharing states, this gives matter its rigidity and every atom a size of about $0.1$ nm.
The same balance, on a vastly larger scale, holds up a white dwarf star. Its electrons are confined to spaces about $10^{-12}$ m across, a hundred times smaller than in an atom, so their momentum spread is about $\hbar/(2 \times 10^{-12}) \approx 5 \times 10^{-23}$ kg m/s — a kinetic energy of order $10$ keV per electron, a thousand times its value in an atom. That motion, forced by confinement rather than heat, supplies the degeneracy pressure that balances gravity, and it is why a star with the Sun's mass can be no smaller than about the size of Earth before it would have to become a neutron star.
The most common misreading of the uncertainty principle is that a measurement disturbs the particle, so we cannot know both position and momentum, though the particle has both. The principle says something stronger: no wave function has a sharp position and a sharp momentum at once. It is derived from the state alone, with no instrument in the argument, and experiments that measure position on some particles and momentum on others, never both on one, still find the product of spreads bounded by $\hbar/2$.
A second trap is to treat the relation as $\sigma_x\sigma_p = \hbar/2$ for every state. It is an inequality; most states exceed the bound, and only the Gaussian meets it. A third is to apply it to quantities that do commute: position along $x$ and momentum along $y$ can both be sharp, as can energy and momentum for a free particle.
An electron's position spread is $0.1$ nm. Write the least momentum spread.
$\sigma_p \geq \dfrac{\hbar}{2\sigma_x}$
Saturating the uncertainty relation.
Put the length in meters.
$\sigma_x = 0.1\ \text{nm} = 1.0 \times 10^{-10}\ \text{m}$
SI units with $\hbar$ in J s.
Substitute the values.
$\sigma_p \geq \dfrac{1.0546 \times 10^{-34}}{2 \times 1.0 \times 10^{-10}} = 5.27 \times 10^{-25}\ \text{kg m/s}$
A tiny number by everyday standards.
Convert to a velocity spread.
$\sigma_v = \dfrac{5.27 \times 10^{-25}}{9.109 \times 10^{-31}} = 5.8 \times 10^{5}\ \text{m/s}$
Divide by the electron's mass.
Interpret the result.
$\sigma_v \approx 580\ \text{km/s}$
The electron's speed is uncertain by hundreds of kilometers a second, comparable with its actual speed in an atom.
For $\psi = \sqrt{30/a^5}\,x(a - x)$, recall the position variance.
$\sigma_x^2 = \dfrac{a^2}{28}$
From $\langle x^2\rangle - \langle x\rangle^2 = \tfrac{2}{7}a^2 - \tfrac{1}{4}a^2$.
Recall the momentum moments.
$\langle p\rangle = 0, \qquad \langle p^2\rangle = \dfrac{10\hbar^2}{a^2}$
Real wave function; $\hbar^2\int(\psi')^2dx$.
Form the momentum variance.
$\sigma_p^2 = \dfrac{10\hbar^2}{a^2}$
The mean is zero.
Multiply the variances.
$\sigma_x^2\sigma_p^2 = \dfrac{a^2}{28} \cdot \dfrac{10\hbar^2}{a^2} = \dfrac{10}{28}\hbar^2$
The length cancels, as it must: the product cannot depend on the size of the box.
Take the root.
$\sigma_x\sigma_p = \sqrt{\dfrac{10}{28}}\,\hbar = 0.598\hbar$
A pure multiple of $\hbar$.
Compare with the bound.
$0.598\hbar \geq 0.5\hbar$
Satisfied, with twenty percent to spare; only a Gaussian would reach exactly $0.5\hbar$.
Write the energy of an electron confined within radius $r$ of a proton.
$E(r) \approx \dfrac{\hbar^2}{2m_er^2} - \dfrac{ke^2}{r}$
Kinetic energy from $p \approx \hbar/r$, plus Coulomb potential energy.
Differentiate with respect to $r$.
$\dfrac{dE}{dr} = -\dfrac{\hbar^2}{m_er^3} + \dfrac{ke^2}{r^2}$
The kinetic term falls with $r$; the potential term rises.
Set the derivative to zero.
$\dfrac{\hbar^2}{m_er^3} = \dfrac{ke^2}{r^2}$
The minimum of the energy.
Solve for the radius.
$r = \dfrac{\hbar^2}{m_eke^2}$
Multiply both sides by $r^3/(ke^2)$.
Evaluate using $\hbar c$ and $ke^2 = 1.440$ eV nm.
$r = \dfrac{(\hbar c)^2}{m_ec^2 \cdot ke^2} = \dfrac{(197.3)^2}{511\,000 \times 1.440}\ \text{nm} = 0.0529\ \text{nm}$
Multiplying top and bottom by $c^2$ gives convenient energy units.
Find the energy at this radius.
$E = \dfrac{\hbar^2}{2m_er^2} - \dfrac{ke^2}{r} = \dfrac{ke^2}{2r} - \dfrac{ke^2}{r} = -\dfrac{ke^2}{2r}$
At the minimum the kinetic energy is half the size of the potential energy.
Evaluate the result.
$E = -\dfrac{1.440}{2 \times 0.0529} = -13.6\ \text{eV}$
The hydrogen ground-state energy.
Say what the estimate shows.
$r \to 0 \quad\Rightarrow\quad E \to +\infty$
The electron cannot fall into the proton: confinement costs more kinetic energy than the attraction gains.
Write the energy-time relation.
$\sigma_E \geq \dfrac{\hbar}{2\tau}$
A finite lifetime means an energy spread.
Substitute the values.
$\sigma_E \geq \dfrac{6.582 \times 10^{-16}\ \text{eV s}}{2 \times 16 \times 10^{-9}\ \text{s}}$
Lifetime in seconds.
Evaluate the result.
An electron's position spread is squeezed from $7$ nm to $3.5$ nm. What happens to the least possible spread in its momentum?
Complete the worked solution: for $\psi = e^{-|x|/a}/\sqrt{a}$ with $a = 4$ nm, find $\sigma_x^2$, $\sigma_p^2$ in units of $\hbar^2$/nm², and the product $(\sigma_x\sigma_p)^2$ in units of $\hbar^2$.
Find the mean square position.
$\sigma_x^2 = \langle x^2\rangle = \dfrac{a^2}{2} =$ x nm²
$\langle x\rangle = 0$, so the variance is the mean square.
Find the mean square momentum.
$\sigma_p^2 = \hbar^2\int(\psi')^2dx = \dfrac{\hbar^2}{a^2} =$ p $\hbar^2$/nm²
$\psi' = \mp\psi/a$, so $(\psi')^2 = \psi^2/a^2$, which integrates to $1/a^2$.
Multiply the two variances.
$(\sigma_x\sigma_p)^2 =$ q $\hbar^2$
The length cancels; the product is $\hbar^2/2$, above the bound $\hbar^2/4$.
Match each statement about quantum spreads to its relation (all for $3$ particle).
| $\sigma_x\sigma_p \geq \hbar/2$ | $\sigma_E\tau \geq \hbar/2$ | $\sigma_x\sigma_p = \hbar/2$ exactly | both can be sharp at once | |
|---|---|---|---|---|
| position and momentum along one axis | ||||
| energy and lifetime | ||||
| a Gaussian wave packet | ||||
| position along $x$ and momentum along $y$ |
An electron is confined with position spread $\sigma_x = 0.2$ nm. Fill in the least momentum spread, in units of $10^{-25}$ kg m/s, and the least velocity spread in m/s. Use $\hbar = 1.0546 \times 10^{-34}$ J s.
| $\sigma_p$ ($10^{-25}$ kg m/s) | $\sigma_v$ (m/s) | |
|---|---|---|
| this electron |
A nucleon inside a nucleus has a position spread of $4$ fm. What is the least spread in its momentum, in MeV/$c$? Use $\hbar c = 197.33$ MeV fm.
Answer: MeV/c
An atom's excited state has a lifetime of $4$ ns before it emits a photon. What is the least spread in the energy of the emitted photons, in neV? Use $\hbar = 6.582 \times 10^{-16}$ eV s.
Answer: neV
Particle accelerators make exotic atoms. For muonic hydrogen, a proton orbited by a muon, the orbiting particle is $206.77$ times as massive as an electron. Use the uncertainty-principle estimate of atomic size to find its radius, in fm. The hydrogen radius is $a_0 = 52\,918$ fm.
Answer: fm
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An electron is confined with position spread $\sigma_x = 0.05$ nm. Fill in the least momentum spread, in units of $10^{-25}$ kg m/s, and the least velocity spread in m/s. Use $\hbar = 1.0546 \times 10^{-34}$ J s.
| $\sigma_p$ ($10^{-25}$ kg m/s) | $\sigma_v$ (m/s) | |
|---|---|---|
| this electron |
You can use the uncertainty principle for estimates and checks. Explain to someone why the electron in a hydrogen atom does not fall into the proton.
16. Your turn: sodium's yellow emission comes from a state with lifetime $16$ ns. What is the least spread in the emitted photons' energy?, step 3
$\sigma_E \geq 2.06 \times 10^{-8}\ \text{eV}$
About one part in $10^{8}$ of the $2.1$ eV photon energy: a very sharp line.