The uncertainty principle

In $1927$, Heisenberg stated the uncertainty principle.

1. a) State it in your own words.
2. b) Which pair of quantities does it relate? Write its mathematical expression.
3. c) Explain what $\Delta x$ and $\Delta p$ mean in that expression: what does each uncertainty represent?

A classmate claims: "The uncertainty principle is due to our measuring instruments being imperfect; with better equipment it would disappear."

1. a) Explain whether this statement is correct: is the principle a technological limitation or an intrinsic property of nature?
2. b) Point out the difference between quantum uncertainty $\Delta x$ and ordinary measurement error with a ruler or stopwatch.

Start from the expression $\Delta x\cdot\Delta p \geq \frac{h}{4\cdot\pi}$ and reason without a calculator:

1. a) If the uncertainty in position is halved, what happens to the minimum uncertainty in momentum?
2. b) What if $\Delta x$ is multiplied by ten?
3. c) Summarise the relationship between both uncertainties: can both be arbitrarily small at the same time?

An electron is confined to a region of size $\Delta x = 10^{-10}\ \text{m}$ (the size of an atom). The uncertainty principle states that $\Delta x\cdot\Delta p \geq \frac{h}{4\cdot\pi}$, where $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$.

1. a) Calculate the minimum uncertainty in its momentum, $\Delta p$.
2. b) If the electron were confined to a region ten times smaller, $\Delta x = 10^{-11}\ \text{m}$, what would $\Delta p$ be? Justify using proportionality.

An electron is confined in an atom of size $\Delta x = 10^{-10}\ \text{m}$. Given: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$ and $m_e = 9,11\cdot10^{-31}\ \text{kg}$.

1. a) Calculate the minimum uncertainty in its momentum, $\Delta p$.
2. b) Calculate the uncertainty in its velocity, $\Delta v = \frac{\Delta p}{m_e}$.
3. c) The typical speed of an electron in an atom is of the order of $10^{6}\ \text{m/s}$ (given): compare $\Delta v$ with that speed and explain why it makes no sense to assign a precise orbit to the electron.

An electron and a proton are each confined in a region of the same size, $\Delta x = 10^{-10}\ \text{m}$. Given: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$, $m_e = 9,11\cdot10^{-31}\ \text{kg}$ and $m_p = 1,67\cdot10^{-27}\ \text{kg}$.

1. a) Which of the two particles has a larger minimum uncertainty in momentum? Justify your answer by looking at the formula: does mass appear in it?
2. b) Calculate the uncertainty in the velocity of each particle.
3. c) Explain why mass does not affect $\Delta p$ but does affect $\Delta v$, and what consequence this has for the heavier particle.

A football moves across the pitch and, in principle, is also subject to the uncertainty principle.

1. a) Explain, using the value of $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$, why the principle has no practical consequences for everyday objects.
2. b) What two characteristics of macroscopic objects (mass and size) make quantum uncertainties negligible compared to ordinary measurements?

A bullet of $10\ \text{g}$ passes through a detector that determines its position with an uncertainty $\Delta x = 10^{-5}\ \text{m}$. Given: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$.

1. a) Calculate the minimum uncertainty in the momentum of the bullet.
2. b) Calculate the uncertainty in its velocity.
3. c) The bullet travels at a speed of $400\ \text{m/s}$: calculate the ratio $\frac{\Delta v}{v}$ and conclude whether quantum uncertainty is noticeable in this measurement.

We confine a dust speck of $10^{-15}\ \text{kg}$ and an electron to the same region, $\Delta x = 10^{-6}\ \text{m}$. Data: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$ and $m_e = 9,11\cdot10^{-31}\ \text{kg}$.

1. a) Calculate the minimum uncertainty in momentum: is it the same for both? Justify your answer.
2. b) Calculate the uncertainty in the speed of the dust speck and that of the electron.
3. c) In light of the results, reason which quantity determines whether quantum effects are relevant for an object: its confinement size, its mass, or both.
4. d) Extrapolate: explain why no laboratory experiment has ever observed noticeable quantum uncertainty in objects visible to the naked eye.

In classical mechanics, a particle follows a trajectory: at every instant it has an exact position and velocity. In quantum mechanics, we speak of probabilities and orbitals.

1. a) Explain, using the uncertainty principle, why the concept of an exact trajectory loses meaning for an electron.
2. b) What information does an orbital provide about the electron, instead of its trajectory?
3. c) Relate the above to the transition from the Bohr model (orbits) to the current quantum model (orbitals).

Before the neutron was discovered, it was thought that the nucleus might contain electrons. Use the uncertainty principle to check what would happen if an electron were confined within a nucleus of size $\Delta x = 10^{-15}\ \text{m}$. Given: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$, $m_e = 9,11\cdot10^{-31}\ \text{kg}$ and $c = 3\cdot10^{8}\ \text{m/s}$.

1. a) Calculate the minimum uncertainty in the momentum of the electron.
2. b) Calculate the uncertainty in its velocity.
3. c) Compare $\Delta v$ with the speed of light and draw a conclusion: can an electron be stably confined within the nucleus?

The uncertainty principle allows us to estimate the minimum energy of a confined particle: if the momentum is of the order of $\Delta p$, its kinetic energy is of the order of $E_c \approx \frac{\Delta p^2}{2\cdot m}$. Data: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$, $m_e = 9,11\cdot10^{-31}\ \text{kg}$, $m_p = 1,67\cdot10^{-27}\ \text{kg}$ and $1\ \text{eV} = 1,6\cdot10^{-19}\ \text{J}$.

1. a) Estimate the minimum kinetic energy of an electron confined in an atom, $\Delta x = 10^{-10}\ \text{m}$, and express it in electronvolts. Compare it with typical chemical bond energies, which are a few electronvolts (given).
2. b) Estimate the minimum kinetic energy of a proton confined in a nucleus, $\Delta x = 10^{-15}\ \text{m}$, and express it in $\text{MeV}$ ($1\ \text{MeV} = 10^{6}\ \text{eV}$).
3. c) Conclude: explain how the uncertainty principle accounts for why nuclear process energies are millions of times greater than those of atomic processes.