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 $\Delta x\times\Delta p \geq \frac{h}{4\times\pi}$ 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?

Heisenberg imagined a microscope capable of locating an electron: to "see" it with a resolution of $\Delta x = 10^{-10}\ \text{m}$, it must be illuminated with photons of comparable wavelength, $\lambda = 10^{-10}\ \text{m}$, whose momentum is $p_{fotón} = \frac{h}{\lambda}$. Data: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $m_e = 9.11\times10^{-31}\ \text{kg}$. Observe the diagram of the thought experiment:

1. a) Calculate the momentum of the photon used.
2. b) If in the collision the photon transfers momentum of that order to the electron, calculate the induced uncertainty in the electron's speed.
3. c) Explain the idea of the thought experiment: the better the position is measured, the more the momentum is disturbed. Then clarify that the uncertainty principle is even deeper than a mere measurement disturbance: what does it state about the nature of the particle prior to measurement?

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 $\Delta x\times\Delta p \geq \frac{h}{4\times\pi}$, where $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.

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.

In an experiment, the momentum of a proton is known with an uncertainty $\Delta p = 10^{-24}\ \text{kg}\times\text{m/s}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.

1. a) Rearrange $\Delta x\times\Delta p \geq \frac{h}{4\times\pi}$ for $\Delta x$ and calculate the minimum uncertainty in its position.
2. b) Compare that size with that of an atom ($10^{-10}\ \text{m}$): is the proton better or worse localized than the size of an atom?

An electron has an uncertainty in its momentum of $\Delta p = 5.3\times10^{-25}\ \text{kg}\times\text{m/s}$. Given: $m_e = 9.11\times10^{-31}\ \text{kg}$.

1. a) Calculate the uncertainty in its velocity, $\Delta v = \frac{\Delta p}{m_e}$.
2. b) Compare the result with the typical speed of an electron in an atom, of the order of $10^{6}\ \text{m/s}$ (given), and comment on what that comparison implies.

An electron is confined in an atom of size $\Delta x = 10^{-10}\ \text{m}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $m_e = 9.11\times10^{-31}\ \text{kg}$.

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.

In an accelerator, a proton moves at $v = 2\times10^{6}\ \text{m/s}$, a velocity measured with an uncertainty $\Delta v = 10^{4}\ \text{m/s}$. Given: $m_p = 1.67\times10^{-27}\ \text{kg}$ and $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.

1. a) Calculate the momentum of the proton, $p = m_p\times v$.
2. b) Calculate the uncertainty in momentum, $\Delta p = m_p\times\Delta v$.
3. c) Apply the uncertainty principle and calculate the minimum uncertainty in its position.
4. d) Calculate the relative uncertainty $\frac{\Delta v}{v}$ as a percentage.

In nanotechnology, electrons are confined in "quantum boxes" of different sizes. Consider two cases: box A, with $\Delta x_A = 1\ \text{nm}$, and box B, with $\Delta x_B = 0.1\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$. The diagram compares the two boxes.

1. a) Calculate the minimum uncertainty in momentum in each box.
2. b) Calculate the ratio $\frac{\Delta p_B}{\Delta p_A}$.
3. c) Explain, using the uncertainty principle formula, the inverse relationship between the confinement size and the uncertainty in momentum.

An electron and a proton are each confined in a region of the same size, $\Delta x = 10^{-10}\ \text{m}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $m_e = 9.11\times10^{-31}\ \text{kg}$ and $m_p = 1.67\times10^{-27}\ \text{kg}$.

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.

In a modern chip, an electron is confined within a transistor of width $10\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $m_e = 9.11\times10^{-31}\ \text{kg}$.

1. a) Calculate the minimum uncertainty in its momentum.
2. b) Calculate the uncertainty in its velocity.
3. c) If industry reduces the scale of the transistor to $1\ \text{nm}$, what happens to $\Delta v$? Comment on what challenge this presents for the miniaturisation of electronics.

A soccer ball moves across the field and, in principle, is also subject to the uncertainty principle.

1. a) Explain, using the value of $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, 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: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.

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.

A speck of dust of mass $10^{-6}\ \text{kg}$ is observed with a microscope that determines its position with an uncertainty $\Delta x = 10^{-6}\ \text{m}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.

1. a) Calculate the minimum uncertainty in its momentum.
2. b) Calculate the uncertainty in its velocity.
3. c) The speck moves at $10^{-3}\ \text{m/s}$: explain whether quantum uncertainty has any relevance for describing its motion.

We confine a dust speck of $10^{-15}\ \text{kg}$ and an electron to the same region, $\Delta x = 10^{-6}\ \text{m}$. Data: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $m_e = 9.11\times10^{-31}\ \text{kg}$.

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.

Imagine a universe where Planck's constant were much larger. In this exercise we will work with a ball of $0.15\ \text{kg}$ located with an uncertainty $\Delta x = 10^{-3}\ \text{m}$. Data: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.

1. a) Calculate, using the actual value of $h$, the minimum uncertainty in the ball's speed.
2. b) Rearrange for $h$ in the expression of the uncertainty principle and calculate what value it should have so that, with the same $\Delta x$, the uncertainty in speed would be noticeable, for example $\Delta v = 0.01\ \text{m/s}$.
3. c) How many orders of magnitude larger than the real constant would that value be?
4. d) Imagine the consequences in that universe: give a reasoned description of how the movement of everyday objects would change if the uncertainty principle were noticeable on a human scale.

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: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $m_e = 9.11\times10^{-31}\ \text{kg}$ and $c = 3\times10^{8}\ \text{m/s}$.

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?

Bohr's model depicts the electron in a circular orbit of radius $r = 5.3\times10^{-11}\ \text{m}$. Suppose we wanted to track this trajectory knowing its position with an uncertainty of $10\%$ of the radius, $\Delta x = 5.3\times10^{-12}\ \text{m}$. Data: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $m_e = 9.11\times10^{-31}\ \text{kg}$. The figure shows the orbit with the assumed uncertainty.

1. a) Calculate the minimum uncertainty in momentum imposed by the uncertainty principle.
2. b) Calculate the uncertainty in speed.
3. c) The speed of the electron in the first Bohr orbit is $2.2\times10^{6}\ \text{m/s}$ (given): compare $\Delta v$ with it and conclude whether the image of an exact orbit is compatible with quantum mechanics.

You can do it on paper and send a photo with your answer.

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 $E_c \approx \frac{\Delta p^2}{2\times m}$. Data: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $m_e = 9.11\times10^{-31}\ \text{kg}$, $m_p = 1.67\times10^{-27}\ \text{kg}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.

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.

A student writes in an exam: "With a perfect measuring instrument I could simultaneously know the position and momentum of an electron with whatever precision I wished; the uncertainty principle only reflects the poor quality of current equipment." Data: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.

1. a) Point out the conceptual error in the statement and correct it.
2. b) Use numbers to support your answer: if the "perfect instrument" improves the position measurement from $\Delta x = 10^{-9}\ \text{m}$ to $\Delta x = 10^{-12}\ \text{m}$, calculate by what factor the minimum uncertainty in momentum increases.
3. c) Is there any value of $\Delta x$ for which $\Delta p$ stops increasing or can become zero? Reason your answer using the expression $\Delta x\times\Delta p \geq \frac{h}{4\times\pi}$.

A beam of electrons with speed $v = 10^{7}\ \text{m/s}$ passes through a narrow slit of width $\Delta y = 10^{-7}\ \text{m}$. As they pass, the vertical position of each electron is determined with that uncertainty, which spreads the beam out. Data: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $m_e = 9.11\times10^{-31}\ \text{kg}$. The figure shows the geometry of the setup.

1. a) Calculate the minimum uncertainty in the vertical component of momentum, $\Delta p_y$.
2. b) Calculate the linear momentum of the electron, $p = m_e\times v$, and estimate the beam's angle of spread, $\theta \approx \frac{\Delta p_y}{p}$, in radians.
3. c) Estimate the width of the beam on a screen placed at $L = 1\ \text{m}$, as $2\times L\times\theta$, and express it in millimeters.
4. d) Predict what happens if the slit is narrowed to $10^{-8}\ \text{m}$ and relate this result to diffraction: why does a smaller aperture spread the beam more?

Newtonian mechanics is deterministic: if at a given instant the position and momentum of all particles in a system are known exactly, the equations of motion predict their entire future. Data: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $m_e = 9.11\times10^{-31}\ \text{kg}$.

1. a) Use the uncertainty principle to argue that this "exact initial state" cannot exist: calculate the minimum uncertainty in momentum for an electron whose position was known to within $\Delta x = 10^{-10}\ \text{m}$ and the corresponding uncertainty in its speed.
2. b) If the initial state cannot be known exactly, what kind of predictions does quantum physics make about the outcome of a measurement?
3. c) Explain why Newtonian mechanics, despite everything, works so well for a ball, a car or a planet, supporting your answer with the value of $h$ and the masses of these objects.