Quantum Mechanics

In the early 1900s physicists performed experiments that probed reality at its fundamental level. They found that matter was made of particles they named atoms. Inside each atom they found the subatomic particles: electrons, protons, and neutrons. They also found that light was made of particles that we now call photons. The strangest result was that all these particles produced wave-like interference patterns.

The double slit experiment is a good example of wave interference. A wave passing through a slit spreads out. It diffracts.

single slit wave diffraction

2 slits produce 2 separate wave diffractions. Once both waves get to their target, they have traveled different distances so they have different phases. When both waves are in the same phase they add constructively. When both waves are half a cycle out of phase they add destructively. The phase differences lead to an alternating constructive and destructive interference pattern.

double slit wave interference
gap separation =

In 1801, Thomas Young produced a double slit interference pattern with beams of light. In 1909, the experiment was repeated with a single photon at a time. After passing through the double slit, each photon was measured at single location, but the distribution of measurements still produced an interference pattern.

The double slit experiment isn't limited to photons. In 1927, the experiment was repeated with electrons. In 2019, an interference pattern was produced by a 2000 atom molecule.

In order to produce a wave interference pattern, particles must be perfectly isolated from their environment. That's why we still can't produce a double slit interference pattern for everyday sized objects, like cats. It's just too hard to isolate large objects long enough to run the experiment.

Simulation: What happens when one slit is blocked? Why?

The interference pattern goes away, because the waves are all originating from the same point, so they have the same phase.

Although, there is still a diffraction pattern related to the width of the slit.

Simulation: When one particle at a time is fired, the distribution of hits still produces an interference pattern. What could the single particle be interfering with?

A single particle goes through both slits. It interferes with itself.

It seems that the things we call particles have wave-like properties.

The overlap between the wave and particle models at the nanometer scale led to many more exciting experiments. Over time, physicists developed and refined a new model, called quantum mechanics. Predictions based on quantum mechanics can only give probabilities, but when applied to repeated experiments they are very accurate.

Photon Energy

In 1899, after investigating the thermal radiation spectrum, Maxwell Planck reluctantly hypothesized that the energy of light is only released in small quantities determined by its frequency. This idea was the beginning of quantum mechanics.

e - e - e - e -

These packets of light were eventually called photons. Photons have a frequency and a wavelength, but no mass. They are produced anytime a charged particle loses energy. Photons spread out propagating through space at the speed of light. They can transfer their energy to a charged particle in a process called absorption.


\(E\) = energy of a photon [J, joules, kg m²/s²]
\(h\) = 6.626 × 10-34 = Planck's constant [J s]
\(f\) = frequency [Hz, 1/s]

f (THz)

668 – 789
606 – 668
526 – 606
508 – 526
484 – 508
400 – 484

λ (nm)

380 – 450
450 – 495
495 – 570
570 – 590
590 – 620
620 – 750
Question: Which has more energy, one red photon or one blue photon?
answer $${\Uparrow \atop E} {\atop = h} {\Uparrow \atop f}$$

Red photons have a lower frequency and a lower energy.

Blue photons have a higher frequency and a higher energy.

Example: What is the energy range for red photons? Calculate the highest and lowest energy possible in the color ranges listed above.
metric prefixes
Name Symbol Factor Power
tera T 1 000 000 000 000 1012
giga G 1 000 000 000 109
mega M 1 000 000 106
kilo k 1 000 103
centi c 0.01 10-2
milli m 0.001 10-3
micro μ 0.000 001 10-6
nano n 0.000 000 001 10-9
pico p 0.000 000 000 001 10-12
solution $$\text{lower range}$$ $$E=hf$$ $$E=(6.626 \times 10^{-34})(400 \times 10^{12})$$ $$E=2.65 \times 10^{-19}J$$
$$\text{upper range}$$ $$E=hf$$ $$E=(6.626 \times 10^{-34})(484 \times 10^{12})$$ $$E=3.21 \times 10^{-19}J$$
Example: The frequency of a photon is 3.6 × 1015 Hz. What color is it? What energy does it have?
electro-magnetic spectrum
region wavelength (m) frequency (Hz)
gamma ray
x-ray 2 × 10-11 1.5 × 1019
ultraviolet 1 × 10-8 3 × 1016
visible light 4 × 10-7 7.5 × 1014
infrared 7.5 × 10-7 4 × 1014
microwave 1 × 10-2 3 × 1010
radio wave 1 3 × 108
solution $$E=hf$$ $$E = (6.626 \times 10^{-34})(3.6 \times 10^{15})$$ $$E = 2.385 \times 10 ^{-18} \, \mathrm{J} $$

The color is not visible, because it is in the ultraviolet.

Example: Find the energy for a photon that has a wavelength of 0.3 m. What part of the E-M spectrum is the photon in?
solution $$c = \lambda f $$ $$f = \frac{c}{\lambda} $$ $$f = \frac{ 3 \times 10^{8}}{ 0.3 } $$ $$f = 10^{9} \, \mathrm{Hz}$$
$$ E = hf $$ $$ E = (6.626 \times 10^{-34})(10^{9}) $$ $$ E = 6.626 \times 10^{-25} \, \mathrm{J}$$

The photon is a microwave.

Example: Find the wavelength of a photon that has 2.65 × 10-19 J of energy.
solution $$ c = f \lambda $$ $$ f = \frac{c}{ \lambda }$$
$$E=hf$$ $$E=\frac{hc}{\lambda}$$ $$\lambda=\frac{hc}{E}$$ $$\lambda=\frac{(6.626 \times 10^{-34})(3 \times 10^8)}{2.65 \times 10^{-19}}$$ $$ \lambda = 7.5 \times 10 ^{-7} \, \mathrm{m} $$ $$ \lambda = 750 \, \mathrm{nm} $$
Example: A helium neon laser outputs 5 mW of optical power. Lookup the wavelength of the light and then calculate the number of photons produced every second.

Wikipedia says the wavelength of each photon is 632.8 nm. Convert the wavelength into energy.

The number of photons is the energy output divided by the energy for each photon. The energy output is 0.005 J every second.

solution $$\lambda = 632.8 \, \mathrm{nm} $$ $$E=\frac{hc}{\lambda}$$ $$E=\frac{(6.626 \times 10^{-34})(3 \times 10^8)}{632.8 \times 10^{-9}}$$ $$E = 3.14 \times 10^{-19} \, \mathrm{J}$$
$$\mathrm{power} = \frac{\mathrm{energy}}{\mathrm{time}} $$ $$\frac{\mathrm{power}}{\mathrm{energy}} = \frac{1}{\mathrm{time}} $$ $$\frac{\mathrm{power}}{\mathrm{energy}} = \mathrm{frequency} $$ $$\frac{0.005\, \mathrm{W}}{3.14 \times 10^{-19} \, \mathrm{J}} = \mathrm{frequency}$$ $$ 1.59 \times 10^{16} \, \mathrm{\tfrac{1}{s}} = \mathrm{frequency}$$
15 900 000 000 000 000 photons/second from a low power laser beam!

The Photoelectric Effect

The photoelectric effect occurs when metals dislodge electrons after being hit by light. Light with a frequency above the visible spectrum is required to produce the effect. Bright red light can't produce the effect, but even dim UV light can.

In 1905, Albert Einstein published an explanation of the photoelectric effect that supported Max Plank's concept of quantized light. Einstein suggested that light is made up of many small packets of energy, and each packet interacts with a single electron. Only high frequency light releases electrons because it has enough energy per packet.

Einstein published an equation that describes the photoelectric effect using conservation of energy. The kinetic energy of a released electron can't exceed the difference between the energy of an incoming photon and the energy needed to dislodge the electron. If the maximum kinetic energy is below zero, an electron is not released.

$$K_{max}=hf - \Phi$$

\(K_{max}\) = maximum kinetic energy of released electron [J, joules]
\(h\) = 6.626 × 10-34 = Planck's constant [J s]
\(f\) = frequency of incoming light [Hz, 1/s]
\(\Phi\) = Work function, the minimum energy to dislodge an electron [J]

Electrons can be thought of as being stuck in a energy well. The work function represents the minimum energy the electrons needs to escape. The work function depends on the material. Metals have a low work function, so it is easier to dislodge an electron from a metal.

Question: Why does dim ultraviolet light produce the photoelectric effect, yet very bright red light doesn't?

Most of the time just one photon is hitting one electron. Low energy photons don't have enough energy to dislodge an electron.

A two photon photoelectric effect can happen, but it's probability is very low at normal light intensities. A high intensity red laser produces enough photons for a two photon effect, but it would also turn the metal into a hot plasma.

Energy at the atomic level is often calculated in electron volts (eV). We can convert between eV and J by multiplying or dividing by the charge of an electron.

$$1 \, \mathrm{eV} = 1.6 \times 10^{-19} \, \mathrm{J} $$ $$32 \times 10^{-19} \, \mathrm{J} \left( \frac{ 1 \, \mathrm{eV}}{ 1.6 \times 10^{-19} \, \mathrm{J}} \right)= 20 \, \mathrm{eV}$$ Example: Find the energy of a 1.69 × 1015 Hz photon in electron-volts.
solution $$E=hf$$ $$E=(6.626 \times 10^{-34}) (1.69 \times 10^{15})$$ $$E = 1.12 \times 10 ^{-18} \, \mathrm{J}$$
$$E = 1.12 \times 10 ^{-18} \, \mathrm{J} \left( \frac{ 1 \, \mathrm{eV}}{ 1.6 \times 10^{-19} \, \mathrm{J}} \right) = 7.0 \, \mathrm{eV} $$
Example: After being hit by light with 7.0 eV per photon, the rare earth metal terbium releases electrons. The electrons have a maximum kinetic energy of 4.0 eV. What is the work function of terbium in electron-volts?
solution $$K_{\text{max}} = hf - \Phi$$ $$\Phi = hf - K_{\text{max}}$$ $$\Phi = 7.0 \, \mathrm{eV} - 4.0 \, \mathrm{eV}$$ $$\Phi = 3.0 \, \mathrm{eV}$$
Example: Find the max kinetic energy of a magnesium electron after being hit by a photon with a frequency of 600 THz. Look up the work function for magnesium and convert it into joules.
metric prefixes
Name Symbol Factor Power
tera T 1 000 000 000 000 1012
giga G 1 000 000 000 109
mega M 1 000 000 106
kilo k 1 000 103
centi c 0.01 10-2
milli m 0.001 10-3
micro μ 0.000 001 10-6
nano n 0.000 000 001 10-9
pico p 0.000 000 000 001 10-12
solution $$\Phi = (3.66 \, \mathrm{eV}) \left( \frac{1.6 \times 10^{-19} \, \mathrm{J}}{1 \, \mathrm{eV}} \right)$$ $$\Phi = 5.856 \times 10^{-19} \, \mathrm{J}$$
$$hf = (6.626 \times 10^{-34})(600 \times 10^{12})$$ $$hf = 3.9756 \times 10^{-19} \, \mathrm{J}$$
$$K_{\text{max}}=hf - \Phi$$ $$K_{\text{max}}= (3.9756 \times 10^{-19} \, \mathrm{J}) - (5.856 \times 10^{-19} \, \mathrm{J}) $$ $$K_{\text{max}}= -1.88 \times 10^{-19} \, \mathrm{J} $$

A negative kinetic energy means the electron will not escape the atom.

Example: A 100 nm photon strikes a lump of magnesium. How much kinetic energy could the released electron have?
solution $$ c = f \lambda $$ $$ f = \frac{c}{\lambda} $$ $$ f = \frac{(3.00 \times 10^{8})}{100 \times 10^{-9}} $$ $$ f = 3.00 \times 10^{15} \, \mathrm{Hz} $$
$$ \Phi = 3.66 \, \mathrm{eV} \left( \frac{1.6 \times 10 ^{-19} \, \mathrm{J}}{1 \, \mathrm{eV}} \right) = 5.856 \times 10^{-19} \, \mathrm{J}$$
$$K_{\text{max}} = hf - \Phi$$ $$K_{\text{max}} = (6.626 \times 10^{-34})(3.00 \times 10^{15}) - 5.856 \times 10^{-19}$$ $$K_{\text{max}} = 19.878 \times 10^{-19} - 5.856 \times 10^{-19}$$ $$K_{\text{max}} = 14.022 \times 10^{-19} \, \mathrm{J}$$

How fast could the electron be moving as it escapes the magnesium atom?
solution $$m_e = 9.10 \times 10^{-31} \, \mathrm{kg} $$ $$ K = \tfrac{1}{2}m_ev^2 $$ $$ v = \sqrt{\frac{2K}{m_e}}$$ $$ v = \sqrt{\frac{2(14.022 \times 10^{-19})}{9.10 \times 10^{-31}}}$$ $$v = \sqrt{3.08 \times 10^{12}}$$ $$ v = 1.76 \times 10 ^6 \, \mathrm{\tfrac{m}{s}}$$
Example: Use the work function table to decide which elements would release electrons from 427 nm wavelength light.

The lowest energy photon that can dislodge electrons will have the same energy as the work function. Convert the photon's wavelength into electronvolts and compare it to the work functions in the table.

solution $$E=\frac{hc}{\lambda}$$ $$E=\frac{(6.626 \times 10^{-34})(3 \times 10^8)}{427 \times 10^{-9}}$$ $$ E = 4.66 \times 10^{-19} \, \mathrm{J}$$ $$ E = 4.66 \times 10^{-19} \, \mathrm{J} \left( \frac{1 eV}{1.602 \times 10^{-19} \, \mathrm{J}} \right)= 2.91 \, \mathrm{eV}$$

Any element with a work function below 2.91 eV. Like: Ce, Gd, Li

Light as Radiation

Radiation is a wave or particle that transmits energy through space. This includes particles with mass (electrons) and massless particles (photons).

A large amount of radiation can hurt living things in a direct way, by increasing their temperature. A toaster and a microwave oven both use radiation to cook food.

Radiation from particles with very high energy can harm living things in a more subtle way. The particles can ionize atoms, which breaks chemical bonds. This can cause cell death and possibly cancer.

Non-ionizing radiation is generally below 1.60 × 10-18 J (10 eV). This safe radiation doesn't have enough energy to break chemical bonds.

Ionizing radiation is generally above 1.60 × 10-18 J (10 eV). This unsafe radiation can potentially break chemical bonds.

region wavelength (m) frequency (Hz) energy (J) energy (eV)
 gamma ray
 x-ray 2 × 10-11 1.5 × 1019 10 × 10-15 62 500
 ultraviolet 1 × 10-8 3 × 1016 2.0 × 10-17 125
visible light 4 × 10-7 7.5 × 1014 5.0 × 10-19 3.1
infrared 7.5 × 10-7 4 × 1014 2.7 × 10-19 1.7
microwave 1 × 10-2 3 × 1010 2.0 × 10-23 0.000 125
radio wave 1 3 × 108 2.0 × 10-25 0.000 001

Try this PhET simulation to see how molecules interact with photons.

Question: Greenhouse gasses need to interact with infrared light. Which molecules from the simulation could be greenhouse gasses?

CO | CO₂ | CH₄ | H₂O | NO₂ | O₃ all could deflect infrared light as it leaves the Earth, preventing the Earth from cooling.

In the last 100 years humans have greatly increased the amount of CO₂. The extra CO₂ has led to a 1° Celsius increase in average global temperature.

Question: Can microwave ovens ionize atoms? Are microwaves dangerous?

Microwave ovens make microwaves. Each microwave photon has an energy around 10-23 J. This is far below the ionization threshold: 1.60 × 10-18 J.

Many studies have been done on the risks of microwaves, but no adverse health effects have been established. Microwaves can heat up things, but they will not directly cause cell damage or cancer. Cell Phones and Wi-Fi also use microwaves to transmit data.

Question: Is sunlight dangerous?

Too much sun can cause sunburns and skin cancer. Sunlight has some ionizing radiation in the UV part of the electromagnetic spectrum. Most of the UV is blocked by the ozone layer, but some still gets through.

Not enough sunlight is also unhealthy. A lack of Sun exposure can lead a vitamin D deficiency. It is recommended that we get at least 15 minutes of direct Sun every day.

Question: Which would damage a person the most: an ultraviolet photon or a gamma-ray photon?

Gamma Rays

UV is less likely to ionize cells. Gamma-rays have much more energy per photon, but gamma-rays also have higher penetration. This means they can pass deeper into the body or not interact at all.

Emission and Absorption

Each electron in an atom can only have an energy that exactly matches an atomic orbital.

quantum mechanics and atomic orbitals

Two bodies can orbit each other if they experience an attractive force. Gravitational attraction keeps the Earth orbiting the Sun every year.

The nucleus of an atom attracts electrons because it has an opposite electric charge. This attraction can also lead to an orbit, but with some quantum mechanical restrictions.

Quantum mechanics tells us that electrons are waves. As the waves try to orbit the nucleus, they destructively interfere with themselves. This limits the wavelengths to values that allow constructive orbital standing waves to form. We say that the allowed wavelengths are quantized. Quantizing means to restrict a variable to specific values.

A 2-D analogy for atomic orbitals is the vibrations of a drum.

An electron in an atom can be described by 4 quantum numbers. Quantizing these numbers adds another restriction. Two electrons can't form the exact same wave due to a type of destructive interference called the Pauli exclusion principle. This means each electron in an atom must have a different combination of the 4 numbers.

This image is a 2-D slice of the allowed 3-D states for electrons in a Hydrogen atom. Each electron in an atom or molecule must occupy a different wave state. This state determines its wavelength, frequency and energy.

All quantum mechanical particles are not required to be quantized. Particles outside an atom are free to have a wide range of values. The quantization of particles comes from electrons in atoms forming standing waves only at specific values. This quantization can spread to photons as they are emitted from electrons in atoms.

If an electron gains the energy difference between two states it can jump to a higher energy level. For example, an electron will transition to a higher energy level if it collides with a photon that has an energy equal to the energy difference between levels.

This process can also run in reverse. An electron can drop to a lower energy level if an atom has an unoccupied energy level. When the electron drops down it emits a photon equal to the difference in energy between each level. This follows the law of conservation of energy.

Photon emission occurs when electrons transition to lower energy levels within an atom. Each electron transition emits a photon with an energy equal to the energy difference between levels.

Hydrogen has a very simple emission spectrum because it doesn't have very many possible energy states and therefore few energy state transitions.

Iron's nucleus has more protons so it has more possible energy transitions.

Photon emission is the working principle behind fluorescent lights. To make light, a tube is filled with various gases. The gases are electrically charged up which brings the electrons to a higher energy level. The electrons are unstable in the higher energy levels. They eventually fall back down to their ground state and emit light. Try looking at fluorescent light reflected off a CD to see the separate bands of color.

Example: Find the frequency of the photon produced when an electron drops from an energy of -3.4 eV to -13.6 eV in a hydrogen atom?
solution $$\Delta E = -13.6 \, \mathrm{eV} + 3.4 \, \mathrm{eV}$$ $$\Delta E = -10.2 \, \mathrm{eV} \left( \frac{1.6 \times 10^{-19} \, \mathrm{J}}{1 \, \mathrm{eV}} \right) = 1.632 \times 10^{-18} \, \mathrm{J}$$
$$E = hf$$ $$f = \frac{E}{h}$$ $$f = \frac{ 1.632 \times 10^{-18} \, \mathrm{J}}{6.626 \times 10^{-34} \, \mathrm{J \, s}}$$ $$f = 2.46 \times 10^{15} \, \mathrm{Hz}$$

The photon is in the ultraviolet spectrum.

Absorption is the reverse of emission. A single photon is absorbed by a single electron in a single atom. This causes the electron to transition to higher energy levels.

If there isn't an energy difference that matches the energy of the colliding photon, the material is transparent to that frequency. A substance may be clear in one range of the spectrum but not in others. For example: glass is mostly transparent to visible light, but it has many absorption frequencies in the infrared and UV.

Exposing an atom to a full range of light will produce an absorption spectrum that matches the energy difference between electron energy levels.

Light absorption occurs at the same energies as emission. Hydrogen's absorption frequencies are the same as its emission frequencies.

Analyzing the spectrum of emission or absorption can actually be used like a fingerprint to identify the elements or molecules being observed. This technique is used in fields like forensics and astronomy.

Sunlight is mostly thermal radiation with large sections of absorption from the molecules in Earth's atmosphere.

250 500 750 1000 1250 1500 1750 2000 2250 2500 Wavelength (nm) Spectrum of Solar Radiation (Earth) 0 0.5 1 1.5 2 2.5 Irradiance (W/m²/nm) 2 H O Atmospheric absorption bands H O 2 H O 2 H O 2 H O 2 CO 2 O 2 O 3 UV Visible Infrared Sunlight without atmospheric absorption 5778K blackbody Sunlight at sea level Question: What molecules are absorbing most of the Sun's infrared rays?

H₂O, CO₂ and O₂ absorb infrared light and then re-emit the light in a random direction.

High H₂O and CO₂ concentrations lead to a warmer Earth. This happens because incoming sunlight is mostly visible and is not absorbed by these gases, but light leaving the Earth is mostly infrared and is absorbed.

Question: What molecule is responsible for absorbing the Sun's UV rays?

O₃ (Ozone) absorbs some of the UV spectrum.
This reduces ionizing radiation.

103 nm n = 1 n = 2 n = 3 n = 4 n = 6 n = 5 434 nm 122 nm Lyman series Balmer series Paschen series 94 nm 410 nm 486 nm 656 nm 1875 nm 1282 nm 1094 nm 97 nm 95 nm

This diagram shows the wavelengths of photons emitted or absorbed when an electron transitions between energy levels. Each wavelength, listed in nanometers, is for a photon with an energy equal to the difference between energy levels.

Question: What wavelength of light could make an electron jump from energy level n = 1 to n = 5?
answer $$\lambda = 95 \, \mathrm{nm}$$
Example: Find the energy of a photon produced as an electron drops from energy level n = 4 to n = 2?
solution $$\lambda = 486 \, \mathrm{nm} = 4.86 \times 10^{-7} \, \mathrm{m}$$ $$E = \frac{hc}{\lambda}$$ $$E = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{4.86 \times 10^{-7}}$$ $$E = 4.09 \times 10^{-19} \, \mathrm{J} $$
$$E = 4.09 \times 10^{-19} \, \mathrm{J} \left(\frac{1 \, \mathrm{eV}}{1.6 \times 10^{-19} \, \mathrm{J}} \right) = 2.56 \, \mathrm{eV}$$
Example: Find three possible light frequencies that could be absorbed by an electron at energy level n = 3.

We can see three wavelengths in the diagram that transition to n = 3:

  • 1875 nm,   1282 nm,   1094 nm.
  • There are more transitions, but the diagram doesn't go past n = 6.

    $$c=f \lambda$$ $$f = \frac{c}{\lambda}$$ $$f = \frac{3\times 10 ^8 }{\lambda}$$ $$f = \frac{3\times 10 ^8 }{1875 \times 10^{-9}} \quad \frac{3\times 10 ^8 }{1282 \times 10^{-9}} \quad \frac{3\times 10 ^8 }{1094 \times 10^{-9}} $$ $$f = 1.6 \times 10^{14} \,\,\,\quad 2.34 \times 10^{14} \,\,\,\quad 2.74 \times 10^{14} \, \mathrm{Hz} $$

    These frequencies are all in the infrared.

    Wavelength and Momentum

    We can calculate the quantum wavelength of a particle in terms of it's momentum with the equation below.

    derivation of particle wavelengths

    Special relativity explains the relationship between energy and momentum. We can use that relationship to find how wavelength and momentum relate.

    $$E^2 = (pc)^2 +(mc^2)^2 $$

    Photons have no mass, so the equation simplifies.

    $$E^2 = (pc)^2$$ $$E = pc$$
    $$E = hf$$ $$E = \frac{hc}{\lambda}$$ $$pc = \frac{hc}{\lambda}$$ $$p = \frac{h}{\lambda}$$ $$\lambda = \frac{h}{p}$$

    $$\lambda = \frac{h}{p}$$

    \(\lambda\) = wavelength [m, meter]
    \(h\) = 6.626 × 10-34 = Planck's constant [J s]
    \(p\) = momentum [kg m/s]

    Photons have no mass, but they they still have momentum. You can actually push a space ship like a sail boat, but with light instead of wind. When photons collide with the light sail, they bounce off pushing the sail forward.

    Example: What is the momentum of a photon of green light? (540 nm)
    solution $$\lambda = \frac{h}{p}$$ $$p = \frac{h}{\lambda}$$ $$p = \frac{6.626 \times 10^{-34}}{540 \times 10^{-9}}$$ $$p = 1.2 \times 10^{-27} \, \mathrm{kg \tfrac{m}{s}}$$

    If we fired photons at a massive object the photons should transfer their momentum because of conservation of momentum.

    Extra Credit: How many green photons would it take to accelerate a 1 kg body from rest to 1 m/s? (assume the green photons reflect off the 1 kg object)

    Use conservation of momentum.

    $$p_{\mathrm{(initial)}} = p_{\mathrm{(final)}}$$ $$p_{\mathrm{(photons)}} + p_{\mathrm{(body)}} = p_{\mathrm{(photons)}} + p_{\mathrm{(body)}}$$

    The body is initially at rest. We can set it to zero.

    $$p_{\mathrm{(photons)}} = p_{\mathrm{(photons)}} + p_{\mathrm{(body)}}$$

    After the photons are reflected they will have the same momentum but in the opposite direction.

    $$p_{\mathrm{(photons)}} = -p_{\mathrm{(photons)}} + p_{\mathrm{(body)}}$$ $$2p_{\mathrm{(photons)}} = p_{\mathrm{(body)}}$$

    We can just add in a variable "n" to represent the number of total photons and scale "n" by the momentum of a single photon.

    $$2np_{\mathrm{(photon)}} = p_{\mathrm{(body)}}$$
    solution $$2np_{\mathrm{(photon)}} = p_{\mathrm{(body)}}$$ $$2np_{\mathrm{(photon)}} = mv$$ $$2np_{\mathrm{(photon)}} = (1)(1)$$ $$np_{\mathrm{(photon)}} = \tfrac{1}{2}$$ $$n = \frac{1}{2p_{\mathrm{(photon)}}}$$
    $$p = 1.2 \times 10^{-27} \, \mathrm{kg \tfrac{m}{s}}$$ $$n = \frac{1}{(2)(1.2 \times 10^{-27})}$$ $$n= 4.16 \times 10^{26}\,\mathrm{photons}$$

    In 1924 Louis de Broglie proposed that particles with mass might have a wavelength. This turned out to be accurate. Electrons, atoms, even large molecules all have a measurable quantum wavelength based on their momentum.

    $$\lambda = \frac{h}{p} \quad \quad \lambda = \frac{h}{mv}$$
    particle mass (kg) speed (m/s) wavelength (m)
    radio photon 0 c ≈ 1
    electron 9.1 × 10-31 1 ≈ 10-4
    visible photon 0 c ≈ 10-7
    oxygen atom 2.7 × 10-26 1 ≈ 10-8
    gamma ray 0 c ≈ 10-12
    electron 9.1 × 10-31 0.5c ≈ 10-12
    a cat? 4 1 ≈ 10-36

    Oxygen is made of smaller particles, so it's odd that it has a single wavelength. Yet, particles made of smaller particles do behave as a single wave. Wave patterns have been verified experimentally for molecules up to around 25,000 protons and neutrons in 2019.

    A cat particle has an extremely small wavelength, but a cat isn't a single quantum particle in a practical sense. The equation begins to lose meaning at the macroscopic scale as particles can no longer be isolated from their environment. This agrees with our everyday observations for cats. They don't have a measurable wave-nature, although they are very sneaky.

    Example: Find the wavelength of a proton moving at 30 m/s.
    Subatomic Particles Data Table
    name symbol charge (e) mass (kg)
    proton \(p^+\) +1 1.6726 × 10-27
    electron \(e^-\) −1 9.1094 × 10-31
    positron \(e^+\) +1 9.1094 × 10-31
    neutron \(n\) 0 ‎1.6749 × 10-27
    neutrino \(v_e\) 0 1.78 × 10-36
    antineutrino \(\bar{v}_e\) 0 1.78 × 10-36
    muon \( \mu^-\) −1 1.8835 × 10-28
    alpha \( \small{}^4_2 \normalsize \alpha\) +2 6.6447 × 10-27
    solution $$\lambda = \frac{h}{mv}$$ $$\lambda = \frac{6.626 \times 10^{-34}}{(1.67 \times 10^{-27})(30)}$$ $$\lambda = 1.32 \times 10^{-8} \, \mathrm{m}$$

    Quantum Mechanics

    Quantum mechanics seems strange. It feels like our everyday experiences are normal, and quantum mechanics is a set of special rules for small things. It's actually the other way around. Quantum mechanics models the fundamental rules of reality, and our everyday experience emerges from those rules.

    Learning quantum mechanics is typically done by studying the underlying math. Building an intuition without the math can be confusing. The grey boxes below have a nonmathematical and inevitably flawed description of quantum mechanics. You may want to skip them.

    how quantum mechanics makes predictions

    The state of a quantum mechanical system is described mathematically by it's wave function. The wave function contains all the measurable information about the system. The evolution of the wave function over time can be calculated for simple situations with the Schrödinger equation. The Schrödinger equation can be applied to one particle, atom, molecule, or possibly the entire universe.

    As a wave function evolves with time it often looks like a field of positive and negative amplitudes oscillating back and forth like a sine wave. It kinda looks like waves and ripples on surface of a pool of water.

    The probability of measuring a system in a particular state doesn't use the Schrödinger equation. It's done by squaring the amplitude of the system's wave function. A larger squared amplitude has a higher probability of of being measured.

    In order to measure the results of an experiment, an observer must interact with the experiment. Interacting is complex because everything is a quantum object, including the observer and the experiment. This means that the wave function of the observer interacts with the wave function of the experiment. The observer and experiment together form a superposition of all possible outcomes.

    The observer can't directly see this superposition, like how sound waves in superposition can't directly effect each other. The observer only experiences one experimental outcome chosen at random. Why can't they see all the outcomes? The answer to that question depends on how you interpret quantum mechanics.

    interpretations of quantum mechanics

    Different interpretations of quantum mechanics suggest suggest strange and exciting things about the fundamental nature of our reality. Yet, each interpretation is just a different way to put math into words. All valid interpretations of quantum mechanics should predict the same experimental outcomes, so it is hard to test which is correct.

    An important feature of quantum mechanical interpretations is how they handle the measurement problem. Measurement of a particle's position can only be done by having a second particle interact with it. For example, you only know the position of this text because photons of light are interacting with the electrons in the text, and in your eyes.

    Before measurement, an electron matches the wave model. It is spread out, oscillating up and down, and forms wave interference patterns in superposition. When measured by a photon, an electron is found at a random position. This position is more likely to be found where it's wave function has a large amplitude.

    This leads to some questions about measurement that different interpretations of quantum mechanics attempt to answer.

  • Is nature made of waves, particles, both, or neither?
  • Are quantum effects transmitted faster than the speed of light?
  • What happens to the wave function during measurement
  • Is measurement random or deterministic?
  • The Many-Worlds Interpretations

    If the mathematics of the Schrödinger equation is interpreted literally, it means that nature is a wave and all possible outcomes of an interaction occur. These different outcomes do not interact with each other, like how sound waves in superposition don't affect each other. The many-worlds interpretation suggests that a version of you exists that observes each possible outcome of a quantum event. No one has devised an experiment to test for these branching worlds. For now their existence is unknowable.

    Having many worlds seems needlessly complicated, but other possible outcomes are already a part of the Schrödinger equation. Many-worlds just doesn't add an extra step to remove unseen outcomes.

  • Nature is modeled as only a wave, with particles that emerge as excited wave states.
  • Quantum effects are transmitted locally.
  • Measurement branches the wave function into many worlds.
  • Measurement isn't fundamentally random. It seems random because a version of you experiences every outcome, and you don't know ahead of time which you you are.
  • The Copenhagen Interpretation

    The Copenhagen interpretation was the first popular explanation of the math behind quantum mechanics. Like the many-worlds interpretation, the evolution of the wave function is still calculated with the Schrödinger equation. It differs from many-worlds during interactions. Instead of suggesting that all possible outcomes of an interaction occur, one random outcome is chosen.

    The unchosen outcomes are said to be collapsed to a single value. In this way, the Copenhagen interpretation imagines a particle during measurement, but treats the particle as a wave of possible outcomes when it is not measured.

  • Nature is modeled as waves that become particles when measured.
  • Quantum effects are transmitted locally.
  • Measurement collapses the wave function to one outcome.
  • Measurement is fundamentally random.
  • Pilot Wave Theory

    The pilot wave theory proposes a deterministic universe made of particles. It also proposes a hidden level of reality for the wave function, which is still calculated with the Schrödinger equation.

    The wave function guides particles down a path based on the particle's position. When the position of the particle gets measured, the wave function doesn't change. Measurement disrupts interference patterns because in order to measure something you have to hit it with another particle, which changes it's velocity.

    The wave function is determined by the instantaneous positions of particles, including particles that are very far away. Knowing these positions instantly can't be used for faster than light communication, because the usable information in the wave function is hidden.

  • Nature is modeled as a wave that guides particles.
  • Quantum effects travel faster than the speed of light.
  • Measurement doesn't change the wave function.
  • Measurement isn't fundamentally random. Measurement appears random because the initial position of a particle isn't known.

  • wave-particle duality

    Is nature made of waves or particles? It depends on how detailed we want to be and how we interpret quantum mechanics. The wave model and the particle model are approximations of nature that are each accurate for different conditions, but if you have to choose, the wave model is probably more accurate at the fundamental level.

    Quantum field theory describes reality with wave functions. In this theory there are several field types spread across all of space and time. Regions of these fields can evolve into localized excited states with consistently high amplitudes. What we think of as particles are a simplification of these localized excited wave states.

    For example, an electron is a region where the electron field has more amplitude, or bigger ripples. The electron field exists everywhere, but electrons are more likely to be measured where the field amplitude is larger.

    Fields interact with other fields, and with themselves. These interactions can be approximated with Feynman diagrams, which treat everything as particles.

    In Feynman diagrams electron field interactions generally model well as particles. This is because the most complex electron interactions are weak enough to be mathematically canceled out. Quarks, the particles inside protons and neutrons, interact through the "strong force". This force has frequent complex interactions that don't cancel out, making the particle model less useful for calculations.

    quantum uncertainty

    Quantum mechanics answers questions I never thought to ask. Like, why do objects with no forces acting on them move in a straight line?

    Calculating how something moves involves adding the amplitudes of every possible path of getting from one place to another. The amplitudes of the possible paths cycle from positive to negative as time moves forward.

    Some paths might take a very long time. They might zigzag or even go backwards. These slow paths tend to have very different times and have dissimilar amplitudes. The randomly dissimilar amplitudes mostly average to zero.

    Paths that are more direct, have similar times and therefore similar amplitudes. These similar amplitudes are constructive, not destructive. The square of the superposition of all the amplitudes predicts the probability of observing a possible outcome, so we are more likely to observe a direct straight path.

    For very short periods of time, this bias towards straight line paths becomes reduced. The less direct paths have less time to develop random phases. This leads to particles moving in a less predictable way.

    There is an uncertainty in the motion of a particle that increases as the total time decreases. Even when a force should act as a barrier under the classical rules of physics, a particle might just quantum tunnel through that barrier if the time scales involved are very short.

    observing quantum phenomena

    Quantum mechanics explains almost all observable phenomena. It explains the color of materials, rigidity of solids, conductivity, superconductivity, magnetism, all of chemistry, and just about everything else in reality. You are observing quantum mechanical phenomena in every moment.

    Quantum mechanics says that small particles, like atoms, have a measurable wave nature. Cats are made of atoms. Why don't we notice cats acting like waves?

    It's because atoms are hard to isolate. They are frequently interacting with their environment. This causes decoherence between the relative phases of atoms. When coherence is lost, destructive and constructive interference patterns between waves become messy and begin to look like a classical system. For large groups of atoms, decoherence happens so frequently that measurable wave interference patterns don't have time to develop.

    A Bose-Einstein condensate is one way to produce particles in superposition at the macroscopic scale. They are made by cooling a class of particles to near absolute zero. Cold particles move slower, and according to quantum mechanics slower particles have longer wavelengths. As the temperature lowers, the wave functions of the particles begin to overlap. This causes them to enter a shared quantum state.

    Quantum mechanics is currently the most accurate theory for how our universe works, but it is still limited. Quantum mechanics doesn't explain large scale phenomena, like gravitation, dark matter, or dark energy. New theories that unite gravity with quantum mechanics are being developed, but so far there aren't any testable predictions.