• Planck derived the blackbody spectrum by assuming that matter and radiation can only exchange energy in discrete chunks (quanta): \[E(\nu) = nh\nu \qquad \text{for} \qquad n=0,1,2,3,\cdots\]
• He obtained: \[\boxed{u(\nu,T)=\frac{8\pi}{c^3} \frac{h \nu^3}{e^{h\nu/kT}-1},}\]
• To fit experimental data, he needed a new constant called Planck's Constant with value \[h = 6.626\times 10^{-34}\,\text{m}^2\text{kg}\,\text{s}^{-1}.\]

By integrating Planck's formula, we obtain the total intensity of radiation emitted at a given temperature: \[\boxed{I = a\sigma T^4},\] where

• $a=1$ for a blackbody,
• $a<1$ otherwise,
• $\sigma = 5.67\times 10^{-8}\,\text{W}\,\text{m}^{-2}\,\text{K}^{-4}$.

The maximum intensity occurs at wavelength \[\boxed{\lambda_{\text{max}} = \frac{b}{T},\] with \[b = \frac{hc}{4.965 k}.\]

By assuming that light is made of particles (photons) with energy $E = h\nu$, Einstein explained the kinetic energy of electrons ejected from a metal in the photoelectric effect:

\[\boxed{h\nu = W + K},\] where

• $W$ is the work function of the metal,
• $K$ is the kinetic energy of each ejected electron.

The wavelength of X-rays scattered off electrons is larger than the wavelength of the incident radiation \[\boxed{\Delta \lambda} = 2\lambda_C \sin^2 \frac{\theta}{2},\] where

• $\theta$ is the scattering angle of the photon and
• $\lambda_C = h/m_ec$ is the Compton wavelength of the electron.

This formula is derived by assuming elastic collisions between photons and electrons.

Just as electromagnetic radiation can behave like a particle, de Broglie proposed that matter can behave like a wave.

The de Broglie wavelength of a particle with momentum $p$ is \[\boxed{\lambda = \frac{h}{p}}.\]

More generally, in three dimensions, momentum $\vec{p}$ and wave vector $\vec{k}$ are related by \[\boxed{\vec{k} = \frac{\vec{p}}{\hbar}},\] where \[\hbar = \frac{h}{2\pi},\] is the modified Planck constant

A particle with definite momentum $\vec{p}$ and energy $E$ is represented by the plane wave wavefunction \[\boxed{\psi(\vec{r},t) = Ae^{i(\vec{p}\cdot\vec{r} - Et)/\hbar}}.\]

In any given experiment, a quantum system either behaves like a wave or like a particle. It is impossible to observe both behaviors at the same time. Bohr called the wave and particle descriptions complementary.

The wavefunction $\psi(x,t)$ of a quantum particle determines the probability density that the particle will be detected at $x$ if its position is measured at time $t$. If we normalize the wavefunction such that \[\int_{-\infty}^{+\infty} \Abs{\psi(x,t)}^2 \,\D x = 1,\] then the Born rule says that $\Abs{\psi(x,t)}^2$ is the probability density, i.e. the probability of detecting the particle in the interval $a<x<b$ is \[\boxed{p(a<x<b) = \int_a^b \Abs{\psi(x,t)}^2\,\D x}.\]

Heisenberg's uncertainty principle states that \[\boxed{\Delta x\Delta p \geq \frac{\hbar}{2}}.\] This has several interpretations. The one we will use is:

• If the probability density for $x$ predicted by the wavefunction $\psi(x,t)$ has standard deviation $\Delta x$ and the probability density for $p$ predicted by the same wavefunction has standard deviation $\Delta p$ then $\Delta x\Delta p \geq \frac{\hbar}{2}$.

There is also an energy-time uncertainty relation $\Delta E \Delta t \geq \frac{\hbar}{2}$, which is more difficult to interpret.

If $\psi_1(x,t)$ and $\psi_2(x,t)$ are solutions to the equations of motion then so is \[\alpha \psi_1(x,t) + \beta \psi_2(x,t),\] for any complex numbers $\alpha$ and $\beta$.

By assuming that the electron in a hydrogen atom can only be in circular orbits with quantized orbital angular momentum \[\boxed{L = n\hbar, \qquad n=1,2,\cdots},\] Bohr derived that the possible energies of the electron are \[\boxed{E_n = -\frac{m_e}{2\hbar^2} \left ( \frac{e^2}{4\pi\epsilon_0}\right )^2 \frac{1}{n^2} = -\frac{\mathcal{R}}{n^2},\qquad $\mathcal{R} = 13.6\,\text{eV}$}.\] An electron may jump between orbitals by emitting or absorbing a photon with energy \[\boxed{h\nu = E_n - E_m}.\]

• $n=1$ is the lowest energy state with $E_1 = -13.6\,\text{eV}$. It is called the ground state of hydrogen.
• States with $n>1$ are called excited states.
• For $E<0$ the electron is bound to the proton and it has a discrete set of possible energies.
• As $n\rightarrow \infty$, the levels become closely spaced and the set of possible energies is continuous for $E > 0$.

The wavefunction $\psi(x,t) = e^{i(kx-\omega t)}$ represents a free particle with definite momentum $p=\hbar k$ abd $E = \hbar \omega$. However, it is spread out over all of space.

The wavefunction $\psi(x) = \delta(x-x_0)$ is localized at $x_0$, but has infinite momentum uncertainty.

We can get more realistic solutions by superposing different momentum states:

\[\psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{+\infty} \phi(p) e^{i(px-Et)/\hbar}\,\D p\]

For the Gaussian wavepacker

\[\psi_0(x) = \left ( \frac{1}{2\pi\sigma_2}\right )^{\frac{1}{4}} e^{-x^2/4\sigma^2},\] we derived $\Delta x = \sigma$, $\Delta p = \hbar/2\sigma$, so these are minimum uncertainty states that saturate the uncertainty relation \[\Delta x \Delta p \geq \frac{\hbar}{2}.\]

They are well-localized in both position and momentum, as illustrated below.

The relationship $\omega(k)$ between angular frequency and wave number is called the dispersion relation, and depends on the equation of motion. If $\omega = c k$ for some constant $c$ then there is no dispersion and the wave packet moves at speed $c$ without changing its shape of spreading.

Generally, the phase velocity is the velocity of the fast oscillations and is given by \[v_p = \frac{\omega}{k}.\]

The group velocity is the velocity of the wave packet envelope and is given by \[v_g = \frac{\D \omega}{\D k}.\]

When there is dispersion, i.e. $\omega \neq ck$, the group and phase velocities differ. In quantum mechanics, for a Hamiltonian \[H = \frac{p^2}{2m} + V,\] with $V$ constant, we have \begin{align*} v_p & = \frac{p}{2m} + \frac{V}{p}, & v_g & = \frac{p}{m}, \end{align*} so the group velocity corresponds to the velocity of a classical particle.

When there is dispersion, wave packets spread in time. For a Gaussian wave packet with initial position uncertainty $\sigma_0$, the uncertainty at time $t$ grows to \[\sigma_t = \sigma_0 \sqrt{1 + \frac{\hbar^2 t^2}{4m^2\sigma_0^4}}.\]