Quantum Mechanics

Quantum mechanics began not with philosophy but with a stubborn experimental fact that classical physics could not explain: the spectrum of light emitted by a hot object. A blackbody is an idealized object that absorbs all radiation falling on it and re-emits a characteristic spectrum that depends only on its temperature, not on what it is made of. A glowing furnace, a star, and a piece of heated iron all approximate this behavior.

The ultraviolet catastrophe

Classical physics treated the radiation in a cavity as a collection of standing electromagnetic waves, each sharing, on average, an energy of k_B T by the equipartition theorem (here k_B is Boltzmann's constant and T the temperature). Because there is no upper limit on how many short-wavelength modes fit inside a cavity, this Rayleigh-Jeans prediction says the emitted energy grows without bound as the wavelength shrinks. The total radiated energy would be infinite - an absurd result known as the ultraviolet catastrophe. Real blackbodies instead show a peak and then a rapid fall-off toward short wavelengths.

Planck's hypothesis

In 1900 Max Planck found a formula that fit the data perfectly, but only by making a radical assumption: the energy of each oscillating mode of frequency f cannot take any value. It is restricted to integer multiples of a fundamental quantum, E = n h f, where n = 0, 1, 2, ... and h is a new constant of nature, Planck's constant, h = 6.626 x 10^-34 joule-seconds. High-frequency modes require a large quantum h f of energy to be excited even once, so at a given temperature they are effectively frozen out. This exponential suppression tames the short-wavelength divergence and reproduces the observed spectrum.

The smallest unit of light energy at frequency f is therefore the quantum E = h f. Using c = f lambda, this is equivalent to E = h c / lambda. It is often convenient to define the reduced Planck constant hbar = h / (2 pi) = 1.055 x 10^-34 J s, so that E = hbar omega where omega = 2 pi f is the angular frequency.

Worked example: energy of a photon

Given: green light with wavelength lambda = 500 nm = 5.00 x 10^-7 m. Take h = 6.626 x 10^-34 J s and c = 3.00 x 10^8 m/s. Find: the energy of one quantum of this light.
Solution: Use E = h c / lambda = (6.626 x 10^-34)(3.00 x 10^8) / (5.00 x 10^-7). The numerator is 1.988 x 10^-25 J m, and dividing by 5.00 x 10^-7 m gives E = 3.97 x 10^-19 J. In electronvolts (dividing by 1.602 x 10^-19 J/eV) that is about 2.48 eV. Visible-light quanta carry a few electronvolts each, which is why they can nudge the outer electrons of atoms.

Planck treated quantization as a property of the oscillating walls. In 1905 Einstein went further: light itself is quantized into particles now called photons, each carrying energy E = h f and momentum p = E / c = h / lambda. He tested this idea against the photoelectric effect, in which light shining on a metal ejects electrons.

Why waves cannot explain it

The classical wave picture predicts that brighter light (more energy) should eject electrons with more kinetic energy, and that even dim light should eventually free electrons after enough time. Experiment flatly contradicts both. Instead: (1) electrons are emitted only if the light frequency exceeds a threshold f_0, no matter how bright; (2) above threshold, the maximum electron energy rises with frequency, not intensity; and (3) emission is essentially instantaneous.

Einstein's photoelectric equation

These facts follow at once if one photon gives all its energy to one electron. To escape the metal the electron must pay an energy cost called the work function W (also written phi). Energy conservation gives the photoelectric equation:

K_max = h f - W

Below the threshold frequency f_0 = W / h, a single photon lacks the energy to free any electron, so no current flows regardless of brightness. Above it, the leftover energy appears as electron kinetic energy. Brightness only sets how many photons arrive, hence how many electrons are freed, not how energetic each one is. This was the work for which Einstein received the Nobel Prize.

Worked example: stopping potential

Given: a metal with work function W = 2.30 eV is illuminated by light of wavelength lambda = 400 nm. Find: the maximum kinetic energy of the ejected electrons and the stopping potential.
Solution: The photon energy is E = h c / lambda. A handy shortcut is h c = 1240 eV nm, so E = 1240 / 400 = 3.10 eV. Then K_max = E - W = 3.10 - 2.30 = 0.80 eV. The stopping potential is the voltage that just halts the fastest electrons, V_stop = K_max / e = 0.80 V. Photons below W = 2.30 eV (wavelengths longer than about 539 nm) would eject nothing.

The Compton effect

Further proof of photon momentum came from Arthur Compton in 1923. When X-rays scatter off electrons, the scattered light has a longer wavelength, shifted by delta lambda = (h / m_e c)(1 - cos theta), where theta is the scattering angle. This is exactly the result of an elastic billiard-ball collision between a photon and an electron, conserving relativistic energy and momentum. Waves alone cannot change wavelength on scattering; particles carrying momentum can.

By 1910 experiments by Rutherford had shown the atom to be mostly empty space, with electrons orbiting a tiny dense nucleus. But classical physics forbade this: an orbiting electron accelerates, and accelerating charges radiate energy, so the electron should spiral into the nucleus in about 10^-11 seconds. Atoms plainly do not collapse. A second puzzle was that atoms emit and absorb light only at sharp, discrete spectral lines, not a continuous rainbow.

The Balmer formula

For hydrogen, the visible emission lines fit a simple empirical pattern found by Balmer and generalized by Rydberg: 1 / lambda = R (1 / n_f squared - 1 / n_i squared), where n_i and n_f are integers with n_i > n_f, and R = 1.097 x 10^7 per meter is the Rydberg constant. That such a clean formula worked cried out for explanation.

Bohr's postulates

In 1913 Niels Bohr proposed a bold fix with three postulates. First, electrons occupy only certain stationary states (allowed orbits) in which, contrary to classical physics, they do not radiate. Second, the allowed orbits are those for which the angular momentum is quantized: L = n hbar, with n = 1, 2, 3, .... Third, light is emitted or absorbed only when an electron jumps between states, with photon energy equal to the energy difference: h f = E_i - E_f.

Combining the quantization condition with the Coulomb force holding the electron in orbit yields the hydrogen energy levels:

E_n = -13.6 eV / n squared, for n = 1, 2, 3, ...

The negative sign means the electron is bound; the lowest state n = 1 (the ground state) sits at -13.6 eV, and it takes 13.6 eV to ionize hydrogen. The levels crowd together toward E = 0 as n grows.

Worked example: a Balmer line

Given: an electron falls from n_i = 3 to n_f = 2 in hydrogen. Find: the photon's energy and wavelength.
Solution: The levels are E_3 = -13.6 / 9 = -1.51 eV and E_2 = -13.6 / 4 = -3.40 eV. The emitted photon carries E = E_3 - E_2 = -1.51 - (-3.40) = 1.89 eV. Its wavelength is lambda = 1240 / 1.89 = 656 nm, the red H-alpha line - exactly what the Balmer formula predicts. The Bohr model, though later superseded by full quantum mechanics, correctly reproduces the hydrogen spectrum and first made quantization concrete.

If light, long thought to be a wave, also behaves as particles, might matter, long thought to be particles, also behave as waves? In 1924 Louis de Broglie proposed exactly this symmetry. He assigned to any particle of momentum p a wavelength:

lambda = h / p

This is the de Broglie wavelength. For everyday objects it is fantastically small and undetectable, but for electrons it lands in the range of atomic spacings, where wave effects become visible.

Electron diffraction

The prediction was confirmed in 1927 when Davisson and Germer fired electrons at a nickel crystal and saw a diffraction pattern - alternating maxima and minima - just as X-rays produce. A stream of particles was interfering like a wave. The regularly spaced atoms of the crystal act as a diffraction grating, and the angles of the bright spots match those predicted from the de Broglie wavelength. Modern electron microscopes exploit this short wavelength to resolve detail far finer than visible light allows.

Wave-particle duality

The lesson is wave-particle duality: quantum objects are neither classical waves nor classical particles but something that shows wave-like or particle-like behavior depending on the experiment. In the famous double-slit experiment, electrons sent one at a time each arrive as a single localized dot (particle-like), yet the accumulated pattern of thousands of dots forms interference fringes (wave-like). Crucially, if you detect which slit each electron passes through, the fringes vanish. The wave describes probabilities, as the next module makes precise.

Worked example: de Broglie wavelength of an electron

Given: an electron (m_e = 9.11 x 10^-31 kg) moving at v = 1.0 x 10^6 m/s. Find: its de Broglie wavelength.
Solution: The momentum is p = m v = (9.11 x 10^-31)(1.0 x 10^6) = 9.11 x 10^-25 kg m/s. Then lambda = h / p = (6.626 x 10^-34) / (9.11 x 10^-25) = 7.3 x 10^-10 m, or about 0.73 nm - comparable to atomic spacings, which is why electron diffraction is observable. For contrast, a 1 kg ball at 1 m/s has lambda = 6.6 x 10^-34 m, unmeasurably small.

In quantum mechanics the complete description of a particle is not a position and a velocity but a single complex-valued function of position and time, the wavefunction, written psi(x, t) (the Greek letter psi). Everything that can be known about the particle is contained in this function. It is generally complex, meaning it has real and imaginary parts, and it is not directly observable.

The Born rule

What is observable is probability. Max Born proposed in 1926 that the square of the magnitude of the wavefunction gives a probability density. In one dimension, the probability of finding the particle between x and x + dx at time t is:

P(x) dx = |psi(x, t)| squared dx = psi* psi dx

Here psi* is the complex conjugate of psi, so psi* psi is real and non-negative, as a probability density must be. This is the Born interpretation, and it is the bridge between the abstract wavefunction and laboratory measurement. The wavefunction does not tell you where the particle is; it tells you the odds of each possible outcome.

Normalization

Because the particle must be found somewhere, the total probability over all space must equal 1:

integral from -infinity to +infinity of |psi(x, t)| squared dx = 1

A wavefunction satisfying this is normalized. Any physically acceptable wavefunction must be square-integrable (the integral of its squared magnitude must be finite) so that it can be scaled to satisfy this condition. To be well-behaved it should also be single-valued, continuous, and (where the potential is finite) have a continuous first derivative. These boundary conditions are what force the discrete energy levels we will find later.

Worked example: normalizing a wavefunction

Given: at a fixed time a particle on the interval 0 <= x <= L (and zero outside) has wavefunction psi(x) = A sin(pi x / L), with A a real constant. Find: the value of A that normalizes it.
Solution: Require integral from 0 to L of A squared sin squared(pi x / L) dx = 1. Using integral from 0 to L of sin squared(pi x / L) dx = L / 2, this becomes A squared (L / 2) = 1, so A squared = 2 / L and A = sqrt(2 / L). The normalized wavefunction is psi(x) = sqrt(2/L) sin(pi x / L). This is exactly the ground state of the particle in a box, which we derive in Module 4.

If the wavefunction is the state of a system, we need a law that tells us how it changes in time - the quantum analog of Newton's second law. That law is the time-dependent Schrodinger equation (TDSE), written down by Erwin Schrodinger in 1926. For a single particle of mass m moving in one dimension under a potential energy V(x, t), it reads:

i hbar (partial psi / partial t) = -(hbar squared / 2m)(partial squared psi / partial x squared) + V(x, t) psi

The imaginary unit i = sqrt(-1) on the left is essential; it is why the wavefunction is complex and why quantum states oscillate in phase rather than simply decaying.

Reading the equation

Each term has a physical meaning. The left side, i hbar (partial psi / partial t), describes how the state evolves in time. On the right, the first term -(hbar squared / 2m)(partial squared psi / partial x squared) is the kinetic energy contribution: the curvature of the wavefunction encodes momentum, since sharper wiggles mean shorter wavelength and (by de Broglie) higher momentum. The second term V psi is the potential energy contribution. Together the right side is the Hamiltonian operator H acting on psi, so the equation can be written compactly as i hbar (partial psi / partial t) = H psi.

What kind of law is this?

The Schrodinger equation is linear, which means that if psi_1 and psi_2 are both solutions, so is any combination a psi_1 + b psi_2. This linearity is the mathematical root of the superposition principle. It is also deterministic: given psi now, the equation fixes psi at all later times exactly. The randomness of quantum mechanics enters only at measurement, through the Born rule, not in the evolution of the wavefunction itself. Finally, it is first order in time, so the wavefunction at one instant (not the wavefunction and its time derivative) is enough to determine the future.

Momentum and energy as operators

The equation embodies a deep dictionary: physical quantities become operators acting on the wavefunction. Momentum becomes p -> -i hbar (partial / partial x) and energy becomes E -> i hbar (partial / partial t). Substituting these into the classical energy relation E = p squared / 2m + V and letting both sides act on psi reproduces the Schrodinger equation exactly. Module 3 develops this operator formalism in full.

Most problems in this course use a powerful simplification. When the potential does not depend on time, V = V(x), we can look for solutions of the time-dependent equation that separate into a product of a space part and a time part: psi(x, t) = f(x) g(t). Substituting this into the TDSE and dividing through, the two sides can be equal for all x and t only if each equals the same constant, which turns out to be the energy E.

The two pieces

The time part solves immediately: g(t) = e^(-i E t / hbar), a pure phase oscillating at angular frequency omega = E / hbar (note the echo of Planck's E = hbar omega). The space part f(x), usually written psi(x), must satisfy the time-independent Schrodinger equation (TISE):

-(hbar squared / 2m)(d squared psi / dx squared) + V(x) psi = E psi

Equivalently, H psi = E psi. This is an eigenvalue equation: we seek functions psi that the Hamiltonian merely multiplies by a number E. The allowed numbers E are the energy eigenvalues, and the corresponding functions are the energy eigenstates. Solving the TISE for a given V(x) is the central task of the next two modules.

Why "stationary"?

A separated solution has the full form psi(x, t) = psi(x) e^(-i E t / hbar). Its probability density is |psi(x, t)| squared = |psi(x)| squared, because the time-dependent phase has magnitude 1 and cancels. The probability distribution does not change in time, so these are called stationary states. An atom in a single energy eigenstate does not radiate, which is precisely what Bohr had to postulate and what the Schrodinger equation now explains.

General solutions by superposition

The stationary states are not merely special cases; they are the building blocks of every solution. Because the TDSE is linear, the general solution is a superposition of stationary states, each carrying its own phase:

psi(x, t) = sum over n of c_n psi_n(x) e^(-i E_n t / hbar)

The constants c_n are fixed by the initial wavefunction. Unlike a single stationary state, such a mixture does evolve, because the different phases e^(-i E_n t / hbar) beat against one another. This is how quantum systems change, oscillate, and show interference in time.

Worked example: two-state beating frequency

Given: a system in an equal superposition of two stationary states with energies E_1 and E_2. Find: the frequency at which the probability density oscillates.
Solution: The two phases are e^(-i E_1 t / hbar) and e^(-i E_2 t / hbar). When we form |psi| squared the cross term oscillates as cos((E_2 - E_1) t / hbar), so the angular frequency of the beating is omega = (E_2 - E_1) / hbar. This is exactly the Bohr transition frequency hbar omega = E_2 - E_1, connecting the mathematics back to spectral lines.

In classical mechanics an observable is just a number you read off - a position, a momentum, an energy. In quantum mechanics each observable is represented by a linear operator that acts on wavefunctions. The key examples in one dimension are position x_hat = x (multiply by x), momentum p_hat = -i hbar (d/dx), and energy, the Hamiltonian H_hat = p_hat squared / 2m + V(x).

Eigenvalues are the possible outcomes

A central postulate ties operators to measurement: the only possible results of measuring an observable are the eigenvalues of its operator. If A_hat psi_a = a psi_a, then a is a value the measurement can return, and psi_a is the state that yields it with certainty. This is why energy came out quantized in Module 2 - the energy eigenvalues are the allowed measured energies. If the system is not in an eigenstate, the measurement still returns one of the eigenvalues, chosen at random with probabilities set by the Born rule.

Why Hermitian?

Measured values are real numbers, so the operators representing observables must have only real eigenvalues. The mathematical property guaranteeing this is that the operator is Hermitian (self-adjoint): it equals its own conjugate transpose, so that integral of psi* (A_hat phi) dx = integral of (A_hat psi)* phi dx for all suitable psi and phi. Hermitian operators have two properties that make them exactly right for physics: their eigenvalues are real, and their eigenfunctions form a complete orthonormal set - any state can be expanded in them. Position, momentum, and the Hamiltonian are all Hermitian.

Orthonormality and expansion

Two eigenstates psi_m and psi_n belonging to different eigenvalues are orthogonal: integral of psi_m* psi_n dx = 0 when m is not equal to n, and equals 1 when they match (if normalized). Completeness means any wavefunction can be written as psi = sum of c_n psi_n, where the expansion coefficient is c_n = integral of psi_n* psi dx. The number |c_n| squared is then the probability that a measurement yields the eigenvalue a_n. This expansion is the practical machinery behind every quantum prediction.

Worked example: is an operator Hermitian?

Given: the operators A = d/dx and B = i (d/dx). Find: which one can represent an observable.
Solution: Integration by parts (with wavefunctions vanishing at infinity) shows integral of psi* (d phi/dx) dx = - integral of (d psi/dx)* phi dx. So d/dx is anti-Hermitian (it picks up a minus sign) and cannot be an observable; its eigenvalues are imaginary. But i (d/dx) gains a compensating factor: the i conjugates to -i, canceling the sign, so i(d/dx) is Hermitian. This is why physical momentum is p_hat = -i hbar (d/dx) and not simply d/dx - the factor of i is required to make it a legitimate, real-valued observable.

Even though individual measurements are random, quantum mechanics makes sharp predictions about averages. The expectation value of an observable A in state psi is the average result of measuring it on many identically prepared systems:

<A> = integral of psi* (A_hat psi) dx

For position this reads <x> = integral of psi* x psi dx = integral of x |psi| squared dx, the mean of the probability distribution. For momentum, <p> = integral of psi* (-i hbar d psi/dx) dx. The expectation value is not usually itself a possible measurement result; it is the mean of many results.

Commutators

Operators, unlike numbers, do not always commute: the order in which you apply them can matter. The commutator of two operators is defined as [A, B] = A B - B A. If [A, B] = 0 the operators commute, and the observables can share eigenstates and be measured simultaneously to arbitrary precision. If [A, B] is nonzero, they are incompatible.

The canonical commutator

The most important commutator in all of quantum mechanics is between position and momentum. Applying [x_hat, p_hat] = x_hat p_hat - p_hat x_hat to a test function and using p_hat = -i hbar (d/dx), the product rule leaves a single surviving term:

[x_hat, p_hat] = i hbar

This nonzero result is the mathematical origin of the uncertainty principle. Position and momentum are incompatible observables: no state can have a definite value of both at once, because they do not share eigenstates.

Worked example: evaluating [x, p]

Given: a test function f(x). Find: [x_hat, p_hat] f and hence the commutator.
Solution: Compute each ordering. x_hat p_hat f = x(-i hbar) f' = -i hbar x f'. And p_hat x_hat f = -i hbar (d/dx)(x f) = -i hbar (f + x f') by the product rule. Subtracting, [x_hat, p_hat] f = -i hbar x f' - (-i hbar)(f + x f') = -i hbar x f' + i hbar f + i hbar x f' = i hbar f. Since this holds for any f, we conclude [x_hat, p_hat] = i hbar.

Worked example: expectation value in a box

Given: the normalized ground state psi(x) = sqrt(2/L) sin(pi x / L) on 0 <= x <= L. Find: the expectation value of position.
Solution: By symmetry the distribution |psi| squared is symmetric about the center x = L/2, so <x> = L/2. (Carrying out integral of x (2/L) sin squared(pi x/L) dx from 0 to L confirms this: the result is exactly L/2.) On average the particle sits at the middle of the box, even though any single measurement can find it anywhere inside.

The nonzero commutator [x, p] = i hbar has a famous physical consequence. Define the uncertainty in an observable as the standard deviation of its measurements, delta A = sqrt(<A squared> - <A> squared). Then for position and momentum the Heisenberg uncertainty principle states:

delta x times delta p >= hbar / 2

You cannot simultaneously know both position and momentum with unlimited precision. Squeezing the wavefunction to pin down position (small delta x) necessarily spreads out the range of momenta (large delta p), and vice versa. This is not a limitation of instruments; it is a property of the quantum state itself, following directly from the fact that x and p do not commute.

A wave picture

The relation makes intuitive sense through waves. A wavefunction with a single, sharply defined wavelength (hence a definite momentum) is a spread-out sinusoid extending everywhere, so its position is completely uncertain. To localize a particle you must superpose many wavelengths into a compact wave packet, but a spread of wavelengths means a spread of momenta. The trade-off is built into the mathematics of waves; Fourier analysis makes it exact.

The general uncertainty relation

For any two observables the bound generalizes to delta A times delta B >= (1/2) |<[A, B]>|. When the commutator is zero the observables can in principle both be sharp; when it is nonzero, as with x and p, a fundamental trade-off appears. There is also an energy-time relation, delta E times delta t >= hbar / 2, which links the lifetime of a state to the sharpness of its energy - short-lived states have broad energy widths.

Worked example: estimating a ground-state energy

Given: a particle of mass m confined to a region of size L. Find: an order-of-magnitude estimate of its minimum kinetic energy.
Solution: Confinement means delta x ~ L, so the uncertainty principle forces delta p ~ hbar / L. The particle cannot sit still: its typical momentum is at least this large, giving a minimum kinetic energy E ~ (delta p) squared / 2m ~ hbar squared / (2 m L squared). This zero-point energy explains why confined quantum particles are never at rest and why matter resists compression. Applied to an electron in a hydrogen-sized region (L ~ 10^-10 m), this estimate lands within a factor of a few of the true 13.6 eV binding energy - remarkable for a one-line argument.

The simplest exactly solvable bound system is the infinite square well, or particle in a box: a particle free to move on 0 <= x <= L but confined by walls of infinite potential, so V = 0 inside and V = infinity outside. The infinite walls force the wavefunction to vanish at the edges - a particle cannot exist where the potential is infinite - giving the boundary conditions psi(0) = 0 and psi(L) = 0.

Solving inside the well

Inside, with V = 0, the TISE is -(hbar squared / 2m) psi'' = E psi, or psi'' = -k squared psi with k squared = 2 m E / hbar squared. The general solution is psi(x) = A sin(k x) + B cos(k x). The condition psi(0) = 0 kills the cosine (B = 0). The condition psi(L) = 0 then requires sin(k L) = 0, so k L = n pi for a positive integer n. Quantization drops out of a boundary condition, exactly as promised.

Energies and wavefunctions

From k = n pi / L and E = hbar squared k squared / 2m, the allowed energies are:

E_n = n squared pi squared hbar squared / (2 m L squared), for n = 1, 2, 3, ...

and the normalized wavefunctions are psi_n(x) = sqrt(2/L) sin(n pi x / L). Three features deserve note. The energies grow as n squared, so levels spread apart at higher energy. The lowest energy is not zero but E_1 = pi squared hbar squared / (2 m L squared), a zero-point energy demanded by the uncertainty principle. And psi_n has n - 1 interior nodes (points where it crosses zero), a pattern that recurs throughout quantum mechanics.

Worked example: an electron in a box

Given: an electron confined to a box of length L = 0.50 nm = 5.0 x 10^-10 m. Use hbar = 1.055 x 10^-34 J s and m_e = 9.11 x 10^-31 kg. Find: the ground-state energy.
Solution: E_1 = pi squared hbar squared / (2 m L squared). Numerically, hbar squared = 1.11 x 10^-68, pi squared = 9.87, and 2 m L squared = 2(9.11 x 10^-31)(2.5 x 10^-19) = 4.56 x 10^-49. So E_1 = (9.87)(1.11 x 10^-68) / (4.56 x 10^-49) = 2.4 x 10^-19 J, about 1.5 eV. Confining an electron to atomic dimensions produces energy-level spacings of a few electronvolts, the scale of atomic and molecular physics.

Real potentials are not infinitely high. The finite square well has walls of finite depth V_0: V = 0 inside a region of width L and V = V_0 outside. This small change has a profound consequence - the wavefunction no longer has to vanish at the walls.

Penetration into the walls

Outside the well, where E < V_0, the Schrodinger equation becomes psi'' = kappa squared psi with kappa = sqrt(2 m (V_0 - E)) / hbar. Its acceptable solution is a decaying exponential psi ~ e^(-kappa |x|). So the wavefunction is nonzero inside the classically forbidden region, decaying over a characteristic length 1/kappa. Classically a particle with E < V_0 could never be there; quantum mechanically it leaks in. Matching the interior sinusoid smoothly to these exterior tails at each wall quantizes the energies, and a finite well always has a finite number of bound states, fewer than the infinite well of the same width.

Quantum tunneling

If instead of a well we have a barrier - a region of high potential V_0 and finite width a - a particle approaching with E < V_0 can pass through and appear on the far side. This is quantum tunneling. The wavefunction decays exponentially inside the barrier but does not reach zero before the far edge, so a reduced-amplitude wave emerges beyond it. A classical ball would simply bounce back; a quantum particle has a nonzero probability of being transmitted.

How the probability scales

For a barrier that is not too thin, the transmission probability falls off exponentially: T ~ e^(-2 kappa a), with kappa = sqrt(2 m (V_0 - E)) / hbar. Tunneling is therefore extremely sensitive to the barrier's width a and height V_0 - E: doubling the width squares an already small number. This exquisite sensitivity is what makes the scanning tunneling microscope able to image individual atoms, since the tunneling current between tip and surface changes sharply with tiny distance changes. Tunneling also explains alpha decay of nuclei and the fusion reactions that power the Sun.

Worked example: effect of doubling the barrier width

Given: a barrier with transmission T = e^(-2 kappa a). Suppose 2 kappa a = 10 for the original width. Find: how the transmission changes if the width is doubled.
Solution: Originally T = e^(-10) = 4.5 x 10^-5. Doubling a doubles the exponent to 20, giving T = e^(-20) = 2.1 x 10^-9. The transmission drops by a factor of e^(-10) = 4.5 x 10^-5, roughly twenty-thousand-fold, for only a doubling of width. This steep dependence is the hallmark of tunneling.

After the box, the most important solvable system is the quantum harmonic oscillator: a particle in the potential V(x) = (1/2) m omega squared x squared, the quantum version of a mass on a spring with classical frequency omega = sqrt(k/m). Its importance is hard to overstate, because any smooth potential looks like a parabola near a minimum. Molecular vibrations, lattice vibrations in solids (phonons), and even the modes of the electromagnetic field are all harmonic oscillators.

The energy spectrum

Solving the TISE for this potential (a classic exercise using either power series or clever raising and lowering operators) gives beautifully simple energy levels:

E_n = (n + 1/2) hbar omega, for n = 0, 1, 2, ...

Two features stand out. First, the levels are equally spaced, separated by hbar omega. This uniform ladder is unique to the harmonic oscillator and is why a vibrating molecule absorbs and emits at a single characteristic frequency. Second, the lowest level is not zero but E_0 = (1/2) hbar omega, the zero-point energy. Even in its ground state the oscillator has energy and its position fluctuates - it can never be perfectly at rest at the bottom of the well, as the uncertainty principle demands.

The wavefunctions

The eigenstates are a Gaussian (bell curve) multiplied by polynomials called Hermite polynomials. The ground state is a simple Gaussian, psi_0(x) proportional to e^(-m omega x squared / 2 hbar), with no nodes. Each higher state adds one node, following the same node-counting rule as the box. Unlike the box wavefunctions, these extend (with exponentially small tails) into the classically forbidden region beyond the turning points, just as the finite well allowed.

Worked example: molecular vibration

Given: a diatomic molecule modeled as a harmonic oscillator with angular frequency omega = 5.0 x 10^14 rad/s. Find: the spacing between adjacent energy levels and the zero-point energy.
Solution: The spacing is delta E = hbar omega = (1.055 x 10^-34)(5.0 x 10^14) = 5.3 x 10^-20 J, about 0.33 eV - in the infrared, which is exactly where molecular vibrational spectra are observed. The zero-point energy is half this, E_0 = (1/2) hbar omega = 2.6 x 10^-20 J, or about 0.16 eV. This residual vibrational energy persists even at absolute zero.

Moving to three dimensions, a new conserved quantity becomes central: angular momentum. Classically L = r x p is a vector with three components. In quantum mechanics these become operators L_x, L_y, L_z, and their commutation relations, such as [L_x, L_y] = i hbar L_z, are nonzero. The immediate consequence is that the three components are incompatible: you cannot know all three at once.

What can be known simultaneously

What does commute is the total magnitude squared L squared = L_x squared + L_y squared + L_z squared with any single component, conventionally chosen as L_z. So a state can have definite values of L squared and L_z together, but not of L_x or L_y. This is why angular momentum is pictured as a vector of fixed length precessing around the z-axis: its z-component is sharp, but its direction in the x-y plane is completely undetermined.

The quantization rules

Solving the eigenvalue problem gives two quantum numbers. The magnitude is set by the orbital quantum number l = 0, 1, 2, ... through L squared = l(l + 1) hbar squared, so the length is |L| = sqrt(l(l + 1)) hbar. The z-component is set by the magnetic quantum number m through L_z = m hbar, where m takes the 2l + 1 integer values from -l to +l. For example, l = 1 allows m = -1, 0, +1, three orientations. This restriction of the allowed directions is called space quantization, and the chemist's s, p, d, f orbitals correspond to l = 0, 1, 2, 3.

Worked example: the l = 2 state

Given: a particle with orbital quantum number l = 2. Find: the magnitude of its angular momentum and the allowed values of L_z.
Solution: The magnitude is |L| = sqrt(l(l+1)) hbar = sqrt(2 x 3) hbar = sqrt(6) hbar = 2.45 hbar. The allowed z-projections are L_z = m hbar with m = -2, -1, 0, +1, +2, that is five values from -2 hbar to +2 hbar. Notice the maximum projection 2 hbar is less than the magnitude 2.45 hbar: the angular momentum vector can never point exactly along the z-axis, because that would make L_x and L_y both exactly zero, violating the uncertainty relations among the components.

The hydrogen atom - a single electron bound to a proton by the Coulomb potential V(r) = -e squared / (4 pi epsilon_0 r) - is the crowning exactly solvable problem of quantum mechanics. Because the potential depends only on the distance r, the three-dimensional Schrodinger equation separates in spherical coordinates into radial and angular parts. The angular part is exactly the angular-momentum problem of the previous lesson, so its solutions are the spherical harmonics labeled by l and m. The radial part introduces a third quantum number.

Three quantum numbers

Each bound state is labeled by three integers:

• the principal quantum number n = 1, 2, 3, ..., which sets the energy;
• the orbital quantum number l = 0, 1, ..., n - 1, which sets the angular momentum magnitude;
• the magnetic quantum number m = -l, ..., +l, which sets the z-component of angular momentum.

The restriction l < n is important: for n = 1 only l = 0 is allowed (the 1s orbital), while n = 2 allows l = 0 (2s) and l = 1 (2p).

The energy levels

Remarkably, solving the Coulomb problem exactly reproduces the Bohr formula:

E_n = -13.6 eV / n squared, for n = 1, 2, 3, ...

The energy depends only on n, not on l or m. This means many distinct states share the same energy, a situation called degeneracy. Counting the allowed (l, m) combinations for a given n gives n squared spatial states at that energy (for example, n = 2 has one 2s plus three 2p states, four in all). This n squared degeneracy, doubled by spin, is the structural reason behind the shape of the periodic table.

The ground state

The lowest state, n = 1, l = 0, m = 0, is the spherically symmetric 1s orbital, psi_100 proportional to e^(-r / a_0), where a_0 = 5.29 x 10^-11 m is the Bohr radius. The probability of finding the electron peaks at a distance of one Bohr radius from the nucleus. Unlike the Bohr model's sharp orbit, the true quantum electron is a diffuse cloud whose density falls off exponentially with distance.

Worked example: counting states

Given: the hydrogen level n = 3. Find: how many spatial states (orbitals) share this energy.
Solution: The allowed l values are 0, 1, 2. Each contributes 2l + 1 states: l = 0 gives 1 (3s), l = 1 gives 3 (3p), l = 2 gives 5 (3d). The total is 1 + 3 + 5 = 9 = 3 squared, confirming the n squared degeneracy. Including the two spin states of the electron doubles this to 18.

Orbital angular momentum is not the whole story. Experiments - most famously the Stern-Gerlach experiment of 1922, in which a beam of silver atoms split into exactly two spots after passing through a nonuniform magnetic field - revealed that the electron carries an additional, intrinsic angular momentum called spin. Spin is not the electron physically rotating; it is a fundamental property, like charge or mass, with no classical picture. It behaves mathematically like angular momentum but with half-integer quantum numbers.

Spin quantum numbers

For the electron the spin quantum number is s = 1/2. The magnitude of the spin angular momentum is |S| = sqrt(s(s+1)) hbar = sqrt(3)/2 hbar, and its z-component takes only two values, S_z = m_s hbar with m_s = +1/2 or -1/2. These are informally called spin-up and spin-down. Because there are exactly two states, the Stern-Gerlach beam splits into two - a result impossible for any integer l, which would give an odd number 2l + 1 of spots.

The complete set of quantum numbers

An electron in an atom is therefore fully specified by four quantum numbers: n, l, m, and m_s. Spin doubles the number of available states at every spatial orbital, so each orbital holds up to two electrons.

The Pauli exclusion principle

The structure of the periodic table follows from one more rule. The Pauli exclusion principle states that no two electrons in an atom can have the same set of all four quantum numbers. Equivalently, at most one electron can occupy each distinct quantum state, so each orbital holds two electrons of opposite spin. This forces electrons to stack into successive shells rather than all collapsing into the 1s ground state, and that shell structure is exactly what determines an element's chemistry. Electrons, and all half-integer-spin particles (fermions), obey this exclusion; integer-spin particles (bosons) do not.

Worked example: filling the n = 2 shell

Given: the n = 2 shell of an atom. Find: the maximum number of electrons it can hold.
Solution: The n = 2 level has n squared = 4 spatial orbitals (one 2s and three 2p). Each holds two electrons of opposite spin by the Pauli principle, for a total of 4 x 2 = 8 electrons. This is the origin of the octet that fills the second row of the periodic table, from lithium to neon.

Because the Schrodinger equation is linear, a quantum system can exist in a superposition of states. If psi_1 and psi_2 are possible states, so is c_1 psi_1 + c_2 psi_2. This is not a statement of ignorance - the system is not secretly in one state or the other. Until measured, it genuinely has no definite value of the observable, and the two components can interfere, as the double-slit experiment shows.

The measurement postulate

Measurement is where the wavefunction meets reality, and quantum mechanics handles it with a distinct postulate. Suppose an observable A has eigenstates psi_n with eigenvalues a_n, and the system is in the superposition psi = sum of c_n psi_n. Then:

• A measurement of A yields one of the eigenvalues a_n, never anything in between.
• The probability of obtaining a_n is |c_n| squared (the Born rule).
• Immediately after the measurement, the state collapses to the corresponding eigenstate psi_n.

This third point is the notorious collapse (or reduction) of the wavefunction. Before measurement the state evolves smoothly and deterministically by the Schrodinger equation; the act of measurement introduces an abrupt, random jump to a single eigenstate. A repeated immediate measurement then gives the same result with certainty, since the system now is in that eigenstate.

Worked example: probabilities from a superposition

Given: a normalized state psi = (1/sqrt(3)) psi_1 + sqrt(2/3) psi_2, where psi_1 and psi_2 are energy eigenstates with energies E_1 and E_2. Find: the probability of each energy outcome and the expectation value of the energy.
Solution: The coefficients are c_1 = 1/sqrt(3) and c_2 = sqrt(2/3). The probabilities are |c_1| squared = 1/3 and |c_2| squared = 2/3, which correctly sum to 1. A measurement returns E_1 one third of the time and E_2 two thirds of the time. The expectation value is the weighted average <E> = (1/3) E_1 + (2/3) E_2 - note this need not equal either eigenvalue.

Decoherence

Why do we never see everyday objects in superposition - a cat both alive and dead, in Schrodinger's famous thought experiment? The modern answer is decoherence: a macroscopic system is constantly entangled with its environment (air molecules, photons, thermal vibrations), which rapidly destroys the delicate phase relationships that make interference visible. Superposition does not vanish in principle, but it becomes practically unobservable for large objects, so the world looks classical at human scales.

Quantum mechanics is the most precisely tested theory in the history of science, yet what it means remains debated. The mathematics is not in question - it predicts experimental results to extraordinary accuracy - but the interpretation of the wavefunction and of collapse is a live question. This lesson surveys the landscape without pretending the matter is settled.

Interpretations

Several major interpretations coexist:

• The Copenhagen interpretation, historically dominant, treats the wavefunction as a tool for predicting measurement probabilities and takes collapse as a basic feature, declining to ask what the system "really" does between measurements.
• The many-worlds interpretation denies collapse entirely: every outcome occurs, each in its own branch of a continually splitting universe, and the appearance of a single result is our experience of one branch.
• Hidden-variable theories, such as the pilot-wave (de Broglie-Bohm) picture, restore determinism by positing additional variables that the wavefunction does not capture.

No experiment to date distinguishes these for ordinary quantum predictions; they agree on all observations and differ only in what they say lies behind them.

Entanglement

The deepest quantum feature is entanglement: two particles can share a joint state that cannot be written as a product of individual states. Measuring one instantly fixes the correlated property of the other, however far apart they are. Einstein, Podolsky, and Rosen argued in 1935 that this "spooky action at a distance" showed the theory was incomplete. Decades later, John Bell devised an inequality that any local-hidden-variable theory must obey, and experiments have repeatedly found it violated, in agreement with quantum mechanics. Nature really is nonlocal in the correlations it permits, though (importantly) entanglement cannot be used to send information faster than light.

The reach of the theory

Whatever its interpretation, quantum mechanics underpins much of modern technology. The transistor and the laser, magnetic resonance imaging, light-emitting diodes, and the entire semiconductor industry rest on quantum principles. Emerging quantum technologies - quantum computing, quantum cryptography, and quantum sensing - aim to harness superposition and entanglement directly. The strange features that troubled the theory's founders are becoming engineering resources. You now have the conceptual and mathematical foundation to go further into any of these directions.