Lesson 2.1 · 2. Quantum Mechanics

QM Foundations

Everything changes here. Classical mechanics describes a world of definite properties: particles have positions and momenta that exist whether we look or not. Quantum mechanics describes something stranger: a world where properties don't exist until measured, where particles are waves, and where observation changes reality.

Why Quantum Mechanics?

Classical physics failed spectacularly in the early 1900s:

Blackbody radiation: Classical theory predicted infinite energy at high frequencies (ultraviolet catastrophe). Planck fixed it by assuming energy comes in discrete packets: $E = h\nu$
Photoelectric effect: Light ejects electrons from metals, but only above a threshold frequency, not intensity. Einstein explained it: light is particles (photons) with energy $E = h\nu$
Atomic stability: Classical electrons orbiting a nucleus should radiate energy and spiral in. Atoms should collapse in $10^{-11}$ seconds. They don't.
Atomic spectra: Atoms emit light only at specific frequencies, not a continuous spectrum. Why?

The resolution required abandoning classical intuitions entirely.

The Wavefunction

Wave Function

The wave function $\psi(x, t)$ is a mathematical function that encodes everything there is to know about a quantum particle. Unlike a classical description that gives a definite position and speed, the wave function spreads out over space like a wave. It is a complex number at each point, and its squared magnitude $|\psi|^2$ at any location gives the probability of finding the particle there. You can think of it as a "probability cloud" that tells you where the particle is likely to be, without pinning it down to a single spot.

In quantum mechanics, a particle is described by the wavefunction $\psi(x, t)$, a complex-valued function of position and time.

Born's Rule

The probability of finding the particle between $x$ and $x + dx$ is:

$$ P(x) \, dx = |\psi(x)|^2 \, dx $$

The wavefunction itself is not directly observable. Only $|\psi|^2$ is.

Since probabilities must sum to 1:

\int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1

The wavefunction ψ(x) and its square |ψ|² giving the probability density.

What IS the wavefunction?

This is the central mystery. Is $\psi$ a real physical field? A representation of our knowledge? Something else? Different interpretations give different answers. We'll explore this in lesson 2.5.

The Schrödinger Equation

The wavefunction evolves according to the Schrödinger equation:

i\hbar \frac{\partial \psi}{\partial t} = \hat{H} \psi

where $\hat{H}$ is the Hamiltonian operator. For a particle in a potential:

\hat{H} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)

So the full equation is:

i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi

Connection: From Hamilton-Jacobi to Schrödinger

The Schrödinger equation is the quantum version of Hamilton-Jacobi. If $\psi = e^{iS/\hbar}$, then in the limit $\hbar \to 0$, you recover classical mechanics. The "quantum" is in the $\hbar$.

Operators and Observables

In quantum mechanics, physical quantities become operators acting on wavefunctions:

Observable	Operator
Position $x$	$\hat{x} = x$ (multiply by $x$)
Momentum $p$	$\hat{p} = -i\hbar \frac{\partial}{\partial x}$
Energy $E$	$\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V$
Angular momentum $L_z$	$\hat{L}_z = -i\hbar \frac{\partial}{\partial \phi}$

Eigenvalues and Measurement

Eigenvalues and Eigenvectors

An eigenvector (or eigenstate) of an operator is a special state that, when the operator acts on it, simply gets multiplied by a number rather than being changed into something completely different. That number is called the eigenvalue. In quantum mechanics, the eigenvalues of an operator are the possible outcomes you can get when you make a measurement, and the eigenvectors are the states the system can be found in after that measurement. For example, the energy eigenvalues of a hydrogen atom are the specific energy levels the electron can occupy.

When you measure an observable $\hat{A}$, you get one of its eigenvalues. If:

\hat{A} |a\rangle = a |a\rangle

then $a$ is a possible measurement result, and $|a\rangle$ is the state with definite value $a$.

The probability of getting result $a$ when measuring state $|\psi\rangle$:

P(a) = |\langle a | \psi \rangle|^2

Wave Function Collapse

Wave function collapse is what happens when you measure a quantum system. Before measurement, the particle's wave function may be spread across many possible outcomes (a superposition). The instant a measurement is made, all those possibilities reduce to a single definite result, and the wave function "collapses" to the eigenstate corresponding to that result. It is as if you had a coin spinning in the air (representing all possibilities), and the act of catching it forces it to land on one side. After measurement yields result $a$, the wavefunction becomes the eigenstate $|a\rangle$. This process is discontinuous, irreversible, and deeply mysterious.

The Uncertainty Principle

The uncertainty principle, discovered by Heisenberg, states that you cannot know both the position and the momentum of a particle with perfect precision at the same time. The more precisely you pin down where a particle is, the less you can know about how fast it is moving, and vice versa. This is not a limitation of our measuring instruments; it is a fundamental property of nature. It arises because position and momentum are linked through the wave nature of quantum particles: a tightly localized wave must be built from many different wavelengths, which means many different momenta.

Position and momentum operators don't commute:

[\hat{x}, \hat{p}] = \hat{x}\hat{p} - \hat{p}\hat{x} = i\hbar

This leads to Heisenberg's uncertainty principle:

\Delta x \cdot \Delta p \geq \frac{\hbar}{2}

where $\Delta x$ and $\Delta p$ are standard deviations of position and momentum.

The uncertainty trade-off: sharper position means fuzzier momentum, and vice versa.

Is uncertainty about knowledge or reality?

Does the particle have definite position and momentum that we just can't know? Or does it genuinely not have definite values? Bell's theorem (lesson 2.4) shows: it's reality, not just knowledge. Particles don't have definite properties until measured.

Superposition

The Superposition Principle

The superposition principle says that if a quantum system can be in state A and it can be in state B, then it can also be in both states at the same time. This is not the same as saying "it is in A or B and we just do not know which." It is genuinely in a combination of both, simultaneously. Only when a measurement is made does the system "choose" one outcome. This is one of the most counterintuitive features of quantum mechanics and has no parallel in everyday experience.

If $|\psi_1\rangle$ and $|\psi_2\rangle$ are valid states, so is any linear combination:

|\psi\rangle = \alpha |\psi_1\rangle + \beta |\psi_2\rangle

This is superposition. The particle is "in both states at once," not in one or the other with some probability, but genuinely in a combination.

A spin can be in a superposition of up and down simultaneously.

Time Evolution

The Schrödinger equation is deterministic. Given $\psi(t=0)$, it uniquely determines $\psi(t)$ for all future times.

For time-independent Hamiltonians, we can find stationary states that only change by a phase:

\hat{H}|\psi_n\rangle = E_n |\psi_n\rangle

These are energy eigenstates. Their time evolution is:

|\psi_n(t)\rangle = e^{-iE_n t/\hbar} |\psi_n(0)\rangle

Any state can be written as a sum of energy eigenstates:

|\psi(t)\rangle = \sum_n c_n e^{-iE_n t/\hbar} |\psi_n\rangle

The Two Rules of Quantum Mechanics

Quantum mechanics has two fundamentally different rules:

Rule 1: Unitary Evolution

When not being measured, the wavefunction evolves according to the Schrödinger equation. This is deterministic, continuous, and reversible.

Rule 2: Collapse

When measured, the wavefunction collapses to an eigenstate of the measured observable. This is probabilistic, discontinuous, and irreversible.

The tension between these two rules is the measurement problem. When exactly does Rule 2 apply? What counts as a "measurement"? This is the deepest unsolved problem in the foundations of physics.

Exercises

Normalize: Find the normalization constant $A$ for $\psi(x) = A e^{-x^2/2\sigma^2}$.
Expectation values: For a normalized wavefunction, show that $\langle x \rangle = \int x |\psi|^2 dx$ and compute it for a Gaussian.
Commutator: Verify that $[\hat{x}, \hat{p}] = i\hbar$ by acting on a test function $f(x)$.
Uncertainty: For a Gaussian wave packet, compute $\Delta x$ and $\Delta p$ and verify the uncertainty principle.

Key Takeaways

Particles are described by wavefunctions; $|\psi|^2$ gives probability density
The Schrödinger equation governs time evolution
Observables are operators; measurements yield eigenvalues
$[\hat{x}, \hat{p}] = i\hbar$ leads to the uncertainty principle
Superposition: particles can be in multiple states simultaneously
Two rules: smooth evolution vs. sudden collapse (the measurement problem)

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