Dirac Sea, Uncertainty & Ehrenfest's Theorem

Part I — The Problem of Negative Energy

Consider a relativistic massless particle (like a photon). The Hamiltonian is $H = cp$. Since momentum $p$ can be positive or negative, the energy can be negative. The plane-wave eigenstates in position space are:

Why is negative energy catastrophic? Suppose a particle occupies a state with energy $E > 0$ (a normal positive-energy state). It could emit a photon (releasing energy $\delta E > 0$) and drop into a state with energy $E - \delta E$. If $E - \delta E < 0$, the particle has fallen below zero energy.

But then it could repeat: emit another photon, fall further negative, emit again, ad infinitum. Each emission releases a photon and pushes the particle deeper into negative energy. The vacuum — empty space — would spontaneously emit an unbounded cascade of radiation. The universe could not exist.

This is not a minor technical issue. It is an existential crisis for relativistic quantum mechanics.

Part II — The Dirac Sea

Paul Dirac's resolution in 1930 was audacious: fill all the negative-energy states.

The key ingredient: Dirac was working with fermions — particles obeying the Pauli Exclusion Principle, which states that no two identical fermions can occupy the same quantum state. Define the vacuum as the state in which every negative-energy level is already occupied. Then:

• A positive-energy fermion cannot fall into a negative-energy state — all of them are full.
• The cascade is blocked. The vacuum is stable.
• The filled sea is invisible — it defines the energy reference level.

Holes = Antiparticles

An energetic photon can promote a negative-energy electron out of the sea, leaving behind a hole — an absence in the sea. This hole behaves like a particle with the same mass as the electron but opposite charge: the positron. Dirac predicted this in 1928; Carl Anderson discovered the positron experimentally in 1932.

Why it fails for bosons: Bosons obey Bose-Einstein statistics — there is no Pauli exclusion. An unlimited number of bosons can pile into the same negative-energy state. The sea cannot be filled and stabilized. The negative-energy catastrophe for relativistic bosons cannot be cured by Dirac's trick; it requires the full machinery of quantum field theory.

Part III — Wave Functions and Representations

For a particle moving in one dimension, the quantum state can be represented in two equivalent ways — as a function of position or of momentum.

Position-Space Wave Function

The wave function in position space is the component of the state along the position eigenstate $|x\rangle$:

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

Momentum-Space Wave Function and Fourier Transforms

In position space, the momentum eigenstates are plane waves: $\langle x|p\rangle = \dfrac{1}{\sqrt{2\pi\hbar}}\,e^{ipx/\hbar}$. The momentum-space wave function $\tilde{\psi}(p) = \langle p|\psi\rangle$ is obtained by the inner product with this basis.

Inserting completeness resolutions, the two representations are related by a Fourier transform pair:

This is a profound duality: position and momentum are Fourier conjugates. A state sharply localized in position (a narrow spike) has a broadly spread momentum distribution, and vice versa. This Fourier duality is the mathematical origin of the uncertainty principle.

Momentum Operator in Position Space

Acting with $\hat{p}$ on the position-space wave function:

Part IV — The Heisenberg Uncertainty Principle: Rigorous Proof

This is one of the most important derivations in quantum mechanics. We prove $\Delta x\,\Delta p \geq \hbar/2$ from first principles using only the inner product and the Cauchy-Schwarz inequality.

Setup

Assume $\langle x\rangle = 0$ (we can always shift the origin). Set $\hbar = 1$ for simplicity. The uncertainties are:

Define two vectors in the function space:

Then $\langle a|a\rangle = (\Delta x)^2$ and $\langle b|b\rangle = (\Delta p)^2$.

Applying Cauchy-Schwarz

The Cauchy-Schwarz inequality for any two vectors states $\langle a|a\rangle\langle b|b\rangle \geq |\langle a|b\rangle|^2$. Therefore:

We need to evaluate $\langle a|b\rangle = \int\psi^*(x)\cdot x\cdot\dfrac{d\psi}{dx}\,dx$. Integrate by parts (boundary term vanishes for normalizable states):

using normalization in the last step. Therefore $|\langle a|b\rangle|^2 = \tfrac{1}{4}$, giving:

Taking square roots and restoring $\hbar$:

Equality holds when $|a\rangle = \lambda|b\rangle$ for some scalar $\lambda$, i.e., when $x\psi = \lambda\,d\psi/dx$. The solution is a Gaussian: $\psi(x) \propto e^{-x^2/4\sigma^2}$. Gaussian wave packets are minimum-uncertainty states.

The same Fourier duality argument gives a time-energy version: $\Delta E\,\Delta t \geq \hbar/2$. A quantum state that exists for time $\Delta t$ has an intrinsic energy uncertainty of at least $\hbar/(2\Delta t)$.

Part V — The Schrödinger Equation for Particles

For a particle of mass $m$ moving in a potential $V(x)$, the Hamiltonian is the kinetic plus potential energy. In position space, the kinetic operator $\hat{p}^2/2m$ becomes $-\hbar^2/2m\cdot\partial^2/\partial x^2$:

The physical interpretation: the right-hand side has two contributions. The kinetic term $-\hbar^2/2m\cdot\partial^2\psi/\partial x^2$ is proportional to the curvature of $\psi$ — high curvature means rapid spatial oscillation, which corresponds to high momentum (large $p$ in $e^{ipx/\hbar}$). The potential term $V(x)\psi$ modulates the wave function according to the local energy landscape.

Since the equation is linear, superpositions of solutions are solutions — the superposition principle holds.

Time-Independent Schrödinger Equation

For a time-independent $V(x)$, separate variables: $\psi(x,t) = \phi(x)\,e^{-iEt/\hbar}$. Substituting:

This is an eigenvalue problem for the Hamiltonian operator. Its solutions $\phi_n(x)$ with eigenvalues $E_n$ are the energy eigenfunctions. The general solution is:

Part VI — Ehrenfest's Theorem

Ehrenfest's theorem makes precise the sense in which quantum mechanics reduces to classical mechanics for expectation values. Starting from the commutator equation of motion $d\langle L\rangle/dt = (i/\hbar)\langle[H,L]\rangle$, with $H = \hat{p}^2/2m + V(\hat{x})$:

These two equations look exactly like Newton's laws — $\dot{x} = p/m$ and $\dot{p} = F$. But there is a critical subtlety:

The caveat: Newton's second law is $dp/dt = F(\langle x\rangle)$ — the force evaluated at the average position. Ehrenfest gives $d\langle p\rangle/dt = \langle F(x)\rangle$ — the average of the force. These are NOT the same in general: $$\langle F(x)\rangle \neq F(\langle x\rangle)$$ unless $F(x)$ is a linear function of $x$. The difference is $O((\Delta x)^2 F''(\langle x\rangle))$ — suppressed when the wave packet is narrow.

The semi-classical approximation is valid when the wave packet width $\Delta x$ is small compared to the spatial scale over which the force varies. In that regime, $\langle F(x)\rangle \approx F(\langle x\rangle)$ and quantum expectation values obey Newton's laws to high accuracy.

Part VII — When Does Classical Mechanics Emerge?

Classical mechanics emerges from quantum mechanics when quantum uncertainty becomes negligible relative to the scales of the problem. Two conditions:

1. Narrow wave packet: $\Delta x \ll \ell$, where $\ell$ is the length scale over which the potential varies significantly (so $\langle F\rangle \approx F(\langle x\rangle)$).
2. Large mass: The uncertainty principle gives $\Delta x \geq \hbar/(2m\Delta v)$. Heavy objects have tiny quantum uncertainties at ordinary velocities.

A bowling ball ($m = 0.1$ kg, $v = 1$ m/s) has a de Broglie wavelength of $\lambda = h/p \approx 6.6\times10^{-33}$ m — far smaller than any atomic structure. The position uncertainty is roughly $10^{-31}$ m, some 16 orders of magnitude below a proton radius. For all practical and theoretical purposes, the bowling ball is a classical point particle.

An electron ($m = 9.1\times10^{-31}$ kg) at the same speed has $\Delta x \sim$ centimeters — genuinely macroscopic quantum uncertainty. It is irreducibly quantum mechanical.