Case Study 1: Position and Momentum — Two Sides of the Same Coin

Introduction

Position and momentum are the two most fundamental observables in quantum mechanics. They are connected by the canonical commutation relation $[\hat{x}, \hat{p}] = i\hbar$ and by the Fourier transform. In this case study, we explore the deep symmetry between them: how position eigenstates look in momentum space and vice versa, how the Fourier transform mediates between representations, and how the uncertainty principle emerges as a direct consequence of their mathematical relationship.

This is not merely an exercise in formalism. The position-momentum duality is the template for every conjugate pair in physics: energy-time, angle-angular momentum, and the generalized coordinates and momenta of Hamiltonian mechanics. Master it here, and you will recognize the pattern everywhere.

Part 1: Two Completeness Relations, One Hilbert Space

The position and momentum eigenstates provide two complete bases for the same Hilbert space $\mathcal{H} = L^2(\mathbb{R})$:

$$\int_{-\infty}^{\infty} |x\rangle\langle x| \, dx = \hat{I}, \qquad \int_{-\infty}^{\infty} |p\rangle\langle p| \, dp = \hat{I}$$

Any state $|\psi\rangle$ can be expanded in either basis:

$$|\psi\rangle = \int \psi(x) |x\rangle \, dx = \int \phi(p) |p\rangle \, dp$$

where $\psi(x) = \langle x|\psi\rangle$ and $\phi(p) = \langle p|\psi\rangle$. The two representations carry the same information — they are two descriptions of the same physical state, written in two different "languages."

The translation between languages is provided by the overlap:

$$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}$$

This single formula — a plane wave with amplitude $(2\pi\hbar)^{-1/2}$ — encodes the entire position-momentum duality.

Part 2: How a Position Eigenstate Looks in Momentum Space

Suppose a particle is (hypothetically) in the position eigenstate $|x_0\rangle$. What does this state look like in momentum space?

$$\phi_{x_0}(p) = \langle p|x_0\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{-ipx_0/\hbar}$$

This is a plane wave of constant amplitude. The momentum probability density is:

$$|\phi_{x_0}(p)|^2 = \frac{1}{2\pi\hbar}$$

Every momentum is equally probable. A particle with perfectly known position has completely unknown momentum. This is the uncertainty principle in its most extreme form.

The inverse. A momentum eigenstate $|p_0\rangle$ in position space:

$$\psi_{p_0}(x) = \langle x|p_0\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ip_0 x/\hbar}$$

Again a plane wave, now of constant amplitude in position. A particle with perfectly known momentum is equally likely to be found anywhere. The duality is perfect.

Part 3: Gaussian Wave Packets — The Optimal Compromise

No physical state can have both perfectly known position and perfectly known momentum. The best compromise is a Gaussian wave packet. Consider:

$$\psi(x) = \left(\frac{1}{2\pi\sigma_x^2}\right)^{1/4} \exp\left(-\frac{(x - x_0)^2}{4\sigma_x^2} + \frac{ip_0 x}{\hbar}\right)$$

This represents a particle centered at $x_0$ with position uncertainty $\sigma_x$ and mean momentum $p_0$.

Position-space properties: $$\langle\hat{x}\rangle = x_0, \qquad \Delta x = \sigma_x$$

Fourier transform to momentum space:

$$\phi(p) = \frac{1}{\sqrt{2\pi\hbar}} \int \psi(x) e^{-ipx/\hbar} \, dx$$

Completing the square in the exponent:

$$\phi(p) = \left(\frac{2\sigma_x^2}{\pi\hbar^2}\right)^{1/4} \exp\left(-\frac{(p - p_0)^2\sigma_x^2}{\hbar^2} - \frac{ip x_0}{\hbar}\right)$$

(The phase factor $e^{-ipx_0/\hbar}$ does not affect probabilities.)

Momentum-space properties: $$\langle\hat{p}\rangle = p_0, \qquad \Delta p = \frac{\hbar}{2\sigma_x}$$

Uncertainty product: $$\Delta x \cdot \Delta p = \sigma_x \cdot \frac{\hbar}{2\sigma_x} = \frac{\hbar}{2}$$

The Gaussian saturates the Heisenberg inequality. It is the minimum-uncertainty state.

The reciprocal width relationship

The position-space width is $\sigma_x$ and the momentum-space width is $\sigma_p = \hbar/(2\sigma_x)$. As one gets narrower, the other gets wider:

This table is a complete summary of the position-momentum uncertainty principle, visualized through the Fourier transform.

Part 4: Operator Symmetry Between Representations

The beautiful symmetry extends to how operators act:

Each operator is "diagonal" (acts by multiplication) in its own eigenbasis and acts as a derivative in the conjugate basis. This is not a coincidence — it is a direct consequence of the commutation relation $[\hat{x}, \hat{p}] = i\hbar$.

Verifying the commutation relation in momentum space. Let $\hat{x} = i\hbar\frac{\partial}{\partial p}$ and $\hat{p} = p$ (multiplication). Then:

$$[\hat{x}, \hat{p}]\phi(p) = i\hbar\frac{\partial}{\partial p}(p\phi) - p \cdot i\hbar\frac{\partial\phi}{\partial p} = i\hbar\left(\phi + p\frac{\partial\phi}{\partial p}\right) - i\hbar p\frac{\partial\phi}{\partial p} = i\hbar\phi(p)$$

So $[\hat{x}, \hat{p}] = i\hbar\hat{I}$, consistent with the position representation.

Part 5: The Schrodinger Equation in Both Representations

Position space (familiar):

$$\left[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)\right]\psi(x) = E\psi(x)$$

Kinetic energy is a second derivative; potential energy is diagonal.

Momentum space (less familiar):

$$\frac{p^2}{2m}\phi(p) + \int \tilde{V}(p - p')\phi(p') \, dp' = E\phi(p)$$

Kinetic energy is diagonal (just $p^2/(2m)$); potential energy becomes a convolution with the Fourier transform $\tilde{V}$ of $V(x)$.

Strategic choice of representation

The best representation depends on the problem:

• QHO ($V = \frac{1}{2}m\omega^2 x^2$): Both representations are equally natural (the QHO is symmetric under the Fourier transform — its eigenstates are Hermite-Gauss functions, which are Fourier transform eigenstates).

• Free particle ($V = 0$): Momentum space is trivially diagonal.

• Coulomb potential ($V = -e^2/r$): Position space is traditional, but momentum-space solutions (Fock, 1935) reveal hidden $SO(4)$ symmetry.

• Periodic potential ($V(x + a) = V(x)$): Momentum space (Bloch's theorem, reciprocal lattice) is standard in solid-state physics.

Part 6: The Fourier Transform as a Unitary Operator

The Fourier transform is not just a computational tool — it is a unitary operator $\hat{F}$ on the Hilbert space. In the position representation:

$$(\hat{F}\psi)(p) = \frac{1}{\sqrt{2\pi\hbar}}\int \psi(x) e^{-ipx/\hbar} \, dx = \phi(p)$$

Its unitarity follows from Parseval's theorem:

$$\int|\phi(p)|^2 dp = \int|\psi(x)|^2 dx \implies \hat{F}^\dagger\hat{F} = \hat{I}$$

The Fourier transform has a remarkable property: $\hat{F}^4 = \hat{I}$. That is, applying the Fourier transform four times returns you to the original function. The eigenvalues of $\hat{F}$ are therefore $\{1, i, -1, -i\}$ — the fourth roots of unity.

QHO eigenstates are Fourier eigenstates. The QHO energy eigenstates $\psi_n(x)$ (Hermite-Gauss functions) satisfy:

$$\hat{F}\psi_n = (-i)^n \psi_n$$

The ground state ($n = 0$) is an eigenstate of $\hat{F}$ with eigenvalue $1$ — its Fourier transform is itself (up to scaling). This is another manifestation of the Gaussian's special role.

Part 7: Physical Implications — Why This Matters

The position-momentum duality has profound physical consequences:

1. Heisenberg's microscope (thought experiment). To determine a particle's position precisely, you must use short-wavelength (high-momentum) photons. These photons transfer large, unpredictable momenta to the particle. The more precisely you determine position, the more you disturb momentum. The Fourier transform makes this quantitative.

2. Tunneling. A particle in a classically forbidden region has imaginary momentum ($p^2 < 0$). In momentum space, the wave function extends beyond the classical turning points. The Fourier relationship ensures that spatial localization near a barrier implies momentum components that can penetrate it.

3. Diffraction. The diffraction pattern of a particle passing through a slit is the Fourier transform of the slit geometry. A narrow slit (small $\Delta x$) produces a wide diffraction pattern (large $\Delta p$), and vice versa. This is the uncertainty principle made visible.

4. Quantum field theory. Fields are naturally decomposed into momentum modes (plane waves). Creation and annihilation operators create and destroy particles of definite momentum. The entire framework of particle physics rests on the Fourier decomposition of quantum fields.

Summary

Position and momentum are not merely two observables that happen to be related by a Fourier transform. They are conjugate variables — dual descriptions of the same degrees of freedom. The Fourier transform is the unitary operator that converts between them, preserving all physical information. The uncertainty principle is not an additional postulate but a mathematical theorem about Fourier transform pairs: a function and its Fourier transform cannot both be arbitrarily narrow. This is as true for sound waves and electrical signals as it is for quantum mechanics — but in quantum mechanics, it has the profound consequence that position and momentum cannot both be simultaneously well-defined.

💡 Key Insight — The relationship $\langle x|p\rangle = (2\pi\hbar)^{-1/2}e^{ipx/\hbar}$ is arguably the most information-dense equation in introductory quantum mechanics. From it, you can derive: the Fourier transform, the uncertainty principle, the position and momentum representations of all operators, the Schrodinger equation in both representations, and the Dirac delta function. Everything in this case study flows from this single overlap.