Case Study 2: From Fourier Transforms to Quantum Mechanics — Position-Momentum Duality

Introduction

The Fourier transform is one of the most important mathematical tools in physics. In classical signal processing, it decomposes a function of time into frequency components. In quantum mechanics, it does something far more profound: it changes the basis of the Hilbert space from position to momentum.

This case study traces the Fourier transform from its classical incarnation through its quantum mechanical meaning, using Dirac notation to reveal the underlying structure. Along the way, we will see that position and momentum are related by a unitary transformation — they are two equally valid ways of looking at the same quantum state.

Part 1: The Classical Fourier Transform — A Quick Review

The Fourier transform pair for a function $f(x)$ and its transform $\tilde{f}(k)$:

$$\tilde{f}(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) \, e^{-ikx} \, dx$$

$$f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \tilde{f}(k) \, e^{ikx} \, dk$$

Key properties: - Parseval's theorem: $\int |f(x)|^2 \, dx = \int |\tilde{f}(k)|^2 \, dk$ - Convolution theorem: The Fourier transform of a product is a convolution. - The Fourier transform is a unitary operator on $L^2(\mathbb{R})$.

In quantum mechanics, we use $p = \hbar k$, so:

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

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

Until now, you may have viewed these as "just formulas." Dirac notation reveals them to be something deeper.

Part 2: The Dirac Derivation — Basis Change in One Line

Position to momentum

Starting from the abstract state $|\psi\rangle$:

$$\phi(p) = \langle p|\psi\rangle$$

Insert the position completeness relation $\hat{I} = \int |x\rangle\langle x| \, dx$:

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

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

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

That is the Fourier transform. In Dirac notation, it is a single completeness insertion.

Momentum to position

Starting from $\psi(x) = \langle x|\psi\rangle$ and inserting momentum completeness $\hat{I} = \int |p\rangle\langle p| \, dp$:

$$\psi(x) = \int \langle x|p\rangle \langle p|\psi\rangle \, dp = \frac{1}{\sqrt{2\pi\hbar}} \int \phi(p) \, e^{ipx/\hbar} \, dp$$

That is the inverse Fourier transform. Same technique, different basis.

The "basis change matrix"

The quantity $\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}$ is the overlap between a position eigenstate and a momentum eigenstate. It is the continuous analogue of the unitary matrix element $U_{mn} = \langle m'|n\rangle$ that connects two discrete bases.

In a finite-dimensional system, the basis change is a matrix multiplication. In the continuous case, it is an integral — but the structure is identical.

Part 3: Parseval's Theorem as Unitarity

Parseval's theorem states $\int |\psi(x)|^2 \, dx = \int |\phi(p)|^2 \, dp$. In Dirac notation:

$$\langle\psi|\psi\rangle = 1$$

This single equation contains Parseval's theorem. The left side can be evaluated in any basis:

Position: $\langle\psi|\psi\rangle = \int \langle\psi|x\rangle\langle x|\psi\rangle \, dx = \int |\psi(x)|^2 \, dx$

Momentum: $\langle\psi|\psi\rangle = \int \langle\psi|p\rangle\langle p|\psi\rangle \, dp = \int |\phi(p)|^2 \, dp$

Both equal 1 because $\langle\psi|\psi\rangle$ is basis-independent. Parseval's theorem is a consequence of unitarity — the basis change preserves the norm.

Part 4: Operators in Both Representations

The position operator

In the position representation:

$$\langle x|\hat{x}|\psi\rangle = x \, \psi(x)$$

In the momentum representation, insert position completeness:

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

But this is the Fourier transform of $x\psi(x)$, which (by the derivative property of Fourier transforms) equals:

$$\langle p|\hat{x}|\psi\rangle = i\hbar \frac{\partial}{\partial p} \phi(p)$$

So in the momentum representation, the position operator is a derivative:

$$\hat{x} \xrightarrow{\text{position rep.}} x \times, \qquad \hat{x} \xrightarrow{\text{momentum rep.}} i\hbar \frac{\partial}{\partial p}$$

The momentum operator

In the position representation:

$$\langle x|\hat{p}|\psi\rangle = -i\hbar \frac{\partial}{\partial x} \psi(x)$$

In the momentum representation:

$$\langle p|\hat{p}|\psi\rangle = p \, \phi(p)$$

So in the momentum representation, the momentum operator is simple multiplication:

$$\hat{p} \xrightarrow{\text{position rep.}} -i\hbar \frac{\partial}{\partial x}, \qquad \hat{p} \xrightarrow{\text{momentum rep.}} p \times$$

The beautiful symmetry

Position and momentum are related by an exact duality: whichever operator is "multiplication" in one representation becomes a "derivative" in the other. This symmetry is a consequence of the Fourier transform being its own inverse (up to sign).

In Dirac notation, the operators $\hat{x}$ and $\hat{p}$ are abstract — they do not have any preferred form. The derivative and multiplication forms are representations, specific to a choice of basis. The abstract canonical commutation relation $[\hat{x}, \hat{p}] = i\hbar$ holds in every representation.

Part 5: The Gaussian Wave Packet — A Worked Example

Setup

Consider the normalized Gaussian wave packet centered at $x_0$ with momentum $p_0$:

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

We will compute its momentum-space representation, expectation values, and uncertainties using both wave mechanics and Dirac notation.

Momentum representation

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

Substituting $\psi(x)$ and completing the square in the exponent (a standard Gaussian integral):

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

This is also a Gaussian, centered at $p_0$ with width $\sigma_p = \hbar/(2\sigma)$.

Expectation values

Position (Dirac method):

$$\langle\hat{x}\rangle = \langle\psi|\hat{x}|\psi\rangle$$

In position representation: $\langle\hat{x}\rangle = \int x |\psi(x)|^2 dx$. By inspection (the probability density $|\psi(x)|^2$ is a Gaussian centered at $x_0$):

$$\langle\hat{x}\rangle = x_0$$

Momentum (Dirac method):

$$\langle\hat{p}\rangle = \langle\psi|\hat{p}|\psi\rangle$$

In momentum representation: $\langle\hat{p}\rangle = \int p |\phi(p)|^2 dp$. Similarly:

$$\langle\hat{p}\rangle = p_0$$

Uncertainties

$$\sigma_x^2 = \langle\hat{x}^2\rangle - \langle\hat{x}\rangle^2 = \sigma^2 \qquad \Rightarrow \qquad \sigma_x = \sigma$$

$$\sigma_p^2 = \langle\hat{p}^2\rangle - \langle\hat{p}\rangle^2 = \frac{\hbar^2}{4\sigma^2} \qquad \Rightarrow \qquad \sigma_p = \frac{\hbar}{2\sigma}$$

The uncertainty product

$$\sigma_x \sigma_p = \sigma \cdot \frac{\hbar}{2\sigma} = \frac{\hbar}{2}$$

The Gaussian wave packet saturates the uncertainty relation $\sigma_x \sigma_p \geq \hbar/2$. It achieves the minimum possible uncertainty product. This is why the Gaussian is called a minimum-uncertainty state.

This result is independent of the choice of $x_0$, $p_0$, and $\sigma$. Every Gaussian saturates the bound. This makes Gaussians the "most classical" quantum states — they come as close as quantum mechanics allows to having definite position and momentum simultaneously. We will meet them again in Chapter 27 as coherent states of the harmonic oscillator.

Part 6: The QHO Energy Eigenstates in Momentum Space

Translating to momentum representation

The QHO energy eigenstates in position space are:

$$\psi_n(x) = \langle x|n\rangle = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \frac{1}{\sqrt{2^n n!}} H_n\!\left(\sqrt{\frac{m\omega}{\hbar}}\,x\right) e^{-m\omega x^2/2\hbar}$$

where $H_n$ are Hermite polynomials. What do they look like in momentum space?

Using $\phi_n(p) = \langle p|n\rangle = \frac{1}{\sqrt{2\pi\hbar}} \int \langle x|n\rangle e^{-ipx/\hbar} dx$, one can show (using properties of Hermite polynomials under Fourier transformation) that:

$$\phi_n(p) = \langle p|n\rangle = \frac{(-i)^n}{\left(\pi m\omega\hbar\right)^{1/4}} \frac{1}{\sqrt{2^n n!}} H_n\!\left(\frac{p}{\sqrt{m\omega\hbar}}\right) e^{-p^2/(2m\omega\hbar)}$$

The momentum-space eigenstates are the same Hermite-Gaussian functions (up to a phase $(−i)^n$ and a rescaling of the argument). This is a remarkable symmetry of the QHO: the Fourier transform maps QHO eigenstates to QHO eigenstates.

Why this happens

In Dirac notation, the Hamiltonian is:

$$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2$$

This is symmetric under the exchange $\hat{x} \leftrightarrow \hat{p}/(m\omega)$ (up to constants). The Fourier transform implements precisely this exchange. So the QHO eigenstates must be eigenstates of the Fourier transform — which means their position and momentum representations have the same functional form.

This is a deep fact that connects the QHO to signal processing (Hermite-Gaussian modes in optics) and to quantum field theory (the vacuum state is a Gaussian).

Matrix representation in the energy basis

In the energy basis $\{|0\rangle, |1\rangle, |2\rangle, \ldots\}$, the Hamiltonian is diagonal by construction:

$$\langle m|\hat{H}|n\rangle = \hbar\omega\left(n + \frac{1}{2}\right)\delta_{mn}$$

The position and momentum operators are off-diagonal (as shown in Section 8.5 of the main chapter). This is the complementary picture: choosing the energy basis makes $\hat{H}$ simple but $\hat{x}$ and $\hat{p}$ complicated. Choosing the position basis makes $\hat{x}$ simple but $\hat{H}$ a differential operator. There is no basis in which all three are simultaneously simple. This is the uncertainty principle at work at the level of representations.

Part 7: Heisenberg Picture — Operators Evolve Instead of States

In the Schrodinger picture, the state evolves: $|\psi(t)\rangle = \hat{U}(t)|\psi(0)\rangle$.

In the Heisenberg picture, the operators evolve: $\hat{A}_H(t) = \hat{U}^\dagger(t)\hat{A}\hat{U}(t)$.

Both pictures give the same expectation values:

$$\langle\psi(t)|\hat{A}|\psi(t)\rangle = \langle\psi(0)|\hat{U}^\dagger(t)\hat{A}\hat{U}(t)|\psi(0)\rangle = \langle\psi(0)|\hat{A}_H(t)|\psi(0)\rangle$$

For the QHO, the Heisenberg equations of motion give:

$$\hat{x}_H(t) = \hat{x}(0)\cos(\omega t) + \frac{\hat{p}(0)}{m\omega}\sin(\omega t)$$

$$\hat{p}_H(t) = \hat{p}(0)\cos(\omega t) - m\omega\hat{x}(0)\sin(\omega t)$$

These are exactly the classical equations of motion for a harmonic oscillator, with $x$ and $p$ replaced by $\hat{x}$ and $\hat{p}$. The operators oscillate sinusoidally, just as classical position and momentum do.

This is Ehrenfest's theorem in action (Chapter 7): quantum expectation values follow classical trajectories for systems where the potential is at most quadratic. The QHO is the archetype of this correspondence.

Summary: What Position-Momentum Duality Teaches Us

1. The Fourier transform is a change of basis in Hilbert space, from position to momentum eigenstates. In Dirac notation, it reduces to inserting a completeness relation.

2. Position and momentum are dual representations of the same physics. Neither is more fundamental. The operator $\hat{x}$ is simple in one basis and complicated in the other; $\hat{p}$ is the reverse.

3. The uncertainty principle is a property of the Fourier transform. A function that is narrow in $x$-space must be broad in $p$-space, and vice versa. The Gaussian saturates this bound.

4. The QHO is self-dual under Fourier transformation. Its eigenstates have the same form in both position and momentum representations. This is a symmetry, not a coincidence.

5. Dirac notation makes all of this transparent. In wave mechanics, these results require page-long Fourier integral calculations. In Dirac notation, the structure is visible in a single line: $\phi(p) = \int \langle p|x\rangle \langle x|\psi\rangle \, dx$.

🔗 Connection — The position-momentum duality explored here will reappear in Chapter 10 (translation symmetry generates momentum conservation), Chapter 27 (coherent states as displaced Gaussians), and Chapter 31 (the path integral formulation, where position and momentum play symmetrical roles in the phase-space path integral).