Fourier Analysis and the Momentum Representation
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You should be comfortable with complex exponentials, basic integration by parts, and the definition of the momentum operator \(\hat{p} = -i\hbar \frac{\partial}{\partial x}\) from the previous unit.
Why this matters
The wavefunction is not just a probability cloud; it is a superposition of plane waves, and this duality is the key to understanding uncertainty.
In classical mechanics, position and momentum are independent variables that can be known simultaneously. In quantum mechanics, the state of a particle is described by a wavefunction \(\psi(x)\) in position space. However, the physical meaning of momentum is tied to the spatial frequency of this wave. By applying Fourier analysis, we can translate the state into a momentum-space representation \(\phi(p)\). This transformation reveals that a state which is sharply defined in position is necessarily broad in momentum, and vice versa. This lecture establishes the mathematical bridge between these two complementary descriptions, providing the foundation for the Heisenberg uncertainty principle and the analysis of wave packet spreading.
The Wave as a Sum of Frequencies
Imagine a complex sound wave, such as a musical chord. It looks complicated in the time domain, but if you analyze it, you find it is simply a sum of pure sine waves, each with a specific frequency and amplitude. Quantum mechanics works the same way. A particle's wavefunction \(\psi(x)\) in position space can be decomposed into a continuous sum of plane waves, \(e^{ikx}\). Each plane wave corresponds to a particle with a definite momentum \(p = \hbar k\). The coefficient of each plane wave tells us the probability amplitude of finding the particle with that specific momentum. Thus, the "shape" of the wave in space is directly determined by the mixture of momenta it contains. A narrow pulse in space requires a wide range of frequencies (momenta) to construct it, while a pure sine wave (infinite extent in space) has a single, precise momentum.
The Fourier Transform Pair
To move between position and momentum representations, we use the Fourier transform. The momentum-space wavefunction \(\phi(p)\) is defined as the integral of the position-space wavefunction multiplied by a complex exponential kernel. The standard normalization convention in quantum mechanics ensures that the probability interpretation remains consistent in both spaces.
The inverse transform allows us to reconstruct the position-space wavefunction from the momentum distribution. This symmetry is crucial: neither representation is "more real" than the other; they are simply different coordinate systems for the same quantum state.
These equations define a unitary transformation. This means that the total probability is preserved. If \(\psi(x)\) is normalized such that \(\int |\psi(x)|^2 dx = 1\), then \(\phi(p)\) is automatically normalized such that \(\int |\phi(p)|^2 dp = 1\). This is a direct consequence of Parseval's theorem in Fourier analysis.
Operators in Momentum Space
One of the most powerful aspects of the momentum representation is the simplification of operators. In position space, the momentum operator involves a derivative, \(\hat{p} = -i\hbar \frac{\partial}{\partial x}\). In momentum space, however, momentum is just a variable. The operator \(\hat{p}\) acts by simple multiplication by \(p\).
Conversely, the position operator becomes a derivative in momentum space. By applying the inverse Fourier transform to the action of the position operator, we can derive that \(\hat{x}\) acts as \(i\hbar \frac{\partial}{\partial p}\). This duality is a manifestation of the canonical commutation relation \([\hat{x}, \hat{p}] = i\hbar\). The operator dictionary from D-141, which established \(\hat{x} = x\) and \(\hat{p} = -i\hbar \partial_x\) in position space, is mirrored perfectly in momentum space as \(\hat{p} = p\) and \(\hat{x} = i\hbar \partial_p\).
Width and the Uncertainty Principle
The relationship between the widths of \(\psi(x)\) and \(\phi(p)\) is governed by the properties of the Fourier transform. A function and its Fourier transform cannot both be sharply localized. If \(\psi(x)\) is a narrow Gaussian with standard deviation \(\sigma_x\), its Fourier transform \(\phi(p)\) is also a Gaussian, but with a standard deviation \(\sigma_p\) that is inversely proportional to \(\sigma_x\).
For a Gaussian wave packet, the product of the standard deviations is minimized. This leads to the Heisenberg uncertainty principle, which states that the product of the standard deviations of position and momentum must be greater than or equal to \(\hbar/2\).
This is not a limitation of measurement technology; it is a fundamental property of the wave nature of matter. The more precisely we define the position (small \(\sigma_x\)), the more high-frequency components (large \(\sigma_p\)) are required to construct the wave, leading to a larger spread in momentum.
The derivation
Expectation Values and the Operator Dictionary
For a normalised state ψ(x), the average value of an observable A over repeated identical measurements is the inner product ⟨A⟩ = ⟨ψ|Â|ψ⟩ = ∫ ψ*(x) Â ψ(x) dx. Imposing the Born rule for position fixes x̂ = x (multiplication), and imposing the de Broglie relation for momentum eigenstates fixes p̂ = −iℏ ∂/∂x. These two representations are the seed of the entire operator dictionary.
Be careful with the sign convention in the Fourier transform. If you use \(e^{-ipx/\hbar}\) for the forward transform, you must use \(e^{+ipx/\hbar}\) for the inverse. Mixing these up will lead to incorrect signs in the operator representations, specifically for \(\hat{x}\) in momentum space. Always check that \(\hat{x}\) acting on a momentum eigenstate yields the correct phase gradient. C
Worked example
Problem. Consider a normalized Gaussian wave packet in position space given by \(\psi(x) = \frac{1}{(2\pi\sigma_x^2)^{1/4}} e^{-x^2/(4\sigma_x^2)}\). Determine the momentum-space wavefunction \(\phi(p)\) and calculate the product \(\sigma_x \sigma_p\).
- Identify the form of the Fourier transform for a Gaussian. The integral \(\int e^{-ax^2} e^{-ikx} dx\) results in another Gaussian. Here, \(a = 1/(4\sigma_x^2)\) and \(k = p/\hbar\).
- Perform the integration. The result is \(\phi(p) = \frac{1}{(2\pi\hbar)^{1/2}} \frac{1}{(2\pi\sigma_x^2)^{1/4}} \int_{-\infty}^{\infty} e^{-x^2/(4\sigma_x^2)} e^{-ipx/\hbar} dx\). Using the standard Gaussian integral \(\int_{-\infty}^{\infty} e^{-ax^2+bx} dx = \sqrt{\frac{\pi}{a}} e^{b^2/4a}\), we substitute \(a = 1/(4\sigma_x^2)\) and \(b = -ip/\hbar\).
- Simplify the exponent. The term \(b^2/4a\) becomes \(\frac{(-ip/\hbar)^2}{4(1/4\sigma_x^2)} = \frac{-p^2/\hbar^2}{1/\sigma_x^2} = -\frac{p^2\sigma_x^2}{\hbar^2}\).
- Determine the normalization constant. The prefactor is \( \frac{1}{\sqrt{2\pi\hbar}} \frac{1}{(2\pi\sigma_x^2)^{1/4}} \sqrt{4\pi\sigma_x^2} \), so \(\phi(p) = \left(\frac{2\sigma_x^2}{\pi\hbar^2}\right)^{1/4} e^{-p^2\sigma_x^2/\hbar^2}\). Then \( |\phi(p)|^2 \propto e^{-2p^2\sigma_x^2/\hbar^2} \) is a Gaussian in \(p\) with variance \(\sigma_p^2 = \hbar^2/(4\sigma_x^2)\), so \(\sigma_p = \hbar/(2\sigma_x)\).
- Calculate the product. \(\sigma_x \sigma_p = \sigma_x \left( \frac{\hbar}{2\sigma_x} \right) = \frac{\hbar}{2}\).
Units check. \(\sigma_x\) has units of \(\mathrm{m}\). \(\sigma_p\) has units of \(\mathrm{kg \cdot m/s}\). The product has units of \(\mathrm{kg \cdot m^2/s}\), which is equivalent to \(\mathrm{J \cdot s}\), the same units as \(\hbar\). The result is dimensionally consistent.
Three quick tests
- If \(\psi(x)\) is a real-valued function, what can you say about the symmetry of \(\phi(p)\)?
Answer
\(\phi(p)\) is symmetric (even) if \(\psi(x)\) is symmetric, but generally, \(\phi(-p) = \phi^*(p)\). For a real \(\psi(x)\), the Fourier transform satisfies \(\phi(-p) = \phi^*(p)\). - How does the momentum operator \(\hat{p}\) act on a state in momentum space?
Answer
It acts by multiplication: \(\hat{p} \phi(p) = p \phi(p)\). - Why is the Fourier transform unitary?
Answer
Because it preserves the inner product (and thus the norm/probability) between the position and momentum representations, as guaranteed by Parseval's theorem.
Problem set for PU-202 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.