Momentum space and the Fourier transform · Quantum Mechanics

Intuition

A state can be written in the momentum basis just as well as in the position basis. Its components there form the momentum-space wavefunction, whose modulus squared is the probability density for momentum. The two descriptions are related by the Fourier transform, and the transform has a built-in trade-off: a wavefunction squeezed narrow in position is necessarily spread wide in momentum.

A chord can be described by the air pressure against time or by the list of notes in it. The two are the same sound, turned into each other by a Fourier transform; a short click contains every note at once.

The momentum density of a particle spread evenly over a box of width $a$ , in units where $\hbar=a=1$ . It vanishes first at $p=\pm2\pi\hbar/a$ : squeezing the box, making $a$ smaller, pushes those zeros out and spreads the momentum.

The Fourier transform between x and p

The momentum-space wavefunction is the component of the state along each momentum eigenstate. Since $\braket{x|p}=e^{ipx/\hbar}/\sqrt{2\pi\hbar}$ , it is the Fourier transform of the position wavefunction, and the position wavefunction is the inverse transform of it.

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

Properties

The inverse transform: $\psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int e^{ipx/\hbar}\tilde\psi(p)\,dp$ .

Momentum eigenfunctions are plane waves

\braket{x|p}=\frac{1}{\sqrt{2\pi\hbar}}\,e^{ipx/\hbar}

Write the eigenvalue equation of momentum in the position representation: it is a first-order differential equation whose solutions are exponentials. The constant in front is fixed by asking the momentum states to be normalised to a delta function, using the Fourier integral $\int e^{iqx}dx=2\pi\delta(q)$ .

Proof steps

-i\hbar\,\frac{d}{dx}\braket{x|p}=p\,\braket{x|p}

The eigenvalue equation $\hat{p}\ket{p}=p\ket{p}$ in the position basis.

\braket{x|p}=C\,e^{ipx/\hbar}

A first-order linear equation with constant coefficients; the Differential Equations course solved these by an exponential.

\braket{p'|p}=|C|^{2}\int e^{i(p-p')x/\hbar}dx=|C|^{2}\,2\pi\hbar\,\delta(p-p')

Insert the position completeness relation and use the Fourier integral of the delta function.

|C|^{2}\,2\pi\hbar=1\ \Rightarrow\ C=\frac{1}{\sqrt{2\pi\hbar}}

Demand $\braket{p'|p}=\delta(p-p')$ , choosing real.

Applications

Practice

Components Along Momentum States

The momentum-space wavefunction is the bracket of each momentum eigenstate with the state: the amplitude for each momentum. It is the Fourier transform of the position wavefunction.

\tilde\psi(p)=\braket{p|\psi}

Try it

What is $|\tilde\psi(p)|^{2}$ ?