Momentum Distribution for a Particle in an Infinite 1D Well

Also read – Particle in a 1-D infinte well  – Energy States

Setup

Infinite well from 0 to a. The position-space eigenstate is

    \[ \psi_n(x)= \begin{cases} \sqrt{\dfrac{2}{a}}\;\sin\!\left(\dfrac{n\pi x}{a}\right), & 0<x<a,\\[6pt] 0, & \text{elsewhere}, \end{cases} \qquad n=1,2,3,\dots \]

Define k_n \equiv \dfrac{n\pi}{a} and p=\hbar k. The momentum-space wavefunction (Fourier transform) is

    \[ \phi_n(p)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\psi_n(x)\,e^{-ipx/\hbar}\,dx =\frac{1}{\sqrt{2\pi\hbar}}\sqrt{\frac{2}{a}}\int_{0}^{a}\sin(k_n x)\,e^{-ikx}\,dx. \]

Compute the Transform

Using \sin(k_n x)=\tfrac{1}{2i}\big(e^{ik_n x}-e^{-ik_n x}\big) and
\int_0^a e^{iqx}\,dx=\dfrac{e^{iqa}-1}{iq} with k_n a=n\pi, one convenient compact form is

    \[ \boxed{\; \phi_n(p)=\sqrt{\frac{2}{a}}\frac{1}{\sqrt{2\pi\hbar}}\, \frac{k_n\left[\,1-(-1)^n e^{-i p a/\hbar}\right]}{k_n^{\,2}-k^{2}} \;} \quad\text{with}\quad k=\frac{p}{\hbar}. \]

(Overall phase is physically irrelevant.)

Momentum Distribution

The probability density in momentum space is

    \[ \boxed{\; |\phi_n(p)|^2=\frac{1}{\pi a\hbar}\; \frac{k_n^{\,2}\,\big|1-(-1)^n e^{-i p a/\hbar}\big|^2}{\big(k_n^{\,2}-k^{2}\big)^2} \;},\qquad k=\frac{p}{\hbar}. \]

Using \big|1-e^{-i\theta}\big|^2=4\sin^2(\theta/2) and
\big|1+e^{-i\theta}\big|^2=4\cos^2(\theta/2) with \theta=\dfrac{p a}{\hbar}, we get

Parity Split

  • n even ((-1)^n=+1):

        \[ |\phi_n(p)|^2=\frac{1}{\pi a\hbar}\; \frac{4k_n^{\,2}\,\sin^2\!\left(\dfrac{p a}{2\hbar}\right)}{\big(k_n^{\,2}-k^{2}\big)^2}. \]

  • n odd ((-1)^n=-1):

        \[ |\phi_n(p)|^2=\frac{1}{\pi a\hbar}\; \frac{4k_n^{\,2}\,\cos^2\!\left(\dfrac{p a}{2\hbar}\right)}{\big(k_n^{\,2}-k^{2}\big)^2}. \]

Interpretation & Checks

  • Two symmetric peaks near p=\pm \hbar k_n=\pm \dfrac{n\pi\hbar}{a}; width \sim \hbar/a. Tails \propto 1/p^2.
  • Parity imprint: for even n, node at p=0; for odd n, maximum at p=0.
  • Normalization: \displaystyle \int_{-\infty}^{\infty}|\phi_n(p)|^2\,dp=1.
  • Moments:

        \[ \langle p\rangle=0,\qquad \langle p^2\rangle=\int p^2|\phi_n(p)|^2\,dp=(\hbar k_n)^2 =\left(\frac{n\pi\hbar}{a}\right)^2, \]

    consistent with 2mE_n where E_n=\dfrac{n^2\pi^2\hbar^2}{2ma^2}.