Fourier transform of an exponential wave function

The function $|\psi|^2=e^{-2|x|/a}/a$ is even, and

$$\int_{\R}|\psi|^2dx=\frac2a\int_0^\infty e^{-2x/a}dx=\frac2a\cdot\frac a2=1 .$$

By parity, $\langle\hat X\rangle=0$. The variance is therefore

$$\sigma_X^2=\langle\hat X^2\rangle=\frac2a\int_0^\infty x^2e^{-2x/a}dx=\frac2a\times2!\Bigl(\frac a2\Bigr)^3=\frac{a^2}{2}, \qquad\sigma_X=\frac{a}{\sqrt2}.$$

We split the integral into two parts according to the sign of $x$, setting $k=p/\hbar$:

$$\begin{aligned} \int_{\R}e^{-ikx}e^{-|x|/a}dx&=\int_0^\infty e^{-(1/a+ik)x}dx+\int_{-\infty}^0e^{(1/a-ik)x}dx\\ &=\frac{1}{1/a+ik}+\frac{1}{1/a-ik}=\frac{2/a}{1/a^2+k^2}=\frac{2a}{1+k^2a^2}. \end{aligned}$$

Both integrals converge because the real parts of the exponents are negative. We obtain

$$\widetilde\psi(p)=\frac{1}{\sqrt{2\pi\hbar}}\cdot\frac{1}{\sqrt a}\cdot\frac{2a}{1+p^2a^2/\hbar^2}=\sqrt{\frac{2a}{\pi\hbar}}\;\frac{1}{1+p^2a^2/\hbar^2}.$$

The momentum density is a squared Lorentzian function centred at $p=0$, with characteristic width $\hbar/a$. To verify the normalisation, set $u=pa/\hbar$, so that $dp=\hbar\,du/a$:

$$\int_{\R}|\widetilde\psi(p)|^2dp=\frac{2a}{\pi\hbar}\cdot\frac\hbar a\int_{\R}\frac{du}{(1+u^2)^2}=\frac2\pi\cdot\frac\pi2=1 .$$

In the momentum-space representation, using the same change of variable,

$$\langle\hat P^2\rangle=\int_{\R}p^2|\widetilde\psi(p)|^2dp=\frac{2a}{\pi\hbar}\Bigl(\frac\hbar a\Bigr)^3\int_{\R}\frac{u^2du}{(1+u^2)^2}=\frac{2\hbar^2}{\pi a^2}\cdot\frac\pi2=\frac{\hbar^2}{a^2}.$$

In the position-space representation, the derivative $\psi'(x)=-\frac{\mathrm{sgn}(x)}{a}\psi(x)$ is defined everywhere except at $x=0$, where it has a jump; we therefore have $|\psi'|^2=|\psi|^2/a^2$ almost everywhere, and

$$\langle\hat P^2\rangle=\hbar^2\int_{\R}|\psi'|^2dx=\frac{\hbar^2}{a^2}\int_{\R}|\psi|^2dx=\frac{\hbar^2}{a^2}.$$

The two methods agree. Since $\widetilde\psi$ is even, $\langle\hat P\rangle=0$, and $\sigma_P=\hbar/a$. The formula $\hbar^2\int|\psi'|^2$ follows from the integration by parts $\braket{\psi}{\hat P^2\psi}=\braket{\hat P\psi}{\hat P\psi}$, which is valid here because $\psi$ is continuous.

The product of the standard deviations is $\sigma_X\sigma_P=\frac{a}{\sqrt2}\cdot\frac\hbar a=\frac{\hbar}{\sqrt2}\simeq0{,}71\,\hbar$, which is greater than $\hbar/2$ as it should be. It is independent of $a$: broadening the wave function in position space reduces the momentum spread by the same proportion. The required probability is

$$P\Bigl(|p|<\frac\hbar a\Bigr)=\frac2\pi\int_{-1}^1\frac{du}{(1+u^2)^2}=\frac4\pi\Bigl[\frac{u}{2(1+u^2)}+\frac12\arctan u\Bigr]_0^1=\frac4\pi\Bigl(\frac14+\frac\pi8\Bigr)=\frac12+\frac1\pi\simeq0{,}82 .$$

We have $\langle\hat P^4\rangle=\frac{2a}{\pi\hbar}\bigl(\frac\hbar a\bigr)^5\int_{\R}\frac{u^4\,du}{(1+u^2)^2}$. As $|u|\to\infty$, the integrand tends to $1$, and the integral diverges. The momentum distribution decreases too slowly, as $1/p^4$, for its fourth-order moment to exist. This behaviour is related to the cusp of $\psi$ at $x=0$: its derivative has a jump there, so the second derivative contains a Dirac delta term, $\psi"=\psi/a^2-\frac{2}{a^{3/2}}\delta(x)$, which is not square-integrable. Thus $\hat P^2\psi$ does not belong to $L^2(\R)$, and $\langle\hat P^4\rangle=\norm{\hat P^2\psi}^2$ is infinite. The less regular a wave function is, the more slowly its Fourier transform decreases and the more high-momentum components it contains.