Inverse Fourier transform of time evolved $e^{-a|x|}$

Question

For a free particle with initial wave function, $$\Psi(x,0) = ae^{-a^2|x|}$$ I am trying to get the time evolved wave function by applying the inverse transform, \begin{equation} \Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dk\,\phi(k)e^{ikx}e^{-iE(k) t/\hbar } \end{equation} where I already found the transform at zero $\phi(k)$ so I have \begin{equation} \Psi(x,t) = \frac{1}{2\pi}\int_{-\infty}^\infty dk\,\frac{2a^3}{a^4 +k^2}e^{-i(\hbar k^2 t/2m-kx)} \end{equation} My question is specifically about how to do this contour integral. I defined a couple constants to have everything dimensionless \begin{equation} \Psi(x,t) = \frac{1}{\pi a}\int_{-\infty}^\infty dz\,\frac{1}{1+z^2}e^{-i( K_2 z^2 -K_1z )} \end{equation} where $K_1 = x a^2$ and $K_2 = \hbar a^4 t/2m >0$, as $t>0$ and $z a^2 = k$. My questions here are:

I am having a hard time justifying to myself that (to recover the $|x|$, I am assuming) I will have to do this over two different contours, the upper semicircle when $x>0$ and the lower semicircle when $x<0$. To me it seems that the behavior of the exponent is actually dominated by the $z^2$ term in both cases, so why shouldn't I just always choose the upper contour. This however, doesn't yield the right answer.

I decided to let $z = a+bi$ and expanded the exponent: \begin{equation} -iK_2(a^2-b^2+2abi)+iK_1(a+bi) = -i(K_2a^2-K_2b^2-K_1 a) + 2abK_2 -bK_1 \end{equation} where I care about what the real part is. So clearly for $x>0$, we have that $e^{-bK_1}$ would require the upper contour to decay. If $x<0$, this term would require the lower contour. But this is of course not the only real part, how do I argue to ignore $ab$ term?

I already computed the residues (assuming the two contour choice) and got \begin{equation} \text{Res}(f,i)=\lim_{z \to i} \frac{e^{-i( K_2 z^2 -K_1z)}}{(i+z)} = \frac{e^{-K_1 +iK_2}}{2i} \end{equation} which plugging back in, yields, \begin{equation} \Psi(x,t) = \frac{e^{-a^2 x} e^{+i\hbar a^4 t/2m}}{a} \text{ if } x>0 \end{equation} and if we do the same analysis for the other contour, we need \begin{equation} \text{Res}(f,-i)=\lim_{z \to -i} \frac{e^{-i( K_2 z^2 -K_1z)}}{(z-i)} = -\frac{e^{K_1 +iK_2}}{2i} \end{equation} which then is \begin{equation} \Psi(x,t) = -\frac{e^{a^2 x} e^{+i\hbar a^4 t/2m}}{a} \text{ if } x<0 \end{equation} so I am inching away from the answer. The absolute value gets recovered but there is also a minus sign out front for the $x<0$ which should not be there. My guess is this is one of those deals where clockwise/anti clockwise contour does something to the sign but I am not sure. Why is there a minus sign here?

I believe, although this is probably a common question, I didn't find anything exactly like it on the site.

Answer 1

An alternative way is to use the propagator. I will set $\hbar^2/m=1$ so that the propagator is: $$ \begin{align} G &= \int e^{-itk^2/2}e^{ikx}\frac{dk}{2\pi} \\ &= \frac{e^{ix^2/2t}}{\sqrt{-i2\pi t}} \end{align} $$ Setting $a=1$, your wave function is: $$ \begin{align} \psi &= \int e^{-|y|}\frac{e^{i(x-y)^2/2t}}{\sqrt{-i2\pi t}}dy \\ &= \int_{-\infty}^x e^{y-x}\frac{e^{iy^2/2t}}{\sqrt{-i2\pi t}}dy+\int_x^{+\infty} e^{x-y}\frac{e^{iy^2/2t}}{\sqrt{-i2\pi t}}dy \\ &= -\frac{e^{-x+it/2}}{\sqrt{-i2\pi t}}\int_{-x}^{+\infty} e^{i(y+it)^2/2t}dy+\frac{e^{x+it/2}}{\sqrt{-i2\pi t}}\int_x^{+\infty} e^{i(y+it)^2/2t}dy \\ &= -\frac{e^{-x+it/2}}{2}\text{erfc}\left(\frac{it-x}{\sqrt{-i2\pi t}}\right)+\frac{e^{x+it/2}}{2}\text{erfc}\left(\frac{it+x}{\sqrt{-i2\pi t}}\right) \end{align} $$

You can recover this formula in Fourier space. However, as Hyperon remarked, your approach was wrong because you are neglecting the contribution of the arc that closes the contour. In your case Jordan's lemma does not apply, and in fact it is precisely this arc that will give you the error function.

Answer 2

Trying to evaluate your integral by using complex integration over a closed contour in the complex plane is doomed to fail (in this case) as the contribution from a large semicircle with radius $R$ will never vanish in the limit $R \to \infty$ because of the presence of the term $\exp(-i K_2 z^2)$ with $K_2 \gt 0$ in the integrand. Parametrizing the large semicircle by $z= R \exp( i \varphi)$ (with $0\le \varphi \le \pi$ for a semicircle chosen in the upper half of the complex plane or $\pi \le \varphi \le 2\pi$ for a semicircle in the lower half plane), the term $$\exp(-i K_2 z^2)= \exp(-i K_2 R^2 \cos 2 \varphi)\exp(K_2 R^2 \sin 2 \varphi)$$ explodes for $K_2 R^2 \sin 2 \varphi\gt 0$, corresponding to $0 \lt \varphi \lt \pi/2 $ (semicircle in the upper half plane) or $\pi \lt \varphi \lt 3\pi/2$ (semicircle in the lower half plane). Note that the subdominant term $\exp(iK_1 z)$ does not play any role in this context. Note also that your integral cannot be expressed in terms of elementary functions (except in the special case $K_2=0$).