Is there any closed formula for the time-dependent gaussian wave packet?

$$\Psi(x,0) = \frac{1}{ \sqrt[4]{ 2 \pi {\sigma_x}^2 }} \exp \left( i k_0 x - \frac{{\left( x - x_0 \right)}^2}{4 {\left(\sigma_x \right)}^2} \right)$$

$$\Psi(x,t) = ?$$

I found some formulas for the case when $x_0=0$

There is also these:

But I am struggling in the algebra for including the $x_0$ in them.

Maybe I should go through all those Fourier integrals, but it would be a nice shortcut if someone could give a final formula.


My goal is to use it for validating numerical results.


Given a wavefunction, $\psi(x,0)$, the corresponding quantity, $\psi(x,t)$, is obtained from the integral: $$ \boxed{\psi(x,t) = \int_{-\infty}^{\infty}dy\,Z[x,t;y,0]\psi(y,0)\,} $$ where $Z[x,t;y,0]$ is a path integral, $\int_{x,0}^{y,t} \mathcal{D}x\,e^{\frac{i}{\hbar}I[x]}$, and $I[x]$ the single particle action of interest. For free non-relativistic particles, $I[x]=\int_0^td\tau\frac{m}{2}(\frac{dx}{d\tau})^2$, and the correctly normalised path integral reads: $$ Z[x,t;y,0] = \sqrt{\frac{m}{2\pi i\hbar t}}\,\exp\Big(i\frac{m(x-y)^2}{2\hbar t}\Big). $$ The initial wave packet is of the form: $$ \psi(y,0)=Ne^{\frac{i}{\hbar}ky}e^{-\frac{1}{2}b(y-x_0)^2}, $$ where you can read off the $b,N$ of interest from your formula, e.g. $b=1/(2\sigma_x^2)$.

The objective is to compute $\psi(x,t)$, which amounts to carrying out the $y$ integral in the above boxed formula. This is a Gaussian and so can easily be carried out. You'll primarily need the result: $$ \int_{-\infty}^{\infty} da\,e^{iza^2} = \sqrt{\frac{\pi i}{z}},\quad {\rm if }\quad {\rm Im}\,z\geq0,\quad z\neq0, $$ which follows from Cauchy's theorem applied to $\oint_C da\,e^{iza^2}=0$, where the contour $C$ is that in the figure: C contour It is convenient to write $z=\rho e^{i\phi}$ (with $0\leq \phi<\pi$), and then given any allowed $\phi$ you choose $\theta$ such that $\phi+2\theta=\pi/2$. This enables you to take $R\rightarrow \infty$ while ensuring the angular contour contributions vanish, and you are left with $\int_{-\infty}^{\infty}da\,e^{iza^2}=\sqrt{i}e^{-i\phi/2}\int_{-\infty}^{\infty}da\,e^{-\rho a^2}$. Evaluating the remaining Gaussian integral and rewriting the result in terms of $z=\rho e^{i\theta}$ leads to the displayed result.

The next step is to redefine your integration variable, $a\rightarrow a'=a-b$ (with $b\in \mathbb{R}$), $$ \int_{-\infty}^{\infty} da\,e^{iz(a+b)^2} = \sqrt{\frac{\pi i}{z}}, $$ which implies: $$ \int_{-\infty}^{\infty} da\,e^{iza^2+2izba} = \sqrt{\frac{\pi i}{z}}\,e^{-izb^2},\quad {\rm if }\quad {\rm Im}\,z\geq0,\quad z\neq0. $$ We are almost there. The only thing remaining to do is to notice that both the left- and right-hand sides of this last expression are analytic in $b$, so we can extend $b$ into the complex plane, $b\rightarrow w=b+ib'$. Finally, we redefine $w\rightarrow zw$. Therefore, for generic complex numbers $z$ (with ${\rm Im}\,z\geq0$, $z\neq0$) and $w$, $$ \boxed{\,\int_{-\infty}^{\infty} da\,e^{iza^2+2iwa} = \sqrt{\frac{\pi i}{z}}\,e^{-iw^2/z},\quad {\rm if }\quad z,w\in \mathbb{C},\quad {\rm and}\quad{\rm Im}\,z\geq0,\quad z\neq0\,\,\,} $$ This is all you need in order to evaluate the above integral that gives you $\psi(x,t)$. You just read off $z,w$ and $a$ as appropriate for your case of interest and apply this result. (Of course the same formula applies for general $x_0$.) The result is: $$ \boxed{\psi(x,t) = \frac{1}{\sqrt{1+i\frac{\hbar b }{m}t}}\exp\Bigg\{\!\!-\frac{i}{\hbar}\frac{k^2t}{2m}\frac{\big[1+i\frac{\hbar b}{k}(x-x_0)\big]^2}{1+i\frac{\hbar b}{m}\,t}\Bigg\}\psi(x,0)\,\,} $$ Recall that $ \psi(x,0)=Ne^{\frac{i}{\hbar}kx}e^{-\frac{1}{2}b(x-x_0)^2}. $

  • $\begingroup$ Wakabaloola your answer is awesome, thank you so much. It is a way more than I need. Thanks. $\endgroup$ – Thiago Melo Feb 16 '18 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.