I am currently studying quantum mechanics and I am struggling to obtain the time evolution of a Gaussian wave packet.
We know that the wave function of a free particle is:
$$\Psi(x,t)=\frac{\sqrt{a}}{(2\pi)^{3/4}}\int e^{-\frac{a^2}{4}(\kappa-\kappa_0)^2}e^{i(\kappa x-\omega t)}d\kappa$$
with $\omega=\frac{\hbar\kappa^2}{2m}$.
Cohen just shows us the result without mentioning how it is performed, so trying to obtain what Cohen obtains is where my question arises.
The exponent of the wave packet is expressed as:
$$-\frac{a^2}{4}(\kappa-\kappa_0)^2+i\kappa x – i\hbar\kappa^2t/2m.$$
Utilizing the completing square method, we can represent it as:
$$-\bigg(\frac{a^2}{4}+\frac{i\hbar t}{2m}\bigg)\bigg[\kappa – \frac{(\frac{\kappa_0a^2}{2}+ix)}{(\frac{a^2}{2}+\frac{i\hbar t}{m})}\bigg]^2-\frac{a^2\kappa_0^2}{4}+\frac{(\frac{\kappa_0a^2}{2}+ix)^2}{(a^2+\frac{2i\hbar t}{m})}$$
I think that one step is to rewrite the term $ -\bigg(\frac{a^2}{4}+\frac{i\hbar t}{2m}\bigg)$ as a complex number, but I do not understand how that solves the problem.
My physicist teacher does this:
Let´s rename $\alpha^2=\bigg(\frac{a^2}{4}+\frac{i\hbar t}{2m}\bigg)$
so we can write
$$\alpha^2=\bigg((\frac{a^2}{4})^2+(\frac{\hbar t}{2m})^2\bigg)^{1/2}e^{2i(\frac{2\hbar t}{ma^2})}$$
At this point, I'm uncertain about the subsequent steps in the calculations. I do not understand why rewriting that term as an exponential, helps me to obtain what Cohen obtains.
For the sake of the argument, the result that I am trying to obtain is the following:
$$
\Psi(x,t)=\bigg(\frac{2a^2}{\pi}\bigg)^{1/4}\frac{e^{i\phi}}{(a^4+\frac{4\hbar^2t^2}{m^2})^{1/4}}e^{i\kappa_0 x}exp\bigg\{-\frac{(x-\frac{\hbar\kappa_0}{m}t)^2}{(a^2+\frac{2i\hbar t}{m})}\bigg\}
$$
where $\phi$ is real and independent of x:
$$\phi=-\theta-\frac{\hbar\kappa_0^2}{2m}t$$
with $\tan 2\theta=\frac{2\hbar t}{ma^2}$
Note:
I have read similar questions like this Time evolution of a Gaussian packet but none of them seem to approach the problem as Claud Cohen-Tannoudji's Volume 1 Compliment G1 page 59.
And I think that now I do not have big mistakes in this question.
Trying to solve the problem I tried to rewrite this term :
$$-\frac{a^2\kappa_0^2}{4}+\frac{(\frac{\kappa_0a^2}{2}+ix)^2}{(a^2+\frac{2i\hbar t}{m})}(1)$$
like this:
$$-\frac{x^2-\kappa a^2ix+\frac{a^2\kappa_0^2\hbar t}{2m}}{(a^2+\frac{2i\hbar t}{m})}$$
but I do not see how I can end up having this term
$$-\frac{(x-\frac{\hbar\kappa_0}{m}t)^2}{(a^2+\frac{2i\hbar t}{m})}$$
Best Answer
This trick works because of an important observation that is left unstated. Choosing symbols to avoid what has been used in your question statement, you should note that the famous Gaußian integral tells us that $$\int_{-\infty}^{+\infty}e^{-\beta^2(x+\gamma)^2}\mathrm dx=\beta^{-1}\sqrt\pi$$ for real values of $\beta$ and $\gamma$.
The part that is particularly unstated, is that this result holds true for complex values of $\beta$ and $\gamma$ too. This is very much a non-trivial argument. The reason why this extension works for complex values of $\beta$ is due to the same kind of contour integration trick as needed to extract the Fresnel integrals from the Gaussian integral. Finally, a constant shift in the complex plane tells us that it works for all complex values of $\gamma$
We are not going to give you the correct form of $\alpha^2$; your mistake is just in the complex phase, and it would give you $\theta$ if you did it correctly.
Your completing the square is correct. You are very close to your required final answer. Just do more work on the constant terms. Everything is going to work out.
The algebraic conversion of the constant terms goes as follows: $$ \begin{align} \tag1-\frac{a^2k_0^2}4+\frac{\left(\frac{k_0a^2}2+ix\right)^2}{a^2+\frac{2i\hslash t}m}&=\frac{-\frac{k_0^2a^4}4-\frac{i\hslash k_0^2a^2t}{2m}+\frac{k_0^2a^4}4+ik_0a^2x-x^2}{a^2+\frac{2i\hslash t}m}\\ \tag2&=\frac{ik_0a^2\left(x-\frac{\hslash k_0}{2m}t\right)-x^2}{a^2+\frac{2i\hslash t}m}\\ \tag3&=\frac{ik_0\left[\left(a^2+\frac{2i\hslash t}m\right)-\frac{2i\hslash t}m\right]\left(x-\frac{\hslash k_0}{2m}t\right)-x^2}{a^2+\frac{2i\hslash t}m}\\ \tag4&=ik_0x-i\frac{\hslash k_0^2}{2m}t-\frac{x^2-ik_0\left(-\frac{2i\hslash t}m\right)\left(x-\frac{\hslash k_0}{2m}t\right)}{a^2+\frac{2i\hslash t}m}\\ \tag5&=ik_0x-i\frac{\hslash k_0^2}{2m}t-\frac{x^2-2\left(\frac{\hslash k_0}mt\right)x+\left(\frac{\hslash k_0}mt\right)^2}{a^2+\frac{2i\hslash t}m}\\ \tag6&=ik_0x-i\frac{\hslash k_0^2}{2m}t-\frac{\left(x-\frac{\hslash k_0}mt\right)^2}{a^2+\frac{2i\hslash t}m} \end {align} $$