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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

$$-\frac{(x-\frac{\hbar\kappa_0}{m}t)^2}{(a^2+\frac{2i\hbar t}{m})}$$

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    $\begingroup$ WP. $\endgroup$ Dec 26, 2023 at 3:32
  • $\begingroup$ The polar form of $\alpha^2$ is incorrect. The correct form will give you the $e^{-i\theta}$ factor. $\endgroup$
    – Ghoster
    Dec 26, 2023 at 6:15
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    $\begingroup$ You haven’t explained whether you understand what the integral over $\kappa$ gives. $\endgroup$
    – Ghoster
    Dec 26, 2023 at 6:24
  • $\begingroup$ Why do you write $\kappa$ and $\kappa_0$ when the textbook writes $k$ and $k_0$? $\endgroup$
    – Ghoster
    Dec 26, 2023 at 6:27

1 Answer 1

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I think that one step is to rewrite the term $−\left(\frac{a^2}4+i\frac{\hslash t}{2m}\right)$ as a complex number, but I do not understand how that solves the problem.

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: $$ \begin{align} \tag1-\frac{a^2k_0^2}4+\frac{\left(\frac{k_0a^2}2+ix\right)^2}{a^2+\frac{2i\hslash t}m}&=\frac{-\frac{k_0^2a^4}4-\frac{i\hslash k_0^2a^2t}{2m}+\frac{k_0^2a^4}4+ik_0a^2x-x^2}{a^2+\frac{2i\hslash t}m}\\ \tag2&=\frac{ik_0a^2\left(x-\frac{\hslash k_0}{2m}t\right)-x^2}{a^2+\frac{2i\hslash t}m}\\ \tag3&=\frac{ik_0\left[\left(a^2+\frac{2i\hslash t}m\right)-\frac{2i\hslash t}m\right]\left(x-\frac{\hslash k_0}{2m}t\right)-x^2}{a^2+\frac{2i\hslash t}m}\\ \tag4&=ik_0x-i\frac{\hslash k_0^2}{2m}t-\frac{x^2-ik_0\left(-\frac{2i\hslash t}m\right)\left(x-\frac{\hslash k_0}{2m}t\right)}{a^2+\frac{2i\hslash t}m}\\ \tag5&=ik_0x-i\frac{\hslash k_0^2}{2m}t-\frac{x^2-2\left(\frac{\hslash k_0}mt\right)x+\left(\frac{\hslash k_0}mt\right)^2}{a^2+\frac{2i\hslash t}m}\\ \tag6&=ik_0x-i\frac{\hslash k_0^2}{2m}t-\frac{\left(x-\frac{\hslash k_0}mt\right)^2}{a^2+\frac{2i\hslash t}m} \end {align} $$

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  • $\begingroup$ Well, I'm glad that finally someone gave me a real answer. I was trying to obtain the phase by myself and I have this. $z=re^{i\theta}$ with $r=( (a^2/4)^2+(\hbar t/2m)^2 )^{1/2}$. So $\cos\theta=a^2/4r$ and $\sin\theta=\hbar t/2mr$ so $\theta=arctan(2\hbar t/ma^2)$. But is not the same $\theta$ that Cohen has. $\endgroup$
    – user353399
    Dec 26, 2023 at 11:38
  • $\begingroup$ I would have to disagree; I think both Cosmas and Ghoster gave good enough answers as comments, just not fully fleshed out. As for the angle, that is correct; Remember, you are defining $\alpha^2=r^2e^{2i\vartheta}$, so, really, you got what Cohen has. It is simply convenient to define it as double angle because we are going to have to take the conventional complex square root later. $\endgroup$ Dec 27, 2023 at 1:55
  • $\begingroup$ Ooo I did not notice that, thx. $\endgroup$
    – user353399
    Dec 27, 2023 at 10:22
  • $\begingroup$ Regarding to rewriting this term $-\frac{a^2\kappa_0^2}{4}+\frac{(\frac{\kappa_0a^2}{2}+ix)^2}{(a^2+\frac{2i\hbar t}{m})}$ as this one $-\frac{x^2-\kappa a^2ix+\frac{a^2\kappa_0^2\hbar t}{2m}}{(a^2+\frac{2i\hbar t}{m})}$ Do you have any suggestions on how can I obtain what Cohen has?. Sorry if the questions are too silly but I have tried a lot to solve this problem, but I can´t. The term that I refer to is in my original question. $\endgroup$
    – user353399
    Dec 27, 2023 at 10:22
  • $\begingroup$ Yes, I thought you might find this difficult. I had actually worked everything out before typing the original answer. Just by carefully brute-forcing the computation. It just is possible to work this out in the end. Note that you are actually wrong in the thinking. The original constant terms will combine to give you something in $\phi$, and then also the $e^{ik_0x}$ term, and then the term that you want to get. It is not that simple. I'm going to give you some time to work it out (mostly because I am busy right now). Ask me again if you have more difficulties. $\endgroup$ Dec 27, 2023 at 10:59

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