Weird integration of gaussian wave packet

I have been learning Fourier transformation of a gaussian wave packet and i don't know how to calculate this integral: In the above integral we try to calculate $\varphi(\alpha)$ where $\alpha$ is a standard deviation, $\alpha^2$ is variance, $x'$ is average for $x$, $p'$ is average for $p$ and:

$$\psi_\alpha = \frac{1}{\sqrt{\sqrt{\pi} \alpha}} \exp \left[ - \frac{(x-x')}{2 \alpha^2} \right]$$

For some reason author of this derivation swaps $p$ with $(p - p')$ (red color) and from $=$ sign (yellow color) forward i am completely lost. Could anyone please explain why did author did what he did? It is weird...

• This question has been asked before in various formats. Answers have been offered there. To do the Gaussian integral shown in the second line of your text, you need to: (i) complete the square so that you get rid of the linear term in the exponent, (ii) Do a shift in your integration variable (which has no effect due to the -infity and +infinity. Take extra care to do the algebra carefully. the extra term must have come from $\psi_\alpha$, as has been said by Lubos. – JKL Mar 3 '13 at 10:52
• Could you please provide any link from where i can read anything about the "more general gaussian". – 71GA Mar 3 '13 at 14:59

I am sure that the author has simply considered a more general Gaussian than yours $$\psi_\alpha = \frac{1}{\sqrt{\sqrt{\pi} \alpha}} \exp \left[ - \frac{(x-x')}{2 \alpha^2} \right] \exp(ip'x/\hbar)$$ which is "shifted" in the momentum direction by $p'$ (it still minimizes the uncertainty relationship's product) and that's the reason why $p$ was "replaced" by $p-p'$.
The last step of the calculation is a simple Gaussian integral. One completes the square by redefining $x\to x+x_S$ for such $x_S$ that the linear terms in $x_S$ get cancelled. This produces some multiple of $x_S^2$ in the exponent – that's in the result – while the rest is calculated via the simple $\int \exp(-K x^2)=\sqrt{\pi/K}$. Note that due to various properties of the complex functions, these operations work even for a complex $x_S$.
• Hi, they appear from completing the squares. Let me be more specific. If the integrand has the exponential of $A(x^2+Bx)$ for some $A,B$, you may write it as $A(x+B/2)^2 - AB^2/4$, please check it. The integral over $x$ is then done by redefining $x = x_N-B/2$ and yields $\sqrt{\pi/-A}$ but the exponential $\exp(-AB^2/4)$ is still there. You must make this exercise yourself if you want to understand it, it's simple but not vacuous. – Luboš Motl Mar 3 '13 at 21:08
• Sorry, I think that what separates your result from the right result are just some elementary errors in the algebra like missing prefactors etc. Also, it's not clear to me why you deliberately changed the notation from $\alpha$ to $\sigma_x$ - is it really the same thing or did you add some prefactors? Your calculation is very incomprehensible for such a simple thing. Can't you just use a symbolic software (Mathematica) to evaluate such integrals if you can't do it right on the paper? – Luboš Motl Mar 5 '13 at 7:58