Completing the square for integration of e^( )

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SUMMARY

The discussion centers on solving the integral ∫e^{-(a*x^2 + b*x)} using the method of completing the square, as outlined in Griffith's "Intro to Quantum Mechanics." The integral can be transformed by defining a new variable y, allowing the expression to be rewritten as e^{-k^2(y^2 - b^2/(4a))}. This leads to separating the exponential and extracting constants, ultimately resulting in the integral ∫e^{-k^2y^2}dy, which is recognized as the Error function (erf(x)). The key takeaway is the importance of variable substitution in simplifying complex integrals.

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Homework Statement


I'm working on problem 2.22 from Griffith's Intro. to Quantum Mechanics (a free particle problem). I am stuck on the final integral from part b. Part a of the problem is normalizing:
A*e-a*x2 which I did. Part b wants the general, time-dependent wave function.

Homework Equations


Griffith says that integrals of the form
∫e-(a*x2+b*x) can be solved by "completing the square." Griffith gives the example of defining y to be (a)(1/2)*(x+(b/(2a)) and using that definition to convert (a*x2+b*x) to y2 - (b2/(4*a)) I presume we are supposed to substitute in the converted expression into the integral and work from there.


The Attempt at a Solution


I have the following integral:
∫e-k2/(4*a)+i*(k*x-(h*k2)/(2*m)*t)
I presume using the "completing the square" method is supposed to make this integral work out; but I don't see how this will help. If I do convert this into something like:
ey + (something about a, h, and m),
how do I integrate with y?

Thanks,
Vance
 
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The exponential, e^{-k^2(y^2- b^2/4a}) can be separated as e^{-k^2y^2}e^{k^2b^2/4} and then the constant, e^{k^2b^2/4a} can be taken out of the integral, leaving \int e^{-k^2y^2}dy. That integral cannot be done in terms of 'elementary' functions (polynomials, fractions, trig functions, exponential, logarithms). That integral is defined to be the "Error function" or "erf(x)".
 
Thank-you so much. The missing part was the change of variables.
Vance
 

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