Complex Substitution and Infinity in Quantum Mechanics Integrals

prismaticcore
Messages
4
Reaction score
0

Homework Statement


In Griffiths' Introduction to Quantum Mechanics problem 2.22 as well as 6.7, I used substitution to complete an integral. The original integral had limits from negative infinity to positive infinity. For my substitution, I had a complex constant term added to the original variable. In computing the new limits of the integral after substitution, I must somehow add a complex number to infinity. Does this imply that the new limits are also negative infinity to infinity? Also, I haven't had analysis nor complex analysis and so I am unsure as to how to appropriately phrase what is going on in computing these new limits.


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
The new integral is over a line shifted by a complex constant from the original line. IF there are no poles in the function between the original line and new line and IF the functions are 'well behaved' at infinity, i.e. go to zero fast enough, then you can argue using the Cauchy Integral theorem that the integral over both lines are equal. Given this is a quantum mechanics problem and not a complex analysis problem I suspect both IF's are probably true. So, yes, you can do that.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
9
Views
2K
Replies
3
Views
1K
Replies
3
Views
2K
Replies
13
Views
2K
Replies
11
Views
2K
Replies
4
Views
941
Back
Top