Understanding Borel transform (sum) in QM

  • Thread starter Karlisbad
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  • #1
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Hello, i have heard tat you can use Borel transform in QM so the divergent perturbation series become convergent..:confused: :confused: i am a bit confused since if you must compute the Borel transform:

[tex] B(x)=\sum_{n=0}^{\infty}\frac{a(n)x^{n}}{n!} [/tex]

But this can only be made for a few cases [tex] a(n)=(-1)^{n} [/tex] and
[tex] a(n)=(-1)^{n} n! [/tex]

But if the a(n) don't follow a known pattern Borel Transform is useless..:grumpy: :grumpy:
 

Answers and Replies

  • #2
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sorry the first "image" (i really hate this f... Latex :mad: :mad:) should be:

[tex] \sum_{n=0}^{\infty}(a(n) x^{n})/n! [/tex]

-the second: [tex]a(n)= (-1)^{n} n! [/tex]

- And the third [tex]a(n)= (-1)^{n} [/tex]
 

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