- #1
Willi Tschau
- 1
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Homework Statement
The story is that I would like to evaluate the coinciding point limit (when ## (x^0, x^1)→(y^0,y^1)##) of these two terms:
\begin{eqnarray*}
&&\frac{1}{2L}e^{\frac{i}{2}eE\left((x ^1)^2-(y^1)^2\right)}\left( im\left( x^0-y^0+ x^1-y^1\right) \right) \sum_{n=0}^{\infty}e^{-ik_n( x^0-y^0+x^1-y^1)}\\
&&\times \frac{1}{L}\int_{-L/2}^{L/2}dx\left( -1\right)^{n}\cos \left( eE\left(\frac{L^2}{4} -x^2\right) +\left( 2n+1\right) \pi \frac{x}{L}\right) \\
&&+\frac{1}{2L}e^{\frac{i}{2}eE\left((x ^1)^2-(y^1)^2\right)}m \sum_{n=0}^{\infty} e^{-ik_n( x^0-y^0+x^1-y^1)}\\
&&\times \frac{1}{L}\int_{-L/2}^{L/2}dx\left(-1 \right)^{n}\left(2x-L \right)\sin\left( eE\left(\frac{L^2}{4}-x^2 \right)+\left(2n+1\right)\pi \frac{x}{L} \right)
\end{eqnarray*}
m, e, E, L are respectively the mass, charge, electric field strength, length of the box, and are constant. The energy levels ##k_n##'s are given by ##k_n=(n+1/2)\frac{\pi}{L}##.
Homework Equations
& 3. The Attempt at a Solution [/B]1. There will be no singularities since another calculation of mine has shown that.
2. When I put ## \frac{1}{L}\int_{-L/2}^{L/2}dx\left( -1\right)^{n}\cos \left( eE\left(\frac{L^2}{4} -x^2\right) +\left( 2n+1\right) \pi \frac{x}{L}\right) ## in Mathematica, I got something that decays like ##\frac{1}{n}##, when ##n## is large.
So I expect the sum will give me ##\log (1-exp...)## (appropriate dissipative term ##\pm i \epsilon## may be needed); of course there will be some residue.
3. My burning question is: is there an analytic way to expand the integral which depends on ##n## as ##\frac{C}{n}+D+...##, for some constant ##C, D##, which are to be found?
Thanks!