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Sokhotskyi-Plemelj formula

The proper method of limiting this point to the real axis is to give the line of integration a small semi-circle bump below the real axis to accommodate the point. One then takes the radius of this semi-circle to zero and achieves the proper integral. It can be shown that the proper form of this integration is given by

$\displaystyle \lim_{y\downarrow 0}\int^{\infty}_{-\infty}\frac{f(x)}{x-x_0-iy}d...
...frac{f(x)}{x-x_0}dx +\lim_{y\downarrow0}\int_S \frac{f(\eta)}{\eta - x_0}d \eta$ (6.1)

The latter is taken over a semicircle which can be shrunk in the limit and thus equals $ \pi i f(x_0)$. Using the postulate of the Dirac delta function, we can write,

$\displaystyle \lim_{y\downarrow 0}\int^{\infty}_{-\infty}\frac{f(x)}{x-x_0-iy}d...
...}_{-\infty}\frac{f(x)}{x-x_0}dx+i\pi \int^{\infty}_{-\infty}\delta(x-x_0)f(x)dx$ (6.2)

This is typically approximated in theoretical physics as,

$\displaystyle \lim_{y\downarrow 0}\frac{1}{x-x_0 - iy} \approx p.v. \frac{1}{x-x_0}+i\pi \delta(x-x_0)$ (6.3)

We note that, with $ \tau=t-t'$, $ d\tau = -dt'$,

$\displaystyle -\int^t_0 \exp(i(\omega-\omega_j)\tau)d\tau = \left(\frac{i}{\omega-\omega_j}\right)[1-\exp(i(\omega-\omega_j)t)]$ (6.4)

The exponential term approximately averages out over the $ \omega$ integration, and we have the approximation,

$\displaystyle \int^t_0 dt' e^{\pm(\omega-\omega_j)(t-t')}=\pi\delta(\omega-\omega_j)\pm iP(\frac{1}{\omega-\omega_j})$ (6.5)

Now looking at the table of calculations on the previous page, we see that many principal value part terms will cancel out, specifically, the B and C terms (i.e. $ \gamma$ cancels $ \gamma^{\dagger}$ whenever all other terms are equal).

For the A D terms, we must do something fancy, that is, use that $ aa^{\dagger}=1+a^{\dagger}a$ to convert all terms to $ a^{\dagger}a$ terms. This will create terms with $ a$ operators, but we notice the symmetry between these terms and that they cancel. We also see the antisymmetry across each row for the A and D terms, but since $ 1+\bar{n}-\bar{n} = 1$, we are left with only one set of terms which retain their principal value part.

We could articulate these details, but we feel that it will profit the reader to examine this by the table presented in this report.

The Dirac delta function will act as we would expect and drop out the $ \omega_j$ terms to $ \omega$, and so we will have our final terms.

next up previous
Next: Result Up: Detailed Derivation of the Previous: The Summation Transform
Timothy Jones 2006-10-11