Delta Function as a limit

Question

In quantum mechanics we often write expression for Green function which have a form similar to

$$ \frac{1}{\omega-\omega_0-i\eta} $$

with $\eta\to0$. Taking the limit one would expect the expression to reduce to:

$$ P.V.\frac{1}{\omega-\omega_0}+i\pi\delta(\omega-\omega_0) $$

where $P.V.$ is the principal value, an $\delta$ is the Dirac delta.

MMA however doesn't no to take this limit apparently, and when taking the limit $\eta\to0$ it doesn't even return a conditional expression which I assumed it would know to return. In fact:

Limit[1/(x - x0 - I \[Eta]), \[Eta] -> 0]

returns the naive:

1/(x - x0)

completely unaware of the possible problem.

Any suggestions on how to improve this behavior?

Answer 1

The result above used in quantum mechanics or QFT is named Sokhotski-Plemelj formula and only makes sense when associated to a certain space $\mathcal{D}$ of test functions, i.e. its a functional expression over generalized functions $\phi(x)$. So, the equality stated by the formula needs an arbitrary function to be applied.

$\displaystyle \lim_{\eta \to 0} \int_{-\infty}^{\infty} {\frac{\phi(x)\,dx}{x\pm i\eta}} = \mathcal{P}\int_{-\infty}^{\infty} \phi(x)\left ({\frac{1}{x}}\right ) dx\mp i\pi\int_{-\infty}^{\infty} \phi(x)\,\delta(x)\,dx\,,\quad \text{since} \,x=\omega-\omega_0$.

Unless this distribution $\phi(x)$ is defined its hard to MMA understands this functional relation and produce a symbolic answer. For example, if we put $\phi(x)=1$ the code will look like this

Limit[Integrate[1/(\[Omega] - \[Omega]0 + I \[Eta]),
   {\[Omega], -\[Infinity], \[Infinity]}, PrincipalValue -> True,
   Assumptions -> Re[\[Eta]] > 0 && Im[\[Eta]] == 0  && 
    \[Omega]0 \[Element] Reals], \[Eta] -> 0, Direction -> -1]

which results

$-i\pi$

Answer 2

You can get Mathematica to produce the desired result in full generality by using its ability to produce distributions as output of Fourier transforms.

Try the following:

spf = Assuming[ω0 ∈ Reals, 
  InverseFourierTransform[
   Limit[
    FourierTransform[
     1/(ω - ω0 - I η), 
    ω, t],
   η -> 0], 
  t, ω]]

$$\frac{1}{\omega -\text{$\omega $0}}+i \pi \delta (\omega -\text{$\omega $0})$$

This is almost what the theorem says. You just have to make sure that you use this result the right way. In particular, since it's meant to be a distribution, it has to be used in an integral. And this integral then must be treated as a principal-value integral. Fortunately, there is an option for this in Integrate:

Take for example a specific choice of function $\phi(\omega)$ and $\phi_0$ to test the result above:

ϕ[ω_] := Sinc[ω - 1]

Block[{ω0 = 0},
 Integrate[ϕ[ω] spf, {ω, -Infinity, Infinity}, PrincipalValue -> True]]

$\pi -\pi \cos (1)$

The trick here was to add the option PrincipalValue -> True. The combination of FourierTransform and its inverse above sandwiches the Limit in which $\eta\to 0$ is taken. This reflects the fact that the naive limit is wrong, and you're intending for that limit to be taken only after the integration.

As mentioned in the comment, it's also important to specify that $\omega0$ is real: I do that in Assuming. The theorem tacitly assumes that our final integration variable $\omega$ is real, and then the result is true only if the pole is on the real axis as well.