# Non-equivalence between $\omega \to \omega \pm i\varepsilon$ and Cauchy principle value

I am looking to gain a more rigorous and deeper understanding as to how an $i\varepsilon$ prescription actually changes the end result of a divergent integral, specifically in regards to Green's functions. I understand the use of $i\varepsilon$ allows one to impose desired boundary conditions. A very helpful Phys.SE answer here suggests that we should think of the $i\varepsilon$ as a contribution due to damping, which all real systems possess. At the end, we take the limit of $\varepsilon \to 0$. Yet I am puzzled by this fact. It seems that if we do not inlcude $i\varepsilon$ at the offset, we will not obtain the same answer (see example below). Both the principle value and the $i\varepsilon$ method both serve as valid Green's functions (with different B.C.'s I suppose), but I am under the impression that the $i\varepsilon$ is fundamentally what we are after.

The closest answer that I have come across comes from Shankar's Prinicples of Quantum Mechanics, which in regards to inverting the Helmholtz operator, he says we must consider a slightly different different operator $\nabla^2 +k^2 +i\varepsilon$, where the eigenfunctions are plane waves of complex wavenumber. He goes on to say that such functions are not part of the space we are restricting oursleves to, namely, the space of functions normalized to unity or the Dirac delta function. Thus $\nabla^2 +k^2 +i\varepsilon$ may be inverted within the physical Hilbert space. And then $i\varepsilon$ is sent to zero at the end of the calculation.

I am curious as to how, by working with functions outside our restricted space, we may thus appropriately invert an operator? I believe this sort of generalization to complex operators (and their associated space they act on) is on the right track.

Here is a basic example on the inequivalence between $i\varepsilon$ and the Cauchy principle value. I have found this result extends to the case with two poles on the real axis.

If we start with a function such as $$f(x) = \int_{-\infty}^\infty dk\ \frac{e^{ik x}}{k}\quad\text{for}\quad x>0$$ and seek to assign a non-divergent value, we may find the principle value by integrating over a contour of $$\text{P.V}[ f(x)] + \int_\delta + \int_{arc} = \oint.$$ Here $\text{P.V}$ is the principle value, $\delta$ represents a small semicircle contribution around the pole, and $arc$ is the contribution from the arc at $+\infty$. With Jordan's lemma, we close above and thus the $arc$ at infinity makes no contribution. We are then left with $$\text{P.V}[ f(x)] = \oint - \int_\delta = \pm i\pi a_{-1} +2\pi i\sum_{res}.$$ Here $a_{-1}$ represents the first term of a Laurent expansion (this is contribution from $\int_\delta$). This term is positive (negative) for a clockwise (counter-clockwise) semi-circle of vanishing radius $\delta$.

To see that principle value is unique, we can make a detour around the pole either clockwise (in which we do enclose the pole) or counter clockwise (no pole enclosed, no residue). Either way $$\text{P.V}[ f(x)] = i\pi.$$

Okay, now lets use the idea of pushing the pole upwards or downwards by $$k \to k\pm i\varepsilon$$ and take the limit as $\varepsilon \to 0$. We may now carry the integration across the real line unhindered and we find, when closing above as we must, $$\lim_{\varepsilon\to 0} \int_{-\infty}^\infty dk\ \frac{e^{i(k\pm i\varepsilon) x}}{k\pm i\varepsilon} = \begin{cases} 0&\quad\text{for}\quad k \to k+ i\varepsilon \\ 2\pi i&\quad\text{for}\quad k \to k-i\varepsilon \end{cases}$$ This last part of obtaining $0$ is useful for retarded/advanced Green's functions.

• This is a long post, but at the end all I take away is that you are struggling with Green's functions/contour integrals/the $\mathrm{i}\epsilon$ prescription. What is the actual question you want answered? – ACuriousMind Feb 19 '16 at 22:35
• @ACuriousMind The question I want answered is denoted by a question mark at the end. – Lone Wolf Feb 19 '16 at 22:39
• You mean "I am curious as to how, by working with functions outside our restricted space, we may thus appropriately invert an operator?"? But...that's precisely what you're doing when you're computing a Green's function - integrating $f$ against the Green's function $G$ of an operator $O$ gives a function $F$ such that $OF = f$. Note that the Green's "function" is often rather a distribution, i.e. not inside your function space. What do you want to know when you ask "how"? – ACuriousMind Feb 19 '16 at 22:46
• A Hilbert space is, by construction, complete, so if you claim that you are doing all of this over a Hilbert space, then the limit lives inside it. If you find yourself constructing something that did not live within the space that you think you are working on, then it's not a Hilbert space by definition. No? A Hilbert space, by the way, is a mathematical, not a physical object. A stone, a metal rod or a magnetic field, those are physical objects and they do not live in your Hilbert space. Be careful with your basic ontology here. All of this is an idealizing description. – CuriousOne Feb 19 '16 at 22:57
• @CuriousOne Is it only possible to obtain the "appropriate" Green's function within say, our Hilbert space, by extending our linear operator outside the Hilbert space, inverting, and then taking a limit such that it does lie within our Hilbert space? – Lone Wolf Feb 19 '16 at 23:14