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In chapter 5 section 9 of Sakurai, 2nd edition, he uses some notation that I am unfamiliar with. This may be suited for Math.se but I figured it could be peculiar physicist notation. Anyways it is equation 5.9.14 and states:

$$\lim_{\varepsilon\to 0}\frac{1}{x+i\varepsilon}={\rm Pr}.\frac1x -i\pi\delta(x).\tag{5.9.14}$$

Could someone please explain what's going on with the Pr./what it means? It seems like it might be some sort of principal value...like the Cauchy principal value.

EDIT

The section is on energy shifts and decay widths from the chapter on perturbation theory. This equation basically arises while doing an expansion on the energy corrections to second order. The second order energy shift is a sum over terms that look like:

$$\lim_{\epsilon\to 0}\frac{1}{x+i\epsilon}.$$

So he does that little trick above to separate the real and imaginary parts of the energy correction.

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This is notation from Distribution Theory in Functional Analysis. The theory of distributions is meant to make things like the Dirac Delta rigorous.

In this context, just to give you one overview, a distribution is a functional on the space of test functions. We define the space of test functions over $\mathbb{R}$ as $\mathcal{D}(\mathbb{R})$ being the space of smooth functions with compact support (that is, the set where they are not zero is bounded and closed).

In that case, the space of distributions is the space of continuous linear functionals over $\mathcal{D}(\mathbb{R})$ and is denoted as $\mathcal{D}'(\mathbb{R})$. If $\eta\in \mathcal{D}'(\mathbb{R})$ and $\phi\in \mathcal{D}(\mathbb{R})$ we usually denote $\eta(\phi)$ by $(\eta,\phi)$. Since distributions are just linear functionals, we say that two distributions $\eta,\zeta$ are equal if $(\eta,\phi)=(\zeta,\phi)$ for all $\phi\in \mathcal{D}(\mathbb{R})$.

The Dirac Delta, for instance, is defined as $\delta\in \mathcal{D}'(\mathbb{R})$ whose action on $\phi\in \mathcal{D}(\mathbb{R})$ is $(\delta,\phi)=\phi(0)$. Now, given $\phi\in\mathcal{D}(\mathbb{R})$ one can always build a distribution associated with it:

$$(\phi,\psi)=\int_{-\infty}^{\infty}\phi(x)\psi(x)dx, \qquad \forall \ \psi\in \mathcal{D}(\mathbb{R}).$$

There are other ways, though, to make one usual function into a distribution, even if the function is not a test function. One of them is the principal value. Consider $f(x) = \frac{1}{x}$. This obviously doesn't have compact support, so $f\notin \mathcal{D}(\mathbb{R})$. We can make $f$ into a distribution, though, by considering the principal value:

$$\left(\operatorname{Pv}\frac{1}{x},\phi\right)=\lim_{\epsilon\to 0^+}\left(\int_{-\infty}^{-\epsilon}\dfrac{\phi(x)}{x}dx+\int_\epsilon^\infty \dfrac{\phi(x)}{x}dx\right).$$

This is what the book means by $\operatorname{Pr}$.

Now, the formula you state is the Sokhotski–Plemelj formula. It should be read in the distributional sense. Saying that:

$$\lim_{\epsilon\to 0}\frac{1}{x+i\epsilon}=\operatorname{Pr}\frac1x -i\pi\delta(x).$$

Really means that for all $\phi\in \mathcal{D}(\mathbb{R})$ we have

$$\lim_{\epsilon\to 0}\left(\frac{1}{x+i\epsilon},\phi\right)=\left(\operatorname{Pr}\frac1x,\phi\right) -i\pi\left(\delta(x),\phi\right),$$

where

$$\left(\frac{1}{x+i\epsilon},\phi\right)=\int_{-\infty}^{\infty}\dfrac{\phi(x)}{x+i\epsilon}dx.$$

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This is not a peculiar physicist notation oddly enough. The notation allows one to interpret $1/x$ as a distribution (which makes sense since it's being added to the delta distribution on the right hand side of the equation). For a suitable test function $\varphi$, one defines this distribution as $$ \mathrm{pv}(1/x)(\varphi) = \lim_{\epsilon\to 0+}\int_{\mathbb R\setminus [-\epsilon, \epsilon]} \frac{\varphi(x)}{x}\, dx $$ As noted by user anon0909, this distribution is called the principal value of $1/x$. Given this definition, the equation that appears in Sakurai should be interpreted as $$ \lim_{\epsilon\to 0^+} \int_{-\infty}^{\infty} \frac{\varphi(x)}{x + i\epsilon} = \lim_{\epsilon\to 0+}\int_{\mathbb R\setminus [-\epsilon, \epsilon]} \frac{\varphi(x)}{x}\, dx - i\pi\varphi(0) $$

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    $\begingroup$ Ah cool so it like you integrated that little chunk around $0$ by bringing the contour into the upper half plane? $\endgroup$ – ClassicStyle May 9 '16 at 1:45
  • $\begingroup$ @TylerHG It seems like you're on the right track trying to get intuition from complex analysis, but I don't quite follow what you mean. $\endgroup$ – joshphysics May 9 '16 at 2:04
  • $\begingroup$ Here's what it looks like to me. You broke the integral up into three pieces. Two that extend to infinity and exclude zero (the principal value part) and a third which is integrated in a $\delta$ neighborhood of $0$. Then looking at the integral around $0$, bc $x$ is small we can Taylor expand $\phi$ and just be left with an integral around $0$ of $\phi (0)$ over the denominator (the first order terms vanishes I believe). Then if we move that little contour to the upper half plane then from the residue theorem we get a $-\pi i \phi (0)$ contribution to the integral...seems valid? $\endgroup$ – ClassicStyle May 9 '16 at 2:15
  • $\begingroup$ @TylerHG Hmm. I don't see how that works. For one thing, the function $\varphi$ isn't defined away from the real line, so I'm not sure how one would integrate it over some contour impinging on the upper-half plane. Also, what would allow you to deform that little piece in the first place? $\endgroup$ – joshphysics May 9 '16 at 2:26
  • $\begingroup$ @TylerHG By the way, I believe the standard way to prove the identity is to multiply the numerator and denominator by $x-i\epsilon$, and then use identities like the one I ask about here: math.stackexchange.com/questions/1156854/… $\endgroup$ – joshphysics May 9 '16 at 2:32

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