# Question on derivation/notation in a report

## Main Question or Discussion Point

This is from "An Essay on the Theory of Collective Spontaneous Emission" by Gross and Haroche (1982).

I was wondering how exactly they were able to separate the operator D in the Heisenberg representation into positive and negative frequencies with an imaginary exponential on only the left side of $D^{\pm}_j$.

I was also wondering what exactly the PP distribution was referring to. I don't recall reading it defined earlier, and am not entirely sure what it is.

Any help with the above would be greatly appreciated.

## Answers and Replies

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vanhees71
Gold Member
2019 Award
Let's set $c=1$ to simplify things and then regularize the integral by introducing a small negative imaginary part to $k$ then you have the integral
$$I(k-k_0)=\int_0^{\infty} \mathrm{d} \tau \exp[-\mathrm{i} \tau (k-k_0-\mathrm{i} 0^+)]=\frac{1}{\mathrm{i}(k-k_0+\mathrm{i} 0^+)}=-\pi \delta(k-k_0) - \mathrm{i} \text{PP} \frac{1}{k-k_0}.$$

Let's set $c=1$ to simplify things and then regularize the integral by introducing a small negative imaginary part to $k$ then you have the integral
$$I(k-k_0)=\int_0^{\infty} \mathrm{d} \tau \exp[-\mathrm{i} \tau (k-k_0-\mathrm{i} 0^+)]=\frac{1}{\mathrm{i}(k-k_0+\mathrm{i} 0^+)}=-\pi \delta(k-k_0) - \mathrm{i} \text{PP} \frac{1}{k-k_0}.$$
Thank you for the response. I think this is fairly basic, but how exactly did you evaluate

$$\frac{e^{-\mathrm{i} \tau (k-k_0-\mathrm{i} 0^+)}}{-\mathrm{i}(k-k_0 - \mathrm{i} 0^+)}$$

in the limit $\tau \rightarrow \infty$?

And I think this is just stemming from my lack of understanding, but how exactly are the two expressions: $\frac{1}{\mathrm{i}(k-k_0+\mathrm{i} 0^+)}$ and $-\pi \delta(k-k_0) - \mathrm{i} \text{PP} \frac{1}{k-k_0}$ equivalent?

George Jones
Staff Emeritus
Gold Member
vanhees71
Gold Member
2019 Award
Thank you for the response. I think this is fairly basic, but how exactly did you evaluate

$$\frac{e^{-\mathrm{i} \tau (k-k_0-\mathrm{i} 0^+)}}{-\mathrm{i}(k-k_0 - \mathrm{i} 0^+)}$$

in the limit $\tau \rightarrow \infty$?

And I think this is just stemming from my lack of understanding, but how exactly are the two expressions: $\frac{1}{\mathrm{i}(k-k_0+\mathrm{i} 0^+)}$ and $-\pi \delta(k-k_0) - \mathrm{i} \text{PP} \frac{1}{k-k_0}$ equivalent?
Concerning the 1st question: That's the point of introducing the $-\mathrm{i} 0^+$. No matter how small you take it, it makes the exponential vanish in the limit $\tau \rightarrow \infty$. That's how you get the final result as the integral only taken at the lower limit $\tau=0$.

Concerning the 2nd question: Take an analytic test function $\phi(k)$ vanishing quickly enough for (complex!) $k$ anywere. Now we want to evaluate the integral
$$\Phi_{\epsilon}(k_0)=\int_{-\infty}^{\infty} \mathrm{d} k \frac{f(k)}{k-k_0+\mathrm{i} \epsilon}.$$
Now we have
$$\frac{1}{k-k_0+\mathrm{i} \epsilon}=\frac{k-k_0-\mathrm{i} \epsilon}{(k-k_0)^2+\epsilon^2}.$$
Thus we can write the integral as
$$\Phi_{\epsilon}(k_0)=\int_{-\infty}^{\infty} \mathrm{d} k \left [\frac{f(k)}{k-k_0} \; \frac{(k-k_0)^2}{(k-k_0)^2+\epsilon^2}-\frac{\mathrm{i} \epsilon}{(k-k_0)^2+\epsilon^2} f(k) \right].$$
For the first term the contributions from $|k-k_0| \gg \epsilon$ the 2nd factor is $\approx 1$ while for $|k-k_0|^2 \ll \epsilon$ the contribution is negligible. For $\epsilon \rightarrow 0^+$ you get thus from this term the principle value, because the 2nd factor is symmetric around $k=k_0$.

In the 2nd term the factor in the integrand is an approximate $\delta$ function, since
$$\int_{-\infty}^{\infty} \mathrm{d} k \frac{\epsilon}{(k-k_0)^2+\epsilon^2}=\arctan \left (\frac{k-k_0}{\epsilon} \right)_{-\infty}^{\infty}=\pi,$$
and for $\epsilon \rightarrow 0^+$ this factor goes to 0 for $k-k_0 \neq 0$ and to $\infty$ for $k-k_0=0$. So finally we have
$$\lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^{\infty} \mathrm{d} k \frac{f(k)}{k-k_0+\mathrm{i} \epsilon} = \text{PP} \int_{-\infty}^{\infty} \mathrm{d} k \frac{f(k)}{k-k_0} -\mathrm{i} \pi f(k_0),$$
and this we wanted to show.

Concerning the 1st question: That's the point of introducing the −i0+−i0+-\mathrm{i} 0^+. No matter how small you take it, it makes the exponential vanish in the limit τ→∞τ→∞\tau \rightarrow \infty. That's how you get the final result as the integral only taken at the lower limit τ=0τ=0\tau=0.
Thank you for the explanation. That was incredibly helpful! I think I'm just missing this initial part: how is it valid to introduce $\mathrm{i} 0^+$ into the expression? If both both limits $\tau \rightarrow \infty$ and $0^+ \rightarrow 0$ are necessary to evaluate the original integral, we just have to ensure $\tau$ is increasing more quickly than $0^+$ is decreasing in order to be able to evaluate the integral only at $\tau = 0$, correct? Regularization of the integral is not something I am too familiar with, but thank you for sharing.

vanhees71
This is just a short-hand notation. Instead of introducing $\mathrm{i} 0^+$ introduce $\mathrm{i} \epsilon$ with $\epsilon>0$. Then you have to integrate first over $\tau \in [0,\infty)$ and then take the limit $\epsilon \rightarrow 0^+$. This limit has to be understood in the weak sense, i.e., the result of the integral is a distribution which can be seen as a functional on the space of test functions (e.g., the $C^{\infty}$ functions with compact support), i.e., you multiply the expression with a finite $\epsilon>0$ with a test function, take the integral and then take $\epsilon \rightarrow 0^+$. For many cases one does this once for all test functions like the Dirac $\delta$ distribution or the expression $1/(k-k_0+\mathrm{i} 0^+)$ which stands for a distribution defined in this kind as a "weak limit" of an integral to be taken together with an arbitrary test function.