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I am reading up about Quantum Field theory and the integral of the following function pops up: $$\frac{1}{\sqrt{z^2 + m^2}}$$The details are actually explained in this question.

The book says that there are two branch cuts, $[-\infty, -im]$ and $[im, +\infty]$.

Now:

1) Since I can write $\frac{1}{\sqrt{z^2 + m^2}} = \frac{1}{\sqrt{z + im}\sqrt{z-im}}$, which branch cut corresponds to which $\sqrt{z\pm im}$ ?

2) Is the branch cut of $\sqrt{z - im}$, $[-\infty, im]$ or $[im, +\infty]$?

3) If it were possible to choose the branch cuts for $\sqrt{z - im}$ to be $[-\infty, im]$ , and for $\sqrt{z + im}$ to be $[-im, +\infty]$, how do I treat the interval $[-im, im]$?

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I didn't look at the link post. I am going off the information giving that the function is $\frac{1}{\sqrt{z^2+m^2}}$.


I don't see why infinite branch cuts are used when we can use a single finite branch cut. If we let $z_1 = z - im = r_1e^{i\theta_1}$ and $z_2 = z + im=r_2e^{i\theta_2}$, we have $$ \frac{1}{\sqrt{z^2+m^2}} = \frac{1}{\sqrt{r_1r_2}\exp(i(\theta_1+\theta_2)/2)} $$ If we wind around each branch point separately, the function is multi-valued since we have $$ \frac{1}{\sqrt{r_1r_2}\exp(i(\theta_1+\theta_2+2\pi)/2)} = -\frac{1}{\sqrt{r_1r_2}\exp(i(\theta_1+\theta_2)/2)} $$ If we wind around both, we have $$ \frac{1}{\sqrt{r_1r_2}\exp(i(\theta_1+\theta_2+4\pi)/2)} = \frac{1}{\sqrt{r_1r_2}\exp(i(\theta_1+\theta_2)/2)} $$ so the function is single valued. Therefore, we can define the branch cut from $[-im, im]$

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  • $\begingroup$ so why are they taking the branch cuts to be $\mathbb{C}\backslash([-im, im].?$ $\endgroup$ – SuperCiocia Dec 31 '14 at 10:05
  • $\begingroup$ @SuperCiocia I have no idea why they chose infinite branch cuts when a finite branch cut will suffice. It could be because they are Physicists not Mathematicians or they have a reason for it. I would suggest asking a Physicist why that was used. With the branch cut I suggested, we can use a dogbone contour. There a few post on how to do so as well. $\endgroup$ – dustin Dec 31 '14 at 16:33
  • $\begingroup$ I'm not sure this works since the argument isn't defined in the image of $\mathbb{C}\backslash [-im,im]$ under the function $z^2+m^2$. $\endgroup$ – hjhjhj57 Dec 31 '14 at 20:23
  • $\begingroup$ @SuperCiocia The more I think about it the more I believe this isn't right. First: It is possible to define a branch of the argument in a region iff it is possible to define a branch of the logarithm in it. Second: The image of $\mathbb{C}\backslash[-im,im]$ (imaginary interval) under $z^2+m^2$ is $\mathbb{C}\backslash[0,m]$ (real interval). Third: A branch of the logarithm (argument) is only defined in a simply connected region of $\dot{\mathbb{C}}$ (punctured plane) and $\mathbb{C}\backslash[0,m]$ isn't a simply connected region of the punctured plane. 1/2 $\endgroup$ – hjhjhj57 Jan 1 '15 at 4:48
  • $\begingroup$ I believe the problem is that from the beginning you assume there is a branch of the argument defined ($\theta_i=\arg z_i$). 2/2 $\endgroup$ – hjhjhj57 Jan 1 '15 at 4:50
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A branch of the square root is only defined in a branch of the logarithm, i.e., in a simply connected region of $\dot{\mathbb{C}}$. So in fact you have an infinite amount of branches: any ray from either $im$ or $-im$ to infinity defines one.

In particular, observe that if $f_{\pm}(z)=z\pm im$, then $$ f_+(-\infty,-im] = (-\infty,0]\quad\text{and}\quad f_-[im,\infty) = [0,\infty), $$ (these intervals are imaginary!) so we can define two different branches of the logarithm (hence two branches of the square root), one in $f_+\left(\mathbb{C}\backslash (-\infty,-im]\right) = \mathbb{C}\backslash(-\infty,0]$ and the other in $f_-\left(\mathbb{C}\backslash[im,\infty)\right) = \mathbb{C}\backslash[0,\infty)$. In order to be able to define a common branch of the logarithm you need a branch that works in both regions (the principal branch works perfectly in this case).

Also observe that in this case the function $f:=\sqrt{f_-(z)f_+(z)}$ will be defined in: $$ \mathbb{C}\backslash(-\infty,-im]\cap\mathbb{C}\backslash[im,\infty). $$ This should've already answered (1) and (2).

For (3) observe that the points in the interval are just as any other point in the plane: $$ \begin{align} f_-(-im,im) = (-2im,0), \\ f_+(-im,im) = (0,2im). \end{align} $$ (these are imaginary intervals!) Which are contained in the region where the square root is defined.

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