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In the appendix B of a physics paper arXiv: 1902.01434, it says $$ \partial_z\frac{1}{\bar{z}}=\partial_{\bar{z}}\frac{1}{z}=2\pi\delta(z)\delta(\bar{z}), $$ same as 2-dimensional delta function (complex plane) and A puzzle with derivative of delta-functions. However, from the definition of Wirtinger derivatives, one can also get $$ \partial_z\frac{1}{\bar{z}}=0, $$ such as What is $\partial_z \frac{1}{\bar{z}}$?. So, my question is, which is the right way to do the calculation? For example, we know $\partial_{z}\bar{z}$ is not differentiable, but we can still have $$ \partial_z\bar{z}=\partial_z \frac{1}{\frac{1}{\bar{z}}}=-2\pi\bar{z}^2\delta(z)\delta(\bar{z}), $$ what is wrong here? What about $\partial_z\frac{\bar{z}-a}{\bar{z}-b}$?

I'm really confused here, thank you for any help.

Daniele Tampieri
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J.-H.
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  • Please do not make the whole of the title a math formula. Include some regular text. – Arturo Magidin Dec 14 '19 at 03:05
  • @Arturo Magidin Sorry for that, I have corrected the title. – J.-H. Dec 14 '19 at 03:15
  • Thank you; there are technical reasons for not having it be all math that are too boring to get into. – Arturo Magidin Dec 14 '19 at 03:20
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    I'm not familiar with the physics literature and would love to be corrected, but the Wirtinger Derivative is defined even if your function is not holomorphic while difference quotients no longer make direct sense. If you try to interpret the difference quotients the best you are going to get is some delta function (it blows up after all) while since the Wirtinger Derivative is no longer required to match the difference quotients (since our input is not analytic) the result can be squashed to zero – Brevan Ellefsen Dec 14 '19 at 03:44
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    @Brevan Ellefsen Thank you. So Wirtinger derivatives is always the right way to do calculation like this one, and the result of delta function is somehow problematic because the input is not analytic. – J.-H. Dec 14 '19 at 06:30
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    The result you get for $\partial_z \bar{z}$ (and the derivative of any polynomial function of $\bar{z}$) is consistent with what you get with Wirtinger derivatives. Note that $\bar{z} \delta(\bar{z}) = 0$ in the distributional sense (see e.g. here ). The Dirac delta terms come from poles. – pregunton Dec 14 '19 at 07:32
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    @pregunton Thanks, this does help a lot! – J.-H. Dec 14 '19 at 07:53

1 Answers1

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What measure are they using in the complex plane when they get the factor $2\pi$?

Changing from $(z, \bar{z})$ to $(x,y)$ and using well-known divergence and rotation of two vector fields I get: $$ \partial_z \frac{1}{\bar{z}} = \partial_z \frac{z}{|z|^2} = \frac12 (\partial_x - i \partial_y) \frac{x+iy}{x^2+y^2} \\ = \frac12 \left[ \left(\partial_x \frac{x}{x^2+y^2} + \partial_y \frac{y}{x^2+y^2} \right) + i \left(\partial_x \frac{y}{x^2+y^2} - \partial_y \frac{x}{x^2+y^2} \right) \right] \\ = \frac12 \left[ 2\pi\,\delta(x,y) + i\,0 \right] = \pi\,\delta(x,y) = \pi\,\delta(x)\,\delta(y) . $$

This is with respect to the area measure $dx \wedge dy = \frac{i}{2} dz \wedge d\bar{z}.$

How do they get a factor $2\pi$?

md2perpe
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  • I don't understand how you go from the second line to the third. If the first parenthesis in the bracket is going to be the Dirac delta in the x-y plane, then surely its limit must be $\infty$ for $(x,y) \to 0$, doesn't it? I find that this limit is however 0. – phenolphthalein Mar 17 '23 at 13:15
  • @dim-doom. If you take the limit of $\delta(x,y)$ as $(x,y) \to (0,0)$ then you also get $0$ since $\delta(x,y)=0$ when $(x,y)\neq(0,0).$ – md2perpe Mar 17 '23 at 17:12
  • I see. But then, how do you verify that it is indeed a delta function, and that the other term, multiplied by $i$, is zero? – phenolphthalein Mar 20 '23 at 12:50
  • The imaginary part can be seen as a 2-dimensional curl of a gradient, or divergence of a curl, both of which vanish. The real part is the 2-dimensional divergence of a field from a point mass. The flow through a closed curve not enclosing origin vanishes, but is $2\pi$ when the curve encloses origin. That's basically the reasons for the terms having those values. – md2perpe Mar 21 '23 at 09:57
  • I know that this question is relatively old, and therefore this is a long shot, but I fail to grasp how you get from line 2 to line 3. I simply do not see the divergence and curl in the equations. Could you perhaps enlighten this still? – Johannes Nauta Jul 28 '23 at 16:45
  • @JohannesNauta. Do you know how divergence and curl are defined in terms of partial derivatives? – md2perpe Jul 28 '23 at 17:32
  • In this case, I believe that the divergence is given by $\frac{\partial f_x(z)}{\partial x} + \frac{\partial f_y(z)}{\partial y}$, but I do not know how the curl is defined in terms of partial derivatives in a two-dimensional space. – Johannes Nauta Jul 29 '23 at 08:07
  • @JohannesNauta. Correct. In two dimensions the curl is $\frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y}.$ If we embed the plane into three dimensions then it's just $\hat z\cdot(\nabla\times\vec f),$ where $\hat z$ is the unit vector in the $z$ direction and $f_z\equiv 0.$ – md2perpe Jul 29 '23 at 09:54
  • Ok so I see that the curl in this case must be $0$. But how does the divergence equal $2\pi\delta(x)\delta(y)$? Also, what does the 'wedge' notation mean, i.e. what does $dx \wedge dy$ mean? – Johannes Nauta Jul 29 '23 at 10:35
  • @JohannesNauta. The wedge notation $dx \wedge $dy$ is a differential form. Often, especially in physics, it's just written $dx,dy,$ but using differential forms has some pros. I suggest that you learn the basics about them; they are really nice. For example, the divergence theorem and other "integral over boundary equals integral over interior of some kind of derivative" can be written $$\oint_{\partial\Omega} \omega = \int_\Omega d\omega,$$ where $\omega$ is some differential form. – md2perpe Jul 29 '23 at 12:11
  • @JohannesNauta. The flow of $(x,y)/(x^2+y^2)$ out of a closed path equals $2\pi$ if the path encloses origin, it equals zero otherwise. Within the language of distributions this means that the divergence equals $2\pi,\delta(x),\delta(y).$ – md2perpe Jul 29 '23 at 12:17