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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.

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  • $\begingroup$ Please do not make the whole of the title a math formula. Include some regular text. $\endgroup$
    – Arturo Magidin
    Commented Dec 14, 2019 at 3:05
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    $\begingroup$ 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 $\endgroup$
    – Brevan Ellefsen
    Commented Dec 14, 2019 at 3:44
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    $\begingroup$ @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. $\endgroup$
    – J.-H.
    Commented Dec 14, 2019 at 6:30
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    $\begingroup$ 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. $\endgroup$
    – pregunton
    Commented Dec 14, 2019 at 7:32
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    $\begingroup$ @pregunton Thanks, this does help a lot! $\endgroup$
    – J.-H.
    Commented Dec 14, 2019 at 7:53

1 Answer 1

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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 $2\pi$?

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  • $\begingroup$ 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. $\endgroup$
    – phenolphthalein
    Commented Mar 17, 2023 at 13:15
  • $\begingroup$ @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 $(x,y)\neq(0,0).$ $\endgroup$
    – md2perpe
    Commented Mar 17, 2023 at 17:12
  • $\begingroup$ 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? $\endgroup$
    – phenolphthalein
    Commented Mar 20, 2023 at 12:50
  • $\begingroup$ 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. $\endgroup$
    – md2perpe
    Commented Mar 21, 2023 at 9:57
  • $\begingroup$ 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? $\endgroup$
    – Johannes Nauta
    Commented Jul 28, 2023 at 16:45

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