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It was shown in the answer of the problem what is the right way to calclulate $\partial_{z}\frac{1}{\bar{z}?}$ that \begin{align} \partial_{z}\frac{1}{\bar{z}}=\pi\delta^{(2)}(\vec{x}), \end{align} but when I tried to calclulate $\partial_{z}\frac{1}{\bar{z}^2}$ further, I found that the answer is depend on the method I used.

Method $1:$ \begin{align} \partial_{z}\frac{1}{\bar{z}^2}&=\partial_{z}(\frac{1}{\bar{z}}\cdot\frac{1}{\bar{z}})\\ &=2\cdot\frac{1}{\bar{z}}\partial_{z}\frac{1}{\bar{z}}\\ &=\frac{2\pi}{\bar{z}}\delta^{(2)}(\vec{x}). \end{align} Method $2:$ \begin{align} \partial_{z}\frac{1}{\bar{z}^2}&=-\partial_{z}\partial_{\bar{z}}\frac{1}{\bar{z}}\\ &=-\partial_{\bar{z}}\partial_{z}\frac{1}{\bar{z}}\\ &=-\pi\partial_{\bar{z}}\delta^{(2)}(\vec{x}). \end{align} I tried to prove that the two answers were equivalent because both methods seem natural, and my starting point is $x\delta'(x)=-\delta(x)$, where prime means derivative with respect to x, but it failed.

comment : It seems physicist prefer the second answer (I'm a physics student and my tutor provides me with method 2, but he can't explain why method 1 is wrong.)

Thank you for any help.

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Hezaraki
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  • I think you have some typos in Method 2 which make it a little difficult to help. – Cameron Williams Mar 16 '20 at 12:19
  • @CameronWilliams Sorry I can't find the typos you said, could you please be more specific? – Hezaraki Mar 16 '20 at 12:35
  • You have a $\partial_{\bar{z}}$ but a second derivative of the Dirac delta. – Cameron Williams Mar 16 '20 at 12:39
  • @CameronWilliams Sorry I still can't get your point, may be I should list the definition $\partial_{\bar{z}}:=\frac{1}{2}(\partial_{x}+i\partial_{y} )$ and $\delta^{(2)}(\vec{x}):=\delta(x)\delta(y)$. – Hezaraki Mar 16 '20 at 13:15
  • Oh whoops. I totally missed that the power changed on $\bar{z}$. My apologies! – Cameron Williams Mar 16 '20 at 13:19
  • @Hezaraki what do you mean by $\partial_z \frac{1}{\overline{z}^2}$ exactly? contrary to $\frac{1}{z}$ or $\frac{1}{\overline{z}}$, the function $\frac{1}{\overline{z}^2}$ is not locally integrable, so it is not a distribution – Albert Mar 16 '20 at 20:04
  • @Glougloubarbaki well thanks, actually I don't know $\frac{1}{\bar{z}^2}$ is not a distribution...The motivation I investigate $\partial_{z}\frac{1}{\bar{z}^2}$ is that I'm not sure whether $\partial_{z}$ and $\partial_{\bar{z}}$ satisfie Leibniz's law or not, when we deal with the formule like it, such as $\partial_{z}\frac{(z+w)(\bar{z}+\bar{w})}{\bar{z}(\bar{z}-\bar{w})}$ , such formula usually occurs in the calculation of conformal field theory, like this – Hezaraki Mar 17 '20 at 02:30

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