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Excuse me, I got one more stupid question in Polchinski's string theory book :( $$\partial \bar{\partial} \ln |z|^2 = 2 \pi \delta^2 (z,\bar{z}) (1) $$ I shall check this equation by integrating both sides over $\int \int d^2z $

The right hand side is obviously $2\pi$. The left hand side is evaluated as following $$ \partial \bar{\partial} \ln |z|^2 = \partial \bar{\partial} \left( \ln z + \ln \bar{z} \right) = \partial \left( \bar{\partial} \ln \bar{z} \right) + \bar{\partial} \left( \partial \ln z \right) (2) $$

with the help of Eq. (2.1.9) in that book, we have $$ \int \int_R d^2 z \left[ \partial \left( \bar{\partial} \ln \bar{z} \right) + \bar{\partial} \left( \partial \ln z \right) \right] = i \oint_{\partial R} \bar{\partial} \ln \bar{z} d \bar z - \partial \ln z d z (3) $$

$$= i \oint_{\partial R} \frac{1}{\bar{z}} d \bar z - \frac{1}{z} d z = 2 \pi + \oint_{\partial R} \frac{1}{\bar{z}} d \bar z $$

Here I have used the contour integral. But is the a remaing term $\oint_{\partial R} \frac{1}{\bar{z}} d \bar z$ zero? Why?

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  • $\begingroup$ In your middle equation (the one after the phrase "evaluated as following"), you don't need the second term on the RHS. You should be able to figure it out from there. $\endgroup$
    – zkf
    Commented Jul 15, 2013 at 22:54
  • $\begingroup$ @zkf Is that because at the singular point (origin), the second-order derivative is not continuous. Therefore $\partial \bar{\partial}$ can not be interchanged. Is that correct? $\endgroup$
    – user26143
    Commented Jul 16, 2013 at 10:40
  • $\begingroup$ @zkf, I see. From the Cauchy-Riemann equation, it can be shown (just a straightforward calculation) the "the Wirtinger derivative of f with respect to the complex conjugate of z is zero" en.wikipedia.org/wiki/Holomorphic_function en.wikipedia.org/wiki/Wirtinger_derivative $\endgroup$
    – user26143
    Commented Jul 16, 2013 at 18:16
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    $\begingroup$ Up to a missing $i$ in the last equality, the calculation above is correct. Polchinski's notation is a bit funny, and it turns out that his $\delta^2(z, \bar z)$ is equal to twice the usual delta function $\delta(\sigma^1, \sigma^2)$ written in Cartesian coordinates. The final result should be $4 \pi$, not $2 \pi$. $\endgroup$
    – user2309840
    Commented Aug 25, 2018 at 22:27

1 Answer 1

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Obviously you are trying to solve problem 2.1 from Polchinski´s book which states:

"Verfify that $\partial\overline{\partial}\ln\left|z\right|^{2}=\partial\frac{1}{\overline{z}}=\overline{\partial}\frac{1}{z}=2\pi\delta^{2}\left(z,\:\overline{z}\right)$

(a) by use of the divergence theorem (2.1.9)

($\int_{R}d^{2}z\left(\partial_{z}v^{z}+\partial_{\overline{z}}v^{\overline{z}}\right)=i\oint_{\partial R}\left(v^{z}d\overline{z}-v^{z}dz\right)$);

(b) by regulating the singularity and then taking the limit."

(a) You have

  1. For holomorphic test functions $f(z)$:

$\int_{R}d^{2}z\partial\overline{\partial}\ln\left|z\right|^{2}f\left(z\right)=\int_{R}d^{2}z\overline{\partial}\frac{1}{z}f\left(z\right)=-i\oint_{\partial R}dz\frac{1}{z}f\left(z\right)=2\pi f\left(0\right)$.

  1. For antiholomorphic Test functions $f\left(\overline{z}\right)$:

$\int_{R}d^{2}z\partial\overline{\partial}\ln\left|z\right|^{2}f\left(\overline{z}\right)=\int_{R}d^{2}z\partial\frac{1}{\overline{z}}f\left(\overline{z}\right)=-i\oint_{\partial R}d\overline{z}\frac{1}{\overline{z}}f\left(\overline{z}\right)=2\pi f\left(0\right)$.

(b) Now comes the second part of the problem: To regulate $\ln\left|z\right|^{2}$ use the good old $\epsilon$-environment trick and rewrite it as $\ln\left(\left|z\right|^{2}+\epsilon\right)$. This regularizes you also $\frac{1}{z}$ and $\frac{1}{\overline{z}}$:

$\partial\overline{\partial}\ln\left(\left|z\right|^{2}+\epsilon\right)=\partial\frac{z}{\left|z\right|^{2}+\epsilon}=\overline{\partial}\frac{\overline{z}}{\left|z\right|^{2}+\epsilon}=\frac{\epsilon}{\left(\left|z\right|^2+\epsilon\right)^{2}}$.

From this point the symmetry of the problem makes the use of polar coordinates more convenient. There consider a general test function $f\left(r,\:\theta\right)$, and define $g\left(r^{2}\right)\equiv\int d\theta f\left(r,\:\theta\right)$ which is assumed to be sufficiently well behaved in the asymptotic cases $0$ and $\infty$, then

$\int d^{2}z\frac{\epsilon}{\left(\left|z\right|^{2}+\epsilon\right)^{2}}f\left(z,\:\overline{z}\right)=\int^{\infty}_{0} du\frac{\epsilon}{\left(u+\epsilon\right)^{2}}g\left(u\right)=\left.\left(-\frac{\epsilon}{u+\epsilon}g\left(u\right)+\epsilon\ln\left(u+\epsilon\right)g^{\prime}\left(u\right)\right)\right|_{0}^{\infty}-\int^{\infty}_{0} du \epsilon\ln\left(u+\epsilon\right)g^{\prime\prime}\left(u\right)=g\left(0\right)=2\pi f\left(0\right).$

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    $\begingroup$ Thank you for your solution. The problem is in my Eq. (2), I have interchanged the partial derivatives $\partial \bar{\partial}$. Is that because at the singular point (origin), the second-order derivative is not continuous. Therefore $\partial \bar{\partial}$ can not be interchanged. Is that correct? $\endgroup$
    – user26143
    Commented Jul 15, 2013 at 23:07
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    $\begingroup$ Dear @user26143, it can be perfectly exchanged. In physics, we almost never deal with pathological situations in which derivatives w.r. to different variables fail to commute. Note that $\partial\bar\partial$ acts on the $\ln(|z|^2)$ here which is completely $z$-$\bar z$-symmetric and the result is symmetric as well so of course the order of the two derivatives can't matter. All the distributions may be imagined as limits of smooth continuous functions for which the derivatives may clearly be exchanged. $\endgroup$
    – Luboš Motl
    Commented Jul 16, 2013 at 7:32
  • $\begingroup$ If you've been educated by math nitpickers "you can't do this, that", referring to identities that work for nicely behaved functions etc., then forget about this whole education because in physics, you can and you should always try to use these identities (exchanging derivatives and many harder ones). Instead of inventing excuses why you shouldn't be doing such elementary operations, you should learn how to do these operations quickly, efficiently, and at the right time. Physics needs it all the time. $\endgroup$
    – Luboš Motl
    Commented Jul 16, 2013 at 7:36
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    $\begingroup$ Dear @Luboš Motl, then what's wrong in the second term in the RHS of Eq.(2) in my post? Why there is an extra term $\oint_{\partial R} \frac{1}{\bar{z}} d \bar{z}$ in the end? $\endgroup$
    – user26143
    Commented Jul 16, 2013 at 10:47
  • $\begingroup$ @Hansenet, in your derivation from holomorphic test functions , $\partial \bar{\partial} \ln |z|^2 f(z) = \bar{\partial} \frac{1}{z} f(z)$, why there is no $\partial f(z)$? $\endgroup$
    – user26143
    Commented Jul 16, 2013 at 11:06

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