In 2 Dimensional Conformal field theory, We have two coordinates $ z=x+i y$ and $\bar{z}=x-i y$. Here $x$ and $y$ are both real. So, one can work directly with $z$ and $\bar{z}$.

I want to define $\partial_z$ and $\partial_{\bar{z}}$. But there is a little subtlety that $\partial_z \frac{1}{\bar{z}}=2 \pi \delta^2(z,\bar{z})$. One can prove this identity in many ways. The action of the derivative operator on a more general function is like $\partial_z \frac{1}{\bar{z}}f[\bar{z}]=2 \pi f[0]$. Here $f[\bar{z}]$ is well-behaved near 0.

Is there a way to define such an operator in Mathematica? Ordinary derivatives $"D"$ won't really work because of subtlety regarding Delta function.

  • 1
    $\begingroup$ It is impossible to implement in Mathematica undefined by you notions $\partial_z$ and $\partial_{\bar{z}}$. $\endgroup$
    – user64494
    Mar 22 at 17:27
  • $\begingroup$ The more I consider this, the more I think perhaps you want polar (rather than holomorphic and antiholomorphic) coordinates. $\endgroup$ Mar 23 at 14:46

2 Answers 2


Hard to say.

The "usual" definitions are like so.

z[x_, y_] := x + I*y
zbar[x_, y_] := x - I*y

dbydz[f_, x_, y_] := 1/2*(D[f, x] - I*D[f, y])

The minus sign is quite intentional. It compensates for the differential form dz behaving as dx+i*dy.

First check dbydz on 1/z.

dbydz[1/z[x, y], x, y]

(* Out[112]= -(1/(x + I y)^2) *)

Now check it on 1/zbar.

dbydz[1/zbar[x, y], x, y]

(* Out[113]= 0 *)

Also not a surprise. Just not what you want. From here I do not see the right way to adapt it.

  • $\begingroup$ How about $\partial_z \frac{1}{\bar{z}}f[\bar{z}]=2 \pi f[0]$? $\endgroup$
    – user64494
    Mar 23 at 9:23

The rule for

  D[1/z,w] ~ Delta[z,w]

for the Wirtinger partial derivatives in the complex "directions" 1 +- I

is a derivative for a distribution, that in full length reads

D[Log[w],w,z] == Laplacian[Log[w]]

This has a meaning as an integral kernel in the Cauchy integral sense, replacing the Cauchy line integral around 0 of 1/z with its value 2 Pi I by an 2d integral with 2-d area form

 dw^dz ~ dx dy 

over the inner of the countour of integration. These identies are consequences of the universal rule

Integral[d f, Volume] = Integral[ f, Boundary[ Volume ]]

where df is the volume n-form of a n-1 surface form f. All that can somehow be broken down to 1-d integrals, at least numerically as

 Integral[ f', {a,b}] = Integral[ f, b] -Integral[ f, a] =f[b]-f[a]

Of course you can knit a Wirtinger calculus for distribution rather straight forward, but better don't use any of the internal D and Derivative algorithms.

Just implement the complete set of caluclus rules from an algorithmic oriented book on functional analysis.

So we conclude, by all the philosophy of working with a CAS-Simplify on localized expressions, the expression

\[Partial]_w (1/z) ==0

evaluates locally to zero and the pole at z = 0 is an exception set of measure 0, difficult to hit by chance in a numeric approximation.


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