What is the intuition behind the Wirtinger derivatives?

Solution 1:

It is natural when dealing with a complex-valued function $f(x,y)$ to think of a change of variables from $(x,y)$ to $(z,\bar z)$, where $z = x+i y$. We need to know how the derivatives transform and we find $$\begin{eqnarray*} \frac{\partial}{\partial x} &=& \frac{\partial z}{\partial x} \frac{\partial}{\partial z} + \frac{\partial \bar z}{\partial x} \frac{\partial}{\partial \bar z} \\ &=& \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar z} \\ \frac{\partial}{\partial y} &=& \frac{\partial z}{\partial y} \frac{\partial}{\partial z} + \frac{\partial \bar z}{\partial y} \frac{\partial}{\partial \bar z} \\ &=& i\left(\frac{\partial}{\partial z} - \frac{\partial}{\partial \bar z}\right). \end{eqnarray*}$$ Solving for $(\partial/\partial z, \partial/\partial \bar z)$ we find the usual expressions for the Wirtinger derivatives. The derivatives act as they should, $$\frac{\partial z}{\partial z} = \frac{\partial \bar z}{\partial \bar z} = 1 \qquad \textrm{and}\qquad \frac{\partial \bar z}{\partial z} = \frac{\partial z}{\partial \bar z} = 0.$$

The derivative $\partial/\partial \bar z$ has the important property that if $$\frac{\partial}{\partial \bar z} f(z,\bar z) = 0$$ then $f$ satisfies the Cauchy-Riemann equations and so is holomorphic (and thus analytic). This is straightforward to verify, $$\begin{eqnarray*} \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right) f(x,y) &=& \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right) (u(x,y)+i v(x,y)) \\ &=& \frac{1}{2}((u_x - v_y) + i(u_y + v_x)), \end{eqnarray*}$$ where subscripts denote partial derivatives. Thus, if $\partial f(z,\bar z)/\partial \bar z = 0$ then the Cauchy-Riemann equations are satisfied, $u_x = v_y$ and $u_y = -v_x$. This gives us a very good intuition indeed. Roughly, if $f$ is not a function of $\bar z$, then $f$ is holomorphic.

Solution 2:

In the first place, as noted above: the Wirtinger partial derivatives "act as they should''. The proof is not trivial, but it is easily seen with simple examples. Not only the two examples above, below the cited phrase, but also somewhat more complicated: $$ \frac{\partial z\overline{z}}{\partial z} = \overline{z} $$ and $$ \frac{\partial z\overline{z}}{\partial \overline{z}} = z.$$ For instance: $$ \frac{\partial z\overline{z}}{\partial z} = \frac{1}{2}\left( \frac{\partial \, x^2+y^2}{\partial x} - i \frac{\partial \, x^2+y^2}{\partial y}\right) = \frac{1}{2}\left(2x- i 2y\right) = x- i y = \overline{z}$$ Of course, the whole point of the Wirtinger partial derivatives is their behavior as ordinary partial derivatives, with a jump immediately from $\partial z\overline{z}/\partial z$ to $\overline{z}$. This is my "mental picture".

In the second place, this mental picture of ordinary partial derivatives can be applied to prove the rule that a function $f$ is complex-differentiable with respect to $z$ if and only if ${\partial f}/{\partial \overline{z}} = 0$, as follows. Start with

\begin{align*} \frac{\mathrm{d} f}{\mathrm{d} z} &= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \, \frac{\mathrm{d} \overline{z}}{\mathrm{d} z}\\ &= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \, \lim_{\Delta z \rightarrow 0} \frac{\Delta \overline{z} }{\,\Delta z \,}\\ &= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \, \lim_{\Delta z \rightarrow 0} \frac{\overline{\Delta z} }{\,\Delta z \,}\\ &= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \, \lim_{\Delta z \rightarrow 0} e^{- i 2\arg(\Delta z )}\\ & = \frac{\partial f}{\partial z} + \lim_{\Delta z \rightarrow 0} \left( \frac{\partial f}{\partial \overline{z}} \, e^{- i 2\arg(\Delta z )} \right). \end{align*}

Then, if ${\partial f}/{\partial \overline{z}} = 0$ holds then we have indeed: $$ \frac{\mathrm{d} f}{\mathrm{d} z} = \frac{\partial f}{\partial z} + \lim_{\Delta z \rightarrow 0} \left( 0 \times e^{- i 2\arg(\Delta z )} \right) = \frac{\partial f}{\partial z} + \lim_{\Delta z \rightarrow 0} 0 = \frac{\partial f}{\partial z} $$ Otherwise, indeed, ${\mathrm{d} f}/{\mathrm{d} z}$ does not exist, because then the $\lim_{\Delta z \rightarrow 0}$ does not exist.