I have a question on the complex scalar field theory.

Say you have a $U(1)$ invariant Lagrangian of the form

$$\int d^4 x |\partial\phi|^2 -m^2 |\phi|^2 + \frac{g}{4}(|\phi|^2)^2.$$

Then if you want to find a vacuum solution then I always find it easiest to rewrite the Lagrangian in terms of real components $\phi=\frac{1}{\sqrt{2}}(\varphi+i\chi)$ then take the functional derivative with respect to $\{\varphi,\chi\}$. However, in some textbook authors use functional derivative with respect to $\phi,\phi^\dagger$.

My question is whether one should check the functional-version of the Cauchy-Riemann equation for complex-differentiability if one wants to do functional derivative with respect to complex scalar fields?


You shouldn't check the Cauchy-Riemann equations as they will not hold - that functional is not holomorphic as it contains explicit $\phi^\dagger$s. This is why you need to impose both $\frac{\delta S}{\delta \phi(x)} = 0$ and $\frac{\delta S}{\delta \phi^\dagger(x)} = 0$. In the holomorphic case the second equation would come for free.

The way I make sense of it is to think of $\frac{\delta}{\delta\phi(x)},\frac{\delta}{\delta\phi^\dagger(x)}$ as shorthand for $\frac{1}{\sqrt{2}}\left(\frac{\delta}{\delta\varphi(x)} \mp i \frac{\delta}{\delta\chi(x)}\right)$. You can reconstruct the real derivatives as: $$ \frac{\delta}{\delta\varphi(x)} = \frac{1}{\sqrt{2}}\left(\frac{\delta}{\delta\phi(x)} + \frac{\delta}{\delta\phi^\dagger(x)}\right), \qquad \frac{\delta}{\delta\chi(x)} = \frac{i}{\sqrt{2}}\left(\frac{\delta}{\delta\phi(x)} - \frac{\delta}{\delta\phi^\dagger(x)}\right). $$ Therefore, the condition for a stationary point, $\frac{\delta S}{\delta\varphi(x)} = \frac{\delta S}{\delta\chi(x)} = 0$ is equivalent to demanding $\frac{\delta S}{\delta\phi(x)} = \frac{\delta S}{\delta\phi^\dagger(x)} = 0$. The Cauchy-Riemann equations would demand that $\frac{\delta S}{\delta\phi^\dagger} = 0$ everywhere.

This is easier notationally with ordinary calculus for functions of two variables, with $z = \frac{x + i y}{\sqrt{2}}$. The same principles apply to the functional case. With a holomorphic function, we can talk about the ordinary derivative $\frac{df}{dz}$. With a nonholomorphic function we can talk about the partial derivatives $\left(\frac{\partial f}{\partial z}\right)_\bar{z}$ and $\left(\frac{\partial f}{\partial \bar{z}}\right)_z$, which are defined as $\frac{1}{\sqrt{2}}\left( \left(\frac{\partial f}{\partial x}\right)_y \mp i \left(\frac{\partial f}{\partial y}\right)_x \right)$. The idea of changing $z$ while holding $\bar{z}$ constant sounds ridiculous, so thinking of them as shorthand like this was the simplest[*] way I could make sense of the concept.

Note that the chain rule still works with these differential operators: $$ df = \left(\frac{\partial f}{\partial x}\right)_y dx + \left(\frac{\partial f}{\partial y}\right)_x dy = \left(\frac{\partial f}{\partial z}\right)_\bar{z} dz + \left(\frac{\partial f}{\partial \bar{z}}\right)_z d\bar{z}. $$ Furthermore, $\left(\frac{\partial z}{\partial z}\right)_\bar{z} = \left(\frac{\partial \bar{z}}{\partial \bar{z}}\right)_z = 1$ and $\left(\frac{\partial \bar{z}}{\partial z}\right)_\bar{z} = \left(\frac{\partial z}{\partial \bar{z}}\right)_z = 0$. This means that use these funny differential operators the same way as partial derivatives with respect to independent variables.

[*]: There is another way of thinking about this that is more complicated, but probably more correct. If one of these makes sense to you, feel free to ignore the other. You say that $x$ and $y$ could be complex numbers, it just so happens that they are both real at this point. If we integrate over $x$ and $y$, we say that we chose a contour that lies along the real axes, but we could have picked a contour anywhere on the complex planes. This means that $\bar{z} = \frac{x - i y}{\sqrt{2}}$ is not necessarily the complex conjugate of $z$, so varying $z$ while holding $\bar{z}$ constant is a legitimate concept.

In that case you would worry about the Cauchy-Riemann equations with respect to the real and imaginary components of $x$, and with respect to the real and imaginary components of $y$, not $z$. But if you define the function away from the real axis by analytic continuation this would be automatic. Trandlating back to the functional language, you would imagine $\varphi$ and $\chi$ being complex, etc.

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  • $\begingroup$ Thank you, Subhaneil Lahiri, for the detailed answer. $\endgroup$ – user239970 Jan 26 at 8:02

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