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I understand, loosely, that for many purposes we can treat $z$ and $z^*$ as independent variables (e.g. while differentiating, and apparently the dynamics of a Lagrangian of 2 real free scalar fields is identical to one of a complex free scalar field), cf. e.g. this Phys.SE post.

However, since unlike utterly independent variables (say, $a$ and $b$), complex numbers have the extra structure that $z^*$ is easily mapped to $z$, it seems like in some (maybe not common) circumstances, this 'lack of utter independence' must be considered.

My guess is that the maths of QFT always uses 'the subset of operations in which you can consider $z$ and $z^*$ as independent' and happens to not include operations where you see their dependence?

Is this true? What examples are there?

my2cts
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Alex Gower
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1 Answers1

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Whenever you have a function of two variables $a$ and $b$, you can plug in $a = (z + z^*) / 2$ and $b = (z - z^*) / 2i$ to get a function of $z$ and $z^*$. If $a$ and $b$ were real, $z$ and $z^*$ will necessarily be complex conjugates. Nothing fancy is going on. The mapping between $z$ and $z^*$ is no more or less special than the mapping between $b$ and $-b$.

You are just talking about a change of variables but there are two other things which might be confusing you.

  1. It is often desirable that when $a$ and $b$ are traded for $z$ and $z^*$, the resulting function depends on only one of them. This indeed requires the function to use only "a subset of operations" so that it satisfies the Cauchy-Riemann equations. Note that $f(a, b) = a^2 + b^2 \Leftrightarrow f(z, z^*) = z z^*$ (which is the QFT example you referenced) does not.

  2. Sometimes people take a function of $z$ and $z^*$ and promote both of them to independent complex variables. This is an analytic continuation which means it can be unique or not depending on the function.

Connor Behan
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