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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 dyanmics of a Lagrangian of 2 real free scalar fields is identical to one of a complex free scalar field).

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.

Is this true? What examples are there?

Alex Gower
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    What are "utterly" independent variables? If $z=a+bi$ then $$\begin{cases}z = a+bi \ z^* = a- bi \end{cases}\iff \begin{cases} a = \frac{1}{2}z + \frac{1}{2}z^* \ b = \frac{1}{2i}z - \frac{1}{2i}z^\end{cases}$$ is an invertible set of equations no? As in, if one set of variables is independent the other must be as well. We could have just as easily written the same equations with the notation $z^ \to w$ and $b\to a^$. Does this mean $a$ and $a^$ are not "utterly" independent variables anymore just because I wrote a suggestive notation? – Ninad Munshi Dec 30 '21 at 13:45
  • I see your point. My main confusion is if we plot say the evolution of some 'utterly' independent variables $a$ and $b$ on some 'phase space real line' their motion would be completely uncorrelated, whereas if we did the same with $z$ and $z^*$ on some 'phase space complex plane' their motion would be correlated. So something (possibly incorrectly) seems 'not independent' to me? – Alex Gower Dec 30 '21 at 13:50
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    That's still not necessarily true. You're making the assumption that if you choose $a$ and $b$ beforehand then $z$ and $z^$ would correlated. But what if you independently chose arbitrary paths for $z$ and $w$ first (again borrowing the replacement notation from my earlier comment). Then $a$ and $a^$ would be correlated. Among the four variables, there are only two degrees of freedom. Arbitrarily set any two and you uniquely determine the rest because you have eaten all of those degrees. For example I could have set $z$ and $a$ ahead of time, "correlating" $w$ and $b$ – Ninad Munshi Dec 30 '21 at 13:53
  • Related, especially the comment noting that ${x, y}$ and ${z, \bar{z}}$ span the same two-dimensional complex vector space if viewed as complex-valued functions. – Andrew D. Hwang Dec 30 '21 at 14:03
  • Hmm but it seems like if I have 2 degrees of freedom, $a$ and $b$, I can package all of that information just into $z$ (because it is a 2 dimensional quantity) and there is no need for $z^*$? – Alex Gower Dec 30 '21 at 14:11
  • Ultimately though, isn't there 'extra structure' that we've added once we introduce complex numbers (with their multiplication rule) rather than 2 independent real numbers. How does this not have an effect in some situation? – Alex Gower Dec 30 '21 at 14:12
  • You're making the same mistake that Andrew Huang addresses in the comments of his answer in his link. Ultimately the conclusions you are trying to draw are incorrect because they were based on a faulty premise of a lack of true independence, and you should make peace with that. Please see his well written answer. – Ninad Munshi Dec 30 '21 at 14:13
  • I think I'm struggling to understand the significance of him highlighting complex-valued functions specifically? – Alex Gower Dec 30 '21 at 14:18
  • The highlighting merely emphasizes that although $x$ and $y$ are real-valued, they span the same vector space as $z$ and $\bar{z}$. ;) <> The overall point is, we have a single two-dimensional vector space, and we have two different bases (${x, y}$ and ${z, \bar{z}}$) of this space. Intuition might suggest we can "discard" $\bar{z}$ and still span the same space, but that intuition (as Ninad explains) is not correct. – Andrew D. Hwang Dec 30 '21 at 15:38

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