I am new to QFT, so I may have some of the terminology incorrect.
Many QFT books provide an example of deriving equations of motion for various free theories. One example is for a complex scalar field:
$$\mathcal{L}_\text{compl scaclar}=(\partial_\mu\phi^*)(\partial^\mu\phi)-m^2\phi^*\phi.$$
The usual "trick" to obtaining the equations of motion is to treat $\phi$ and $\phi^*$ as separate fields. Even after this trick, authors choose to treat them as separate fields in their terminology. This is done sometimes before imposing second quantization on the commutation relations, so that $\phi$ is not (yet) a field of operators. (In particular, I am following the formulation of QFT in this book by Robert D. Klauber, "Student Friendly Quantum Field Theory".)
What is the motivation for this method of treating the two fields as separate? I intuitively want to treat $\phi^*$ as simply the complex conjugate of $\phi,$ not as a separate field, and work exclusively with $\phi$.
Is it simply a shortcut to obtaining the equations of motion
$$(\square +m^2)\phi=0\\
(\square + m^2)\phi^*=0~?$$
I also understand that one could write $\phi=\phi_1+i\phi_2$ where the two subscripted fields are real, as is done here; perhaps this addresses my question in a way that I don't understand.
Best Answer
TL;DR: Yes, it is just a short-cut. The main point is that the complexified map
$$\tag{A} \begin{pmatrix} \phi \\ \phi^{*} \end{pmatrix} ~=~ \begin{pmatrix} 1 & i\\ 1 &-i \end{pmatrix} \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} $$
is a bijective map :$\mathbb{C}^2 \to \mathbb{C}^2 $.
Notation in this answer: In this answer, let $\phi,\phi^{*}\in \mathbb{C}$ denote two independent complex fields. Let $\overline{\phi}$ denote the complex conjugate of $\phi$.
I) Let us start with the beginning. Imagine that we consider a field theory of a complex scalar field $\phi$. We are given a Lagrangian density
$$\tag{B} {\cal L}~=~{\cal L}(\phi,\overline{\phi},\partial\phi, \partial\overline{\phi})$$
that is a polynomial in $\phi$, $\overline{\phi}$, and spacetime derivatives thereof. We can always decompose a complex field in real and imaginary parts
$$\tag{C} \phi~\equiv~\phi_1+ i \phi_2 ,$$
where $\phi_1,\phi_2 \in \mathbb{R}$. Hence we can rewrite the Lagrangian density (B) as a theory of two real fields
$$\tag{D}{\cal L}~=~{\cal L}(\phi_1,\phi_2,\partial\phi_1, \partial\phi_2).$$
II) We can continue in at least three ways:
Vary the action wrt. the two independent real variables $\phi_1,\phi_2 \in \mathbb{R}$.
Originally $\phi_1,\phi_2 \in \mathbb{R}$ are of course two real fields. But we can complexify them, vary the action wrt. the two independent complex variables $\phi_1,\phi_2 \in \mathbb{C}$, if we at the end of the calculation impose the two real conditions $$\tag{E} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2). $$
Or equivalently, we can replace the complex conjugate field $\overline{\phi}\to \phi^{*}$ in the Lagrangian density (B) with an independent new complex variable $\phi^{*}$, i.e. treat $\phi$ and $\phi^{*}$ as two independent complex variables, vary the action wrt. the two independent complex variables $\phi,\phi^{*} \in \mathbb{C}$, if we at the end of the calculation impose the complex condition $$\tag{F} \phi^{*} ~=~ \overline{\phi}. $$
III) The Euler-Lagrange equations that we derive via the two methods (1) and (2) will obviously be exactly the same. The Euler-Lagrange equations that we derive via the two methods (2) and (3) will be just linear combinations of each other with coefficients given by the constant matrix from eq. (A).
IV) We mention for completeness that the complexified theory [i.e. the theory we would get if we do not impose condition (E), or equivalently, condition (F)] is typically not unitary, and therefore ill-defined as a QFT. Recall for starter that we usually demand that the Lagrangian density is real.
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