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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:

1. Vary the action wrt. the two independent real variables $\phi_1,\phi_2 \in \mathbb{R}$.

2. 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 $$\tag{E} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2). $$

3. 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.

References:

1. Sidney Coleman, QFT notes; p. 56-57.

TL;DR: Yes, it is just a shortcut. 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 into 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 $(\text{B})$ 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:

1. Vary the action wrt. the two independent real variables $\phi_1,\phi_2 \in \mathbb{R}$.

2. 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 $$\tag{E} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2). $$

3. Or equivalently, we can replace the complex conjugate field $\overline{\phi}\to \phi^{*}$ in the Lagrangian density $(\text{B})$ 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. $(\text{A})$.

IV) We mention for completeness that the complexified theory [i.e. the theory we would get if we do not impose condition $(\text{E})$, or equivalently, condition $(\text{F})$] is typically not unitary, and therefore ill-defined as a QFT. Recall, for starters, that we usually demand that the Lagrangian density is real.

References:

1. Sidney Coleman, QFT notes; p. 56-57.

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:

1. Vary the action wrt. the two independent real variables $\phi_1,\phi_2 \in \mathbb{R}$.

2. 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 $$\tag{E} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2). $$

3. 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.

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:

1. Vary the action wrt. the two independent real variables $\phi_1,\phi_2 \in \mathbb{R}$.

2. 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 $$\tag{E} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2). $$

3. 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.

References:

1. Sidney Coleman, QFT notes; p. 56-57.

Yes, it is just a short-cut. The point is that the complexified map

$$\tag{1} \begin{pmatrix} \phi \\ \phi^{*} \end{pmatrix} ~=~ \begin{pmatrix} 1 & i\\ 1 &-i \end{pmatrix} \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} $$

is a bijective map :$\mathbb{C}^2 \to \mathbb{C}^2 $.

1. 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 $$\tag{2} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2). $$

2. Or equivalently, we can 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

$$\tag{3} \phi^{*} ~=~ \text{complex conjugate of }\phi. $$

The Euler-Lagrange equations that we derive via the two methods (1) and (2) will be just linear combinations of each other given by the constant matrix from eq. (1).

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:

1. Vary the action wrt. the two independent real variables $\phi_1,\phi_2 \in \mathbb{R}$.

2. 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 $$\tag{E} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2). $$

3. 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.