# Why treat complex scalar field and its complex conjugate as two different fields?

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.

• They are separate fields. They are linearly independent, the vector space they form is of dimension 2. – auxsvr Jun 26 '16 at 11:34

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.

I would like to make a comment, which may clarify and simplify the things a little bit.

In complex analysis [see e.g. Introduction to Complex Analysis" by B.V. Shabat] by definition derivatives over the complex variables $z$ and $\bar z$ are given by: $$\mbox{def:} \quad \partial_z \equiv \frac{1}{2} \left(\partial_{\rm a} - i \partial_{b}\right) \quad \partial_{\bar z} \equiv \frac{1}{2} \left(\partial_{\rm a} + i \partial_{b}\right),$$ where $a$ and $b$ stand for real and imaginary parts of $z$ correspondingly. The equalities $$\partial_{ z} \bar z = 0 \quad \mbox{and} \quad \partial_{\bar z} z = 0$$ imply, that the variations over z and $\bar z$ are independent, while the variables $z$ and $\bar z$ (being mutually complex conjugated) are not independent. There is no any doubling of degrees of freedom, but one can vary over the field and its conjugated considering them as independent.

• the conjugates are independent in the sense that there is no analytic function that relates them. Moreover to get the underlying (possibly independent) real numbers one needs both $z$ and its conjugate. This makes them (functionaly) independent. One cannot be derived from the other analyticaly – Nikos M. Jan 21 '16 at 21:25
• @ Nikos M. : What do you mean when you say that one needs z and its conjugate if one wants to obtain the underlying real numbers? If I am given a complex number, won't it always be a just a superposition of "1" and "i" ? – Quantumwhisp Feb 1 '16 at 8:46
• @NikosM. The normal prescription to get the complex conjugate is to simply replace $i \rightarrow -i$ everywhere in the equations. Are there other prescriptions that give inequivalent results, but that are still compatible with analysis? – Andrea Jan 21 '19 at 11:51
• @Andrea, sure you are right. What I was trying to say is that the complex conjugate of a complex number is analyticaly and functionaly independent of the complex number itself..In other words there is no analytic function of $z$ which can derive the complex conjugate. Also the replacement of $i$ to $-i$ is not an analytic function. Therefore since to derive the real and imaginary parts one needs both $z$ and its complex conjugate (through the known identities) and since real and imaginary part can be independent ..contd – Nikos M. Jan 21 '19 at 12:30
• @Andrea contd.. this leads to complex conjugate being independent from $z$ as well (certainly functionaly indpependent) and can be treated as an independent complex feild in itself as far as the variations are concerned – Nikos M. Jan 21 '19 at 12:30

Of course @QMechanic's answer is correct.

i would like to show a very simple reason why this is so (and also point to possible generalisations)

First of all, any complex number $z=a+bi$, is 2-dimensional and each part (the real part $a$ or the imaginary part $bi$) can be completely independent of each other. As a result a complex number can represent in condensed form 2 numbers. Moreover this also means that a complex number to be completely determined each of the dimensions needs to be determined as well.

On the other hand, from every complex number $z=a+bi$ (along with its complex conjugate $\bar{z}=a-bi$), one can calculate 2 real numbers ($a$, $b$) as:

$$a = (z + \bar{z})/2$$

$$b = (z - \bar{z})/{2i}$$

Since $a$ and $b$ can be completely independent of each other, so can $z$ and $\bar{z}$.

There is a complete symmetry of representation (if such a term can be used).

This means that in QFT (for example), instead of doing variations on the $a$, $b$ real fields, one can equivalently (by the same token) do variations on the $z$, $\bar{z}$ complex fields and so on.

UPDATE:

To get into the abstract mathematics a little more.

Complex conjugation is (the natural) automorphism of the field of complex numbers. Furthermore the complex conjugate of a complex number $z$ can not be derived from any analytic function of $z$ (roughly meaning rational functions of $z$ and power series). This further makes the complex conjugate $\bar{z}$ natural candidate for treating as separate field.

Quiz: How many components are needed to compute the velocity $v=dx/dt$ of an object having position $x$, and can these be considered independent? Or in other words knowing position $x$ (at a given time $t$), can we also know velocity $v$ (at the same given time)??

• Thanks Nikos, I like this perspective. As a follow up question: A general complex differential equation, like the Schrodinger equation, is enough to define the evolutionary characteristics of a system. What makes the KG equation different? Why are two equations needed here whereas only one is needed for the Schrodinger equation? – BMS Aug 17 '14 at 18:46
• @BMS, a little textbook on relativistic QFT i had, says, they simple reproduce different energy-momentum relations. Schrodinger equation reproduces the non-relativistic relation $E=p^2/2m +V$ whereas Klein-Gordon reproduces the relativistic $E^2=p^2+(mc^2)^2$ This equation is second order so this means it can describe bosons (simple way to see it is the conditions that make the total energy bounded), whereas the Dirac equation also reproduces the relativistic e-m relation but in first order which bounds the total energy suitable for fermions (spin-statistics theorem et al) – Nikos M. Aug 17 '14 at 18:54

(This post is an answer to the marked as duplicate question there : Independent fields and the Lagrange Density of Schrodinger equation)

# The Lagrangian Density of the Schr$$\ddot{\bf o}$$dinger equation

The necessity to treat the complex fields $$\:\psi,\psi^{\boldsymbol{*}}\:$$ as independent will be clear in the following effort to build an accepted Lagrangian Density for the Schr$$\ddot{\rm o}$$dinger equation.

With real potential $$V$$ the Schr$$\ddot{\rm o}$$dinger equation and its complex conjugate are \begin{align} &\hphantom{--}\!i\hbar \overset{\:\:\centerdot}{\psi}\:\boldsymbol{+}\dfrac{\hbar^2}{2m}\nabla^2\psi\:\:\boldsymbol{-}V\left(\mathbf{x},t\right)\psi\, \boldsymbol{=} 0\,,\quad \,\psi\,\left(\mathbf{x},t\right) \in \mathbb{C}\,, \quad \overset{\:\:\centerdot}{\psi}\,\boldsymbol{\equiv} \dfrac{\partial \psi}{\partial t} \tag{C-01.1}\label{eqC-01.1}\\ &\boldsymbol{-}i\hbar \overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\!\boldsymbol{+}\dfrac{\hbar^2}{2m}\nabla^2\psi^{\boldsymbol{*}}\boldsymbol{-}V\left(\mathbf{x},t\right)\psi^{\boldsymbol{*}}\! \boldsymbol{=} 0\,,\quad \psi^{\boldsymbol{*}}\!\left(\mathbf{x},t\right) \in \mathbb{C}\,, \quad \overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\!\!\boldsymbol{\equiv} \dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t} \tag{C-01.2}\label{eqC-01.2} \end{align}

To find a Lagrange density we change from the complex fields $$\psi,\psi^{\boldsymbol{*}}$$ to the real fields $$\psi_1,\psi_2$$-the real and imaginary parts of $$\psi$$ $$$$\left. \begin{cases} \psi \:\boldsymbol{=} \psi_1 \boldsymbol{+}\mathrm i\, \psi_2\\ \psi^{\boldsymbol{*}} \! \boldsymbol{=} \psi_1 \boldsymbol{-}\mathrm i\, \psi_2 \end{cases}\! \right\} \qquad \psi_1,\psi_2 \in \mathbb{R} \tag{C-03}\label{eqC-03}$$$$

Adding \eqref{eqC-01.1}$$\boldsymbol{+}$$\eqref{eqC-01.2} $$\boldsymbol{\Longrightarrow}$$

$$$$\mathrm i\hbar\left(\overset{\:\:\centerdot}{\psi}\boldsymbol{-} \overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\right)\boldsymbol{+}\dfrac{\hbar^2}{2m}\nabla^2\left(\psi\boldsymbol{+} \psi^{\boldsymbol{*}}\right)\boldsymbol{-}V\left(\psi\boldsymbol{+} \psi^{\boldsymbol{*}}\right)\boldsymbol{=} 0 \nonumber$$$$

$$$$\boxed{\:\: \boldsymbol{-}\hbar\overset{\:\:\centerdot}{\psi}_2\boldsymbol{+}\dfrac{\hbar^2}{2m}\nabla^2\psi_1\boldsymbol{-}V\psi_1\boldsymbol{=} 0\:\:} \tag{C-04}\label{eqC-04}$$$$

Subtracting \eqref{eqC-01.1}$$\boldsymbol{-}$$\eqref{eqC-01.2} $$\boldsymbol{\Longrightarrow}$$

$$$$\mathrm i\hbar\left(\overset{\:\:\centerdot}{\psi}\boldsymbol{+} \overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\right)\boldsymbol{+}\dfrac{\hbar^2}{2m}\nabla^2\left(\psi\boldsymbol{-} \psi^{\boldsymbol{*}}\right)\boldsymbol{-}V\left(\psi\boldsymbol{-} \psi^{\boldsymbol{*}}\right)\boldsymbol{=} 0 \nonumber$$$$

$$$$\boxed{\:\:\hphantom{\boldsymbol{-}}\hbar\overset{\:\:\centerdot}{\psi}_1\boldsymbol{+}\dfrac{\hbar^2}{2m}\nabla^2\psi_2\boldsymbol{-}V\psi_2\boldsymbol{=} 0\:\:} \tag{C-05}\label{eqC-05}$$$$ Equations \eqref{eqC-04},\eqref{eqC-05} are independent with respect to the real fields $$\psi_1,\psi_2$$. So we must treat these fields as independent. These two equations are candidates as the Euler-Lagrange equations of the Schr$$\ddot{\rm o}$$dinger equation.

So consider the Lagrangian density as function of these fields and their space-time derivatives
$$$$\mathcal{L}\left(\psi_1,\boldsymbol{\nabla}\psi_1,\overset{\:\:\centerdot}{\psi}_1,\psi_2,\boldsymbol{\nabla}\psi_2, \overset{\:\:\centerdot}{\psi}_2\right) \tag{C-06}\label{eqC-06}$$$$ The Euler-Lagrange equations are $$$$\dfrac{\partial }{\partial t}\left(\dfrac{\partial \mathcal{L}}{\partial \overset{\:\:\centerdot}{\psi}_k}\right)\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_k\right)}\right]\boldsymbol{-}\dfrac{\partial \mathcal{L}}{\partial \psi_k}\boldsymbol{=}0\,, \quad k=1,2 \tag{C-07}\label{eqC-07}$$$$ that is \begin{align} \dfrac{\partial }{\partial t}\left(\dfrac{\partial \mathcal{L}}{\partial \overset{\:\:\centerdot}{\psi}_1}\right)\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_1\right)}\right]\boldsymbol{-}\dfrac{\partial \mathcal{L}}{\partial \psi_1} & \boldsymbol{=}0 \tag{C-08.1}\label{eqC-08.1}\\ \dfrac{\partial }{\partial t}\left(\dfrac{\partial \mathcal{L}}{\partial \overset{\:\:\centerdot}{\psi}_2}\right)\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_2\right)}\right]\boldsymbol{-}\dfrac{\partial \mathcal{L}}{\partial \psi_2} & \boldsymbol{=}0 \tag{C-08.2}\label{eqC-08.2} \end{align} Expressing equations \eqref{eqC-04} and \eqref{eqC-05} in forms similar to \eqref{eqC-07} we have \begin{align} \dfrac{\partial }{\partial t}\left(\boldsymbol{-}\hbar\psi_2\right)\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\biggl[\dfrac{\hbar^2}{2m}\boldsymbol{\nabla}\psi_1\biggr]\boldsymbol{-}V\psi_1 &\boldsymbol{=} 0 \tag{C-09.1}\label{eqC-09.1}\\ \dfrac{\partial }{\partial t}\left(\boldsymbol{+}\hbar\psi_1\right)\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\biggl[\dfrac{\hbar^2}{2m}\boldsymbol{\nabla}\psi_2\biggr]\boldsymbol{-}V\psi_2 &\boldsymbol{=} 0 \tag{C-09.2}\label{eqC-09.2} \end{align} If we suppose that \eqref{eqC-09.1} and \eqref{eqC-09.2} are produced from \eqref{eqC-08.1} and \eqref{eqC-08.2} respectively then we have good reasons to guess the following $$$$\left. \begin{cases} \left(\dfrac{\partial \mathcal{L}}{\partial \overset{\:\:\centerdot}{\psi}_2}\right)\stackrel{\text {to give}}{-\!\!\!-\!\!\!\longrightarrow} \alpha\,\hbar\,\psi_1\\ \left(\dfrac{\partial \mathcal{L}}{\partial \overset{\:\:\centerdot}{\psi}_1}\right)\stackrel{\text {to give}}{-\!\!\!-\!\!\!\longrightarrow} \beta\,\hbar\,\psi_2 \end{cases} \right\} \Longrightarrow \left. \begin{cases} \alpha\,\hbar\,\psi_1\overset{\:\:\centerdot}{\psi}_2 \in \mathcal{L}\vphantom{\left(\dfrac{\partial \mathcal{L}}{\partial \overset{\:\:\centerdot}{\psi}_2}\right)}\\ \beta\,\hbar\,\overset{\:\:\centerdot}{\psi}_1\psi_2 \in \mathcal{L}\vphantom{\left(\dfrac{\partial \mathcal{L}}{\partial \overset{\:\:\centerdot}{\psi}_2}\right)} \end{cases} \right\} \tag{C-10}\label{eqC-10}$$$$

$$$$\left. \begin{cases} \left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_1\right)}\right]\stackrel{\text {to give}}{-\!\!\!-\!\!\!\longrightarrow} \gamma\,\dfrac{\hbar^2}{2m}\,\boldsymbol{\nabla}\psi_1\\ \left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_2\right)}\right]\stackrel{\text {to give}}{-\!\!\!-\!\!\!\longrightarrow} \delta\,\dfrac{\hbar^2}{2m}\,\boldsymbol{\nabla}\psi_2 \end{cases} \right\} \Longrightarrow \left. \begin{cases} \gamma\,\dfrac{\hbar^2}{4m}\,\Vert\boldsymbol{\nabla}\psi_1\Vert^2 \in \mathcal{L}\vphantom{\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_1\right)}\right]}\\ \delta\,\dfrac{\hbar^2}{4m}\,\Vert\boldsymbol{\nabla}\psi_2\Vert^2 \in \mathcal{L}\vphantom{\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_1\right)}\right]} \end{cases} \right\} \tag{C-11}\label{eqC-11}$$$$

$$$$\left. \begin{cases} \dfrac{\partial \mathcal{L}}{\partial \psi_1}\stackrel{\text {to give}}{-\!\!\!-\!\!\!\longrightarrow} \zeta\,V\psi_1\vphantom{\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_1\right)}\right]}\\ \dfrac{\partial \mathcal{L}}{\partial \psi_2}\stackrel{\text {to give}}{-\!\!\!-\!\!\!\longrightarrow} \eta\,V\psi_2\vphantom{\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_1\right)}\right]} \end{cases} \right\} \Longrightarrow \left. \begin{cases} \zeta\,V\psi^2_1 \in \mathcal{L}\vphantom{\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_1\right)}\right]}\\ \eta\,V\psi^2_2 \in \mathcal{L}\vphantom{\left[\dfrac{\partial \mathcal{L}}{\partial \left(\boldsymbol{\nabla}\psi_1\right)}\right]} \end{cases} \right\} \tag{C-12}\label{eqC-12}$$$$ From equations \eqref{eqC-10},\eqref{eqC-11} and \eqref{eqC-12} we conclude that the Lagrangian density of \eqref{eqC-06} must have the general form \begin{align} &\mathcal{L}\left(\psi_1,\boldsymbol{\nabla}\psi_1,\overset{\:\:\centerdot}{\psi}_1,\psi_2,\boldsymbol{\nabla}\psi_2, \overset{\:\:\centerdot}{\psi}_2\right)\boldsymbol{=} \nonumber\\ &\alpha\,\hbar\,\psi_1\overset{\:\:\centerdot}{\psi}_2 \boldsymbol{+}\beta\,\hbar\,\overset{\:\:\centerdot}{\psi}_1\psi_2 \boldsymbol{+}\gamma\,\dfrac{\hbar^2}{4m}\,\Vert\boldsymbol{\nabla}\psi_1\Vert^2\boldsymbol{+}\delta\,\dfrac{\hbar^2}{4m}\,\Vert\boldsymbol{\nabla}\psi_2\Vert^2\boldsymbol{+}\zeta V\psi_1^2\boldsymbol{+}\eta V\psi^2_2 \tag{C-13}\label{eqC-13} \end{align} where $$\:\alpha,\beta,\gamma,\delta,\zeta,\eta \:$$ real coefficients to be determined.

Inserting this expression of $$\;\mathcal{L}\;$$ in \eqref{eqC-08.1},\eqref{eqC-08.2} we have respectively \begin{align} \dfrac{\partial }{\partial t}\biggl[\left(\beta\boldsymbol{-}\alpha \right)\hbar\,\psi_2\biggr]\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\biggl[\gamma\,\dfrac{\hbar^2}{2m}\,\boldsymbol{\nabla}\psi_1\biggr]\boldsymbol{-}2\zeta V\psi_1 & \boldsymbol{=}0 \tag{C-14.1}\label{eqC-14.1}\\ \dfrac{\partial }{\partial t}\biggl[\left(\alpha\boldsymbol{-}\beta \right)\hbar\,\psi_1\biggr]\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\biggl[\delta\,\dfrac{\hbar^2}{2m}\,\boldsymbol{\nabla}\psi_2\biggr]\boldsymbol{-}2\eta V\psi_2 & \boldsymbol{=}0 \tag{C-14.2}\label{eqC-14.2} \end{align} Comparing \eqref{eqC-14.1},\eqref{eqC-14.2} with \eqref{eqC-09.1},\eqref{eqC-09.2} we must have $$$$\dfrac{\alpha\boldsymbol{-}\beta}{1}=\dfrac{\beta\boldsymbol{-}\alpha}{\boldsymbol{-}1}=\dfrac{\gamma}{1}=\dfrac{\delta}{1}=\dfrac{2\zeta}{1}=\dfrac{2\eta}{1}=\lambda \tag{C-15}\label{eqC-15}$$$$ Setting the common free factor $$\;\lambda=\boldsymbol{-}2\;$$ we have $$\beta=\alpha+2,\,\gamma=\delta=-2,\, \zeta=\eta=-1$$ and equation \eqref{eqC-13} yields \begin{align} &\mathcal{L}\left(\psi_1,\boldsymbol{\nabla}\psi_1,\overset{\:\:\centerdot}{\psi}_1,\psi_2,\boldsymbol{\nabla}\psi_2, \overset{\:\:\centerdot}{\psi}_2\right)\boldsymbol{=} \nonumber\\ &\alpha\,\hbar\,\psi_1\overset{\:\:\centerdot}{\psi}_2 \boldsymbol{+}\left(\alpha\boldsymbol{+}2\right)\hbar\,\overset{\:\:\centerdot}{\psi}_1\psi_2\boldsymbol{-}\dfrac{\hbar^2}{2m}\left(\Vert\boldsymbol{\nabla}\psi_1\Vert^2\boldsymbol{+}\Vert\boldsymbol{\nabla}\psi_2\Vert^2\right)\boldsymbol{-}V\left(\psi^2_1\boldsymbol{+}\psi^2_2\right) \tag{C-16}\label{eqC-16} \end{align} We return now from the real fields $$\psi_1,\psi_2$$ to the complex fields $$\psi,\psi^{\boldsymbol{*}}$$ replacing in \eqref{eqC-16} $$$$\left. \begin{cases} \psi_1 \boldsymbol{=} \dfrac{\psi\boldsymbol{+}\psi^{\boldsymbol{*}}}{2}\\ \psi_2 \boldsymbol{=} \mathrm i \dfrac{\psi^{\boldsymbol{*}}\boldsymbol{-}\psi}{2} \end{cases} \right\} \tag{C-17}\label{eqC-17}$$$$ Now \begin{align} \alpha\,\hbar\,\psi_1\overset{\:\:\centerdot}{\psi}_2 & \boldsymbol{=}\mathrm i\,\alpha\,\hbar\,\left(\dfrac{\psi\boldsymbol{+}\psi^{\boldsymbol{*}}}{2}\vphantom{\dfrac{\overset{\:\:\centerdot}{\psi}}{2}}\right)\left(\dfrac{\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\boldsymbol{-}\overset{\:\:\centerdot}{\psi}}{2}\right) \nonumber\\ &\boldsymbol{=}\mathrm i\,\alpha\,\hbar\, \left(\dfrac{\psi\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\boldsymbol{-}\psi\overset{\:\:\centerdot}{\psi}\boldsymbol{+}\psi^{\boldsymbol{*}}\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\boldsymbol{-}\overset{\:\:\centerdot}{\psi}\psi^{\boldsymbol{*}}}{4}\right) \tag{C-18.1}\label{eqC-18.1}\\ \left(\alpha\boldsymbol{+}2\right)\hbar\,\overset{\:\:\centerdot}{\psi}_1\psi_2 &\boldsymbol{=}\mathrm i \left(\alpha\boldsymbol{+}2\right)\hbar\,\left(\dfrac{\overset{\:\:\centerdot}{\psi}\boldsymbol{+}\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}}{2}\right)\left(\dfrac{\psi^{\boldsymbol{*}}\boldsymbol{-}\psi}{2}\vphantom{\dfrac{\dot{\psi}}{2}}\right) \nonumber\\ &\boldsymbol{=}\mathrm i \left(\alpha\boldsymbol{+}2\right)\hbar\,\left(\dfrac{\overset{\:\:\centerdot}{\psi}\psi^{\boldsymbol{*}}\boldsymbol{-}\psi\overset{\:\:\centerdot}{\psi}\boldsymbol{+}\psi^{\boldsymbol{*}}\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\boldsymbol{-}\psi\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}}{4}\right) \tag{C-18.2}\label{eqC-18.2} \end{align} so $$$$\alpha\,\hbar\,\psi_1\overset{\:\:\centerdot}{\psi}_2 \boldsymbol{+}\left(\alpha\boldsymbol{+}2\right)\hbar\,\overset{\:\:\centerdot}{\psi}_1\psi_2\boldsymbol{=}\mathrm i\,\hbar\,\left(\dfrac{\overset{\:\:\centerdot}{\psi}\psi^{\boldsymbol{*}}\boldsymbol{-}\psi\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}}{2}\right) \boldsymbol{+}\mathrm i\,\hbar\,\left(\alpha\boldsymbol{+}1\right)\left(\dfrac{\psi^{\boldsymbol{*}}\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\boldsymbol{-}\psi\overset{\:\:\centerdot}{\psi}}{2}\right) \tag{C-19}\label{eqC-19}$$$$ Also $$$$\Vert\boldsymbol{\nabla}\psi_1\Vert^2\boldsymbol{+}\Vert\boldsymbol{\nabla}\psi_2\Vert^2\boldsymbol{=}\left(\boldsymbol{\nabla}\psi_1\boldsymbol{+}\mathrm i\boldsymbol{\nabla}\psi_2\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\psi_1\boldsymbol{-}\mathrm i\boldsymbol{\nabla}\psi_2\right)\boldsymbol{=}\boldsymbol{\nabla}\psi\boldsymbol{\cdot}\boldsymbol{\nabla}\psi^{\boldsymbol{*}} \tag{C-20}\label{eqC-20}$$$$ and $$$$\psi_1^2\boldsymbol{+}\psi_2^2\boldsymbol{=}\left(\psi_1\boldsymbol{+}\mathrm i\psi_2\right)\left(\psi_1\boldsymbol{-}\mathrm i\psi_2\right)\boldsymbol{=}\psi\psi^{\boldsymbol{*}} \tag{C-21}\label{eqC-21}$$$$ Inserting the expressions \eqref{eqC-19}, \eqref{eqC-20} and \eqref{eqC-21} in \eqref{eqC-16} we have finally

\begin{align} &\mathcal{L}\left(\psi,\boldsymbol{\nabla}\psi,\overset{\:\:\centerdot}{\psi},\psi^{\boldsymbol{*}},\boldsymbol{\nabla}\psi^{\boldsymbol{*}}, \overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\right)\boldsymbol{=} \nonumber\\ &\mathrm i\,\hbar\,\left(\dfrac{\overset{\:\:\centerdot}{\psi}\psi^{\boldsymbol{*}}\boldsymbol{-}\psi\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}}{2}\right) \boldsymbol{+}\mathrm i\,\hbar\,\left(\alpha\boldsymbol{+}1\right)\left(\dfrac{\psi^{\boldsymbol{*}}\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\boldsymbol{-}\psi\overset{\:\:\centerdot}{\psi}}{2}\right)\boldsymbol{-}\dfrac{\hbar^2}{2m}\boldsymbol{\nabla}\psi\boldsymbol{\cdot}\boldsymbol{\nabla}\psi^{\boldsymbol{*}} \boldsymbol{-}V\psi\psi^{\boldsymbol{*}}\:\:\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}b}} \tag{C-22}\label{eqC-22} \end{align} It's not difficult to verify that the Euler-Lagrange equations of above Lagrangian Density with respect to $$\:\psi^{\boldsymbol{*}}\:$$ and $$\:\psi\:$$ are the Schr$$\ddot{\rm o}$$dinger equation \eqref{eqC-01.1} and its complex conjugate \eqref{eqC-01.2} respectively. This is valid for any value of the parameter $$\:\alpha$$.

Now, the Lagrangian Density we meet in many textbooks $$$$\mathcal{L}\left(\psi,\boldsymbol{\nabla}\psi,\overset{\:\:\centerdot}{\psi},\psi^{\boldsymbol{*}},\boldsymbol{\nabla}\psi^{\boldsymbol{*}}, \overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\right)=\mathrm i\hbar\psi^{\boldsymbol{*}}\overset{\:\:\centerdot}{\psi}\!\boldsymbol{-}\dfrac{\hbar^2}{2m}\boldsymbol{\nabla}\psi\!\boldsymbol{\cdot}\!\boldsymbol{\nabla}\psi^{\boldsymbol{*}} \!\boldsymbol{-}\!V\psi\psi^{\boldsymbol{*}} \tag{C-22a}\label{eqC-22a}$$$$ could not be reached from \eqref{eqC-22} for any value of the parameter $$\:\alpha$$. To do this we will find a more general Lagrangian Density. The basic idea comes from the Lagrangian Mechanics of discrete systems. We know that therein the Euler-Lagrange equations are invariant under the addition to the Lagrange function $$\:L\left(q_{i},\overset{\!\centerdot}{q}_{i},t\right)\:$$ of the total differential of a function $$\:F\left(q_{i}\right)\:$$ of the generalized coordinates. Extending this idea herein we note that the Euler-Lagrange equations will be invariant if to the Lagrangian Density \eqref{eqC-22} we add the total differential of a function $$\:F\left(\psi,\psi^{\boldsymbol{*}}\right)\:$$ of the complex fields $$\:\psi,\psi^{\boldsymbol{*}}$$ so that $$$$\mathcal{L'}\boldsymbol{=}\mathcal{L}\boldsymbol{+}\dfrac{\partial F\left(\psi,\psi^{\boldsymbol{*}}\right)}{\partial t}\boldsymbol{=}\mathcal{L}\boldsymbol{+}\dfrac{\partial F}{\partial \psi}\overset{\:\:\centerdot}{\psi}\boldsymbol{+}\dfrac{\partial F}{\partial \psi^{\boldsymbol{*}}}\overset{\!\centerdot}{\psi^{\boldsymbol{*}}} \tag{C-23}\label{eqC-23}$$$$ We use two of the most simple functions \begin{align} F_1\left(\psi,\psi^{\boldsymbol{*}}\right) & \boldsymbol{=} \mathrm i\,\hbar\,\dfrac{\rho\,\psi\,\psi^{\boldsymbol{*}} }{2} \quad \Longrightarrow \quad \dfrac{\partial F_1\left(\psi,\psi^{\boldsymbol{*}}\right)}{\partial t}\boldsymbol{=}\mathrm i\,\hbar\,\left(\dfrac{\rho\,\overset{\:\:\centerdot}{\psi}\psi^{\boldsymbol{*}}\boldsymbol{+}\rho\,\psi\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}}{2}\right) \tag{C-24.1}\label{eqC-24.1}\\ F_2\left(\psi,\psi^{\boldsymbol{*}}\right) & \boldsymbol{=} \mathrm i\,\hbar\,\dfrac{\sigma \left(\psi^{\boldsymbol{*}2}\boldsymbol{+}\psi^2\right)}{4} \quad \Longrightarrow \quad \dfrac{\partial F_2\left(\psi,\psi^{\boldsymbol{*}}\right)}{\partial t}\boldsymbol{=}\mathrm i\,\hbar\,\left(\dfrac{\sigma\,\psi^{\boldsymbol{*}}\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\boldsymbol{+}\sigma\,\psi\overset{\:\:\centerdot}{\psi}}{2}\right) \tag{C-24.2}\label{eqC-24.2} \end{align} so that $$$$\mathcal{L'}\boldsymbol{=}\mathcal{L}\boldsymbol{+}\dfrac{\partial F_1\left(\psi,\psi^{\boldsymbol{*}}\right)}{\partial t}\boldsymbol{+}\dfrac{\partial F_2\left(\psi,\psi^{\boldsymbol{*}}\right)}{\partial t} \tag{C-25}\label{eqC-25}$$$$ With $$\:\chi\equiv\alpha\boldsymbol{+}1\:$$ the new more general Lagrangian Density is \begin{align} &\mathcal{L}\left(\psi,\boldsymbol{\nabla}\psi,\overset{\:\:\centerdot}{\psi},\psi^{\boldsymbol{*}},\boldsymbol{\nabla}\psi^{\boldsymbol{*}}, \overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\right)\boldsymbol{=} \nonumber\\ &\mathrm i\hbar\left[\dfrac{\left(1\!\boldsymbol{+}\!\rho\right)\overset{\:\:\centerdot}{\psi}\psi^{\boldsymbol{*}}\!\boldsymbol{-}\!\left(1\!\boldsymbol{-}\!\rho\right)\psi\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}}{2}\right]\!\boldsymbol{+}\!\mathrm i\hbar\left[\dfrac{\left(\chi\!\boldsymbol{+}\!\sigma\right)\psi^{\boldsymbol{*}}\overset{\!\centerdot}{\psi^{\boldsymbol{*}}}\!\boldsymbol{-}\!\left(\chi\!\boldsymbol{-}\!\sigma\right)\psi\overset{\:\:\centerdot}{\psi}}{2}\right]\!\boldsymbol{-}\!\dfrac{\hbar^2}{2m}\boldsymbol{\nabla}\psi\!\boldsymbol{\cdot}\!\boldsymbol{\nabla}\psi^{\boldsymbol{*}} \!\boldsymbol{-}\!V\psi\psi^{\boldsymbol{*}}\:\:\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}b}} \tag{C-26}\label{eqC-26} \end{align} Again we could verify that the Euler-Lagrange equations of above Lagrangian Density with respect to $$\:\psi^{\boldsymbol{*}}\:$$ and $$\:\psi\:$$ are the Schr$$\ddot{\rm o}$$dinger equation \eqref{eqC-01.1} and its complex conjugate \eqref{eqC-01.2} respectively. This is valid for any values of the parameters $$\:\chi,\rho,\sigma$$. But especially $$$$\left. \begin{cases} \chi=0\\ \rho=1\\ \sigma=0 \end{cases} \right\} \Longrightarrow \mathcal{L}=\mathrm i\hbar\psi^{\boldsymbol{*}}\overset{\:\:\centerdot}{\psi}\!\boldsymbol{-}\dfrac{\hbar^2}{2m}\boldsymbol{\nabla}\psi\!\boldsymbol{\cdot}\!\boldsymbol{\nabla}\psi^{\boldsymbol{*}} \!\boldsymbol{-}\!V\psi\psi^{\boldsymbol{*}} \tag{C-27}\label{eqC-27}$$$$ that is the Lagrangian Density \eqref{eqC-22a}.