# 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.

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

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.

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Although I agree that in the context described by the OP, its usually used as a mathematical trick, it's also that case that one could consider an extended field theory of independent, complex scalar fields if one so desired. Also is it not also important to make some remark about how, complexification + EL equations + imposing conjugate condition will yield the same equations of motion as not having used the trick in the first place? – joshphysics Dec 5 '13 at 0:02
See also Sidney Coleman's QFT notes; p. 56-57. – Qmechanic Jul 24 '15 at 12:12

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)??

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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

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.

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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 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 at 8:46