# Treating phi and phi-dagger as independent

In treating the complex scalar field in QFT, $\phi$ and $\phi^\dagger$ are treated as independent variables. I'd like to make sure I understand what this actually means:

Is this the same idea as treating $\phi$ and $\partial \phi$ as independent variables in the Lagrangian? Formally, we view the Lagrangian as a function of two variables, and we can differentiate with respect to the two slots as we please, but at the end of the day, we will plug $\phi (t)$ and $\partial \phi (t)$ into the slots. Physicists abbreviate this procedure as treating $\phi$ and $\partial \phi$ as independent variables.

Is this correct?

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No. A complex scalar field is actually two real scalar fields that obey SO(2) symmetry. Since SO(2) ~ U(1), these two fields are equivalent to one complex scalar field. You should think of the real and imaginary parts of the complex scalar field as independent variables.

No. A complex scalar field is actually two real scalar fields that obey SO(2) symmetry. Since SO(2) ~ U(1), these two fields are equivalent to one complex scalar field. You should think of the real and imaginary parts of the complex scalar field as independent variables.
A couple questions:
What precisely does it mean to "obey a symmetry?"

QFT in a Nutshell and Peskin and Schroeder (If I recall correctly -- I don't have the latter on hand right now) adopt the view that $\phi$ and $\phi ^\dagger$ are independent variables, writing expressions such as $\phi^\dagger \phi$ in the lagrangian and differentiating with respect to them. What do you make of, for example, differentiating with respect to $\phi^\dagger$, as this is seemingly incompatible with your assertion.

Well, we know that:

$$\begin{array}{rcl} \phi & = & \frac{\phi_{R} + i \, \phi_{I}}{\sqrt{2}} \\ \phi^{\dagger} & = & \frac{ \phi_{R} - i \, \phi_{I}}{\sqrt{2}} \end{array} \Rightarrow \begin{array}{rcl} \phi_{R} & = & \frac{\phi + \phi^{\dagger}}{\sqrt{2}} \\ \phi_{I} & = & \frac{\phi - \phi^{\dagger}}{\sqrt{2} \, i} \\ \end{array}$$
where $\phi_{R}, \phi_{I}$ are real scalar fields and the factor $2^{-1/2}$ is introduced to make the Jacobian of the transformation have modulo 1.

Then, one can write the expression $\phi^{\dagger} \, \phi$ as:
$$\frac{1}{2} \, \left(\phi^{2}_{R} + \phi^{2}_{I} + i \, [\phi_{R}, \phi_{I}]\right)$$
and the last term drops out assumng the two real fields commute between each other, which is the same as:
$$[\phi_{R}, \phi_{I}] = \left[\frac{\phi + \phi^{\dagger}}{\sqrt{2}}, \frac{\phi - \phi^{\dagger}}{\sqrt{2} \, i}\right] = i [\phi, \phi^{\dagger}] = 0$$
Then, following the chain rule for partial differentiation:
$$\frac{\partial}{\partial \phi^{\dagger}} = \frac{\partial \phi_{R}}{\partial \phi^{\dagger}} \, \frac{\partial}{\partial \phi_{R}} + \frac{\partial \phi_{I}}{\partial \phi^{\dagger}} \, \frac{\partial}{\partial \phi_{I}} = \frac{1}{\sqrt{2}} \, \left(\frac{\partial}{\partial \phi_{R}} + i \, \frac{\partial}{\partial \phi_{I}}\right)$$
The partial derivatives w.r.t. to the real and imaginary part are:
$$\frac{\partial}{\partial \phi_{R}} \, \left(\phi^{\dagger} \, \phi\right) = \frac{\partial}{\partial \phi_{R}} \, \left[\frac{1}{2} \, \left(\phi^{2}_{R} + \phi^{2}_{I}\right)\right] = \phi_{R}$$
$$\frac{\partial}{\partial \phi_{I}} \, \left(\phi^{\dagger} \, \phi\right) = \frac{\partial}{\partial \phi_{I}} \, \left[\frac{1}{2} \, \left(\phi^{2}_{R} + \phi^{2}_{I}\right)\right] = \phi_{I}$$
and we finally have:
$$\frac{\partial}{\partial \phi^{\dagger}} \, \left(\phi^{\dagger} \, \phi\right) = \frac{\phi_{R} + i \, \phi_{I}}{\sqrt{2}} = \phi$$
which is exactly as it should be.

The symmetry SO(2) means that we have a liberty of choosing what we call $\phi_{R}$ and $\phi_{I}$. Namely, we can perform a 'rotation':

$$\left(\begin{array}{c} \phi'_{R} \\ \phi'_{I}\end{array}\right) = \left(\begin{array}{cc} \cos{\alpha} & \sin{\alpha} \\ -\sin{\alpha} & \cos{\alpha} \end{array}\right) \cdot \left(\begin{array}{c} \phi_{R} \\ \phi_{I}\end{array}\right)$$
The matrix that achieves this rotation is the general form of a 2x2 matrix that is orthogonal (meaning $O \cdot O^{T} = 1$), hence the term $O(2)$ and has a determinant one (hence the prefix special).

If one translates what this means for the complex scalar field, one gets:
$$\phi' = \frac{\phi'_{R} + i \, \phi'_{I}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \, \left[(\cos{\alpha} \, \phi_{R} + \sin{\alpha} \, \phi_{I}) + i \, (-\sin{\alpha} \, \phi_{R} + \cos{\alpha} \, \phi_{I})\right] = (\cos{\alpha} - i \, \sin{\alpha}) \, \frac{\phi_{R} + i \, \phi_{I}}{\sqrt{2}} = e^{-i \, \alpha} \, \phi$$
and similarly:
$$\phi'^{\dagger} = \frac{\phi'_{R} - i \, \phi'_{I}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \, \left[(\cos{\alpha} \, \phi_{R} + \sin{\alpha} \, \phi_{I}) - i \, (-\sin{\alpha} \, \phi_{R} + \cos{\alpha} \, \phi_{I})\right] = (\cos{\alpha} + i \, \sin{\alpha}) \, \frac{\phi_{R} - i \, \phi_{I}}{\sqrt{2}} = e^{i \, \alpha} \, \phi^{\dagger}$$
This is what is called $U(1)$ transfromation meaning a unitary 1x1 matrix (i.e. a complex number with modulo 1).

The term $\phi^{\dagger} \, \phi$ remains unchanged under the $U(1)$ transformation and so are the other terms in the lagrangian. Hence, the Lagrangian has $U(1)$ symmetry. You can rewrite everything in term of the real and imaginary parts and see that the Lagrangian has $SO(2)$ symmetry. Hence, these two symmetries are isomorphic.

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