Treating phi and phi-dagger as independent

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  • #1
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In treating the complex scalar field in QFT, [itex]\phi[/itex] and [itex]\phi^\dagger[/itex] are treated as independent variables. I'd like to make sure I understand what this actually means:

Is this the same idea as treating [itex]\phi[/itex] and [itex]\partial \phi[/itex] 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 [itex]\phi (t)[/itex] and [itex] \partial \phi (t) [/itex] into the slots. Physicists abbreviate this procedure as treating [itex]\phi[/itex] and [itex]\partial \phi[/itex] as independent variables.

Is this correct?
 

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  • #2
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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.
 
  • #3
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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.
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 [itex]\phi[/itex] and [itex]\phi ^\dagger[/itex] are independent variables, writing expressions such as [itex] \phi^\dagger \phi [/itex] in the lagrangian and differentiating with respect to them. What do you make of, for example, differentiating with respect to [itex]\phi^\dagger[/itex], as this is seemingly incompatible with your assertion.
 
  • #4
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Well, we know that:

[tex]
\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}
[/tex]
where [itex]\phi_{R}, \phi_{I}[/itex] are real scalar fields and the factor [itex]2^{-1/2}[/itex] is introduced to make the Jacobian of the transformation have modulo 1.

Then, one can write the expression [itex]\phi^{\dagger} \, \phi[/itex] as:
[tex]
\frac{1}{2} \, \left(\phi^{2}_{R} + \phi^{2}_{I} + i \, [\phi_{R}, \phi_{I}]\right)
[/tex]
and the last term drops out assumng the two real fields commute between each other, which is the same as:
[tex]
[\phi_{R}, \phi_{I}] = \left[\frac{\phi + \phi^{\dagger}}{\sqrt{2}}, \frac{\phi - \phi^{\dagger}}{\sqrt{2} \, i}\right] = i [\phi, \phi^{\dagger}] = 0
[/tex]
Then, following the chain rule for partial differentiation:
[tex]
\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)
[/tex]
The partial derivatives w.r.t. to the real and imaginary part are:
[tex]
\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}
[/tex]
[tex]
\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}
[/tex]
and we finally have:
[tex]
\frac{\partial}{\partial \phi^{\dagger}} \, \left(\phi^{\dagger} \, \phi\right) = \frac{\phi_{R} + i \, \phi_{I}}{\sqrt{2}} = \phi
[/tex]
which is exactly as it should be.
 
  • #5
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The symmetry SO(2) means that we have a liberty of choosing what we call [itex]\phi_{R}[/itex] and [itex]\phi_{I}[/itex]. Namely, we can perform a 'rotation':

[tex]
\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)
[/tex]
The matrix that achieves this rotation is the general form of a 2x2 matrix that is orthogonal (meaning [itex]O \cdot O^{T} = 1[/itex]), hence the term [itex]O(2)[/itex] and has a determinant one (hence the prefix special).

If one translates what this means for the complex scalar field, one gets:
[tex]
\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
[/tex]
and similarly:
[tex]
\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}
[/tex]
This is what is called [itex]U(1)[/itex] transfromation meaning a unitary 1x1 matrix (i.e. a complex number with modulo 1).

The term [itex]\phi^{\dagger} \, \phi[/itex] remains unchanged under the [itex]U(1)[/itex] transformation and so are the other terms in the lagrangian. Hence, the Lagrangian has [itex]U(1)[/itex] symmetry. You can rewrite everything in term of the real and imaginary parts and see that the Lagrangian has [itex]SO(2)[/itex] symmetry. Hence, these two symmetries are isomorphic.
 
Last edited:
  • #6
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Thank you! This was helpful.

I am still confused about the symmetry business: you claim that because SO(2) ~ U(1) that two real scalar fields (endowed with SO(2) symmetry) behave as a single complex field. I understand what you mean by possession of a symmetry now, but I don't understand why isomorphic symmetries implies equivalent fields.
 

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