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Treating phi and phi-dagger as independent

  1. Jul 19, 2011 #1
    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?
  2. jcsd
  3. Jul 19, 2011 #2
    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.
  4. Jul 19, 2011 #3
    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.
  5. Jul 19, 2011 #4
    Well, we know that:

    \phi & = & \frac{\phi_{R} + i \, \phi_{I}}{\sqrt{2}} \\

    \phi^{\dagger} & = & \frac{ \phi_{R} - i \, \phi_{I}}{\sqrt{2}}
    \end{array} \Rightarrow
    \phi_{R} & = & \frac{\phi + \phi^{\dagger}}{\sqrt{2}} \\

    \phi_{I} & = & \frac{\phi - \phi^{\dagger}}{\sqrt{2} \, i} \\
    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:
    \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.
  6. Jul 19, 2011 #5
    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':

    \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} \\

    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:
    \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 [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: Jul 19, 2011
  7. Jul 19, 2011 #6
    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.
  8. Jul 20, 2011 #7


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