Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Field and conjugate independent?

  1. May 17, 2013 #1
    I'm having trouble figuring out how a field and it's conjugate are independent quantities. How can they be, when they are related by conjugation?

    Suppose you have real fields x and y, and form fields: L=(x+iy)/sqrt2 R=(x-iy)/sqrt2

    In a path integral, you'd have .5(∂x∂x+∂y∂y) in your Lagrangian. Changing variables to L and R you'd have ∂L∂R. How can L and R be independent in the path integral? At each spacetime point, the quantity in the Lagrangian is of the form ∂L∂R=number*(its conjugate). If L and R vary independently, then your Lagrangian might not even be real, since only a number times its conjugate is real.
  2. jcsd
  3. May 18, 2013 #2
    If you deal with x and y, you only have two (real) field degrees of freedom. Hence, if you pass to some new fields L and R, you must somehow get the same number of field degrees of freedom. This could in principle be done by letting L and R be independent complex fields (4 field d.o.f.) with two constraint equations, or you can treat L and R as independent real-valued fields. This way you'll always get a real Lagrangian.

    (Peskin and Schroeder seem to use the latter convention, for example. However, they are rather cryptic about it, only saying that L and R are treated as the new dynamical variables.)

    In addition, and perhaps more importantly, it is quite simple to show that both methods lead to the same equations of motion (i.e. physics), by just considering the variation of the action:
    [tex]\delta S =\int d^4 x \left( F \delta L + G\delta R \right) \quad \Rightarrow F=G=0[/tex] or, in the constituent fields, [tex]\delta S =\int d^4 x \left[ \left( F+G \right) \delta x + i \left( F-G \right) \delta y \right] \quad \Rightarrow F+G=F-G=0[/tex]
    (F and G are some functions, which necessarily depend on the exact nature of S.)
  4. May 18, 2013 #3
    How can L and R be two (independent) real fields? Is some type of analytic continuation made? In the path integral when integrating over the complex fields, it's often just treated as if they were real for some reason.

    I think that argument can be made for any transformation. It's a general property that Lagrange's equations are transformation invariant.
  5. May 18, 2013 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    A field and its complex conjugate are obviously not independent, but in some situations, it makes sense to pretend that they are. I don't think I can explain it specifically for path integrals, since I haven't looked at those in a long time. But I think that what I'm about to say will always be part of the explanation. (Thanks to Avodyne for showing me this in another thread).

    Given a function ##f:\mathbb C\to\mathbb R##, we can define a function ##R:\mathbb R^2\to\mathbb R## by
    $$R(x,y)=f(x+iy)$$ for all ##x,y\in\mathbb R##, and a function ##C:\mathbb C^2\to\mathbb R## by
    $$C(z,w)=R\left(\frac{z+w^*}{2},\frac{z-w^*}{2i}\right).$$ Let ##z\in\mathbb C## be arbitrary and define ##x,y\in\mathbb R## by ##z=x+iy##. We have
    $$f(z)=f(x+iy)=R(x,y)=R\left(\frac{z+z^*}{2},\frac{z-z^*}{2i}\right)=C(z,z^*)=C(x+iy,x-iy).$$ If you see a notation like
    $$\frac{\partial f}{\partial z^*}$$ it should be interpreted as ##D_2C(z,z^*)##.

    There's a simple relationship between the partial derivatives of R and the partial derivatives of C.
    D_1R(x,y) &=\frac{\partial}{\partial x}C(x+iy,x-iy)=D_1C(z,z^*)+D_2C(z,z^*)\\
    D_2R(x,y) &=\frac{\partial}{\partial y}C(x+iy,x-iy)=iD_1C(z,z^*)-iD_2C(z,z^*).
    \end{align} These results imply that
    Results like these can be used to motivate notations like
    $$\frac{\partial f}{\partial z}=\frac 1 2\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)f.$$ I haven't thought about how to apply this sort of stuff to path integrals, but maybe this can help you figure it out on your own.
  6. May 20, 2013 #5
    If the transformation were missing the factor of i, everything would be fine. Then you would have z=x+y and z*=x-y which are two perfectly independent variables and you can chain rule everything.

    Formally, transformations z=x+iy and z*=x-iy have nonzero determinant, so the two variables are independent in the sense that (x,y) and (z,z*) are equally good descriptions. But because you are introducing complex variables, the i tells you something, making (x,y) and (z) equally good descriptions. So I think for manipulations, because of the nonzero of the determinant, everything is fine, like using a (x+y,x-y) description with chain rule.

    The trouble is how do you integrate with complex variables? What exactly does ∫ dz mean? Is it a path in the complex plane, if so, what path!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook