Complex conjugate an independent variable?

[Chen]

It's very commong to use z and z* as two independent variables, differentiating with respect to one while keeping the other constant. Can you please give me some intuitive insight into this method, and why it works so well? Because every time I see this my first thought is that z and z* are NOT independent, there is a very clear transformation from one to the other...

Thanks,
Chen

 

[Chen]

And also, when I see an integration being done over both z and z*, with the integrals being one-dimensional (as far as I can tell, no boundaries are given), am I to understand that this is simply a formal way to say that the integration is done over the whole complex space?

 

[Chen]

To give a concrete example to my last question, take a look here, at the second identity:
http://www.pact.cpes.sussex.ac.uk/~markh/Teaching/RQF3/node22.html
(this is actually exactly what I'm dealing with)

 

[OrderOfThings]

It's very commong to use z and z* as two independent variables, differentiating with respect to one while keeping the other constant. Can you please give me some intuitive insight into this method, and why it works so well? Because every time I see this my first thought is that z and z* are NOT independent, there is a very clear

Indeed, z and \bar{z} are not independent, but they are linear independent. The partial derivatives \partial /\partial z and \partial /\partial\bar{z} measures the orientation-preservering and orientation-reversing parts of a complex function. They can be defined as

\frac{\partial f}{\partial z} = \lim_{r\rightarrow 0}\frac{1}{2\pi}\int_0^{2\pi}\frac{f(z+re^{it})-f(z)}{re^{it}}dt

\frac{\partial f}{\partial\bar{z}} = \lim_{r\rightarrow 0}\frac{1}{2\pi}\int_0^{2\pi}\frac{f(z+re^{it})-f(z)}{re^{-it}}dt

 

[OrderOfThings]

And also, when I see an integration being done over both z and z*, with the integrals being one-dimensional (as far as I can tell, no boundaries are given), am I to understand that this is simply a formal way to say that the integration is done over the whole complex space?

Yes. dzd\bar{z} is nothing but -2i times the usual area element. The calculation goes like this:

dzd\bar{z} = dz\wedge d\bar{z} = (dx+idy)\wedge(dx-idy) = dx\wedge dx -2i\, dx\wedge dy + dy\wedge dy = -2i\, dx\wedge dy = -2i\, dxdy[/itex]

 

[Chen]

Hmm, pardon my ignorance, but what does that wedge stand for?

Thank you

 

[OrderOfThings]

It is anticommutative multiplication, a\wedge b = -b\wedge a. It is usually called exterior product.

 

[MathematicalPhysicist]

not necessarily, the sign depends on the degrees of the differential forms, it's positive if it deg(da)*deg(db)=even and else is minus.

 

[OrderOfThings]

It depends on how you think about it. Sure, we have

a\wedge (b\wedge c) = (b\wedge c)\wedge a

But that is only because

a\wedge b\wedge c = -b\wedge a\wedge c = b\wedge c\wedge a

The reason to have commuting higher order forms is that such an operation is composed of an even number of anticommuting operations on one-forms.