Differentiating the complex scalar field

[Avodyne]

My problem is not so much with "way 2", but with the way it's presented in physics books. If they had said something like

In this Lagrangian, the symbol * doesn't denote complex conjugation, and z* is just another variable. We determine the equations of motion for these two functions, and find a) that they're exactly the same, and b) that the complex conjugate of any solution is a solution. This means that if we set z* equal to the complex conjugate of z after we have determined the equation satisfied by both, we obtain a theory of a single complex-valued function instead of a theory of two.​

I would have been OK with it.

They don't say this because that's not what's done.

Suppose z=x+iy is a complex field, and z*=x-iy is its complex conjugate. We can express the lagrangian as a function of x and y, or as a function of z and z*. I will call the first function R and the second function C. (R is to remind us of "real" and C of "complex".) R is a function from ℝ² into ℝ, and C is a function from ℂ² into ℂ. These functions are related by

R(x,y)=C(x+iy,x-iy).

From this, it follows immediately (using the chain rule) that the derivatives of these functions are related by (using your non-standard notation)

D₁R = (D₁ + D₂)C

D₂R = (1/i)(D₁ - D₂)C.

We can now solve for D₁C and D₂C, with the result

D₁C = (1/2)(D₁ + iD₂)R

D₂C = (1/2)(D₁ - iD₂)R.

(These are the formulas that you call "strange and unnecessary".)

The equations of motion are

D₁R = 0

D₂R = 0.

Using the "strange and unnecessary" formulas, we find that

D₁C = 0

D₂C = 0.

Now it so happens that it is generally easier to compute D₁C and D₂C than it is to compute D₁R and D₂R. One reason for this is that C has the property that C(x+iy,x-iy) must be real, and this implies that D₂C=(D₁C)^* when evaluated at (x+iy,x-iy). Thus we only need to compute D₁C.

Note that it is never necessary to say that "the symbol * doesn't denote complex conjugation".

 

[Fredrik]

Note that it is never necessary to say that "the symbol * doesn't denote complex conjugation".

OK, you're right. I have no objections to anything you did in this post.

your non-standard notation

I think all of these notations are standard, but perhaps I picked the least popular one:

D_1f(x,y)=\partial_1 f(x,y)=f_{,1}(x,y)=\frac{\partial f(x,y)}{\partial x}=\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x_1}\bigg|_{(x,y)}f(x_1,x_2)

Edit: I think you actually made the best post in the entire thread. It explains a lot, without any weird terminology or strange definitions. One of the things you proved is that there's no need to take the formulas I called "strange and unnecessary" as definitions, because the result

\frac{\partial C(z,z^*)}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)C(x+iy,x-iy)

follows from the standard definition of a partial derivative, assuming of course that the function C is defined and analytic in an open subset of ℂ² that includes the point (z,z*).

 

[Avodyne]

OK, you're right. I have no objections to anything you did in this post.

!

I think you actually made the best post in the entire thread.

! !

I think all of these notations are standard, but perhaps I picked the least popular one:

Well, I haven't seen it before (except in Mathematica). But if the arguments were called (x_1,x_2), then I would know what \partial_1 meant.

Incidentally, the property that C(x+iy,x-iy) must be real is better expressed as the statement that C must be a symmetric function on ℂ².

 

[A. Neumaier]

OK, you're right. I have no objections to anything you did in this post.I think all of these notations are standard, but perhaps I picked the least popular one:

D_1f(x,y)=\partial_1 f(x,y)=f_{,1}(x,y)=\frac{\partial f(x,y)}{\partial x}=\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x_1}\bigg|_{(x,y)}f(x_1,x_2)

Edit: I think you actually made the best post in the entire thread. It explains a lot, without any weird terminology or strange definitions. One of the things you proved is that there's no need to take the formulas I called "strange and unnecessary" as definitions, because the result

\frac{\partial C(z,z^*)}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)C(x+iy,x-iy)

follows from the standard definition of a partial derivative, assuming of course that the function C is defined and analytic in an open subset of ℂ² that includes the point (z,z*).

The point of the Wirtinger calculus is that this assumption (which I made for didactical reasons) is not needed when you work in terms of his definitions. Because in general you only have a function f(z) that is continuous in z and continuously differentiable in Re z and I am z (i.e., considered as a function in R^2), such as F(z)=|Re z|^3-|Im z|^3. One cannot apply Avodyne's recipe in that case, but the Wirtinger derivative 9and the second derivatives) are still well-defined.

It is a bit like the difference between a real analytic function and a real differentiable function. To get the latter, you need more careful definitions than to get the former.

 

[A. Neumaier]

Incidentally, the property that C(x+iy,x-iy) must be real is better expressed as the statement that C must be a symmetric function on ℂ².

But this is not a requirement for the calculus to work.

 

[Fredrik]

!

! !

Do I seen so stubborn and grumpy that I can't give someone else credit for something good? Maybe it's the avatar.

Well, I haven't seen it before (except in Mathematica).

You made me wonder if the book that I picked up the D_if notation from (a little known Swedish book) was using a non-standard notation, but I just checked my copy of the infamous "baby Rudin" (Principles of mathematical analysis), and it's the default notation there too.

I actually like the notation f_{,\,i} a lot, because it makes the chain rule look so nice, e.g.

(f\circ g)_{,i}(x)=f_{,j}(g(x))g_{j,i}(x)

 

[A. Neumaier]

I actually like the notation f_{,\,i} a lot, because it makes the chain rule look so nice, e.g.
(f\circ g)_{,i}(x)=f_{,j}(g(x))g_{j,i}(x)

The notation without indices (regarding D as a vector with components D_i) looks even nicer: D(f(g(x)) = Df(g(x))Dg(x)

 

[Avodyne]

Arrggh! I somehow managed to screw up some minus signs in my big post above, and now I'm not seeing the "edit" button.

Correct versions are

D₁R = (D₁ + D₂)C

D₂R = i(D₁ - D₂)C.

We can now solve for D₁C and D₂C, with the result

D₁C = (1/2)(D₁ - iD₂)R

D₂C = (1/2)(D₁ + iD₂)R.

This implies

{\partial\over\partial z}C(z,z^*)={1\over2}\left({\partial\over\partial x}-i{\partial\over\partial y}\right)C(x+iy,x-iy)

which is the same formula given by A.Neumaier in post #17.

 

[Avodyne]

in general you only have a function f(z) that is continuous in z and continuously differentiable in Re z and I am z (i.e., considered as a function in R^2), such as F(z)=|Re z|^3-|Im z|^3. One cannot apply Avodyne's recipe in that case

I don't see the problem ... Is it because of the absolute-value signs?

 

[Avodyne]

Do I seen so stubborn and grumpy that I can't give someone else credit for something good? Maybe it's the avatar.

I thought my math would be too "lowbrow" for you, though I did make an effort to specify the domain and range of my functions, and to avoid sloppy language, such as referring to the function R as R(x,y).

 

[A. Neumaier]

I don't see the problem ... Is it because of the absolute-value signs?

Yes. The absolute value is not an analytic function. i.e., you can't expand it into a power series.