In QFT how to take this derivative?

1. Sep 2, 2010

IFNT

In QFT given a Lagrangian $L=\partial_a \phi^* \partial^a\phi$, how do you take this derivative $$\frac{\partial L}{\partial_a \phi}$$?

2. Sep 2, 2010

LAHLH

The start of Sean Carroll's GR book has a nice explaination of this if you can access it.

$$L=\partial_a \phi^* \partial^a\phi=\eta^{ab}\partial_a \phi^* \partial_b\phi$$

The trick is to make sure index placement is compatible, so if you have lower indices on thing being differentiated wrt, you should only have lower indices in thing being differentiated etc. So now all indices are lowered in the above. Now the rule is $$\frac{\partial V_{\alpha}}{\partial V_{\beta}}=\delta^{\beta}_{\alpha}$$

Now $$\frac{\partial L}{\partial(\partial_c \phi)}=\frac{\partial(\eta^{ab}\partial_a \phi^* \partial_b\phi)}{\partial(\partial_c \phi)}=\eta^{ab}\partial_a \phi^* \delta^{c}_{b} =\eta^{ac}\partial_a \phi^*$$

Note if you didn't have a complex scalar field, but a real one and were differentiating: $$\frac{\partial(\eta^{ab}\partial_a \phi \partial_b\phi)}{\partial(\partial_c \phi)}=\eta^{ab}[\partial_a \phi \delta^c_b+\delta^{c}_{a}\partial_b \phi ]=\eta^{ac}\partial_a \phi +\eta^{cb}\partial_b \phi =\eta^{ca}\partial_a \phi +\eta^{ca}\partial_a \phi =2\eta^{ca}\partial_a \phi$$

3. Sep 2, 2010

IFNT

What is the difference between $$\eta ^{ab}$$ and $$\eta _{ab}$$?
Is this true? $$\partial_a \phi^* \partial^a \phi = \eta^{ab} \partial_a \phi^* \partial_b \phi= \eta _{ab} \partial^a \phi^* \partial ^b \phi$$ and is $$\eta^{ab} = \eta _{ab}$$ ?

4. Sep 2, 2010

xepma

The statement:
$$\partial_a \phi^* \partial^a \phi = \eta^{ab} \partial_a \phi^* \partial_b \phi= \eta _{ab} \partial^a \phi^* \partial ^b \phi$$

is true, but

$$\eta^{ab} = \eta _{ab}$$

is not. Rather, the two are related as each others inverse, so a contraction leads to the Kronecker delta (the "identity matrix")

$$\eta^{ab} \eta _{bc} = \delta_c^a$$

5. Sep 2, 2010

IFNT

But in the Minkowski space, is
$$\eta^{ab} = \eta _{ab}$$
?

6. Sep 2, 2010

LAHLH

They are equal in terms of their matrix entries, i.e. they are both the matrix diag(-1,1,1,1) say, and it just so happens that this matrix is its own inverse. However coceptually they are different.

7. Sep 2, 2010

nicksauce

This question doesn't make sense. For a tensorial equality to be valid, the indices have to be the same on both sides (same number of top and bottom indices).

8. Sep 2, 2010

IFNT

Hmm okay... I just started a QFT course and the lecturer just starts throwing phrases like "tensor", "covariance" , and so on without defining them properly.

9. Sep 2, 2010

nicksauce

I'd suggest looking over the first chapter or two of Carroll's GR book for a brief introduction to these ideas.

10. Sep 2, 2010

cesiumfrog

Why is
$$\frac{\partial(\partial_a \phi^*)}{\partial (\partial_b \phi)}=0$$
?

(It seems particularly odd in the case where $$\phi$$ may have zero imaginary part.)

Last edited: Sep 2, 2010
11. Sep 2, 2010

cesiumfrog

I think my question is addressed here:
http://www.dfcd.net/articles/fieldtheory/complexder.pdf

Although the cause doesn't feel entirely clear to me yet (perhaps someone would explain it differently?) apparently it shows that treating complex conjugates "as if" they were not dependent does turn out to lead to the correct partial derivatives.

12. Sep 3, 2010

strangerep

The point is that you can either consider a function of a complex variable f(z) as being
a function of independent real and imaginary variables f(x, iy), or you can consider it as a
function of z and z*. If the function is complex-analytic (obeys Cauchy-Riemann), this
can be written as
$$\frac{\partial f}{\partial z^*} ~=~ 0 ~.$$

Similarly, if one has a Lagrangian made from a complex field variable $\phi$, one can
consider either the field's real & imaginary parts as independent, or one can consider
$\phi$ and $\phi^*$ as independent. This is because they're just different
(orthogonal) linear combinations of the independent variables.

If (as you suggested in a previous post), the imaginary part is 0, then the imaginary part
is no longer an independent _variable_ -- so you only have a single independent variable
in that case.

13. Sep 3, 2010

Fredrik

Staff Emeritus
You need to think of the Lagrangian as a polynomial in several variables (in this case eight). The $\partial/\partial(\partial_a\phi)$ notation really just means $D_i$ (take the ith partial derivative) for some i. This is an operator that takes a function to a function, so it doesn't matter at all what the variables are (what symbols you use to represent members of the sets in the cartesian product that's the domain of the polynomial). You compute the derivatives first, the way you compute the derivatives of any polynomial in 8 variables, and then you can start thinking about the values of the variables.

14. Sep 3, 2010

cesiumfrog

What do you mean by orthogonal? (In what space? And under what metric? Do you possibly only mean "not generally linearly dependent"?) Surely x+iy and x-iy can only be independent if x and y are complex (rather than real, as is the relevent case); how is it mathematically possible for a complex number to change without the conjugate of that number also changing?

To me that makes sense iff the conjugates are indeed treated as separate parameters in the derivation of the Euler-Lagrange equation. (But by symmetry, I'd expect this to add duplicate terms equivalent in effect to the factor of 2 that arises in the real case. Haven't checked yet whether this is the case.)

15. Sep 4, 2010

strangerep

Another way of saying it is that if x and y are two independent variables,
then u:=(x+iy) and w:=(x-iy) are also two independent variables:

$$\frac{du}{dw} ~=~ \frac{\partial u}{\partial x} \frac{\partial x}{\partial w} ~+~ \frac{\partial u}{\partial y} \frac{\partial y}{\partial w} ~=~ \frac{1}{2} - i \frac{1}{2i} ~=~ 0$$

People sometimes (ab)use the word "orthogonal" to describe the relationship between
(x+y) and (x-y). The terminology originates from ordinary vectors in the plane,
where "x" and "y" are considered as two orthogonal vectors (in the standard
Euclidean metric sense).

HTH.