- #1
IFNT
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In QFT given a Lagrangian [itex]L=\partial_a \phi^* \partial^a\phi[/itex], how do you take this derivative [tex]\frac{\partial L}{\partial_a \phi}[/tex]?
IFNT said:But in the Minkowski space, is
[tex]
\eta^{ab} = \eta _{ab}
[/tex]
?
Why isLAHLH said:Now [tex] \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^*
[/tex]
Note if you didn't have a complex scalar field, but a real one and were differentiating: [tex] \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
[/tex]
cesiumfrog said: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.
You need to think of the Lagrangian as a polynomial in several variables (in this case eight). The [itex]\partial/\partial(\partial_a\phi)[/itex] notation really just means [itex]D_i[/itex] (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.cesiumfrog said:Why is
[tex]\frac{\partial(\partial_a \phi^*)}{\partial (\partial_b \phi)}=0[/tex]
?
(It seems particularly odd in the case where [tex]\phi[/tex] may have zero imaginary part.)
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 relevant case); how is it mathematically possible for a complex number to change without the conjugate of that number also changing?strangerep said:Similarly, if one has a Lagrangian made from a complex field variable [itex]\phi[/itex], one can
consider either the field's real & imaginary parts as independent, or one can consider
[itex]\phi[/itex] and [itex]\phi^*[/itex] as independent. This is because they're just different
(orthogonal) linear combinations of the independent variables.
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.)Fredrik said: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.
cesiumfrog said: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 relevant case);
In QFT, the process of taking a derivative is similar to that in classical mechanics. You can use the usual calculus rules, such as the chain rule and product rule, to take derivatives of functions.
A partial derivative is taken with respect to a specific variable, while a covariant derivative is taken with respect to an arbitrary direction in space-time. In QFT, the covariant derivative is used to account for the curvature of space-time.
Yes, the chain rule can be used in QFT to take derivatives. However, in some cases, the chain rule may be modified to account for the non-commutativity of operators in QFT.
To take higher-order derivatives in QFT, you can use the same techniques as in classical mechanics, such as repeated application of the chain rule. However, the non-commutativity of operators in QFT may require some modifications to the usual calculus rules.
There is no one general formula for taking derivatives in QFT as it depends on the specific field theory being studied. However, the usual calculus rules and techniques, as well as the non-commutativity of operators, can be applied to take derivatives in most cases.