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Mathematics
Linear and Abstract Algebra
Second derivative of a complex matrix
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[QUOTE="idmena, post: 5916325, member: 447078"] Hi all I am trying to reproduce some results from a paper, but I'm not sure how to proceed. I have the following: ##\phi## is a complex matrix and can be decomposed into real and imaginary parts: $$\phi=\frac{\phi_R +i\phi_I}{\sqrt{2}}$$ so that $$\phi^\dagger\phi=\frac{\phi_R^2 +\phi_I^2}{2}$$ Using this matrix I can construct invariants like: $$Tr(\phi^\dagger\phi)\\ Tr\left(\phi^\dagger\phi\phi^\dagger\phi\right)\\ Tr(\phi^\dagger\phi)Tr(\phi^\dagger\phi)$$ and so on. I am interested in the second derivative of this invariants for a specific configuration, where ##\phi## is diagonal and real. I am having trouble with the second one, I hope you can help me. First I have the derivative with respect to the real part: $$\frac{\delta}{\delta\phi_{Rcd}} Tr\left(\phi^\dagger\phi\phi^\dagger\phi\right) = Tr\left(2\phi_{Rcd}\,\phi^\dagger\phi\right)$$ For the case that ##\phi## is real, ##\phi_I = 0## and ##\phi_R \to \sqrt{2}\phi##. Now, writing down the indices I have been omitting, we have: $$= 2\sqrt{2}Tr\left(\phi_{cd}\phi_{xy}\phi_{xy}\right)\\ = 2\sqrt{2}\delta_y^c\delta_d^x\delta_x^y\phi_{cd}\phi_{xy}\phi_{xy}\\ = 2\sqrt{2}\delta_d^c\,\phi_{cd}^3$$ where in the second line I used the trace operator to contract the indices. This result matches the paper I am reproducing. Now, for the second derivative I have: $$\frac{\delta}{\delta\phi_{Rab}\delta\phi_{Rcd}} Tr\left(\phi^\dagger\phi\phi^\dagger\phi\right) = \frac{\delta}{\delta\phi_{Rcd}}Tr\left(2\phi_{Rcd}\,\phi^\dagger\phi\right) = 2\,Tr\left(\delta_c^a\delta_d^b\,\phi^\dagger\phi +\phi_{Rcd}\,\phi_{Rab}\right)$$ I am unsure on how to treat the first term here. Writing down the indices explicitly like I did before, I have: $$Tr\left(\delta_c^a\delta_d^b\,\phi^\dagger\phi\right) = Tr\left(\delta_c^a\delta_d^b\,\phi_{xy}\phi_{xy}\right)$$ I know the trace contracts the outer indices, but how do I treat the deltas? Which indices are contracting with which in this case? Simplifying the second term yields: $$=2\,Tr\left(\delta_c^a\delta_d^b\,\phi^\dagger\phi\right) +2\delta_d^a \delta_c^b \,\phi_{Rba}\,\phi_{Rab}\\ =2\,Tr\left(\delta_c^a\delta_d^b\,\phi^\dagger\phi\right) +4\delta_d^a \delta_c^b \,\phi_{ba}\,\phi_{ab}$$ where in the second line I assumed that ##\phi## is real and swapped ##\phi_R \to \sqrt{2}\phi## I was supposed to get: $$=2\delta_c^a\delta_d^b \left(\phi_a^2 +\phi_b^2\right) +2\delta_d^a\delta_c^b \phi_a \phi_b$$ Where do the two ##\phi_a^2 +\phi_b^2## come from? And how come I got a factor of 2 extra in the second term? Can anyone help me on how to fill the gap? Thank you all very much! [/QUOTE]
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Second derivative of a complex matrix
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