- #1

idmena

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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!