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So if I define $y(U)=|Tr(U^*V)|^2$

If I do the Gateaux derivative:

$y_U[\tilde{U}] =\frac{d}{d\epsilon}|Tr((U^*+\epsilon\tilde{U})V)|^2 $,

Here is where I get confused because of how things are nested - the trace has a real and imaginary part and I'm not sure how to handle this

My intuition says that the derivative is like $y_U[\tilde{U}] = 2Tr(V^*)$ with the variation somewhere in here, but I can't figure out how to get that if it should be the case. Some tips would be much appreciated

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$y(U) = \overline{\operatorname{tr} (U^*V)} \operatorname{tr} (U^*V)$.

\begin{eqnarray} y(U+tH) -y(U) &=& \overline{\operatorname{tr} ((U+tH)^*V)} \operatorname{tr} ((U+tH)^*V) - \overline{\operatorname{tr} (U^*V)} \operatorname{tr} (U^*V) \\ &=& (\overline{\operatorname{tr} (U^*V)} + t\overline{\operatorname{tr}(H^*V)} ) ( \operatorname{tr} (U^*V) + t\operatorname{tr} (H^*V) ) - \overline{\operatorname{tr} (U^*V)} \operatorname{tr} (U^*V) \\ &=& 2 t \operatorname{re} ( \overline{\operatorname{tr} (U^*V)} \operatorname{tr} (H^*V) ) + O(t^2) \end{eqnarray}

Hence $d y(U,H) = 2 \operatorname{re} ( \overline{\operatorname{tr} (U^*V)} \operatorname{tr} (H^*V) ) $.

Note that there are many different notations for the Gateaux derivative.

Note, if $H$ is real, with $\operatorname{tr} (U^*V) = a+ib$ and $V= V_r+i V_i$ then $d y(U,H) = 2 \operatorname{re} ( (a-ib) ( \operatorname{tr} (V_i^TH) -i \operatorname{tr} (V_r^TH)) = 2 \operatorname{tr}((aV_r-bV_i)^T H)$, and similarly, $d y(U,iH) = 2 \operatorname{re} ( (a-ib) ( \operatorname{tr} (V_i^TH) -i \operatorname{tr} (V_r^TH)) = 2 \operatorname{tr}((bV_r+aV_i)^T H)$

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  • $\begingroup$ Thanks copper, is $\frac{dy}{dU}$ sitting here somewhere? For a scalar variation I would expect like $\frac{dy}{du}H$, so I could just see what the derivative/gradient was. In this case where the variation is a matrix. something is different that I can't put my finger on. I'm really after $\frac{dy}{dU}$ $\endgroup$
    – Vogtster
    Jun 30, 2020 at 22:47
  • $\begingroup$ I'm not sure what you mean by derivative, $y$ is a map from a complex space into the reals, so it cannot have a (Frechet) derivative in the usual sense. $\endgroup$
    – copper.hat
    Jun 30, 2020 at 22:50
  • $\begingroup$ If I have $U = u+iv$, does $\frac{dy}{du}$ and $\frac{dy}{dv}$ make sense in terms of Frechet? $\endgroup$
    – Vogtster
    Jun 30, 2020 at 23:41
  • $\begingroup$ Yes, then the $H$ above would be purely real or purely imaginary and the resulting map is Frechet differentiable. $\endgroup$
    – copper.hat
    Jun 30, 2020 at 23:45
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    $\begingroup$ Got it. Thank you so much for taking the time to help me get my head straight on this! $\endgroup$
    – Vogtster
    Jul 1, 2020 at 1:46

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