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Given a functional which depends on a function (ket), and its complex conjugate (bra), e.g. $$F[\varphi] = \langle \varphi|\hat{F}|\varphi\rangle = \int \varphi^{*}(\mathbf{r}) \hat{F} \varphi(\mathbf{r}) \, \mathrm{d}\mathbf{r} $$ I have been told that we can vary the bra and ket independently, i.e. the first variation of $F$ in the bra is given by $$\delta F = \int \frac{\delta F}{\delta \varphi^{*}} \eta(\mathbf{r}) \, \mathrm{d}\mathbf{r} = \frac{\mathrm{d}}{\mathrm{d}\epsilon}\left[ \int (\varphi^{*}(\mathbf{r})+\epsilon\eta(\mathbf{r}))(\mathbf{r}) \hat{F} \varphi(\mathbf{r}) \mathrm{d}\mathbf{r}\right]_{\epsilon = 0},$$ and not $$\delta F = \int \frac{\delta F}{\delta \varphi^{*}} \eta(\mathbf{r}) \, \mathrm{d}\mathbf{r} = \frac{\mathrm{d}}{\mathrm{d}\epsilon}\left[ \int (\varphi^{*}(\mathbf{r})+\epsilon\eta(\mathbf{r}))(\mathbf{r}) \hat{F} (\varphi(\mathbf{r})+\epsilon\eta(\mathbf{r})) \mathrm{d}\mathbf{r}\right]_{\epsilon = 0},$$ as one might expect.

If the above is correct, how can it be shown that the bra and the ket can be independently varied?

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    $\begingroup$ the argument boils down to the fact that your ket represents complex numbers, where the real part and imaginary part can be varied independently. This gets translated into that the complex number and it's complex conjugate can be viewed as independent variables. $\endgroup$ – Mikael Fremling Feb 4 '16 at 18:04
  • $\begingroup$ Related: physics.stackexchange.com/q/89002/2451 $\endgroup$ – Qmechanic Feb 4 '16 at 20:30
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This has nothing to do with "bras" or "kets" and more with the elementary observation that a complex number has two real degrees of freedom, and that derivatives are with respect to one real degree of freedom.

The $\frac{\partial}{\partial\phi}$ and $\frac{\partial}{\partial\phi^\ast}$ are the Wirtinger derivatives, which in particular fulfill $\frac{\partial\phi^\ast}{\partial\phi} = 0$, i.e. the derivative of something with respect to its conjugate is zero.

This naturally generalizes to the functional derivatives with respect to a complex function.

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  • $\begingroup$ Thanks for your helpful answer. When I posted the question, I realized that it was more fundamental than the bra-ket notation. However, this was the way I originally conceived the question (before researching the issue) and I thought that other people with the same question might find this quicker by relating it directly to bra-ket notation. $\endgroup$ – James Womack Feb 5 '16 at 9:07
  • $\begingroup$ It would be a great help if you could elaborate on the generalization to functional derivatives. There are many resources online discussing this result in terms of partial derivatives, but I have yet to find one that clearly casts the result in terms of functional derivatives. $\endgroup$ – James Womack Feb 5 '16 at 9:08
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    $\begingroup$ @JamesWomack: You can, in full analogy, define $\frac{\partial}{\partial \phi_i}$ as the functional derivatives w.r.t. the real parts, and then define the complex derivatives exactly like the Wirtinger derivatives. Since the functional derivative fulfills the same chain rule as the ordinary derivative, and the reason why the Wirtinger derivatives are the "correct" complex derivatives is just the chain rule, it doesn't matter at all that the derivatives are functional. $\endgroup$ – ACuriousMind Feb 5 '16 at 14:20

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