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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?

Qmechanic
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James Womack
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1 Answers1

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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.

ACuriousMind
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