# Why can the bra and ket be varied independently?

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?

• 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. – Mikael Fremling Feb 4 '16 at 18:04
• – Qmechanic Feb 4 '16 at 20:30

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.
• @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. – ACuriousMind Feb 5 '16 at 14:20