I have learnt that to get the functional derivative, we must carry out the variation. The functional derivative is the thing next to the direction the variation is taken. For example for some real functions and functionals: $$ F[n] = \int V(\vec{r}) n(\vec{r}) \ d \vec{r} $$ we have the variation $$ \delta F[n; f] = \frac{\partial}{\partial t} \int V(\vec{r}) (n(\vec{r}) + t f(\vec{r})) \ d \vec{r} \Big|_{ 0 } = \int V(\vec{r}) f(\vec{r}) \ d \vec{r} $$ Therefore we have the functional derivative $$ \frac{\partial F}{\partial n} = V, $$ But how does this extend to complex functions? I want to find the variation (and also the functional derivative) of the following functional $$ F[\psi,\psi^*] = \int \psi^* \psi \ d \vec{r} $$ My attempt: \begin{align*} \delta F[\psi,\psi^*; h,h^*] &= \frac{\partial}{\partial t} \int (\psi^* + t h^*)^* (\psi + t h) \psi \ d \vec{r} \Big|_{ 0 } \\ &= \int \frac{\partial}{\partial t} (\psi + t h) (\psi + t h) \ d \vec{r} \Big|_{ 0 } = \int h \psi + \psi h \ d \vec{r} \end{align*} Therefore $$ \frac{\delta F}{\delta \psi} = \psi. $$ My feeling is that this is the wrong anwser, I should get the conjugate of that instead. For the other functional derivative I should get $$ \frac{\delta F}{\delta \psi^*} = \psi, $$ according to my lecture notes, but there is no term of the form $h^* \psi$ inside the integral! So I cant seem to get it. Is my variation wrong, or is something else wrong? I put a pretty high bounty on this one because I feel it's such an important question. Hoping for a speedy clarification.


What you've done seems fine, it's just that you added an extra conjugate by accident. Using your notation, if $$F[\psi, \psi^*] = \int \psi^* \psi \, d \vec{r}$$ then we have $$F[\psi + t h, \psi^* + t h^*] = \int (\psi^* + t h^*) (\psi + t h) \, d \vec{r}.$$ Taking the derivative with respect to $t$ at $t = 0$ gives $$\frac{\partial F[\psi + t h, \psi^* + t h^*]}{\partial t}\bigg|_{t = 0} = \int h^* \psi + \psi^* h \, d \vec{r}$$ from which we read off the answer, $$\frac{\delta F}{\delta \psi} = \psi^*, \quad \frac{\delta F}{\delta \psi^*} = \psi.$$ The only difference is that in your derivation, you conjugated $(\psi^* + t h^*)$ an extra time.

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  • $\begingroup$ I hope to giving you the bounty in 21 hours. That's the first time I am allowed $\endgroup$ – Mikkel Rev Jan 22 at 17:16
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    $\begingroup$ @MikkelRev No problem, and also feel free to wait longer, in case more comprehensive answers appear! $\endgroup$ – knzhou Jan 22 at 17:16

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