In a Lagrangian like the one that follows:
$$\mathcal{L} = {i} \Psi^*\dot{\Psi} – \frac{1}{2m} \nabla{\Psi} ^* \nabla \Psi.\tag{1}$$
How can I apply the Euler-Lagrangian equation, shown in $(2)$, to obtain an equation like $(3)$. I don't understand how I can deal with all the different fields?
$$ \frac{d}{dt}\left( \frac{\partial \mathcal{L}}{\partial \dot{\Psi}}\right)= \frac{\partial \mathcal{L}}{\partial \Psi} \tag{2}$$
$$i \frac{\partial \Psi}{\partial t} = – \frac{\nabla^2}{2m} \Psi \tag{3}$$
I have seen in some notes that I must treat $\Psi$ and $\Psi^*$ as different fields and must first calculate the different derivatives, shown in $(4)$:
$$\frac{\partial \mathcal{L}}{\partial \dot{\Psi^*}}= i\Psi \hspace{5mm}; \hspace{5mm} \frac{\partial \mathcal{L}}{\partial \nabla{\Psi^*}}= – \frac{1}{2m} \nabla \Psi \hspace{5mm} ; \hspace{5mm} \frac{\partial \mathcal{L}}{\partial {\Psi^*}}= i\dot{\Psi} \tag{4}$$
But how I can apply $(2)$ to these to obtain $(3)$? And do I not need the $\frac{\partial \mathcal{L}}{\partial \nabla\Psi^*}$ ?
How can I use $(2)$ if I do not have a $\Psi$ variable or a $\dot\Psi^*$ to complete the equation?
Best Answer
I'll try to give a slightly more precise view which might be confusing at first but will pay off later.
You are minimizing the action $S[\psi,\psi^*]$ where the square brackets indicate that $S$ is a functional: an object which takes a function as argument and spits out a number. Here $\psi$ and $\psi^*$ are still considered completely independent functions. In the following equations I will drop the dependence on $\psi^*$ since otherwise it would be too long but you can just as easily put them back in. The action then looks like $$S[\psi]=\int\mathrm dx\,\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)\tag{1}$$ where $x$ is four vector so it includes time as well. Here I emphasize that $\mathcal L$ is just a regular function: it takes in 4 arguments and spits out a number. I could also fill in $a,b,c,d$ instead of $\psi,\dot\psi,\nabla\psi,\nabla\dot\psi$ in the Lagrangian (I will use $\psi$ instead of $\psi^*$ for brevity). $$\mathcal L(a,b,c,d)=\frac i 2ab-\frac 1{2m}dc\tag{2}$$ Now to minimize the action we will have to calculate the functional derivative. The functional derivative, $\frac{\delta S}{\delta\psi}( x)$, is formally defined as $$\int\mathrm d x\frac{\delta S}{\delta \psi}( x)\eta(x)=\lim_{\epsilon\rightarrow\infty}\frac{S[\psi+\epsilon\eta]-S[\psi]}{\epsilon}\tag{3}$$ where $\eta( x)$ is an arbitrary test function that satisfies the right boundary conditions, in this case the condition that $\eta$ goes to zero as $ x\rightarrow\infty$. $\eta$ is often written as $\delta\psi$. We can now calculate $S[\psi+\epsilon\eta]$ by Taylor expanding the Lagrangian: \begin{align} S[\psi+\epsilon\eta]=&\int\mathrm d x\,\mathcal L(\psi+\epsilon\eta,\dot\psi+\epsilon\dot\eta,\nabla\psi+\epsilon\nabla\eta,\nabla\dot\psi+\epsilon\nabla\dot\eta)\\ &=\int\mathrm dx\left[\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\ \epsilon\,\eta(x)\frac{\partial}{\partial\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\ \epsilon\,\dot\eta(x)\frac{\partial}{\partial\dot\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\ \epsilon\nabla\eta(x)\frac{\partial}{\partial\nabla\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\ \epsilon\nabla\dot\eta(x)\frac{\partial}{\partial\nabla\dot\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi) \right]\tag{4} \end{align} This almost has the form of the LHS of (2) but $\eta$ still has derivatives acting on it. We can fix this by integrating by parts to move the derivative from $\eta$ to the other factor at the cost of introducing a minus sign. Plugging this into (3) gives \begin{align}\int\mathrm d x\frac{\delta S}{\delta \psi}( x)\eta(\mathbf x)&=\int\mathrm dx\left[\frac{\partial\mathcal L}{\partial\psi}-\frac{\partial}{\partial t}\frac{\partial\mathcal L}{\partial\dot\psi}-\nabla\cdot\left(\frac{\partial\mathcal L}{\partial\nabla\psi}\right)\pm\frac{\partial}{\partial t}\nabla\cdot\left(\frac{\partial\mathcal L}{\partial\nabla\dot\psi}\right)\right]\eta(x) \end{align} We recognize the term between brackets as the functional derivative and setting it to zero will give us the EL equations. I'm not so sure if the $\pm$ should be plus or minus but I think it's $+$. If we had included $\psi^*$ as a separate field from the start we would have got a similar equation but with $\psi^*$ instead of $\psi$.
I believe there's an error in your Lagrangian and that $\nabla\dot\psi$ should be $\nabla\psi$. The factor $\frac i 2\psi^*\dot\psi$ should either be $i\psi^*\dot\psi$ or $\frac i 2(\psi^*\dot\psi-\psi\dot\psi^*)$. See this question or this question. If you now calculate the EL you get the desired formula.