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This is found at the bottom of page 9 of David Tong's QFT lectures. The Euler-Lagrange equations for the complex scalar field:

$$\mathcal L=\frac{i}{2}(\psi^*\dot\psi-\dot{\psi^*}\psi)-\nabla\psi^*\cdot\nabla\psi-m\psi^*\psi \tag{1.15}.$$

However, to obtain the equation of motion for $\psi^*$, we need the following derivatives of the Lagrangian density:

$$\partial_\mu\left(\frac{\partial\mathcal L}{\partial(\partial_\mu\psi^*)}\right),\quad \frac{\partial\mathcal L}{\partial\psi^*}.$$

In the notes, the following derivatives are instead used:

$$\frac{\partial\mathcal L}{\partial\psi^*},\quad \frac{\partial\mathcal L}{\partial\dot{\psi^*}}, \quad \frac{\partial\mathcal L}{\partial\nabla\psi^*}. \tag{1.16}$$

I don't see why we can use these instead (ie. take the derivative of $\mathcal L$ wrt. the time derivative and the gradient of $\psi^*$ separately and then combine them, which I think is what's being done here).

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3 Answers 3

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The double $\mu$ index is being summed over, so if you write it out, you get

$$\partial_\mu\left(\frac{\partial\mathcal L}{\partial(\partial_\mu\psi^*)}\right) =\sum_{\mu=0}^3\partial_\mu\left(\frac{\partial\mathcal L}{\partial(\partial_\mu\psi^*)}\right) = \partial_0\left(\frac{\partial\mathcal L}{\partial(\partial_0\psi^*)}\right)+\sum_{i=1}^3\partial_i\left(\frac{\partial\mathcal L}{\partial(\partial_i\psi^*)}\right) =\partial_t\left(\frac{\partial\mathcal L}{\partial(\dot{\psi^*})}\right)+\nabla\left(\frac{\partial\mathcal L}{\partial(\nabla\psi^*)}\right)$$

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  • $\begingroup$ Ah of course, thank you! $\endgroup$
    – Charlie
    Aug 31, 2020 at 10:24
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If you take the Euler-Lagrange equation you have the following term $$\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi^*)}\right)\tag{1}$$ in which the $\mu$ index is summed over. This means that the quantity $(1)$ has to be thought as $$\partial_0\left(\frac{\partial\mathcal{L}}{\partial(\partial_0\psi^*)}\right)+\partial_1\left(\frac{\partial\mathcal{L}}{\partial(\partial_1\psi^*)}\right)+\partial_2\left(\frac{\partial\mathcal{L}}{\partial(\partial_2\psi^*)}\right)+\partial_3\left(\frac{\partial\mathcal{L}}{\partial(\partial_3\psi^*)}\right)\tag{2}$$ but since $0\equiv t$, $1\equiv x$, $2\equiv y$, $3\equiv z$, the last three elements of $(2)$ are just the gradient of the derivative of the lagrangian wrt the gradient of $\psi$ $$\nabla\left(\frac{\partial\mathcal{L}}{\partial\nabla\psi^*}\right)$$ Moreover, given that $\partial_0\equiv\partial_t$ the same can be said to the first term of $(2)$ which is going to be $$\partial_t\left(\frac{\partial\mathcal{L}}{\partial\dot{\psi}^*}\right)$$ With this you get your result

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  • $\begingroup$ Sorry for the redundant answer, didn't see the first one! $\endgroup$
    – Quiver
    Aug 31, 2020 at 10:26
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It seems to me that it is better to place the Lagrangian of the complex field in a covariant form and with the covariant derivatives in relation to the field and its derivatives it will be more direct and universal

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