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I am a mathematician who is taking a quantum field theory course without much prior pyhsics. We have had the term "up to a total derivative" a few times, yet every time I asked what it meant I didn't really grasp it.

As an example, for our last tutorial we were given the Lagrangian $$ \mathcal{L} = i\psi^*\partial_0\psi - \frac{1}{2m}\nabla\psi^*\cdot\nabla\psi,\tag{1} $$ but then immediately in the tutorial it was given that this is equivalent (up to a total derivative) to $$ \mathcal{L} = \frac{i}{2}(\psi^*\partial_0\psi - (\partial_0\psi^*)\psi) - \frac{1}{2m}\nabla\psi^*\cdot\nabla\psi.\tag{2} $$

The things I really don't understand are:

  • how exactly are these things the same? (/what does "up to total derivative" mean)

  • how do I know when I should try to convert something to another thing through a total derivative?

Qmechanic
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2 Answers2

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  1. The Euler-Lagrange (EL) equations are not affected by total derivative terms, cf. e.g. this Phys.SE post.

  2. In OP's concrete example the Lagrangian density (2) is preferred as it is manifestly real. See also this related Phys.SE post.

Qmechanic
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Since the Lagrangian density (which is confusingly also referred as a Lagrangian) is defined as a function which is integrated on, we may always think the $\mathcal{L}$ in inside a 4D integral, since the action is defined via $$ S = \int d^4x \mathcal{L}. $$ Now, we may decompose the term to two identical parts and integrating one of them by parts, \begin{align*} \int dt i\psi^*\partial_0\psi &= \int dt\frac{1}{2}i\psi^*\partial_0\psi + \int dt\frac{1}{2}i\psi^*\partial_0\psi\\ &=\int dt\frac{1}{2}i\psi^*\partial_0\psi + \left. \frac{1}{2}i\psi^*\psi \right|_{\pm \infty} - \int dt\frac{1}{2}i\partial_0(\psi^*)\psi \\ &= \int dt\frac{i}{2}(\psi^*\partial_0\psi - (\partial_0\psi^*)\psi) \end{align*} where the substitution term vanishes, since the field values are considered to vanish at infinity: $\psi(\pm \infty) \rightarrow 0$. This is the total derivative term.

Zeick
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