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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?

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Sandstar
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    Up to total derivative simply means “up to terms that do not contribute to the action IF I neglect boundary terms”. It is usual to assume that fields vanish on boundary of space time so any boundary terms vanish. This assumption does not always apply and for some purposes boundary terms are important. Based on your question, you are at a level where you can always just “integrate by parts” and drop boundary terms without worry. – Prahar Mar 18 '20 at 12:51
  • Okay so the second definition can be regarded as the same as the first because they induce the same action? So if I was to integrate both of these over spacetime I should get the same results? And then how would I be able to instantly see that I can make such a substituion in future, should I remember previous ones or guess that I might need one and then use trial and error to find one? – Sandstar Mar 18 '20 at 12:56
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    Yes to your first question. If you integrate you’ll get same result BECAUSE the difference between them is a total derivative so integrating this will give you a boundary term (Stokes theorem) and boundary terms are zero. There’s no need to remember anything, you can always just do it. Whether it’s a useful thing to do or not is a separate matter. – Prahar Mar 18 '20 at 13:00
  • @Prahar Comments are for requesting clarifications or suggesting improvements. If you have an answer, please post it as one. – BioPhysicist Mar 18 '20 at 13:00
  • @Aaron It appears evident Prahar 's conversation is precisely focussing the question into something meaningful and coherent by clarifying language the OP is clearly unaware of. It is all in the "really"... – Cosmas Zachos Mar 18 '20 at 13:48
  • @CosmasZachos I don't really see anywhere that Prahar is trying to focus the question. It looks like Prahar is answering the question to me. Prahar is not suggesting certain edits to be made to the question, and they are not asking the OP to clarify something that is confusing in the post. – BioPhysicist Mar 18 '20 at 13:59
  • @Aaron Oh, OK, I do... One need not be brutally direct. By clarifying language and getting communication and culture problems out of the way, P is steering the otherwise sophisticated OP into cohering the question into something meaningful, worthy of the extant answer. – Cosmas Zachos Mar 18 '20 at 14:04
  • In light of this comment thread - Prahar (intentionally or not) has indeed fully answered my question. But I cannot upvote them or accept their answer. – Sandstar Mar 18 '20 at 14:07
  • @CosmasZachos The problem the OP is currently having is the one I was trying to avoid in my first comment. Prahar has indeed answered the OP's question, but since it was in a comment there is nothing more to be done. – BioPhysicist Mar 18 '20 at 14:12
  • @Aaron I see I misread the question... but ipso facto, it should have been clearer and Prahar helped... It is not an uncommon plight to see a bevy of answers answering different questions the stated question did not take care to exclude and distance itself from. (Some OPs get so frustrated that add "I am not asking xxx".) – Cosmas Zachos Mar 18 '20 at 14:19

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.

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