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This notation can be found on page 254 of Victor Stenger's Comprehensible Cosmos and in David Tong's Lectures on QFT (Equation 2.4 http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf), and in

EDIT: on page 254 of Stenger's Comprehensible Cosmos, the Lagrangian is written with a ${\partial^i}{\partial_i\phi}$ instead of the usual ${\partial^i {\phi}}{\partial_i{\phi}}$ (that David Tong uses).

Why is ${\partial^i}{\partial_i\phi}$ = ${\partial^i {\phi}}{\partial_i{\phi}}$ in QFT ? This fact is used to calculate the Lagrange Equations of Motion (The Klein Gordon Equation) from the Lagrange Density for a Scalar Field.

This clearly isn't true for elementary functions like $y^2$ because ${\partial_y}{\partial_y\ ({y^2})}$ =/= $ {\partial_y {y^2}}{\partial_y{y^2}}$

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It's not. That relation isn't even dimensionally correct. Can you elaborate? –  DJBunk Jul 6 '12 at 18:08
I've seen the Lagrangian written ${\partial^i}{\partial_i\phi}$ or ${\partial^i {\phi}}{\partial_i{\phi}}$ or $({\partial^i {\phi}})^2$ Is this just a notational choice? –  MadScientist Jul 6 '12 at 18:15
Where did you see that? One thing you can do is partial integration, but then you are missing at least one $\phi$. In deriving the Euler-Lagrange equations, you use partial integration. –  orbifold Jul 6 '12 at 18:21
@BB1 Equations 2.4 and 1.13 of David's lectures are just two copies of the KG equation: $\partial^\mu\partial_\mu\phi+m^2\phi=0$. The scalar field lagrangian is equation 1.7 of the same lectures: $\mathcal{L} = \frac{1}{2}\partial^\mu\phi\partial_\mu\phi-\frac{1}{2}m^2\phi^2$. I don't see any place where he is equating $\partial^\mu\phi\partial_\mu\phi$ and $\partial^\mu\partial_\mu\phi$. In section 1.1.1 he is just applying the ordinary process to get the equations of motion from the Lagrangian. –  mmc Jul 7 '12 at 0:54
@BB1 The KG equation is always written as $(\partial^\mu\partial_\mu + m^2)\phi = 0$, $(\Box+m^2)\phi = 0$ or equivalent representations. See eq. 7 in this lecture notes or eq. 1.1.4 in Weinberg's QFT book. $\partial^\mu\phi\partial_\mu\phi$ is the kinetic term of the KG Lagrangian. –  mmc Jul 7 '12 at 19:16

2 Answers 2

up vote 7 down vote accepted

You have overlooked a letter. The kinetic term for the Klein-Gordon field is usually written as $$ {\mathcal L} = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi. $$ However, the equations of motion (and the action, assuming vanishing of the fields at infinity) don't change if we add a total derivative (or divergence) to the Lagrangian. So we may subtract $${\mathcal L}\to {\mathcal L}' = {\mathcal L} - \partial_\mu \left(\frac 12\phi \partial^\mu \phi\right). $$ Using the Leibniz rule (i.e. $(uv)'=u'v+uv'$), this modified Lagrangian is easily seen to be $$ {\mathcal L}' = -\frac{1}{2} \phi \cdot \partial_\mu \partial^\mu \phi $$ which is essentially what you wrote except that you omitted the factor of $-\phi$.

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But David Tong and Victor Stenger both write it without the extra $\phi$. See the link I posted above to David Tong's Lectures. –  MadScientist Jul 6 '12 at 18:39
OK, so David forgot the factor. Please send my best regards to David, I know him quite well. ... It's a somewhat usual, although not the most frequent, way of writing the Lagrangian in terms of the d'Alembertian (and it makes the derivation of the equations of motion somewhat more straightforward, by milliseconds), but the extra $\phi$ really but really can't be forgotten. –  Luboš Motl Jul 6 '12 at 18:41
I'm surprised he forgot something. I thought he was using a weird notation. You're lucky to know him. His lectures were very entertaining. One of the best lecturers I've seen. –  MadScientist Jul 6 '12 at 18:43
There's one more issue. If you have a field redefinition of the kind $\phi=\exp(a)$ or $\ln(b)$, then the non-derivative prefactors may change in various ways. ... David is a very smart and amusing guy. ... I don't understand how you could say that this mistake is a "different notation". It's like saying that $12\times 12=14$ in a different notation where I may omit the extra $4$ in the result. –  Luboš Motl Jul 6 '12 at 18:45
Is it possible that the $\phi$ doesn't contribute to the action because of the standard boundary condtions? –  MadScientist Jul 6 '12 at 18:45

OK just to be clear the lagrangian for a scalar field theory for scalar field is written as

$ \mathcal{L} = \frac{1}{2} \partial^\mu \phi \partial_\mu \phi -\frac{1}{2} m^2 \phi^2 $

sometimes this is written as

$ \mathcal{L} = \frac{1}{2} ( \partial^\mu \phi)^2 -\frac{1}{2} m^2 \phi^2 $


$( \partial^\mu \phi)^2 = \partial^\mu \phi \partial_\mu \phi$

the latter notation is a bit sloppy, but widespread and standard and usually clear from the context. Also watch out for stuff like

$ (F^{\mu \nu})^2 = F^{\mu \nu} F_{\mu \nu} $

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