From the Wikipedia page on the d'Alembert operator it is stated that equivalent ways of writing the d'Alembert operator are as follows,
$$\begin{align}\Box =\eta^{\mu\nu} \partial_\nu\partial_\mu = \partial^\mu \partial_\mu=\partial^2\end{align}$$
But is there even more notation for this operator?
I ask this as in QFT I have seen the Lagrangian density for the Klein-Gordon equation written as
$$\mathcal{L}=\frac{1}{2}\partial^\mu\phi\partial_\mu\phi – \frac{1}{2}m^2\phi^2$$
and equivalently as
$$\mathcal{L}=\frac{1}{2}\left(\partial_\mu\phi\right)^2 – \frac{1}{2}m^2\phi^2$$
So is it also true that
$$\mathcal{L}=\frac{1}{2}\partial_\mu \partial_\mu\phi – \frac{1}{2}m^2\phi^2?$$
If all the above is true, then there are at least $7$ different ways of denoting the d'Alembert operator,
$$\Box = \eta^{\mu\nu} \partial_\nu\partial_\mu = \partial^\mu \partial_\mu =\partial^2=\partial^\mu\phi\partial_\mu\phi$$
$$=\left(\partial_\mu\phi\right)^2 =\partial_\mu \partial_\mu\phi\tag{1}$$
and similarly for the contravariant versions in the last equality above, $\left(\partial^\mu\phi\right)^2=\partial^\mu \partial^\mu\phi$.
In summary, do all these notations mean the exact same thing?
Best Answer
No. We do not get the Klein-Gordon equation from $\mathcal{L}=\frac{1}{2}\partial_\mu \partial_\mu\phi - \frac{1}{2}m^2\phi^2$. As far as I know, the notation $\left(\partial_\mu\phi\right)^2$ simply means $\partial_\mu\phi\partial^\mu\phi$.
Additionally, the object $\partial_\mu \partial_\mu\phi$ is not Lorentz-invariant. Note that \begin{align}\partial_\mu \partial_\mu\phi&=\partial_0 \partial_0\phi+\partial_1 \partial_1\phi+\partial_2 \partial_2\phi+\partial_3\partial_3\phi\\ &=\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}+\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2}. \ \ \ \ \ (1)\end{align} On the other hand, \begin{align}\partial_\mu \partial^\mu\phi&=\eta^{\mu\nu}\partial_\mu \partial_\nu\phi=\partial_0 \partial_0\phi-\partial_1 \partial_1\phi-\partial_2 \partial_2\phi-\partial_3\partial_3\phi\\ &=\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}-\frac{\partial^2\phi}{\partial x^2}-\frac{\partial^2\phi}{\partial y^2}-\frac{\partial^2\phi}{\partial z^2}. \ \ \ \ \ \ \ (2)\end{align} It is possible to show that Eq. $(2)$ is Lorentz-invariant, but Eq. $(1)$ is not.