Conserved current of quartic interaction QFT ($φ⁴$-Theory) The Lagrangian of the real massless $φ⁴$-theory is
\begin{align}
L=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\lambda\phi^4
\end{align}
Therefore the action integral has the global symmetry
\begin{align}
x_\mu \rightarrow x'_\mu=e^{-\alpha}x_\mu \\
\phi(x) \rightarrow \phi'(x') = e^{\alpha}\phi(x)
\end{align}
It is straight forward to calculate the corresponding conserved noether current
\begin{align}
J^\mu = \phi\partial^\mu\phi+x^\nu\partial_\nu\phi\partial^\mu\phi-\frac{1}{2}x^\mu\partial_\nu\phi\partial^\nu\phi+\lambda x^\mu\phi^4
\end{align}
My problem is, that I can't see, why this current is conserved when looking at its derivative
\begin{align}
\partial_\mu J^\mu = \frac{3}{2}\partial_\mu\phi\partial^\mu\phi+\phi\partial_\mu\partial^\mu\phi+x^\nu(\partial_\mu\partial_\nu\phi\partial^\mu\phi+\partial_\nu\phi\partial_\mu\partial^\mu\phi)-\frac{1}{2}(\partial_\mu\partial_\nu\phi\partial^\nu\phi+\partial_\nu\phi\partial_\mu\partial^\nu\phi)+\lambda\phi^4+4\lambda x^\mu\phi^3\partial_\mu\phi
\end{align}
Usually one would use the euler-lagrange equations
\begin{align}
4\lambda\phi^3=-\partial_\mu\partial^\mu\phi
\end{align}
to see, that $\partial_\mu J^\mu=0$. But for example the $\phi^4$-term in the derivative above can in no way cancel out, so how can this current be conserved?
 A: You took incorrectly some derivatives, because $ \partial_\mu x^\mu = 4 \neq 1 $.
If I take the derivatives one by one, the correct divergence of the current is
$$ \partial_\mu J^\mu = \partial_\mu \phi \partial^\mu \phi + \phi \partial_\mu  \partial^\mu \phi + \partial_\mu \phi \partial^\mu \phi + x^\nu \partial_\mu \partial_\nu \phi \partial^\mu \phi + x^\nu \partial_\nu \phi \partial_\mu \partial^\mu \phi - \frac{1}{2} 4 \partial_\nu \phi \partial^\nu \phi - x^\mu \partial_\mu \partial_\nu \phi \partial^\nu \phi + 4 \lambda \phi^4 + 4 \lambda \phi^3 x^\mu \partial_\mu \phi \\ = \phi \partial_\mu  \partial^\mu \phi + x^\nu \partial_\nu \phi \partial_\mu \partial^\mu \phi + 4 \lambda \phi^4 + 4 \lambda \phi^3 x^\mu \partial_\mu \phi = 0$$
Note that the $\partial_\mu \phi \partial^\mu \phi $ and $x^\mu \partial_\mu \partial_\nu \phi \partial^\nu \phi$ terms automatically cancel out, and using the equations of motion (euler-langrange equations), the $4 \lambda \phi^4$ term cancels with $\phi \partial_\mu \partial^\mu \phi$ and the $x^\nu \partial_\nu \phi \partial_\mu \partial^\mu \phi$ term cancels with $4 \lambda \phi^3 x^\mu \partial_\mu \phi$
