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?


1 Answer 1


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 $4 \lambda \phi^3 x^\mu \partial_\mu \phi$


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