For my field theory class I have the following Lagrangian density


Where $\eta^{\mu\nu}$ is the metric tensor (+--- convention) and * denotes complex conjugation. The Lagrangian is invariant under $\phi\rightarrow e^{i\alpha}\phi$ and if we thus let $\alpha$ be of infinitesimal size then we have the following expansion of the transformation $\phi\rightarrow \phi + i\alpha\phi$. From Noether's theorem I know that the conserved currents for s parametric symmetry transformations is given by

$$J^k_n=-\frac{\partial\mathscr{L}}{\partial(\partial_k\phi_I)}(\Phi_{I,n}-\partial_m\phi_I X^m_n)-\mathscr{L}X^k_n,$$

and $n=1,...,s$, for a transformation

$$x^i\rightarrow x'^i= x^i+\delta x^i, \;\;\;i=1,...,d,$$

$$\phi_I(x)\rightarrow \phi_I'(x')=\phi_I(x)+\delta\phi_I(x).$$

With $\phi_I$ being the fields in $\mathscr{L}$. Where $X$ and $\Phi$ are given by the following way

$$\delta x^i=\sum_{1\leq n\leq s}X^i_n\delta\omega_n, \;\;\;\;\;\; \delta\phi_I(x)=\sum_{1\leq n\leq s}\Phi_{I,n}\delta\omega_n$$

Now in the above Lagrangian density we have that $\delta\omega=i\alpha$, $X^i=0$ and $\Phi=\phi$. Now when I try to calculate the conserved current I kind of get stuck here


Which according to my professor should equal $\frac{1}{2}(\partial^k\phi\phi^*-\partial^k\phi^*\phi)$. I have no idea how he arrives at that result from my above $J^k$.


1 Answer 1


Assuming no quantum gravity, $\eta^{\mu\nu}$ is a constant and can be pulled out of the derivative and what remains looks like a $\delta^k_{\mu}$ or $\delta^k_{\nu}$-type expression (in the sense of a Kronecker $\delta$), pulling the $k$ into the $\partial^\mu$ or $\partial^\nu$ respectively.

If you are confused about where the minus sign comes from, I believe the correct way to think about this is to think about the variable $\partial_\mu \phi^*$ as being formally independent from $\partial_\mu \phi$. This means that you need to add derivatives for Noether's theorem to be happy. So you only have half of the expression; if we instead start with:

$$ \delta\mathscr{L} = \frac{\partial\mathscr{L}}{\partial \phi} \delta \phi + \frac{\partial\mathscr{L}}{\partial \phi^*} \delta \phi^* + \eta^{\mu\nu} \left( \frac{\partial\mathscr{L}}{\partial (\partial_\mu \phi)} \partial_\nu \delta \phi + \frac{\partial\mathscr{L}}{\partial (\partial_\mu \phi^*)} \partial_\nu \delta \phi^* \right)$$

Using the Euler equations the left two terms are derivatives of the Lagrangian which combine into a total derivative, giving a current

$$ \delta \mathscr{L} ~\propto~ \frac{\partial\mathscr{L}}{\partial (\partial_\mu \phi)} \delta \phi + \frac{\partial\mathscr{L}}{\partial (\partial_\mu \phi^*)} \delta \phi^*$$

and if your variation $\delta\phi$ is pure-imaginary, then we'll get a $-$ sign between these two expressions.

  • $\begingroup$ Yes you're right, I put them in the derivative to easier hold track of which upper/subscripts are replaced. I have the most trouble proving that the term $\frac{\partial(\eta^{\mu\nu}\partial_\mu\phi^*)}{\partial(\partial_k\phi)}\partial_\nu\phi\phi=-\partial^k\phi\phi^*$. $\endgroup$ Mar 3, 2015 at 21:45
  • $\begingroup$ @Gehaktmolen Given that you have this more-specific problem, I have updated my answer to indicate where I think it comes from. I don't know your professor's exact formalism but you should be able to use those techniques to make the minus sign appear. $\endgroup$
    – CR Drost
    Mar 3, 2015 at 22:46
  • $\begingroup$ Yes I do believe that I understand what you mean, in my case by treating them as seperate fields would mean that instead of choosing $\Phi=\phi$ I have that $\Phi_1=\phi$ and $\Phi_2=-\phi^*$. Which would give me $J^k=-\frac{\partial\mathscr{L}}{\partial(\partial_k\phi)}\phi+\frac{\partial\mathscr{L}}{\partial(\partial_k\phi^*)}\phi^*$. Is this right? $\endgroup$ Mar 3, 2015 at 22:51
  • $\begingroup$ I think so, yes. $\endgroup$
    – CR Drost
    Mar 3, 2015 at 22:55

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