If I have two interacting fields

$$ \mathcal{L} = \frac{1}{2}(\partial_\mu \phi_1)^2 - \frac{1}{2}m^2\phi_2^2 + \frac{1}{2}(\partial_\mu \phi_2)^2 - \frac{1}{2}m^2\phi_2^2 - g^2(\phi_1^2 + \phi_2^2)^2, $$

then this lagrangian density is invariant under $\operatorname{SO}(2)$ rotations. An infinitesimal rotation produces $D\phi_j$: $D\phi_1 = -\phi_2$ and $D(\phi_2)=\phi_1$. The corresponding momenta are $\Pi^0_{\phi_1} = \partial^0\phi_1$ and $\Pi^0_{\phi_2} = \partial^0\phi_2$. Then the Noether charge is

$$ \hat{Q}_N = \int d^3x (-(\partial^0\hat{\phi_1}))\hat{\phi_2} + (\partial^0\hat{\phi_2}))\hat{\phi_1}. $$

To calculate this quantity I'm using the mode expansions:

\begin{split} \hat{\phi}_1 &= \int\frac{d^3p}{(2\pi)^{3/2}(2E_{\vec{p}})^{1/2}}(\hat{a}_\vec{p} e^{-ip\cdot x} + \hat{a}_\vec{p}^\dagger e^{ip\cdot x}),\\ \hat{\phi}_2 &= \int\frac{d^3p}{(2\pi)^{3/2}(2E_{\vec{p}})^{1/2}}(\hat{b}_\vec{p} e^{-ip\cdot x} + \hat{b}_\vec{p}^\dagger e^{ip\cdot x}), \end{split}

where $[\hat{a}_\vec{p},\hat{a}_\vec{q}^\dagger] = [\hat{b}_\vec{p},\hat{b}_\vec{q}^\dagger]=\delta(\vec{p}-\vec{q})$ and all other commutators are zero.

My question is the following. The charge operator $\hat{Q}_N$ must be the number of particles of the first field plus (or minus) the number of particles on the second field. If I substitute the mode expansions in this integral, I get combination of operators like $\hat{a}_\vec{p}\hat{b}_\vec{q}^\dagger$, but I will never get a number operator. Did I made a mistake by writing different mode operators $\hat{a}_\vec{p},\hat{b}_\vec{p}$ for the two fields? Where I am wrong in my reasoning?

  • $\begingroup$ There is nothing "infinitesimal" about your infinitesimal rotation as written $\endgroup$
    – nox
    May 11, 2019 at 7:16
  • 1
    $\begingroup$ The infinitesimal rotation (with infinitesimal parameter ε) is φ -> φ + ε D(φ) $\endgroup$ May 11, 2019 at 18:57

1 Answer 1


If your Noether charge was

$$ Q = Q_1 -Q_2 \ , $$

where $Q_i$ are the particle numbers of field $\phi_i$, then a symmetry transformation on a field would be

$$ e^{i t Q} \phi_1 e^{-itQ}= e^{i t} \phi_1$$

and similarily for the $\phi_2$ field. That is not your desired transformation.

Another argument: the Noether charge maps 1-particles to 2-particles. If there would be no $ab$ terms, this would be impossible.

  • $\begingroup$ Then, what is the conserved charge? Because if I make the proper substitution I can work with the complex scalar field, and there I have the conserved charge as the difference in the number of particles - antiparticles. $\endgroup$ May 11, 2019 at 18:52
  • $\begingroup$ Would it be possible if you give at least what $Q$ is expected for my case? $\endgroup$ May 11, 2019 at 18:53
  • $\begingroup$ You calculated a charge, right? Did you check whether it gives the correct transformation behaviour? $\endgroup$ May 11, 2019 at 18:56
  • $\begingroup$ The particle is not created by the real (or imaginary) part of the field though, but by the real part + i imaginary part. Following this analogy i guess you can derive your formula from what you now for the complex field. I could try doing this later. $\endgroup$ May 11, 2019 at 19:01
  • $\begingroup$ I think my first question is, if I have the complex scalar field, then the Noether charge is the difference in number of particles and antiparticles. If I have the two scalar interacting scalar fields, should I expect the same Noether charge? $\endgroup$ May 13, 2019 at 16:09

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