# How to obtain the Noether charge for two interacting fields. Correct mode expansion for field operators

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?

• There is nothing "infinitesimal" about your infinitesimal rotation as written – lux May 11 '19 at 7:16
• The infinitesimal rotation (with infinitesimal parameter ε) is φ -> φ + ε D(φ) – Lorenz Mayer May 11 '19 at 18:57

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.

• 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. – user2820579 May 11 '19 at 18:52
• Would it be possible if you give at least what $Q$ is expected for my case? – user2820579 May 11 '19 at 18:53
• You calculated a charge, right? Did you check whether it gives the correct transformation behaviour? – Lorenz Mayer May 11 '19 at 18:56
• 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. – Lorenz Mayer May 11 '19 at 19:01
• 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? – user2820579 May 13 '19 at 16:09