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In the equation, where the $\vec{\Psi}$'s are particle states,

$$ \sum_{l,m} C_{lm}\Big\{\Big[\frac{n^{2}}{c^{2}}\dfrac{\partial^2}{\partial t^2}\vec{\Psi}_l(\vec{r}_{1},t)-\nabla^{2}_{1}\vec{\Psi}_l(\vec{r}_{1},t)+\vec{f}_{l}(\vec{r}_{1},t)\Big]\otimes\vec{\Psi}_m(\vec{r}_{2},t)\\ +\vec{\Psi}_l(\vec{r}_{1},t)\otimes\Big[\frac{n^{2}}{c^{2}}\dfrac{\partial^2}{\partial t^2}\vec{\Psi}_m(\vec{r}_{2},t)-\nabla^{2}_{2}\vec{\Psi}_m(\vec{r}_{2},t)+\vec{f}_{m}(\vec{r}_{2},t)\Big]\Big\}=0 $$

can I say that for it to be true, given the states $\vec{\Psi}$'s are not zero, the terms in square brackets have to be zero each one separetely? If it is the case, how can I prove it?

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  • $\begingroup$ does Anyone know? $\endgroup$ – Alfredo Guima Jul 7 at 23:00

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