1. Oct 16, 2016

So I just read a paper called Fermion masses, mixings and proton decay in a Randall–Sundrum model (it's in Physics Letters B 498(3-4):256–262, but you can also find it at arXiv:hep-ph/0010195v2). Anyways, there us an equation in it [Pg. 8, 4.14] $$\int \,dx^4 \int \,dy \sqrt{-g} \frac{1}{M_5^3} \bar{\Psi_i}^{(0)}\Psi_j^{(0)} \bar{\Psi_k}^{(0)} \Psi_l^{(0)} \equiv \int \,dx^4 \frac{1}{M_4^2} \bar{\Psi_i}^{(0)}\Psi_j^{(0)} \bar{\Psi_k}^{(0)} \Psi_l^{(0)}$$ So, if we were to make this dependent on the field decomposition equation [Pg. 2, 2.4] $$\Psi(x^\mu, y) = \frac{1}{\sqrt{2\pi R}} \sum_{n = 0}^ \infty \Psi^{(n)}(x^\mu) f_n(y))$$ and the Hamiltonian of a system (H), couldn't the more realistic equation be? $$\Psi(x^\mu, y) \int \,dx^4 \int \,dy \sqrt{-g} \frac{1}{M_5^3} \bar{\Psi_i}^{(0)}\Psi_j^{(0)} \bar{\Psi_k}^{(0)} \Psi_l^{(0)} \equiv \int \,dx^4 \Psi(x^\mu, y) H \frac{1}{M_4^2} \bar{\Psi_i}^{(0)}\Psi_j^{(0)} \bar{\Psi_k}^{(0)} \Psi_l^{(0)}$$

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2. Oct 21, 2016