I need to find multiplication of two exponential contain vectors as below,
$$e^{i\textbf{p}.\textbf{R}_{A}}e^{-i\textbf{p}.\textbf{R}_{B}}$$ here $\textbf{p},\textbf{R}_{A},\textbf{R}_{B}$ are vectors, therefore $\textbf{p}\cdot \textbf{R}_{A},\textbf{p}\cdot \textbf{R}_{B}$ denote dot product. Moreover, $$\textbf{R}=\textbf{R}_{A}-\textbf{R}_{B}$$
I need to show $$e^{i\textbf{p}\cdot \textbf{R}_{A}}e^{-i\textbf{p}\cdot\textbf{R}_{B}}=e^{i\textbf{p}.\textbf{R}}$$
I first thought this is too simple and tried like this,
$$e^{i\textbf{p}\cdot\textbf{R}_{A}}e^{-i\textbf{p}\cdot\textbf{R}_{B}}=e^{ipR_{A}\cos(\theta_{A})}.e^{-ipR_{B}\cos(\theta_{B})}$$
$$e^{ipR_{A}\cos(\theta_{A})}.e^{-ipR_{B}\cos(\theta_{B})}=e^{ip(R_{A}\cos(\theta_{A})-R_{B}\cos(\theta_{B}))}$$
But, I am stuck here. In order to prove this I need to take that $R_{A}\cos(\theta_{A})-R_{B}\cos(\theta_{B})=R\cos(\theta)$, where $\cos(\theta)$ is the angle between $\textbf{p}$ and $\textbf{R}$. But I don't know how to continue after that.
Best Answer
This is trivial and doesn't need any kind of decomposition with angles. $i(p\cdot R_A)$ is a complex number, so the usual rules for exponentiation hold. Also note that the dot product is linear in the second argument:
$$e^{i(p\cdot R_A)} e^{-i(p\cdot R_B)} = e^{i(p \cdot R_A - p \cdot R_B)} = e^{i(p\cdot(R_A - R_B))} = e^{i(p\cdot R)}$$