For the state $\left|\vec{p}\right> = a_{\vec{p}}^{\dagger}\left|0\right>$ we have the energy
$H\left|\vec{p}\right>=E_{\vec{p}}\left|\vec{p}\right>$ $\space\space\space\space\space\space\space$ (1)
with the usual commutation relations $\left[H,a_{\vec{p}}^{\dagger}\right]=E_{\vec{p}}a_{\vec{p}}^{\dagger}$ etc.
The classical total momentum $\vec{P}$ acts on $\left|\vec{p}\right>$ as
$\vec{P}\left|\vec{p}\right>=\vec{p}\left|\vec{p}\right>$. $\space\space\space\space\space\space\space$ (2)
My problem is both intuition and calculation. Other than the words total, state and 3-momentum respectively, what is the intuitive idea\difference behind $\vec{P}$, $\left|\vec{p}\right>$ and $\vec{p}$? As in, what does equation (2) tell us physically? In fact, how has equation (2) been derived using the integral equation of $\vec{P}$?
Let $\mathrm A$ be an $n \times n$ matrix. An eigenvalue of $\mathrm A$ is a scalar $\lambda$ such that there exists a non-zero vector $v$ satisfying
$$ \mathrm A v = \lambda v.$$
Now, as you study Physics, we'll do some abuse of notation and i'll call
$a$ my eigenvalue $\lambda$;
$\left|a\right\rangle$ my eigenvector $v$ associated to $a$.
Now my equation goes like this
$$ A \left|a\right\rangle = a \left|a \right\rangle.$$
The nice thing with this notation is that you keep the information of which eigenvalue $v$ is associated to.
There is no more than notations here…