Express state as eigenkets This is very basic but I just suddenly got confused. Any state can be expressed as complete set of eigenkets with discrete eigenvalues:
$$|P\rangle = \sum^n c_n |p_n\rangle$$
I understand the above. But when going from there to a continuous eigenvalue why is
$$|P⟩ = \int\mathrm dp \: f(p)|p⟩$$
What is the meaning of time a $dp$ then sum over? Does that mean with continuous eigenvalue with range $(a,b)$,
$$|P\rangle = \frac{b-a}{\infty}\sum^\infty c_n |p_n\rangle~?$$
(Discrete and continuous kets are orthogonal to each other and form a complete set for the state.)
 A: Your second equation isn't quite right. If you have a continuous complete set of states $\{|p⟩\}$, then the correct expansion of a given arbitrary state $|P⟩$ in that basis is of the form
$$
|P⟩ = \int\mathrm dp \: f(p)|p⟩,
\tag 1
$$
with a single arbitrary function $f(p)$ over the indexing variable $p$ as a (continuous) coefficient. Here the $dp$ denotes integration over $p$ as usual.
The question of what that integral actually means, and how it is defined, is not a particularly simple topic mathematically. In essence, you are taking a function that takes $p\mapsto f(p)|p⟩$, so that it takes real numbers into state vectors, $\mathbb R\to\mathcal H$, and integrating vector-valued functions is not particularly easy when the vector space is some huge space like $L_2(\mathbb R)$. If you want to do this properly, you need some pretty chunky functional analysis and measure theory chops to get it right.
The cool part is that the answer mostly comes down to "just do it component-by-component". More specifically, say you want to provide a good definition for $(1)$. Then you first choose some basis for the space, say, the position representation $\{|x⟩\}$, and then you project both sides to get the component along $|x⟩$:
$$
⟨x|P⟩ = \int\mathrm dp \: f(p)⟨x|p⟩.
\tag 2
$$
This is actually a much easier integral to define, because you simply have a complex-valued function of a real variable, $p\mapsto f(p)⟨x|p⟩$, where $⟨x|p⟩$ is some known function, so the integral in $(2)$ folds into the integrals we already know how to define. (In practice, you want to be using the Lebesgue integral rather than Riemann sums, but that's for when you care about the measure theoretic considerations.) If you've done everything correctly and all your functions are nice enough, this will work, and give you the same vector $|P⟩$ regardless of what representation you use to calculate it.
A: Ok I think I miss a really important thing. in discrete express. I use primed for discrete and umprimed for continuous$$|P'\rangle = \sum^n |p'_n\rangle$$
$|p'\rangle$ is a different from continuous one. in case of position, the discrete eigenket is probability but for continuous, its probability density. for continuous $|p\rangle dp$ is analogy to $|p'\rangle$ in discrete. So If write the express of continuous one in terms of sum
$$|P⟩ = \int\mathrm dp \:|p⟩ = \sum^\infty |p\rangle dp \rightarrow \lim_{n\to \infty} \sum^n |p'_n\rangle$$
it just like create discrete delta function from continuous 
$$\delta_{mn} = \delta(m-n)da$$
with $\delta(m-n) = \frac{\delta_{mn}}{da}$, which is "large" and is analogy to probablity density ket. $\delta_{mn}$ is analogy to probablity ket.
I am really sorry to brought this up, I automatically assume that both eigenkets have same physical meaning which is not. (people tend not to change the letter when they express the continuous one.) So this is really all about the idea of taking a limit to construct an integration and symbol abuse. Thank you all the help and sorry again for such trival question.
