What is meant by normalized projection operator? What is meant by normalized projection operator? What is its physical meaning in quantum mechanics? I am pretty confused regarding the physical interpretation of both projection operator and normalized projection operator.
 A: Suppose we have arbitrary vectors $|\alpha\rangle$ and $|\beta\rangle$ that are not necessarily aligned with one another. We can determine the component of $|\beta\rangle$ that lies along the direction of $|\alpha\rangle$ by defining an operator
$$
\hat{P}=|\alpha\rangle\langle\alpha|
$$
which we call the projection operator. Note that $\hat{P}$ is considered normalized if 
$$\sum\hat{P}=\sum|\alpha\rangle\langle\alpha|=1$$
which is then the normalized projection operator.
We can then use $\hat{P}$ as,
$$
\hat{P}|\beta\rangle=|\alpha\rangle\langle\alpha|\beta\rangle=\langle\alpha|\beta\rangle|\alpha\rangle
$$
If, however, $\hat{P}$ is not normalized, then we must use
$$
\hat{P}|\beta\rangle=\frac{\langle\alpha|\beta\rangle|\alpha\rangle}{\sum|\alpha\rangle\langle\alpha|}
$$


As an aside, let's look at $\hat{P}^2$:
$$
\hat{P}^2=\left(|\alpha\rangle\langle\alpha|\right)\left(|\alpha\rangle\langle\alpha|\right)=|\alpha\rangle\langle\alpha|\alpha\rangle\langle\alpha|=|\alpha\rangle\langle\alpha|=\hat{P}
$$
where we used the fact that $\langle\alpha|\alpha\rangle=1$. This property is called idempotence and is a consequence of the fact that once you've projected a vector onto another vector, projecting it a second time gives you the same projection.
Using this property on a vector $|a\rangle$,
$$
\hat{P}^2|a\rangle=\hat{P}|a\rangle
\\
\left(\hat{P}^2-\hat{P}\right)|a\rangle=0
\\
\left(p^2-p\right)|a\rangle=0
$$
Where $p$ is the eigenvalue $\hat{P}$. The solution is either $p=0$ or $p=1$.
The eigenvectors for the appropriate eigenvalues must satisfy
$$
\hat{P}|a\rangle=|\alpha\rangle\langle\alpha|a\rangle=|a\rangle\quad p=1
\\
\hat{P}|a\rangle=|\alpha\rangle\langle\alpha|a\rangle=0\qquad p=0
$$
For $p=1$, the appropriate eigenvector is $A|\alpha\rangle$ for constant $A$ (i.e., any vector parallel to $|\alpha\rangle$ is an eigenvector). For $p=0$, the appropriate eigenvector is 0 (i.e., it is orthogonal).
