# 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.

-
It's sort of like casting a shadow (that's why they named it a "projection" operator). –  DumpsterDoofus Apr 12 '14 at 21:58

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).

-
I don't think the eigenvalue part makes much sense, since even if $|\beta\rangle$ is not an eigenvector of $\hat P$, we always have $(\hat{P}^2-\hat P)|\beta\rangle=0$, since $\hat{P}^2=\hat P$. –  Adam Apr 12 '14 at 22:52
@Adam: The key is not to look at the $\hat{P}^2-\hat{P}$ part but to look at $\hat{P}|\beta\rangle$. This will either be $\mathbb 1|\beta\rangle$ or $\mathbb 0|\beta\rangle$ because it is either parallel or orthogonal to basis $|\alpha\rangle$. –  Kyle Kanos Apr 13 '14 at 0:19
I think you have a problem of notation then. You said $|\beta\rangle$ is not necessarily orthogonal to $|\alpha\rangle$, so you have in general $0\leq\langle\alpha|\beta\rangle\leq 1$, and therefore $\hat P|\beta\rangle=a|\alpha\rangle$ with $0\leq a\leq 1$ (that's the definition of a projector). –  Adam Apr 13 '14 at 1:01
@Adam: The aside is separate from the initial answer (before the thin solid line). I'll clean it up in a few minutes to assert that more clearly. –  Kyle Kanos Apr 13 '14 at 1:17
@Adam: Aside cleaned up. Let me know if this is still confusing. –  Kyle Kanos Apr 13 '14 at 1:41