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.
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asked Apr 12, 2014 at 20:49
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$\begingroup$ It's sort of like casting a shadow (that's why they named it a "projection" operator). $\endgroup$– DumpsterDoofusCommented Apr 12, 2014 at 21:58
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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).
answered Apr 12, 2014 at 21:08
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$\begingroup$ 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 $\hat{P}^2=\hat P$. $\endgroup$– AdamCommented Apr 12, 2014 at 22:52
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$\begingroup$ @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 $|\alpha\rangle$. $\endgroup$ Commented Apr 13, 2014 at 0:19
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$\begingroup$ 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). $\endgroup$– AdamCommented Apr 13, 2014 at 1:01
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$\begingroup$ @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. $\endgroup$ Commented Apr 13, 2014 at 1:17
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1$\begingroup$ You are still assuming that $|a\rangle$ is an eigenvector of $\hat P$, which is not necessarily the case for a general vector. There are plenty of $|a\rangle$ such that $\hat P |a\rangle= c|\alpha\rangle$ with $0\leq c\leq1$. Btw, I don't think your fourth equation make much sense too, since you have $\sum |\alpha\rangle\langle\alpha|$ in the numerator, which is certainly not a normalization. $\endgroup$– AdamCommented Apr 13, 2014 at 7:45