If $\sum_{x=1}^{|A|} |\psi_x\rangle\langle \psi_x|=I$, then $|\psi_x\rangle\langle \psi_x|$ is a basis projectoer.

Question

Let $\left\{E_{x}\right\}_{x=1}^{|A|}$ is a rank 1 matrix in $\operatorname{Herm}(A)$ which $\sum_{x=1}^{|A|} E_{x}=I^A$. $A$ is our Hilbert space and $I^A$ is the identity matrix.Then I want to show that $E_{x}=p_x|y\rangle\langle y|$ which $\left\{|y\rangle\right\}_{x=1}^{|A|}$ are the basis of Hilbert space $A$ and $p_x$ is a number.

$\textbf {Note}:$

All of them are in the Dirac notation. It means that $|y\rangle$ is a column vector and $\langle y|$ is hermitian conjugate of $|y\rangle$.

$\textbf {Attempt}:$

I know $E_x$ is rank-$1$, so it has a form like $|\psi_x\rangle\langle \psi_x|$. We assumed that $\sum_{x=1}^{|A|} E_{x}=I^A$, So $\sum_{x=1}^{|A|} |\psi_x\rangle\langle \psi_x|=I^A$. Then I wrote $|\psi_x\rangle=\sum_{y=1}^{|A|} a_{x,y} |y\rangle$, So $\sum_{x=1}^{|A|}|\psi_x\rangle\langle \psi_x|= \sum_{x=1}^{|A|} \sum_{y=1}^{|A|} \sum_{z=1}^{|A|} a_{x,y}a_{x,z}^* |y\rangle\langle z|=I^A$. On the other hand we know that $\sum_{y=1}^{|A|} |y\rangle\langle y|=I^A$. But now I have no idea to conect them with each other to conclude that $E_{x}=p_x|y\rangle\langle y|$. Could you please help me ?

Answer 1

First off, you mean $E_x=p_x|x\rangle\langle x|$, not $p_x|y\rangle\langle y|$. Second off, you're misinterpreting the problem.

If the basis $\{|x\rangle\}$ is already given, you can't say $E_x=p_x|x\rangle\langle x|$, because that's certainly not true! Rather we need to say there exists a basis $\{|x\rangle\}$ for which the formulas for $E_x$ are true. Plus we can go further and show it is an orthogonal basis, or even an orthonormal basis (so the $p_x$s are $1$).

Here is a version of the proposition that ought to have been given:

If $\{E_x\}$ is a set of $d$ rank-$1$ Hermitian matrices (on a Hilbert space of dimension $d<\infty$) for which the sum $\sum_x E_x$ is the identity matrix $I$, then $E_x=|x\rangle\langle x|$ for some orthonormal basis $\{|x\rangle\}$.

First off, we already know each $E_x$ may be written as $|\psi_x\rangle\langle\psi_x|$ for some vectors $|\psi_x\rangle$, or by factoring out the square magnitude $p_x$ of $|\psi\rangle$ we could write these as $E_x=p_x|x\rangle\langle x|$ for some unit vectors $|x\rangle$. Now we want to show $\{|x\rangle\}$ is an orthonormal basis and the $p_x$s are $1$.

To see that it's a basis, note that the image of $I=\sum_x p_x|x\rangle\langle x|$ is the whole space and it is also contained in the span of $\{|x\rangle\}$, so this is a spanning set, and since it's the same number of elements as the Hilbert space's dimension (which is finite) this must be a basis.

Second, if we apply $I$ to a specific $|y\rangle$ from the basis we get

$$ |y\rangle=\sum_x p_x|x\rangle \langle x|y\rangle $$

Equating coefficients gives $1=p_y$ and $\langle x,y\rangle=0$ for $x\ne y$. Since $y$ is arbitrary, that means all $p_y$s are $1$, and the unit vectors $\{|x\rangle\}$ are orthogonal so it's an orthonormal basis.