Prove that the square root measurement $\Lambda_y=\frac14(\rho_{B^3})^{-\frac12}|\psi_y\rangle\langle\psi_y|(\rho_{B^{3}})^{-\frac{1}{2}}$ is a POVM

Question

Consider $\textit{X}\sim \mathrm{Unif}([0,1,2,3]), |\mathcal{Y}|=|\mathcal{X}|=4$. Also for every random variable realization {\it x} we use three parallel quantum channels like the one employed before such that:

\begin{equation} \displaystyle \rho_{XB^{3}}=\sum_{x}p_{X}(x)|x\rangle\langle x|_{X}\otimes|\psi_{x}\rangle\langle\psi_{x}|_{B^{3}}, \end{equation}

Prove that the Square Root Measurement: $$ \Lambda_{y}=\frac{1}{4}(\rho_{B^{3}})^{-\frac{1}{2}}|\psi_{y}\rangle\langle\psi_{y}|(\rho_{B^{3}})^{-\frac{1}{2}},$$ for $ y\in[0,1,2,3], $

is a positive operator-valued measure.

A positive operator-valued measure (POVM) is a set of operators $\{\Lambda_j\}$ that satify: \begin{align*} \Lambda_j&\succeq 0\\ \sum_j \Lambda_j &=I. \end{align*}

I have proved the first property (I think):

For any state $|\phi\rangle$, we need to prove that $\langle \phi | \Lambda | \phi \rangle \geq 0$: \begin{align*} \langle \phi | \Lambda_{y} | \phi \rangle &= \frac{1}{4} \langle \phi | (\rho_{B^{3}})^{-\frac{1}{2}}|\psi_{y}\rangle\langle\psi_{y}|(\rho_{B^{3}})^{-\frac{1}{2}} | \phi \rangle \\ &= \frac{1}{4} | (\rho_{B^{3}})^{-\frac{1}{2}}|\phi \rangle |^2 \\ &\geq 0 \end{align*}

I am having trouble to prove the second property:

I have to prove the following equality: $$\sum_{y} \langle \phi | \Lambda_{y} | \phi \rangle = \langle \phi | \phi \rangle$$

For that what I have is:

\begin{align*} \sum_{y} \langle \phi | \Lambda_{y} | \phi \rangle &= \sum_{y} \frac{1}{4} \langle \phi | (\rho_{B^{3}})^{-\frac{1}{2}}|\psi_{y}\rangle\langle\psi_{y}|(\rho_{B^{3}})^{-\frac{1}{2}} | \phi \rangle \\ &= \frac{1}{4} \langle \phi | (\rho_{B^{3}})^{-\frac{1}{2}}\left(\sum_{y} |\psi_{y}\rangle\langle\psi_{y}|\right)(\rho_{B^{3}})^{-\frac{1}{2}} | \phi \rangle \\ \end{align*}

But I don't know how to follow.

Answer 1

Thanks to all people in the comments section that have helped me to arrive to this conclusion:

\begin{align*} \sum_y \Lambda_y &= \sum_y \frac{1}{4}(\rho_{B^{3}})^{-\frac{1}{2}}|\psi_{y}\rangle\langle\psi_{y}|(\rho_{B^{3}})^{-\frac{1}{2}}\\ &=(\rho_{B^{3}})^{-\frac{1}{2}}\sum_y \left(\frac{1}{4}|\psi_{y}\rangle\langle\psi_{y}|\right)(\rho_{B^{3}})^{-\frac{1}{2}}\\ &=(\rho_{B^{3}})^{-\frac{1}{2}}\rho_{B^3}(\rho_{B^{3}})^{-\frac{1}{2}}\\ &=I \end{align*}

Where we have used that $X\sim \mathrm{Unif}([0,1,2,3])$, so the probabilities in the spectral decomposition are $\frac{1}{4}.$