# Rank of a measurement

I am thinking about the following question:

Assuming that we have some given state $\rho$ and we perform a measurement with $k$ outcomes on this state. Then we can describe the measurement in outcomes as eigenvalues of the measurable, i.e., the Hermitian operator that I denote by $D$, with probabilities $\mathrm{Tr}[D_i\rho]$, where $D_i$ are the projectors in the $i^{th}$ eigenspace of $D$, i.e. for the eigendecomposition $D = \sum_i \lambda_i s_i s_i^T = \sum_i \lambda_i D_i$.

I was wondering if my assumption is true. If the number of (distinguishable?) outcomes for any Hermitian operator is given by $k$ i.e. then we have only $k$ non-zero eigenvalues and hence $D$ must be of rank $\leq k$?

• Welcome to quantum computing SE! When you say "probabilities $\mathrm{Tr}[P_i\rho]$, where $D_i$ are the projectors ...", do you mean $\mathrm{Tr}[D_i\rho]$? – Mithrandir24601 May 30 '18 at 18:45
• Sorry, yes exactly. It should be that $D_i$ are the projectors indeed! – LeoW. May 31 '18 at 16:54

You are implicitly making a specific assumption here: that the $\{D_i\}$ are rank 1 projectors.
If your $\{D_i\}$ are rank-1 projectors, i.e. taking the form $D_i=s_is_i^T$, then because there is a completeness relation for measurement operators, $$\sum_iD_i=\mathbb{I},$$ then you must have a number of outcomes equal to the dimension of the Hilbert space you're measuring. Call that $k$. Now, if you define $D=\sum_i\lambda_iD_i$ where the $\lambda_i$ are distinct, then $D$ must have rank either $k$ or $k-1$: if one of the $\lambda_i$ is 0, then the number of non-zero values (which is equivalent to the rank) is $k-1$.
Now, strictly, the $D_i$ could be projectors, but not have rank 1 (in fact, they don't even have to be projectors, but we won't go there...), but instead a rank $r_i=\text{Tr}(D_i)$. In this case, either $D$ is full rank (which we'll still call $k$) or, if a particular $\lambda_j=0$, then it has rank $k-r_j$, because that's the number of non-zero eigenvalues $D$ has. Here, the number of distinguishable outcomes is potentially much smaller than the rank of $D$. All you really know is that $\text{rk}(D)\geq |\{D_i\}|-1$ (i.e. the number of measurement operators minus 1, in case one of the eigenvalues is 0). But that could be a very loose bound in some circumstances (and the bound is the opposite way round to what you were asking).