6
$\begingroup$

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 $\leq k$?

$\endgroup$
  • $\begingroup$ Welcome to quantum computing SE! When you say "probabilities $\mathrm{Tr}[P_i\rho]$, where $D_i$ are the projectors ...", do you mean $\mathrm{Tr}[D_i\rho]$? $\endgroup$ – Mithrandir24601 May 30 '18 at 18:45
  • $\begingroup$ Sorry, yes exactly. It should be that $D_i$ are the projectors indeed! $\endgroup$ – LeoW. May 31 '18 at 16:54
3
$\begingroup$

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).

Overall, the answer is that the number of distinguishable measurement outcomes is not equal to the rank of the measurement operator.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.