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We define the tight operator frame as a set of operators $\{E_i\}_{i=1}^{n}$ satisfying \begin{equation} \sum_{i=1}^n \vert \langle \langle E_i \vert X \rangle \rangle \vert^2 = C \Vert V \Vert_2^2, \forall X \in L(H), \end{equation} where $X$ is any linear operator, $C$ is some constant, $\Vert \cdot \Vert_2$ denotes Fronbenius norm.

Is there a tight operator frame $\{E_i\}_{i=1}^{n}$ that also forms a POVM, i.e., $\sum_{i=1}^{n} E_i = I$, where $E_i$ is positive semidefinite?

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    $\begingroup$ How do you define $V$? Have you tried anything? $\endgroup$
    – Rammus
    Commented Aug 29, 2023 at 11:25

1 Answer 1

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Tight frames in general

A general set of vectors $v_k\in V$ is a frame if there are constants $A,B\in\mathbb{R}$ such that $$A\|v\|^2 \le \sum_k |\langle v_k,v\rangle|^2 \le B \|v\|^2$$ for all $v\in V$. The frame is said to be tight if $A=B$, in which case you have the condition analogous to the one you mentioned (assuming a typo on the right-hand side of the equation): $$\sum_k |\langle v_k,v\rangle|^2 = A \|v\|^2.$$ If the frame is made up of linear operators in some Hilbert space, the definition doesn't change, although you have to take care of what inner product is used in the definition. A standard choice is to use the $L_2$ inner product, meaning that a set of operators $\{E_k\}$ is a (operator) frame iff $$ A\|X\|^2 \le \sum_k |\langle E_k,X\rangle|^2 \le B \|X\|^2 $$ for any linear operator $X$, with $\langle E_k,X\rangle\equiv \operatorname{tr}(E_k^\dagger X)$.

An equivalent condition for a frame being tight is formulated in terms of its "frame operator". A generic frame $\{v_k\}$ is tight iff the operator $S\equiv \sum_k \mathbb{P}(v_k)$, $\mathbb{P}(v_k)\equiv |v_k\rangle\!\langle v_k|$, is a multiple of the identity, i.e. $S=\alpha I$ for some $\alpha\in\mathbb{R}$. This is true regardless of the nature of the vectors, and thus also if $v_k$ are linear operators.

Tight (operator) frames of IC-POVMs

Now, you can certainly consider frames made up of POVMs. There is however a subtlety associated with the definition of tightness. Denote the POVM elements as $\{\mu_b\}\subseteq \operatorname{Pos}(\mathbb{C}^n)$, where I assumed they act on an $n$-dimensional space. Restricting our attention to finite frames in finite dimensions, asking $\{\mu_b\}$ to be a frame is equivalent to asking it to be IC. As per the definition above, the frame operator reads $S=\sum_b \mathbb{P}(\mu_b)$. This is also often referred to as the frame superoperator, it being a linear operator acting on a space of linear operator. Explicitly, this notation means it's the operator defined by the action $$S (X) = \sum_b \mu_b \langle \mu_b,X\rangle\equiv \sum_b \mu_b \operatorname{tr}(\mu_b X),$$ for any $X\in\operatorname{Lin}(\mathbb{C}^n)$. For $\{\mu_b\}$ to be a tight frame, we'd need $S$ to be a multiple of the identity. The "identity" in this context means the identity quantum map $\operatorname{Id}$. In other words, the condition $S = \alpha\operatorname{Id}$ for some $\alpha\in\mathbb{R}$.

This condition, however, can't in general be satisfied for POVMs. To see this, observe that $S=\alpha \operatorname{Id}$ implies $$\sum_b \mu_b \operatorname{tr}(\mu_b) = S(I) = \alpha \operatorname{Id}(I) = \alpha I.$$ Tracing out both sides of this equation we'd get $\sum_b \operatorname{tr}(\mu_b)^2 = \alpha n$. At the same time, if we instead take the (superoperator) trace of the expression $S=\alpha \operatorname{Id}$, we get $\sum_b \operatorname{tr}(\mu_b^2)=\alpha n^2$ (remember $\operatorname{Id}$ is the identity quantum map acting on the subspace of Hermitian linear operators, which thus has dimension $n^2$).

But for any positive semidefinite operator $A$ the inequality $\operatorname{tr}(A^2)\le \operatorname{tr}(A)^2$ holds. Thus we'd get $$\alpha n^2 = \sum_b \operatorname{tr}(\mu_b^2) \le \sum_b \operatorname{tr}(\mu_b)^2 = \alpha n.$$ This is clearly a contradiction for all $n>1$, hence we conclude that an IC-POVM cannot be tight in the standard sense. In fact, this proves that more generally a frame of positive semidefinite operators (thought of as spanning the subspace of Hermitian operators) cannot be tight.

Tight measurement frames

People nevertheless talk about tight measurement frames, which have been shown to be "optimal" for state estimation purposes, for a few definitions of optimal. In this context, you define a rescaled frame, for example as $\mu_b'\equiv\mu_b/\sqrt{\operatorname{tr}(\mu_b)}$ (note that this choice is not unique; in fact, a multiplicative constant is sometimes included in this rescaling for various reasons). If $\{\mu_b\}$ is a frame, so is $\{\mu_b/\sqrt{\operatorname{tr}(\mu_b)}\}$, so you can always do this. The corresponding frame superoperator then reads $$S' = \sum_b \frac{\mathbb{P}(\mu_b)}{\operatorname{tr}(\mu_b)}.$$ You can readily verify that this now verifies $S'(I)=I$, meaning $I$ spans a subspace left invariant by $S'$. Being $S'$ Hermitian (as a superoperator), its orthogonal subspace is also an invariant. The subspace orthogonal to $I$ is the subspace of traceless Hermitian operators. All this tells us that we can decompose $S'$ as $$S' = \mathbb{P}(I) + \tilde S,$$ where $\tilde S(I)=0$, meaning $\tilde S$ acts on the subspace of traceless Hermitian operators. This writing effectively allows to "neglect" the part of the action of $S'$ which is due to the intrinsic redundancies of POVMs that come from the normalisation condition. With this we can finally define a "tight measurement frame", by requiring $\tilde S$ to be a multiple of the identity (superoperator). In other words, a POVM $\{\mu_b\}$ is a "tight measurement frame" if $S'$ is "as close" to be a multiple of the identity as it can be (i.e. if $\tilde S$ is a multiple of the identity superoperator, on the space of traceless Hermitian operators).

References discussing these topics at length include quant-ph/0604049, quant-ph/0310075, 1404.3453, 2301.13229. Some of the notation I used here is from the latter one.

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  • $\begingroup$ Thank you glS for this clear answer! $\endgroup$ Commented Aug 30, 2023 at 10:47

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