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As I understand, a set of effects $\mathcal{E}\subset\mathcal{E}(\mathcal{H})$ are called informational complete if given $\sigma \in \mathcal{T}\mathcal(H)$, $\operatorname{tr}[\sigma E]=0$ for all $E\in\mathcal{E}$, then $\sigma=0$.

This is also equivalent to: if $\operatorname{tr}[\sigma_1 E]=\operatorname{tr}[\sigma_2 E]$ for all $E\in\mathcal{E}$ then $\sigma_1=\sigma_2$.

However, if we are performing state tomography, then our state is not a general operator in the trace class. Rather, it has trace one.

In other words, now $\rho\in\mathcal{S}(\mathcal{H})\subset \mathcal{T}(\mathcal{H})$, where $\mathcal{S}(\mathcal{H}):=\{\rho\in \mathcal{T}(\mathcal{H})\,|\,\rho\ge 0,\,\operatorname{tr}[\rho]=1\}$.

Therefore, for the definition of tomographical completeness, can we replace $\mathcal{T}(\mathcal{H})$ with $\mathcal{S}(\mathcal{H})$ in the definition of informational completeness?

In other words, can we now define tomographical completeness as: given $\rho_1, \rho_2 \in \mathcal{S}(\mathcal{H})$ if $\operatorname{tr}[\rho_1 E]=\operatorname{tr}[\rho_2 E]$ for all $E\in\mathcal{E}$ then $\rho_1=\rho_2$?

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Yes it's the same thing. If a set of effects $\{E_i\}_i$ linearly spans the set of all density matrices (i.e. the unit-trace positive semidefinite operators) then it also spans all operators. Trivially because any operator can be written as a linear combination of density matrices.

In practice, the terms "tomographically complete" and "informationally complete" are typically used as synonyms in the literature.

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  • $\begingroup$ Do you mean that whether it is $\sigma \in \mathcal{T}(H)$ or $\sigma \in \mathcal{S}(\mathcal{H})$ does not make any difference? My gut feeling is that it does make a difference as $\sigma \in \mathcal{S}(\mathcal{H})$ has $d^2 - 1$ degrees of freedom while $\sigma \in \mathcal{T}(\mathcal{H})$ has $d^2$ degrees of freedom, where $d$ is the dimension of the Hilbert space. Hence, tomographical completeness differs from informational completeness in that it requires one less effect. $\endgroup$
    – Godfly666
    Commented Apr 22 at 11:20
  • $\begingroup$ I mean that (at least in finite dimensions) a set of effects spans all density matrices iff it spans all operators. Think of a qubit: you can't span all qubit states without spanning $I,X,Y,Z$, and thus all Hermitian operators. Formally, any operator decomposes uniquely in terms of 2 Hermitians, any Hermitian decomposes as $H=P-Q$ with $P,Q\ge0$, and any PSD is a scalar multiple of a density matrix. Thus any linear operator is a linear combination of density matrices $\endgroup$
    – glS
    Commented Apr 22 at 12:15
  • $\begingroup$ A qubit has the density operator $\rho = \frac{I + xX + yY + zZ}{2}$ and hence has only three degrees of freedom. So, aren't the effects $\{\Pi_1, \Pi_2, \Pi_3\} = \{\frac{I+X}{2}, \frac{I+Y}{2}, \frac{I+Z}{2}\}$ tomographically complete? Solving the equations $\operatorname{tr}[\rho \Pi_k] = p_k, \quad k=1,2,3$ can uniquely determine the qubit $\rho$. However, informational completeness requires four effects. $\endgroup$
    – Godfly666
    Commented Apr 23 at 9:45
  • $\begingroup$ @Godfly666 that's true, though that set of effects wouldn't represent a physical measurement due to the lack of normalisation: you don't have $\sum_i \Pi_i=I$. I guess the proper way to state this is then that there is no difference between informational and tomographic completeness as you defined them, for any set of effects that forms a POVM. But if you're just going for the mathematical abstraction, sure, you can have a set of effects which span a basis of traceless Hermitians for the space, but not the identity $\endgroup$
    – glS
    Commented Apr 24 at 7:25
  • $\begingroup$ In a typical physical experiment, we don't perform a single-setting measurement where there is a single POVM. Instead, we use multiple measurement settings, each of which is a POVM. For example, to measure a qubit, we usually don't use the SIC-POVM with four elements. Instead, we use three measurement settings $\{\frac{1\pm X}{2}\}$, $\{\frac{1\pm Y}{2}\}$, $\{\frac{1\pm Z}{2}\}$. That's where my doubt comes from. $\endgroup$
    – Godfly666
    Commented Apr 24 at 8:38

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