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If we have a two qubit system, can we find a convenient set $\{\hat e_i\}$ for the $4 \times 4$ density matrix $\hat \rho$, such that $\rho$ can be writen as a linear combination of the elements in the set, and each element of the set a physical state, i.e. the $\mathrm{Tr} \{ \hat e_i\} = 1$, $\hat e^{\dagger} = \hat e$ and $\hat e_i \ge 0$?

For example, two qubits can be expanded as: $$\hat \rho = \frac{1}{4} \sum_{i,j = 0}^3 S_{ij} \sigma_i \otimes \sigma_j $$ Where $\sigma_i$ belongs to the vector $\vec \sigma = \{1_2, \sigma_1, \sigma_s, \sigma_3 \}$, i.e. the $2 \times 2$ identity matrix followed by the Pauli matrices. However, not all of the 16 terms in this expansion are physical states. Most of them have trace zero.

We can also expand $\hat \rho$ as: $$ \hat \rho = \sum_{i, j, k, l = 0}^{1} A_{ijkl} | i \rangle \langle j | \otimes | k \rangle \langle l | $$

But again, the basis states in this expansion are not physical, most of them have trace 0.

I'm looking for an expansion $\hat \rho = \sum_{i=0}^{15} \hat e_i$ where all of $\hat e_i$'s are physical. Is it possible, or is there any resource out there which talks about this?

EDIT

A single qubit can be written as a linear combination of $\frac{1_2 + \sigma_i}{2}$ (of course, there'll be conditions on the coefficients) where $i$ runs from 1 to 3. All three of these are physical.I'm looking for something similar for two qubits.

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    $\begingroup$ Well, I don't know if this helps, but the space of density matrices is convex with the pure density matrices as extreme elements. Thus, every density matrix admits at least one (possibly trivial), possibly infinitely many, convex decomposition(s). In particular, every density opertor admits a spectral decomposition, i.e. we can write $ \rho = \sum\limits_{k=1}^{\mathrm{dim}H} p_k\, |k\rangle\langle k|$, which clearly is a convex combination of pure density matrices. $\endgroup$
    – Tobias Fünke
    Nov 1, 2022 at 9:23
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    $\begingroup$ Part of the problem is that the space of density matrices is decidedly not a vector space (it's not closed under addition or multiplication by scalars), so it won't have a basis in the usual sense. The idea of a convex combination (as noted in a previous comment) is the closest analogy, and any density matrix can be written as a convex combination of extremal states (which are the pure states). However, I don't know whether there is a "basis" of extremal states in the sense that they are in some sense independent and any density matrix can be written as a convex combination of them. $\endgroup$
    – march
    Nov 1, 2022 at 15:48
  • $\begingroup$ @Bard Do you require positive weights in your expansion? $\endgroup$
    – Norbert Schuch
    Nov 1, 2022 at 17:44
  • $\begingroup$ @march There is no such basis -- no pure state can be expressed in a basis which does not contain said state. $\endgroup$
    – Norbert Schuch
    Nov 1, 2022 at 17:45
  • $\begingroup$ @NorbertSchuch No, for example in the single qubit case I mentioned the weights need not be positive. $\endgroup$
    – Bard
    Nov 1, 2022 at 17:47

2 Answers 2

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There is several options, as long as you don't require positive weights:

  1. Choose a SIC-POVM.

  2. Take your favorite hermitian basis -- for instance, the tensor products of the Pauli matrices (including the identity), and add the identity of them until the resulting operators are positive semidefinite. (In case you chose the Pauli products $P_i$, this means your basis has e.g. entries $I+P_i$, together with the identity matrix.) If you want that they have trace 1: Divide by the trace.

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  • $\begingroup$ In the second method, won't there be an extra condition that the sum of the coefficients is 1? Which is not true in general. $\endgroup$
    – Bard
    Nov 5, 2022 at 17:05
  • $\begingroup$ What do you mean? I am just specifying basis of positive semi-definite operators, in which you can expand your state (or, for that matter, any operator). The coefficients will depend on the state $\rho$. They will without doubt satisfy some condition which imply that the state you expand in this basis is positive semidefinite and has trace one, but this will be true for whatever basis you specify. (In particular, the positivity constraint will always be non-trivial.) $\endgroup$
    – Norbert Schuch
    Nov 5, 2022 at 18:25
  • $\begingroup$ I get it now, thanks for the answer. $\endgroup$
    – Bard
    Nov 6, 2022 at 13:44
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An interesting choice is to use so-called measurement frames. In brief, the idea is that given any informationally-complete POVM with elements $\Pi_i$, you can find another set of operators, call them $F_i$, such that any state can be decomposed as $\rho=\sum_i \operatorname{Tr}(\Pi_i \rho) F_i$. This makes the coefficients "physical" because they just become the output probabilities. This is useful in tomography applications and many other things. A good reference is (A. J. Scott 2006).

This probably also works for not informationally-complete POVMs, but some modifications in the formalism might be required.

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  • $\begingroup$ Isn't that just the notion of dual bases? And will those F_i be all positive (I don't think so, otherwise those wouldn't be bi-orthogonal -- but maybe bi-orthogonality is not needed/wanted here?)? $\endgroup$
    – Norbert Schuch
    Nov 2, 2022 at 11:26
  • $\begingroup$ yes, it's pretty much that. Well, slightly more general because it works also when the vectors are not bases but might contain more elements than the space dimension. It's more generally an application of the formalism of frame theory to compute "dual frames", which reduces to dual bases when applied to, well, bases. It's nice because it works for general sets/bases, regardless of their orthogonality. Eg an IC-POVM can't have mutually orthogonal elements, hence you need to apply this kind of formalism to work out the "dual basis" (or "dual frame", in this context) $\endgroup$
    – glS
    Nov 2, 2022 at 12:31
  • $\begingroup$ just to further clarify: this is completely general. In finite-dimensional spaces, a finite set of vectors is a "frame" iff it spans the space, and you can then work out its dual frame etc. See eg this recent related question of mine. The result about POVMs etc is just an application of this framework to the vector (Hilbert) space of Hermitian operators (and more specifically, using as frames sets of PSD operators) $\endgroup$
    – glS
    Nov 2, 2022 at 12:38
  • $\begingroup$ Interesting. -- But then again: If the $\Pi$ are positive semidefinite, the $F_i$ cannot all be positive semidefinite (since the weights $\mathrm{tr}(\Pi_i\rho)\ge0$). So in that sense it does not address the OPs question. $\endgroup$
    – Norbert Schuch
    Nov 2, 2022 at 13:17
  • $\begingroup$ @NorbertSchuch that's true. I guess it depends whether one wants "physical coefficients" or that the operators used in the decomposition are physical. Interestingly though, the dual of the dual is the frame itself. Meaning that if $\rho=\sum_i \operatorname{Tr}(\Pi_i \rho)F_i$ for all Hermitian $\rho$, then also $\rho=\sum_i \operatorname{Tr}(F_i \rho)\Pi_i$. So one can get a decomposition in terms of the elements of the POVM. But then it's a bit harder to interpret physically the expansion coefficients $\endgroup$
    – glS
    Nov 2, 2022 at 14:27

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