I've been trying to decompose a 3x3 density matrix with 3-dimensional Pauli matrices but it doesn't work for all matrices.

For example, the density matrix of the state $|0\rangle + |1\rangle + |2\rangle$ can be decomposed by obtaining the coefficients of the equation $\rho = a_{X}X + a_{Y}Y + a_{Z}Z + a_{V}V + a_{X^2}X^2 + a_{Y^2}Y^2 + a_{Z^2}Z^2 + a_{V^2}V^2 + a_{I}I $ using trace, e.g. $a_X = Tr(\rho X)/3 $. Here, $Y=XZ, V=XZ^2$, and X and Z are $3 \times 3$ Pauli matrices shown in https://en.wikipedia.org/wiki/Generalizations_of_Pauli_matrices.

On the other hand, when I tried to decompose the state $|0\rangle + |2\rangle$ using the above equation, the result was not the same as $|0\rangle + |2\rangle$.

Is it impossible to decompose a high-dimensional matrix using Pauli matrices?

Cross-posted on physics.SE

  • $\begingroup$ You are talking about state tomography. Nielsen's book, chap 8, process tomography(starting with state tomography) might be helpful. $\endgroup$
    – narip
    Jul 1, 2021 at 1:56
  • 3
    $\begingroup$ @James The space of linear operators $\mathcal{L}(\mathbb{C}^{3}) = \{ A ~|~ A: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3}, A \text{ is linear} \}$ is a $9$-dimensional vector space. So any $9$ linearly independent matrices will span this space and give a unique decomposition for every matrix. This has nothing to do with Pauli matrices, they are just a convenient choice to work with. $\endgroup$ Jul 1, 2021 at 6:25
  • 2
    $\begingroup$ @James Moreover, $\mathcal{L}(\mathbb{C}^{3})$ is itself a Hilbert space with respect to the Hilbert-Schmidt inner product, defined as $\left\langle A,B \right\rangle = \operatorname{Tr}\left[ A^{\dagger}B \right]$. Notice, this is an inner product on the space of operators (and not vectors). The Gell-Mann matrices are orthogonal with respect to this inner product, just like Pauli matrices. I'd suggest rechecking your calculations or sharing some of your work. $\endgroup$ Jul 1, 2021 at 6:29
  • $\begingroup$ see Is the Pauli group for $n$-qubits a basis for $\mathbb{C}^{2^n\times 2^n}$? and links therein $\endgroup$
    – glS
    Jul 1, 2021 at 8:32


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