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I am currently on a set of lecture notes which says that for a state vector $| \psi \rangle_{AB}$ describing a tensor product state, its density operator $| \psi \rangle \langle \psi |_{AB}$ can be expanded in the basis of traceless Hermitian Pauli operators $\delta_{0} \otimes \delta_{0}, \delta_{0} \otimes \delta_{1}, ..., \delta_{3} \otimes \delta_{3}$ = ${ \delta_{i} \otimes \delta_{j} }_{i, j = 0}^{3}$

This way:

$$| \psi \rangle \langle \psi |_{AB} = \frac{1}{4}\left[I \otimes I + \vec{a}.\vec{\sigma} \otimes I + I \otimes \vec{b}.\vec{\sigma} + \sum_{i,j = 0}^{3} T_{i,j} \sigma_{i} \otimes \sigma_{j}\right],\tag{1}$$

in which,

$$T_{i,j} = Tr\left[\sigma_{i} \otimes \sigma_{j}| \psi \rangle \langle \psi |\,\right]\,.\tag{2}$$

How can I show that the above expression is true?

Any help is appreciated.

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    $\begingroup$ any operator can be decomposed as a sum of products of Pauli matrices, see eg quantumcomputing.stackexchange.com/q/8725/55 and links therein for how this is done. Does that answer the question? $\endgroup$
    – glS
    Commented Sep 17, 2023 at 17:05

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