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In my QM lecture it was claimed that if you have a system with degrees freedom $\vec{s}$ and its surroundings which have degrees of freedom $\vec{u}$ then every density matrix for the combined system can be expressed as

$$ \hat\rho = \sum_{\vec{u}} \sum_{\vec{s}} p_{\vec{u}, \vec{s}} \left|\vec{u}, \vec{s} \right> \left< \vec{u}, \vec{s}\right|$$

(I suppose that the sum is should range over orthonormal bases $\{\left| u \right> \}$ and $\{\left| s \right> \}$.)

To me it seems that this is not general enough to express all possible density matrices. Is that right or am I missing something?

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To me it seems that this is not general enough to express all possible density matrices. Is that right or am I missing something?

You are correct. The most general density matrix should read $$ \hat\rho = \sum_{\vec{u},\vec{s},\vec{u}',\vec{s}'} p_{\vec{u}, \vec{s},\vec{u}',\vec{s}'} \left|\vec{u}, \vec{s} \middle> \middle< \vec{u}', \vec{s}'\right|, $$ with $|\vec u⟩$ and $|\vec s⟩$ running over orthonormal bases for the environment and the system, respectively, with the requirements that $$ p_{\vec{u}, \vec{s},\vec{u}',\vec{s}'} = p_{\vec{u}', \vec{s}',\vec{u},\vec{s}}^{\ \ast} $$ for hermiticity, $$ \sum_{\vec{u},\vec{s}} p_{\vec{u}, \vec{s},\vec{u},\vec{s}} = 1 $$ for unit trace, as well as positive-semidefiniteness on the matrix.

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    $\begingroup$ Almost. You can impose some conditions on that parameter since $\rho$ is hermitian, positive semidefinite and has unit trace. $\endgroup$ – Gabriel Golfetti Sep 22 '18 at 19:54
  • $\begingroup$ @both I thought that went without saying, but you're both correct. $\endgroup$ – Emilio Pisanty Sep 22 '18 at 19:57
  • $\begingroup$ I think this still isn't enough for positive-semidefiniteness, $\mathrm{diag}(1, 1, 0, -1)$ satisfies your conditions $\endgroup$ – 0x539 Sep 22 '18 at 19:59
  • $\begingroup$ @0x539 The matrix needs to be positive-semidefinite, which is already explicitly called for in this answer. Your example does not fall into the class described here. $\endgroup$ – Emilio Pisanty Sep 22 '18 at 20:02
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    $\begingroup$ @ZeroTheHero that would only be the case if every eigenvector of every density matrix were a product state, which isn't true at all $\endgroup$ – Gabriel Golfetti Sep 22 '18 at 21:28

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