# Density matrix for system and surroundings

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?

## 1 Answer

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.

• Almost. You can impose some conditions on that parameter since $\rho$ is hermitian, positive semidefinite and has unit trace. Sep 22, 2018 at 19:54
• @both I thought that went without saying, but you're both correct. Sep 22, 2018 at 19:57
• I think this still isn't enough for positive-semidefiniteness, $\mathrm{diag}(1, 1, 0, -1)$ satisfies your conditions Sep 22, 2018 at 19:59
• @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. Sep 22, 2018 at 20:02
• @ZeroTheHero that would only be the case if every eigenvector of every density matrix were a product state, which isn't true at all Sep 22, 2018 at 21:28