# Reduced density matrix: Derive or postulate?

Usually this expression is neither *postulated* nor *derived*, but rather a serves as a *definition* of the reduced density matrix.

For more context let me add:

Reduced density matrix typically appear in certain contexts, notably, when you consider a system coupled to a bath, and want to trace out the degrees of freedom of the bath. One then typically shows that such a reduced density matrix possesses all the properties of a true density matrix, necessary for the complete description of the system of interest.

As indicated in the answer by @Vadim the expression you give is a definition of the reduced density matrix. From this definition, the Born rule and some probability you can **prove** that the reduced density matrix carries information about the marginal probability distribution for the subsystem which it describes.

In addition to the existing answers I will try to add some context and elaborate on the condition that the reduced density matrix must fulfill. Consider a measurable $\hat Q$ that only acts on the system of interest. We know from the postulates (e.g., Postulate 3 in Nielsen and Chuang) that the expectation value is $$ \langle\hat Q\rangle = \mathrm{Tr}\{\hat Q \hat \rho_\mathrm{S}\}. $$ Now, we demand that the measurable $\hat Q \otimes \hat I_\mathrm{E}$ (that acts on the complete system) must yield the same expectation value $$ \langle\hat Q \otimes \hat I_\mathrm{E}\rangle = \mathrm{Tr}\{(\hat Q \otimes \hat I_\mathrm{E}) \hat \rho\} $$

This derivation requires some juggling around with tensor and scalar products, but will ultimately deliver an opportunity to define the reduced density matrix. Personally, I found these notes very helpful.

**Edit: Added derivation**

Let $\hat Q = \hat Q_\mathrm{S} \otimes \hat I_\mathrm{E}$. We need to verify that $$ \mathrm{Tr}\{\hat Q_\mathrm{S} \hat \rho_\mathrm{S} \} = \mathrm{Tr}\{\hat Q \hat \rho \} $$ or, using the property of the partial trace $\mathrm{Tr}\{\cdot\} = \mathrm{Tr}_\mathrm{S}\{\mathrm{Tr}_\mathrm{E}\{\cdot\}\}$, that $$ \mathrm{Tr}\{\hat Q_\mathrm{S} \hat \rho_\mathrm{S} \} = \mathrm{Tr}_\mathrm{S}\{\mathrm{Tr}_\mathrm{E}\{\hat Q \hat \rho\} \}. $$ We note that the trace operation on the left hand side is actually equivalent to the trace over the system S. Therefore, we aim to verify the equivalence of the arguments. Since any operator on the complete system can be extended using the basis operators $\hat \alpha_i$ (of the reduced system) and $\hat \beta_j$ (of the environment), we can write $$ \mathrm{Tr}_\mathrm{E}\{\hat Q \hat \rho\} = \mathrm{Tr}_\mathrm{E}\{(\hat Q_\mathrm{S} \otimes \hat I_\mathrm{E}) \hat \rho \} = \mathrm{Tr}_\mathrm{E}\left\{(\hat Q_\mathrm{S} \otimes \hat I_\mathrm{E}) \left( \sum_{i,j} \hat \alpha_i \otimes \hat \beta_j \right) \right\} $$ and, by exploiting the property $(\hat A \otimes \hat B)(\hat C \otimes \hat D) = \hat A \hat C \otimes \hat B \hat D$ of the tensor product, $$ \mathrm{Tr}_\mathrm{E}\{\hat Q \hat \rho\} = \mathrm{Tr}_\mathrm{E}\left\{ \sum_{i,j} \hat Q_\mathrm{S} \hat \alpha_i \otimes \hat \beta_j \right\}. $$ Now, since the partial trace $\mathrm{Tr}_\mathrm{E}\{\hat Q_\mathrm{S} \hat \alpha_i \otimes \hat \beta_j\} = \hat Q_\mathrm{S} \hat \alpha_i \mathrm{Tr}\{\hat \beta_j\} = \hat Q_\mathrm{S} \mathrm{Tr}_\mathrm{E}\{\hat \alpha_i \otimes \hat \beta_j\}$, we can pull the measurable $\hat Q_\mathrm{S}$ out of the partial trace and write $$ \mathrm{Tr}_\mathrm{E}\{\hat Q \hat \rho\} = \hat Q_\mathrm{S} \mathrm{Tr}_\mathrm{E}\left\{ \sum_{i,j} \hat \alpha_i \otimes \hat \beta_j \right\} = \hat Q_\mathrm{S} \mathrm{Tr}_\mathrm{E}\left\{ \hat \rho \right\} = \hat Q_\mathrm{S} \hat \rho_\mathrm{S}. $$