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Let us consider a quantum mechanical system of interest S that interacts with the environment E. Then, the reduced density matrix $$ \hat \rho_\mathrm{S} = \mathrm{Tr}_\mathrm{E} \{ \hat \rho \} $$ is the partial trace over the environment, where $\hat \rho$ denotes the density matrix of the complete system (S + E).

Is this expression postulated or can it be derived (from basic postulates, such as the set given by Nielsen and Chuang for example)?

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  • $\begingroup$ If you want to derive the formula, you first need to give an independent definition of both sides of the equation. What is your independent definition of ${\hat \rho}_\text{S}$? $\endgroup$ – Prahar Oct 27 at 14:25
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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.

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  • $\begingroup$ OK, but then there should be some derivation which is typically carried out before (at some point) the definition is introduced, right? $\endgroup$ – carlosvalderrama Oct 27 at 14:59
  • $\begingroup$ I am not sure what you mean... 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 the density matrix, necessary for the complete description of the system of interest. $\endgroup$ – Vadim Oct 27 at 15:03
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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.

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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 $$ \mathrm{Tr}_\mathrm{E}\{\hat Q \hat \rho\} = \hat Q_\mathrm{S} \mathrm{Tr}_\mathrm{E}\left\{ \sum_{i,j} \hat \alpha_i \otimes \hat \beta_j \right\} = \hat Q_\mathrm{S} \mathrm{Tr}_\mathrm{E}\left\{ \hat \rho \right\} = \hat Q_\mathrm{S} \hat \rho_\mathrm{S}. $$

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  • $\begingroup$ Thanks, that helps a lot. Also thanks for the pointer to the notes. The derivation seems to be quite complicated, though. Is there a way to simplify that a bit? $\endgroup$ – carlosvalderrama Oct 28 at 12:00
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    $\begingroup$ I have edited my answer and added my approach. Not completely sure whether everything is correct, though -- it seems to be too simple to be true ;-) $\endgroup$ – lucky_luke Oct 28 at 12:33
  • $\begingroup$ Cool, thanks :-) $\endgroup$ – carlosvalderrama Oct 28 at 13:37

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