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Let $A$ and $B$ be 2 subsystems of a quantum mechanical system, so a state of the whole system is a vector in $A \otimes B$. As far as I understand, a density operator $ \rho $ in general can't be written as a tensor product of the density operators of its subsystems. If $\rho_A$ and $\rho_B$ are the density operators of two independent systems, then you can write: $$ \rho = \rho_A \otimes \rho_B. $$ Otherwise (the two systems are entangled), to get an expression that contains the same information as $\rho_A$, you would have to take the partial trace of the operator over the subspace $B$: $$ \tilde{\rho}_A = \mathrm{tr}_B \rho. $$

Now we know that in thermodynamic equilibrium, the von Neumann entropy of the system (of $\rho$) is at its maximum. Can we derive from that that the von Neumann entropy of the reduced density matrix $\tilde{\rho}_A$ is also at its maximum?

In the case of no entanglement, $$ S[\rho_A \otimes \rho_B] = S[ \rho] = S[\rho_A] + S[\rho_B] $$ holds, and since all expressions are bigger than zero, one could argue that for $S[\rho]$ to me maximized, one needs also to maximize $S[\rho_A]$ and $S[\rho_B]$. This doesn't work anymore for an entangled system, because here we don't have additivity of the entropy, but instead only subadditivity for $\tilde{\rho}_A = \mathrm{tr}_B \rho$ and $\tilde{\rho}_B = \mathrm{tr}_A \rho$. Is $S[\mathrm{tr}_B \rho]$ still maximized?

Let $A$ and $B$ be 2 subsystems of a quantum mechanical system, so a state of the whole system is a vector in $A \otimes B$. As far as I understand, a density operator $ \rho $ in general can't be written as a tensor product of the density operators of its subsystems. If $\rho_A$ and $\rho_B$ are the density operators of two independent systems, then you can write: $$ \rho = \rho_A \otimes \rho_B. $$ Otherwise (the two systems are entangled), to get an expression that contains the same information as $\rho_A$, you would have to take the partial trace of the operator over the subspace $B$: $$ \tilde{\rho}_A = \mathrm{tr}_B \rho. $$

Now we know that in thermodynamic equilibrium, the von Neumann entropy of the system (of $\rho$) is at its maximum. Can we derive from that that the von Neumann entropy of the reduced density matrix $\tilde{\rho}_A$ is also at its maximum?

In the case of no entanglement, $$ S[\rho_A \otimes \rho_B] = S[ \rho] = S[\rho_A] + S[\rho_B] $$ holds, and since all expressions are bigger than zero, one could argue that for $S[\rho]$ to me maximized, one needs also to maximize $S[\rho_A]$ and $S[\rho_B]$. This doesn't work anymore for an entangled system, because here we don't have additivity of the entropy, but instead only subadditivity for $\tilde{\rho}_A = \mathrm{tr}_B \rho$ and $\tilde{\rho}_B = \mathrm{tr}_A \rho$. Is $S[\mathrm{tr}_B \rho]$ still maximized?

Edit: To give a reason, why I'm asking this. The question I was thinking about originally was: If a quantum system is in thermodynamic equilibrium, are the sub systems also in thermodynamic equilibrium? My naive answer to that is "Yes, they should be", but I'm not sure about that, and I can't give a proper reason, why they should.

Let $A$ and $B$ be 2 Subsystems of a quantum mechanical System (so a state of the whole system is a vector from $A \otimes B$. As much as I understand, a density operator $ \rho $ in general can't be written as a tensor product of the density operators of it's subsystems. If $\rho_A$ and $\rho_B$ are the density operators of two independend systems, then you can write: $$ \rho = \rho_A \otimes \rho_B $$ Elseways (the two systems are entangled), to get an expression that contains the same information as $\rho_A$, you would have to take the partial trace of the operator over the subspace $B$: $$ \tilde{\rho}_A = \mathrm{tr}_B \rho $$

Now we know that in thermodynamic equilibrium, the von Neumann entropy of the system (of $\rho$) is at its maximum. Can we derive from that that the von Neumann entropy of the reduced densitiy matrix $\tilde{\rho}_A$ is also at its maximum?

In the case of no entanglement, $$ S[\rho_A \otimes \rho_B] = S[ \rho] = S[\rho_A] + S[\rho_B] $$ holds, and since all expressions are bigger than zero, one could argue that for $S[\rho]$ to me maximized, one needs also to maximize $S[\rho_A]$ and $S[\rho_B]$. This doesn't work anymore an entangled system, because here we don't have additivity of the entropy, but instead only subadditivity for $\tilde{\rho}_A = \mathrm{tr}_B \rho$ and $\tilde{\rho}_B = \mathrm{tr}_A \rho$. Is $S[\mathrm{tr}_B \rho]$ still maximized?

Let $A$ and $B$ be 2 subsystems of a quantum mechanical system, so a state of the whole system is a vector in $A \otimes B$. As far as I understand, a density operator $ \rho $ in general can't be written as a tensor product of the density operators of its subsystems. If $\rho_A$ and $\rho_B$ are the density operators of two independent systems, then you can write: $$ \rho = \rho_A \otimes \rho_B. $$ Otherwise (the two systems are entangled), to get an expression that contains the same information as $\rho_A$, you would have to take the partial trace of the operator over the subspace $B$: $$ \tilde{\rho}_A = \mathrm{tr}_B \rho. $$

Now we know that in thermodynamic equilibrium, the von Neumann entropy of the system (of $\rho$) is at its maximum. Can we derive from that that the von Neumann entropy of the reduced density matrix $\tilde{\rho}_A$ is also at its maximum?

In the case of no entanglement, $$ S[\rho_A \otimes \rho_B] = S[ \rho] = S[\rho_A] + S[\rho_B] $$ holds, and since all expressions are bigger than zero, one could argue that for $S[\rho]$ to me maximized, one needs also to maximize $S[\rho_A]$ and $S[\rho_B]$. This doesn't work anymore for an entangled system, because here we don't have additivity of the entropy, but instead only subadditivity for $\tilde{\rho}_A = \mathrm{tr}_B \rho$ and $\tilde{\rho}_B = \mathrm{tr}_A \rho$. Is $S[\mathrm{tr}_B \rho]$ still maximized?

Let $A$ and $B$ be 2 Subsystems of a quantum mechanical System (so a state of the whole system is a vector from $A \otimes B$. As much as I understand, a density operator $ \rho $ in general can't be written as a tensor product of the density operators of it's subsystems. If $\rho_A$ and $\rho_B$ are the density operators of two independend systems, then you can write: $$ \rho = \rho_A \otimes \rho_B $$ Elseways (the two systems are entangled), to get an expression that contains the same information as $\rho_A$, you would have to take the partial trace of the operator over the subspace $B$: $$ \tilde{\rho}_A = \mathrm{tr}_B \rho $$

Now we know that in thermodynamic equilibrium, the von Neumann entropy of the system (of $\rho$) is at its maximum. Can we derive from that that the von Neumann entropy of the reduced densitiy matrix $\tilde{\rho}_A$ is also at its maximum?

In the case of no entanglement, $$ S[\rho_A \otimes \rho_B] = S[ \rho] = S[\rho_A] + S[\rho_B] $$ holds, and since all expressions are bigger than zero, one could argue that for $S[\rho]$ to me maximized, one needs also to maximize $S[\rho_A]$ and $S[\rho_B]$. This doesn't work anymore an entangled system, because here we don't have additivity of the entropy, but instead only subadditivity.

Let $A$ and $B$ be 2 Subsystems of a quantum mechanical System (so a state of the whole system is a vector from $A \otimes B$. As much as I understand, a density operator $ \rho $ in general can't be written as a tensor product of the density operators of it's subsystems. If $\rho_A$ and $\rho_B$ are the density operators of two independend systems, then you can write: $$ \rho = \rho_A \otimes \rho_B $$ Elseways (the two systems are entangled), to get an expression that contains the same information as $\rho_A$, you would have to take the partial trace of the operator over the subspace $B$: $$ \tilde{\rho}_A = \mathrm{tr}_B \rho $$

Now we know that in thermodynamic equilibrium, the von Neumann entropy of the system (of $\rho$) is at its maximum. Can we derive from that that the von Neumann entropy of the reduced densitiy matrix $\tilde{\rho}_A$ is also at its maximum?

In the case of no entanglement, $$ S[\rho_A \otimes \rho_B] = S[ \rho] = S[\rho_A] + S[\rho_B] $$ holds, and since all expressions are bigger than zero, one could argue that for $S[\rho]$ to me maximized, one needs also to maximize $S[\rho_A]$ and $S[\rho_B]$. This doesn't work anymore an entangled system, because here we don't have additivity of the entropy, but instead only subadditivity for $\tilde{\rho}_A = \mathrm{tr}_B \rho$ and $\tilde{\rho}_B = \mathrm{tr}_A \rho$. Is $S[\mathrm{tr}_B \rho]$ still maximized?