# Is the trace of a reduced density operator always equal to 1?

From my understanding, we define a reduced density operator $$\rho_A$$ of an operator $$\rho_{AB} = |a_1⟩⟨a_2|\otimes |b_1⟩⟨b_2|$$, as: $$ρ_A=Tr_a(ρ_{AB})=|a_1⟩⟨a_2|Tr(|b_1⟩⟨b_2|)$$

I also know that $$Tr(\rho)$$ must always be 1. Does this hold also for reduced density operators?

Yes, because $$\mathrm{Tr}_B\circ\mathrm{Tr}_A=\mathrm{Tr}$$, that is, the composition of the partial trace operations gives the standard trace.

For this we only need to focus on the main diagonal elements. The full operator has diagnoal elements $$(\hat{\rho}_{AB})_{i, j} := \rho_{i,j} |a_i\rangle\langle a_i| \otimes |b_j\rangle\langle b_j|$$

The reduced operator will have diagonal elements $$(\hat{\rho}_{A})_i = \text{Tr}_B(\hat{\rho}_{AB})_i = \sum_j \rho_{i,j}|a_i\rangle\langle a_i|$$

Taking the trace of the reduced operator we see that $$\text{Tr}(\hat{\rho}_A) = \sum_i (\hat{\rho}_A)_i = \sum_{i,j} \rho_{i,j} = \text{Tr}_{A,B}\hat{\rho}_{AB} = 1$$

So if the full operator has unity trace, so does the reduced form of it.

Just to add on to what was said by there other answers, it is trivially true that for a composite system in $$H^{A \times B}$$ composed of systems in $$H^{A}$$ and $$H^{B}$$, not every $$\hat{\rho}_{AB}$$ can be written as a $$\hat{\rho}_{A} \times \hat{\rho}_{B}$$, but let us consider what the effects of the composite subsystems entangled-ness will be on the subsystems density matrices?

First, embedded in your question is the fact that: if $$\hat{\rho}_{AB} = \hat{\rho}_{A} \times \hat{\rho}_{B}$$, then $$\hat{\rho}_{A}$$ and $$\hat{\rho}_{B}$$ are both pure systems. This is because their density matrices can be written as $$|\psi \rangle \langle \psi |$$ and $$|\phi \rangle \langle \phi |$$ where $$|\psi \rangle \in H^A$$ and $$|\phi \rangle \in H^B$$ respectively.

So we can say "Composite System Unentangled $$\rightarrow$$ Subsystems Pure"

Now lets go the other way, what happens if we assume the Subsystems are pure? Without loss of generality lets assume $$\hat{\rho}_{A} = |a_0 \rangle \langle a_0 |$$ since we can just define a basis where pure state is a basis vector:

$$\hat{\rho}_{A} = |a_0 \rangle \langle a_0 |$$ $$=\text{Tr}_B(\hat{\rho}_{AB})$$ $$=\text{Tr}_B(|\gamma \rangle \langle \gamma |)$$ $$=\text{Tr}_B(\sum_{w,x,y,z} \gamma_{w,x} \gamma_{y,z} |a_w b_x \rangle \langle a_y b_z | )$$ $$=\sum_{k}\langle b_k |(\sum_{w,x,y,z} \gamma_{w,x} \gamma_{y,z} |a_w b_x \rangle \langle a_y b_z | )|b_k \rangle$$ $$=\sum_{w,y,k} \gamma_{w,k} \gamma_{y,k} |a_w \rangle \langle a_y|$$

But we know this has to be: $$=|a_0 \rangle \langle a_0 |$$ From the first line, therefore: $$\gamma_{w,k}=\delta_{w,0} \sigma_k$$ $$|\sigma \rangle := \sum_{k} \sigma_k |b_k\rangle$$ Which shows that $$|\gamma \rangle = |a_0 \rangle |\sigma \rangle$$, which is unentangled. So we can also say that "Composite System Unentangled $$\leftarrow$$ Subsystems Pure"

Combining both results gives the fact that "Composite System Unentangled $$\leftrightarrow$$ Subsystems Pure".

Of course this does not affect the trace of the total density matrix or even that of the subsystems, but it does affect the trace of the square of each of the density matrices. The result is also quite powerful in that it allows us to combine entangled-ness and mixed-ness of quantum composite systems, which is far easier both computationally and heuristically than involving the Von Neumann Entropies. It also allows for quick cheats on tests if one is asked to prove a state is entangled or not.

For further reading see Schmidt Decomposition and Singular value decomposition.