### Density operator of a subsystem

Let us consider a closed quantum system $$S$$ consisting of two subsystems $$S_1$$ and $$S_2$$. The Hilbert space $$\mathcal{H}$$ of system $$S$$ is a tensor product of Hilbert spaces $$\mathcal{H}_1$$ and $$\mathcal{H}_2$$ of subsystems $$S_1$$ and $$S_2$$, respectively, i.e., $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2 \,.$ Let $$\{ | a \rangle \}$$ and $$\{ | b \rangle \}$$ be complete orthonormal bases in $$\mathcal{H}_1$$ and $$\mathcal{H}_2$$, respectively, so that $\langle a | a' \rangle = \delta_{aa'} \,, \qquad \langle b | b' \rangle = \delta_{bb'} \,,$ and any state $$| \Psi \rangle$$ of $$S$$ can be written as $| \Psi \rangle = \sum_{a,b} \Psi_{ab} | a b \rangle \,, \qquad | a b \rangle \equiv | a \rangle \otimes | b \rangle \,.$ We take $$| \Psi \rangle$$ to be normalized to one: $\langle \Psi | \Psi \rangle = \sum_{a,b} |\Psi_{ab}|^2 = 1 \,.$ For notational simplicity, we treat sets $$\{ | a \rangle \}$$ and $$\{ | b \rangle \}$$ as being discrete; a generalization to the case of continuous (or mixed) basis sets is straightforward. Also, indexes $$a$$ and $$b$$ may in general denote sets of quantum numbers.

Let us also consider an observable $$A$$ of the subsystem $$S_1$$. In the full Hilbert space $$H$$, the observable is describe by the operator $A = A_1 \otimes I_2 \,,$ where $$A_1$$ is the corresponding operator in $$\mathcal{H}_1$$, and $$I_2$$ is the identity operator in $$\mathcal{H}_2$$. The expectation value of the observable is given by $\langle A \rangle \equiv \langle \Psi | A | \Psi \rangle = \sum_{a,a',b,b'} \Psi_{ab}^* \Psi_{a'b'} \langle a b | A| a' b' \rangle \,,$ or $\langle A \rangle = \sum_{a,a',b} \Psi_{ab}^* \Psi_{a'b} \langle a | A_1 | a' \rangle = \sum_{a,a'} \sigma_{a'a} \langle a | A_1 | a' \rangle \,,$ where $\sigma_{a'a} = \sum_b \Psi_{ab}^* \Psi_{a'b} \,.$ This expectation value can be expressed in terms of a (reduced) density operator of the subsystem $$S_1$$, defined as $\rho_1 = \sum_{a,a'} \sigma_{a'a} | a' \rangle \langle a | \,.$ The density operator is normalized to one: $\operatorname{Tr} \rho_1 = \sum_a \sigma_{aa} = \sum_{a,b} |\Psi_{ab}|^2 = 1 \,.$ In terms of $$\rho_1$$, the expectation value of $$A$$ can be written as $\langle A \rangle = \operatorname{Tr} \rho_1 A_1 \,.$ Indeed, $\operatorname{Tr} \rho_1 A_1 = \sum_{a'} \langle a' | \rho_1 A_1 | a' \rangle = \sum_{a,a'} \langle a' | \rho_1 | a \rangle \langle a | A_1 | a' \rangle = \sum_{a,a'} \sigma_{a'a} \langle a | A_1 | a' \rangle = \langle A \rangle \,.$

The density operator has the following properties:

1. All eigenvalues of $$\rho_1$$ are real and lie inside the interval $$[0,1]$$.

Proof: Since $\sigma_{aa'} = \sum_b \Psi_{a'b}^* \Psi_{ab} = \left( \sum_b \Psi_{ab}^* \Psi_{a'b} \right)^* = \sigma_{a'a}^* \,,$ the density operator is Hermitian, $$\rho_1^{\dagger} = \rho_1$$. This implies that all eigenvalues of $$\rho_1$$ are real. The fact that all eigenvalues of $$\rho_1$$ lie between 0 and 1 can be established as follows. For any state $$| \varphi \rangle$$ in $$\mathcal{H}_1$$, we have $\langle \varphi | \rho_1 | \varphi \rangle = \sum_{aa'} \sigma_{a'a} \langle \varphi | a' \rangle \langle a | \varphi \rangle = \sum_{a,a',b} \langle \Psi | a b \rangle \langle a | \varphi \rangle \langle \varphi | a' \rangle \langle a' b | \Psi \rangle \,,$ or $\langle \varphi | \rho_1 | \varphi \rangle = \langle \Psi | P_{\varphi} | \Psi \rangle \,, \qquad P_{\varphi} = | \varphi \rangle \langle \varphi | \otimes I_2 \,.$ Here, $$P_{\varphi}$$ is the projection operator that projects any state of subsystem $$S_1$$ onto $$| \varphi \rangle$$, and its expectation value must lie in the interval $$[0,1]$$.

2. The density matrix satisfies $\operatorname{Tr} \rho_1^2 \le 1 \,.$
Proof: It follows from the previous property that $$\rho_1$$ can be diagonalized: $\rho_1 = \sum_{n} p_n | n \rangle \langle n | \,,$ where $$\{ | n \rangle \}$$ is an orthonormal basis in $$\mathcal{H}_1$$, and eigenvalues $$p_n$$ are such that $$0 \le p_n \le 1$$ and $\operatorname{Tr} \rho_1 = \sum_n p_n = 1 \,.$ Then, $\operatorname{Tr} \rho_1^2 = \sum_n p_n^2 \le 1 \,.$

The subsystem $$S_1$$ is in a mixed state if $$\operatorname{Tr} \rho_1^2 < 1$$, and in a pure state if $$\operatorname{Tr} \rho_1^2 = 1$$. The latter is the case if and only if the state $$| \Psi \rangle$$ of the whole system $$S$$ is a product state, $$| \Psi \rangle = | \psi_1 \rangle \otimes | \psi_2 \rangle$$ with $$| \psi_1 \rangle \in \mathcal{H}_1$$ and $$| \psi_2 \rangle \in \mathcal{H}_2$$.