# How to discern that a subsystem of entangled system is in a mixed state

We have a an entangled system, composed of two subsystems, $A$ and $B$, in a pure state $|\Psi\rangle$, in a basis $|ab\rangle$, described by wave function $\Psi(a,b)$. Density matrices are written as $\rho$. There is also some observable $O$ that does nothing to state $B$ and operates only on $A$. By $X_{a'b',ab}$ I mean the matrix $\langle a'b' | X | ab \rangle$. Since

$${\langle}O{\rangle}=Tr({\rho}O) \tag{1}$$

then

$$\langle O \rangle = \sum_{a,a'} O_{a',a} {\rho}_{a,a'} \, . \tag{2}$$

So the expectation value of our observable is

$$\langle O \rangle = \sum_{ab,a'b'} \Psi^*(a',b') O_{a'b',ab}{\Psi}(a,b) \, . \tag{3}$$

However, since $O$ acts only on $A$, we can write this as

$$\langle O \rangle = \sum\limits_{a,b,a'} \Psi^*(a',b) O_{a',a} \Psi(a,b) \, . \tag{4}$$

First we can simplify the last equation as a one similar to Eq. $(2)$, with the equivalent of $\rho_{a,a'}$ being $\sum_b \Psi^*(a',b) \Psi(a,b)$. This somehow implies mixed state, but I don't understand why.

• Have you read en.wikipedia.org/wiki/Partial_trace? Sep 16, 2015 at 20:19
• No, but this shouldn't be solved using this thing. My book said nothing about partial traces yet, so it should be possible to be done without them. Sep 16, 2015 at 20:21
• What do you mean? The state of system A alone -- which describes any measurement on A alone -- is obtained by tracing out system B (i.e. performing a partial trace). Since the initial state on AB is entangled, you end up with a mixed state on A alone. Not coincidentally, the formula for the partial trace is exactly given by the sum over B. Sep 16, 2015 at 20:22
• Since you updated your comment: You just did solve it without partial traces, didn't you? So there is nothing wrong! (And you certainly never specified that an answer had to fit your book!) Sep 16, 2015 at 20:23
• You should read up on partial traces and reduced density matrices. Sep 16, 2015 at 20:54

Let $\rho_{aa'} = \langle a|\rho_A|a'\rangle = \sum_{b}{\Psi(a,b) \Psi^*(a',b)}$ as you noticed already.
We can tell if the density matrix $\rho_A$ corresponds to a pure state or not by looking at $\rho_A^2$:
If $\rho_A$ is a pure state, then $\rho_A^2 = \rho_A$ and $Tr\rho_A^2 = Tr\rho_A = 1$.
If $\rho_A$ is a mixed state, then $\rho_A^2 \neq \rho_A$ and $Tr\rho_A^2 < Tr\rho_A = 1$.
In your case, $$Tr \rho_A^2 = \sum_{aa'}{\rho_{aa'}\rho_{a'a}} = \sum_{aa'}{|\rho_{aa'}|^2} = \\ = \sum_{aa'}{\left|\sum_{b}{\Psi(a,b) \Psi^*(a',b)}\right|^2} \le \sum_{aa'}{\left(\sum_b{\left|\Psi(a,b) \right|^2}\right)\left(\sum_{b'}{\left|\Psi(a',b') \right|^2}\right)} = \\ =\sum_{aa'}{\rho_{aa}\rho_{a'a'}} = \left( Tr\rho_A \right)^2 = 1$$ where the inequality is an application of the Cauchy–Schwarz inequality (Wikipedia).
In other words, for the reduced state you have $Tr \rho_A^2 \le 1$, which implies that $\rho_A$ is generally mixed. To have a pure state $\rho_A$, the inequality must become equality, which the Cauchy-Schwartz inequality tells us it happens only when $\Psi(a,b)/\Psi(a',b) = \text{const.}$, $\forall b$. But this implies that $\Psi(a,b) = \Psi(a)\Psi(b)$ and so $|\Psi\rangle$ must be separable: $|\Psi\rangle = |\Psi_A\rangle\otimes|\Psi_B\rangle$, $\rho_A = |\Psi_A\rangle \langle \Psi_A|$.