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

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


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

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.

  • $\begingroup$ Have you read en.wikipedia.org/wiki/Partial_trace? $\endgroup$ Sep 16, 2015 at 20:19
  • $\begingroup$ 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. $\endgroup$
    – Qwedfsf
    Sep 16, 2015 at 20:21
  • $\begingroup$ 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. $\endgroup$ Sep 16, 2015 at 20:22
  • $\begingroup$ 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!) $\endgroup$ Sep 16, 2015 at 20:23
  • 1
    $\begingroup$ You should read up on partial traces and reduced density matrices. $\endgroup$ Sep 16, 2015 at 20:54

1 Answer 1


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: $|\Psi\rangle = |\Psi_A\rangle\otimes|\Psi_B\rangle$, $\rho_A = |\Psi_A\rangle \langle \Psi_A|$.


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