So for the combined state of a pair of two-level atoms, A and B, with a density matrix

$$\rho =\frac{1}{2}\lvert g_A, g_B \rangle\langle g_A, g_B \rvert + \frac{1}{2}\lvert g_A, e_B \rangle\langle g_A, e_B \rvert \tag{1}$$

Where g and e denotes the ground and excited state, respectively. I have calculated the reduced matrix operator for system A by taking the partial trace with respect to B

$$ \rho_A = Tr_B(\rho)=\lvert g_A \rangle \langle g_A \rvert \tag{2}$$

Assuming I have calculated this correctly, I would like to ask the following:

(1) Is the purity of system A simply equal to 1 as it only contains one pair of a kat-bra?

(2) Regarding the state of the two atoms, how can I establish if the combined state is entangled or a product state from using the reduced density matrices of each system?

Thank you.

EDIT: Clarified my second question. Also, here is the reduced matrix operator I calculated for system B in case it is relevant to my second question

$$ \rho_B = Tr_A(\rho)=\frac{1}{2}(\lvert g_B \rangle \langle g_B \rvert +\lvert e_B \rangle \langle e_B \rvert) \tag{3}$$

  • 2
    $\begingroup$ The answer to question 1 is yes, by definition of pure state. But regarding question 2, I think we can write $ \rho = \left| g_A \right> \left< g_A \right| \otimes \frac{1}{2} \left( \left| g_B \right> \left< g_B \right| + \left| e_B \right> \left< e_B \right| \right) $. So it looks like equation 1 can be separated. Am I wrong? $\endgroup$
    – DoeJohn
    Commented Nov 7, 2019 at 18:47
  • $\begingroup$ Thank you for your comment. Hmmm that isn't exactly what I mean by separated but if you know a different definition/technique for determining the nature of the combined state, that would be very helpful! $\endgroup$
    – Coconut
    Commented Nov 7, 2019 at 23:19
  • $\begingroup$ What do you mean by separated? $\endgroup$
    – DoeJohn
    Commented Nov 8, 2019 at 8:41
  • $\begingroup$ This wikipedia article describes it better than I can understand it en.wikipedia.org/wiki/Separable_state $\endgroup$
    – Coconut
    Commented Nov 8, 2019 at 12:34
  • $\begingroup$ Edited my second question to provide a bit more clarity about what I'm trying to determine $\endgroup$
    – Coconut
    Commented Nov 8, 2019 at 13:12

1 Answer 1


Question (1)

Yes, $A$ is in a pure state. By definition, a state $\rho$ is pure if and only if it is a projector, that is $\rho^2 = \rho$. This is equivalent to $\mathrm{Tr}(\rho^2) = 1$. From this, it is clear that $\rho_A$ is a pure state.

Question (2)

I suppose you want an explanation of the definitions behind the notion of entanglement. If $\rho$ is the state of a composite system made of two subsystems $A$ and $B$, we distinguish three types of correlation that may exist between the two subsystem, which correspond to three types of states $\rho$: factorized states (1), separable states (2) and, finally, entangled states (3). In the following explanation we will use $O_A$ and $O_B$ to designate generic observables of system $A$ and $B$ respectively.

  1. Factorized states are also called product states or simply separable states. We say that $\rho$ is factorized if it is the tensor product of two states: $$ \rho = \rho_A \otimes \rho_B . $$ In this case the expected value of the product observable $O_AO_B$ is simply $$ \left< O_A O_B \right>_\rho = \left< O_A \right>_{\rho_A} \left< O_B \right>_{\rho_B} . $$ That is, the two subsystems are statistically independent. You can easily check if a state $\rho$ is factorized: just compute $\rho_A = \mathrm{Tr}_B(\rho)$ and $\rho_B = \mathrm{Tr}_A(\rho)$, then calculate $\rho' = \rho_A \otimes \rho_B$. If $\rho' = \rho$, then $\rho$ is factorized. Note that the $\rho$ of your example is factorized. Therefore, as we said, the subsystems are statistically independent. In this case there is no correlation between $A$ and $B$, so there is obviously no entanglement.

  2. Separable states are also called classically correlated states. A state $\rho$ is said to be separable if it can be written as a convex combination of factorized states, that is $$ \rho = \sum_i p_i \rho_A^{(i)} \otimes \rho_B^{(i)} , $$ with $p_i>0$ and $\sum_i p_i = 1$. In this case the expected value of $O_AO_B$ is $$ \left< O_A O_B \right>_\rho = \sum_i p_i \left< O_A \right>_{\rho_A^{(i)}} \left< O_B \right>_{\rho_B^{(i)}} . $$ This time the two subsystems are correlated, but we can say that this correlation is "classical" in some sense: the statistics of the system can be interpreted as if we knew that

    • with probability $p_1$, system $A$ is in state $\rho_A^{(1)}$ and at the same time system $B$ is in $\rho_B^{(1)}$;
    • with probability $p_2$, system $A$ is in state $\rho_A^{(2)}$ and at the same time system $B$ is in $\rho_B^{(2)}$;
    • with probability $p_3$, system $A$ is in state $\rho_A^{(3)}$ and at the same time system $B$ is in $\rho_B^{(3)}$;
    • ... (and so on).

    So there is nothing inherently quantum mechanical in the correlations between $A$ and $B$. They can be understood with classical reasoning. Of course, we are only talking about the correlations existing between $A$ and $B$ as long as the state is $\rho$. The inner workings of $A$ and $B$, as well as the future evolution of the composite system are still quantum mechanical in nature.

  3. All the states that do not correspond to the last two definitions are said to be entangled. In this case, the correlations between $A$ and $B$ cannot be interpreted with "classical reasoning".

  • $\begingroup$ Thank you very much for your comprehensive answer! You cleared up my confusion between the nature of the different composite states, and I have an understanding now of how to determine this :) $\endgroup$
    – Coconut
    Commented Nov 8, 2019 at 17:38

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