# Confusion about mixed states and pure states

Suppose I have a system composed of two subsystems (each is a 2-state system). I understand, that there exist two types of such systems: separable, and entangled. A separable system can be written as $$|\psi\rangle = (a|0\rangle_A+b|1\rangle_A)\otimes(c|0\rangle_B-d|1\rangle_B)$$ Then I measure the subsystem $$A$$ and find out that it is in state $$|0\rangle$$. After the measurement the system becomes

$$|\psi'\rangle = |0\rangle_A\otimes(c|0\rangle_B-d|1\rangle_B)$$ So, after the measurement of subsystem $$A$$, I do not get any information about the state of subsystem $$B$$. An entangled state would be $$|\phi\rangle = \frac{1}{\sqrt 2}(|0\rangle_A\otimes|0\rangle_B +|1\rangle_A\otimes|1\rangle_B)$$ In this case, if I measure the subsystem $$A$$ and get $$|0\rangle_A$$ then I can be 100% sure, that $$B$$ is in state $$|0\rangle_B$$, and the wavefunction of the full system collapses into $$|\phi'\rangle = |0\rangle_A\otimes|0\rangle_B$$. Now my questions are:

• As far as I know, state $$|\psi\rangle$$ is a superposition of the four basis vectors ($$|00\rangle$$, $$|01\rangle$$, $$|10\rangle$$, $$|11\rangle$$), but it is considered a pure state.
• In contrary, $$|\psi'\rangle$$ is said to be a classical statistical mixture, so it is not a pure state. Why? Isn't it also a simple superposition of the basis vectors?
• Why is it called a classical statistical mixture?
• The state $$|\phi'\rangle$$ is also a pure state. Does this mean that entangled subsystems can not be in a mixed state?
• How can I decide if a state is a mixture, or a pure state?
• Where did you see $|\psi'\rangle$ called a "classical statistical mixture"? There is no mixed state/mixture anywhere in your post. Jun 6, 2019 at 13:02
• To add to @NorbertSchuch 's comment, a ket vector such as $|\psi\rangle$ can only represent pure states. In order o represent a statistical mixture (or any other form of mixed state) you need to move to a density matrix foramlism. Jun 6, 2019 at 13:06

All the states you listed are indeed pure states. I think you are confusing the notions of pure and separable states. It is more convenient to work with density matrices.

A state $$\rho$$ is said to be a pure state if $$\rho=|\psi\rangle\langle \psi|$$, or more generally if $$\mathrm{Tr}(\rho^2)=\mathrm{Tr}(\rho)$$, thus all states that you can express as a simple vector $$|\psi\rangle$$ are indeed pure states. A pure state can be entangled, and a non pure state can be non entangled. In contrast a classical mixture is a state of the form

$$\rho=\sum_k \lambda_k |\psi_k\rangle\langle\psi_k|$$

With at least two linearly independent $$|\psi\rangle$$. You can see that in this case we cannot bring $$\rho$$ to the form $$|\phi\rangle\langle\phi|$$ for some $$\phi$$ because the rank of $$\rho$$ is greater than $$1$$, and indeed $$\mathrm{Tr}(\rho^2)=\mathrm{Tr}(\rho)$$ does not hold. In this case you can interpret the state as a classical probability distribution over the pure states $$|\psi_k\rangle\langle\psi_k|$$.

This in general has nothing to do with entanglement. A state on $$AB$$ is called separable if you can write it as

$$\rho=\sum_{k} p_k \rho^A_k\otimes\rho^B_k$$ with $$p_k>0$$ s.t. $$\sum_k p_k=1$$. In particular a pure state is separable if $$\rho^{AB}=\rho^A\otimes\rho^B$$. Notice that this implies if $$\rho^{AB}=|\psi\rangle\langle \psi|$$ that $$|\psi\rangle=|\psi\rangle^A \otimes |\psi\rangle^B$$. A non separable state is called entangled, in particular a pure state can very well be entangled, for example $$\rho^{AB}=|\psi\rangle\langle \psi|$$ with $$|\psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$

• In case it is really a classical mixture, as in the definition I provided, it is because you can interpret it as the system being in a random state $$\rho_k$$ with probability $$p_k$$
• A state $$\rho$$ is pure if and only if $$\mathrm{Tr}(\rho^2)=\mathrm{Tr}(\rho)$$, otherwise it's mixed
• For example, if you have an entangled pair of photons in the pure state$|\psi\rangle\langle \psi|$ with $|\psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ and lose one of the two photons, the remaining one will be in the maximally mixed state $\frac{1}{2}(|0\rangle\langle 0| + |1\rangle\langle 1|)$. This is very common when dealing with photons and it's one of the many sources of decoherence. Jun 7, 2019 at 15:12