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?
  • $\begingroup$ Where did you see $|\psi'\rangle$ called a "classical statistical mixture"? There is no mixed state/mixture anywhere in your post. $\endgroup$ Commented Jun 6, 2019 at 13:02
  • $\begingroup$ 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. $\endgroup$ Commented Jun 6, 2019 at 13:06

1 Answer 1


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 $$|\psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) $$

To answer point by point:

  • It is a pure state because any state that can be expressed as a vector is pure.
  • On the contrary, it is pure too. A non pure state is not one that is not in a superposition of basis vectors, because as you say all vectors are in some basis, but rather one that is in a (convex) superposition of density matrices.
  • 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
  • $\begingroup$ But how can a pair of two-state systems (e.g. two 1/2-spins) be in a mixed state? Can I somehow prepare a mixed state in a laboratory? $\endgroup$
    – dnnagy
    Commented Jun 7, 2019 at 14:11
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Jun 7, 2019 at 15:12
  • 1
    $\begingroup$ I realized that maybe you meant two particles that together are in a mixed state, same principle, take three entangled particles and lose one. $\endgroup$ Commented Jun 7, 2019 at 15:28

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