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I am currently trying to establish a clear picture of pure/mixed/entangled/separable/superposed states. In the following I will always assume a basis of $|1\rangle$ and $|0\rangle$ for my quantum systems. This is what I have so far:

  • superposed: A superposition of two states which a system $A$ can occupy, so $\frac{1}{\sqrt{2}}(|1\rangle_A+|1\rangle_A)$
  • seperable: $|1\rangle_A|0\rangle_B$ A state is called separable, if its an element of the (tensor)product basis of system $A$ and $B$ (for all possible choices of bases)
  • entangled: $\frac{1}{\sqrt{2}}(|0\rangle_A|1\rangle_B+|1\rangle_A|0\rangle_B)$ is not a state within the product basis (again for all possible bases).
  • mixed state: Is a statistical mixture, so for instance $|1\rangle$ with probability $1/2$ and $|0\rangle$ with probability $1/2$
  • pure state: Not a mixed state, no statistical mixture

I hope that the above examples and classifications are correct. If not it would be great if you could correct me. Or add further cases, if this list is incomplete.

On wikipedia I read about quantum entanglement

Another way to say this is that while the von Neumann entropy of the whole state is zero (as it is for any pure state), the entropy of the subsystems is greater than zero.

which is perfectly fine. However I also read on wikipedia a criterion for mixed states:

Another, equivalent, criterion is that the von Neumann entropy is 0 for a pure state, and strictly positive for a mixed state.

So does this imply that if I look at the subsystems of an entangled state, that they are in a mixed state? Sounds strange... What would be the statistical mixture in that case?

Moreover I also wanted to ask, whether you had further illustrative examples for the different states I tried to describe above. Or any dangerous cases, where one might think a state of one kind to be the other?

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i think trace remains same even after a unitary transformation.. so if its true in schmidt basis then its true in other basis also. –  user31383 Oct 20 '13 at 7:49

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up vote 4 down vote accepted

Yes, subsystems of an entangled state – if this subsystem is entangled with the rest – is always in a mixed state or "statistical mixture" which is used as a synonym in your discussion (or elsewhere).

If we're only interested in predictions for a subsystem $A$ in a system composed of $A,B$, then $A$ is described by a density matrix $\rho_A$ calculable by "tracing over" the indices of the Hilbert space for $B$: $$\rho_A = {\rm Tr}_{i_b} \rho_{AB}$$ Note that if the whole system $AB$ is in a pure state, $$\rho_{AB}= |\psi_{AB}\rangle\langle \psi_{AB}| $$ If $\psi_{AB}$ is an entangled i.e. not separable state, i.e. if it cannot be written as $|\psi_A\rangle\otimes |\psi_B\rangle$ for any states $|\psi_A\rangle$ and $|\psi_B\rangle$, then the tracing over has the effect of picking all the terms in $|\psi_{AB}\rangle$, forgetting about their dependence on the $B$ degrees of freedom, and writing their probabilities on the diagonal of $\rho_{AB}$. That's why the von Neumann entropy will be nonzero – the density matrix will be a diagonal one in a basis and there will be at least two entries that are neither $0$ nor $1$.

Take a system of two qubits. We have qubit $A$ and qubit $B$. There are 4 natural basis vectors for the two qubits, $|00\rangle$, $|01\rangle$, $|10\rangle$, and $|11\rangle$ where the first digit refers to the value of $A$ and the second digit to $B$. A general pure state is a superposition of these four states with four coefficients $\alpha_{AB}$ where $A,B$ are $0,1$, matched to the corresponding values.

If $\alpha_{AB}$ may be written as $\beta_A\gamma_B$ i.e. factorized in this way, the pure state is separable. $|01\rangle$ is separable, for example. If it is not, then it is entangled. For example, $|00\rangle+|11\rangle$ is not separable so it is entangled.

The mixed state is a more general state than a pure state. In this case, it is given by a $4\times 4$ Hermitian matrix $\rho$. The matrix entries are $\rho_{AB,A'B'}$ where the unprimed and primed indices refer to the values of qubits $AB$ in the bra and ket vectors, respectively. If these matrix entries may be factorized to $$\rho_{AB,A'B'} = \alpha^*_{AB}\alpha_{A'B'}$$ for some coefficients $\alpha_{A'B'}$ and their complex conjugates that specify a pure state $|\psi_{AB}\rangle$, then the density matrix $\rho$ is equivalent to the pure state $|\psi_{AB}\rangle$ and we say that the system is in a pure state. In the more general case, $\rho$ can't be written as this factorized product but only as a sum of similar products. If you need at least two terms like that to write $\rho$, then the state is mixed and the von Neumann entropy is therefore nonzero.

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thanks a lot for the clarifications:-) –  ftiaronsem Jul 10 '13 at 13:30
    
If ψAB is an entangled i.e. not separable state, i.e. if it cannot be written as |ψA⟩⊗|ψB⟩ for any states |ψA⟩ and |ψB⟩, then the tracing over has the effect of picking all the terms in |ψAB⟩, forgetting about their dependence on the B degrees of freedom, and writing their probabilities on the diagonal of ρA.- Isn't that only true in the (Schmidt decomposition for ψAB) basis? –  Max M Sep 27 '13 at 23:28

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