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A Reduced Density Operator and Entanglement

  1. Apr 4, 2017 #1
    I'm having a little bit of trouble getting my head around the idea of the reduced density operator being used to tell us about the entanglement of a state.

    I understand that if you take the reduced density operator of any of the Bell states, you get a reduced density operator proportional to the identity, and this is defined as being a maximally entangled state.

    But why is this a proof that the state is entangled?

    I have written problem that asks to solve the reduced density operator for a particular state. I solve for the reduced density operator and I get a 2x2 matrix with elements in each slot. It satisfies the conditions to be a density operator, but what does it tell me about the entanglement of the state?

    Any clarification would be much appreciated!
  2. jcsd
  3. Apr 4, 2017 #2


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    You could calculate the purity. If the reduced state is less pure than the full state, you have entanglement.
  4. Apr 4, 2017 #3


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    Let's stick to the very simplest type of wave function, which is a 2-component spinor. If we have two spin-1/2 particles, then the most general composite state is:

    [itex]|\Psi\rangle = C_{uu} |u\rangle |u\rangle + C_{ud} |u\rangle |d\rangle + C_{du} |d\rangle |u\rangle + C_{dd} |d\rangle |d\rangle[/itex]

    where [itex]|u\rangle |u\rangle[/itex] means both particles are spin-up in the z-direction, [itex]|u\rangle |d\rangle[/itex] means the first is spin-up, the second is spin-down, etc.

    The composite state is "entangled" if it cannot be written as a product [itex]|\psi\rangle |\phi\rangle[/itex]. If I did the calculation correctly, the state above is "entangled" unless [itex]C_{uu} C_{dd} = C_{ud} C_{du}[/itex]. For example, the following is a maximally entangled state:
    [itex]\frac{1}{\sqrt{2}} (|u\rangle |d\rangle - |d\rangle |u\rangle)[/itex]

    A density matrix [itex]\rho[/itex] is "pure" if it can be written in the form [itex]|\psi\rangle \langle \psi\rangle[/itex] for some state [itex]|\psi\rangle[/itex]. Otherwise, it is mixed.

    So the connection between mixed states and entanglement is this:
    • Start with an entangled pure state [itex]|\Psi\rangle[/itex]
    • Form the corresponding density matrix.
    • "Trace out" the degrees of freedom of one of the particles.
    • Then what is left will be a mixed reduced density matrix.
    So tracing out turns a pure state into a mixed state, if the original pure state is entangled.
  5. Apr 4, 2017 #4
    Ok, great! Thanks for the help kith and stevendaryl. I've created a couple of simple states for examples and played around with them and I think I'm getting to grips with it.

    I understand then that the trace of the square of the density operator tells us about how "pure" the state is - ie if it's 1 then it's a pure state, otherwise it's mixed.
    So returning to the Bell states again, I can see that if I calculate the the reduced density operator I get one half of the identity. If I further calculate the purity, I get a value of one half. Since the Bell states are "maximally entangled" does this mean the smallest possible value I can get for the purity (of any density matrix) is one half?

  6. Apr 5, 2017 #5


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    For a two-level system, yes. In general, the lowest purity is [itex]1/d[/itex] where [itex]d[/itex] is the number of degrees of freedom of the system.

    The purity isn't necessarily associated with subsystems. You can also look at a single system. Let's say we have a system with two possible energies. If you have the system in a well-prepared state (one of the energy eigenstates or a superposition of them), the purity is zero. If you have a thermal state at a given temperature, the density matrix is given by the Boltzmann distribution and the purity depends on the temperature. In the high-temeperature limit, the density matrix approaches one half times the identity and the purity is minimal.
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