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Density matrix for bell states

  1. Mar 1, 2013 #1
    I have three states (I believe bell states) and want to find the density matrix, am I right in thinking:
    1) [itex] \frac{|00> + |11>}{\sqrt{2}} \rightarrow \rho = \left( \begin{array}{cc}
    \frac{1}{\sqrt{2}} & 0 \\
    0 & \frac{1}{\sqrt{2}} \\
    \end{array} \right) [/itex] (because it is pure)

    2) [itex] \frac{|00> - |11>}{\sqrt{2}} \rightarrow \rho = \left( \begin{array}{cc}
    \frac{1}{\sqrt{2}} & 0 \\
    0 & \frac{1}{\sqrt{2}} \\
    \end{array} \right) [/itex] (because it is pure)

    3) [itex] \frac{|01> + |01>}{\sqrt{2}} \rightarrow \rho = \left( \begin{array}{cc}
    \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
    \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
    \end{array} \right) [/itex] (because it is mixed)
  2. jcsd
  3. Mar 2, 2013 #2
    Shouldn't the density matrix be 4x4? The first one should be something like

    1 & 0 & 0 & 1\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    1 & 0 & 0 & 1

    The others then analogously, but a 100% I am not certain, because looking at the Schmidt-coefficients, they are not all the same (2 are actually 0) as required for a maximally entangled state. Anyone can solve this issue?
  4. Mar 2, 2013 #3


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    You get the density matrix of the composite system by writing out ρ= |ψ><ψ| for your states. This is a 4x4 matrix.

    If you want to get the reduced density matrix of one subsystem,you have to trace over the other system. You then get a 2x2 matrix.

    PS: Why do you think that the third state is not pure? Every ket corresponds to a pure state. You seem to have some fundamental misunderstanding. You need to give more details about your calculations.
    Last edited: Mar 2, 2013
  5. Mar 2, 2013 #4


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    You need only as many coefficents as the dimension of the Hilbert space of the smaller subsystem. Here, we have two 2-dimensional Hilbert spaces, so we need two coefficients.
  6. Mar 2, 2013 #5
    Please could you describe the difference between a pure and a mixed state?
    I am also unclear as to how to calculate the density matrix, could you give me a simple example, I can't see why [itex] \sum_i |\psi_i><\psi_i|[/itex] would give a 4x4 matrix.

    Would you have to decompose [itex] |00>[/itex] into |0>|0> and then into,
    (A|0> + B|1>)Tensor-product(C|0>+D|1>)? I am a bit confused
  7. Mar 3, 2013 #6


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    A pure state can be a linear combination of other pure states. This is the quantum mechanical superposition.

    For a pure state |ψ> we simply chose this state |ψ> to construct the density matrix

    ρ = |ψ><ψ|

    As you can see you have ρ2 = ρ, so ρ is a projector (here we do not care whether there are other states, e.g. u and v, which can be combined via a linear superposition to get ψ).

    A mixed state is always described via a density matrix which is not a projector.

    For a mixed state, in a basis where the density matrix is diagonal, you have two or more states with non-vanishing probability. So for two orthogonal states contributing to a density matrix you have

    ρ = p|u><u| + (1-p)|v><v|

    Note that for a diagonal density matrix the trace of the probailities must be = 1.

    Testing whether ρ is a projector you find

    ρ2 = p2|u><u| + (1-p)2|v><v| ≠ ρ

    The probabilities p and (1-p) in ρ are classical probabilities, not related to the quantum mechanical superpositions. So if you would construct ψ from u and v using quantum mechanical superpositions you get the pure state as described above.
  8. Mar 3, 2013 #7


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    Have you calculated [itex]\rho = |\psi><\psi| = \frac{1}{2}(|00> + |11>)(<00|+<11|)[/itex]? (First example from you)

    ρ has 16 entries which are the coefficients in front of the 16 possible combinations |ij><kl|. Most of them are zero, but 4 are non-zero. If you arrange them in matrix form, you are done.

    No decomposing of |ij> is required, because we are now dealing with the combined state space of two quibits. There are states which can't be decomposed and these are the entangled states.
    Last edited: Mar 3, 2013
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