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Non-entangled 2 qubit states

  1. Aug 1, 2014 #1

    I'm going through some course notes for QM.
    A state for a 2 qubit system is called non-entangled (or separable) if it can be decomposed in a tensorproduct of 2 single qubit states.

    If we write a general state as
    [tex]|\psi> = a_{00}|00>+a_{01}|01>+a_{10}|10>+a_{11}|11>[/tex]

    A theorem states the following:
    A two-qubit state is non-entangled [itex]\Leftrightarrow a_{00}a_{11}=a_{10}a_{01}[/itex]

    Getting the right side from the left side is easy, trivial even. By just writing out the tensor product you automatically get this condition.

    Proving it the other way around has been difficult for me.
    I tried using the fact that the expectation of [itex]A\otimes B[/itex] would give the product of expectations on the one-qubit states. This didn't work as far as I could see.
    Then I tried using contraposition i.e. prove that if the state is in fact entangled, the equation would be negated. Here I got stuck since I'm not sure how to do work with a general entangled state. Or would an example, like a Bell state, be enough to conclude the proof? (prolonged staring really makes one doubt his capabilities)

  2. jcsd
  3. Aug 1, 2014 #2
    Given that [itex]a_{00}a_{11}=a_{10}a_{01}[/itex], one can rearrange (with assumption that denominators are non-zero)
    \frac{a_{00}}{a_{01}}=\frac{a_{10}}{a_{11}}=\alpha \ldots(\text{say})

    |\psi\rangle = a_{00}|00\rangle + a_{01}|01\rangle + a_{10}|10\rangle + a_{11}|11\rangle \\
    = a_{01}|0\rangle \otimes (\alpha |0\rangle + |1\rangle) + a_{11}|1\rangle\otimes(\alpha |0\rangle + |1\rangle)\\= (a_{01}|0\rangle + a_{11}|1\rangle)\otimes (\alpha |0\rangle + |1\rangle).
    Last edited: Aug 1, 2014
  4. Aug 1, 2014 #3
    Thanks, exactly what I needed. How could I have missed that :rolleyes:
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