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The Schmidt Decomposition: Looking for some intuition

  1. Feb 16, 2013 #1
    Hi,

    I finished reading about the Schmidt decomposition from Preskill's notes today. I understand and follow his derivation but it still seems completely non intuitive to me. We have
    [tex]
    \mid\psi\rangle_{AB}=\sum_{i,u}a_{iu}\mid i\rangle_{A}\mid u\rangle_{B}=\sum_{i}\mid i\rangle_{A}\mid\tilde{i}\rangle_{B}
    [/tex]
    where we have
    [tex]
    \mid\tilde{i}\rangle_{B}=\sum_{i,u}a_{iu}\mid u\rangle_{B}
    [/tex]

    At this stage, the system B is represented by [itex]\mid\tilde{i}\rangle_{B}[/itex] which are not orthonormal. Then, one chooses a basis for the first subsystem such that the partial trace [itex]\rho_{a}=Tr_{B}\rho_{AB}[/itex] is diagonal. Somehow, this automatically makes the vectors of system B orthonormal. I can see that it's true but I have no clue why. If anyone has some intuition for what is going on, I'd be very grateful.

    Thank you.
     
  2. jcsd
  3. Jun 13, 2013 #2

    naima

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    Doing so you can associate to each vector in HA a vector in HB when you have entanglement. (iA -> iB)
    If one takes Va as an eigenvector for an operator O: O Va = a Va can we say that the associate vector Vb verifies:
    O Vb = b Vb? (a measurement result on A would always be associated to a definite result on B with the same measurement)?
    Is it what it means physically?
     
  4. Jun 15, 2013 #3

    naima

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    I think the correct sentence is:
    if the Schmidt decomposition has equal [itex]\lambda_i[/itex] the results of measurements on each subsystem are perfectly correlated:
    Measuring an observable on A makes sure the measurement result of the same observable on subsystem B. the
    choice of the measured observable can be done after the systems have finished to interact.

    Could anyone help me to prove it?
    thanks
     
  5. Jun 15, 2013 #4

    kith

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    That's an interesting way to proof the Schmidt decomposition. The sources I know use the singular value decomposition which is not very intuitive. I don't know an answer to your question right away. I will have a look at this proof if I find the time.

    But you don't lack intuition regarding what the Schmidt decomposition states, do you? It seperates a state as much as possible and therefore quantifies entanglement.

    For maximum entanglement/correlation, the number of the coefficients also has to be the same as the dimension of the smaller Hilbert space. Note that in general, you can't measure the same observable on both systems because the Hilbert spaces can be different.

    /edit: I've just seen, that this thread is rather old.
     
    Last edited: Jun 15, 2013
  6. Jun 15, 2013 #5

    kith

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    /edit: please delete, I posted in the wrong thread
     
  7. Jun 15, 2013 #6

    naima

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    It may be true in mathematics but not in physics.
    you can measure the abscissa of a point on a line and that of a point in a plane containing the line.
     
  8. Jun 15, 2013 #7

    kith

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    Not all observables exist for all kinds of systems. There is no position operator for the electromagnetic field, for example.

    I have thought a bit more about this and I think it doesn't hold in general. For example, we can entangle different spin components of two particles. |+z>|+x> + |-z>|-x> is a counter example to your statement.
     
  9. Jun 15, 2013 #8

    naima

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    What I wrote comes from french 2012 co-Nobel prize Serge Haroche
    It is in french. I google translated page 3:

    If an entangled state has equal Schmidt decomposition (λi) it can be expressed in
    different orthonormal bases associated with non-compatible observables states:
    (he gives two examples) then he writes:
    The results of measurements on each subsystem are random, but perfectly correlated:
    measuring system A makes sure the measurement result of the same observable on B. the
    choice of the measured observable can be done after the systems have come to interact.

    I do not understand if what he says after the examples is only valid when the λi are equal.
     
  10. Jun 16, 2013 #9

    kith

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    If the λi are not equal, you get perfect correlations only in a single set of bases (it consists of the eigenstates of the reduced density matrices). So it is a necessary condition for his statement that the λi are equal.

    But as I have mentioned above, I don't think it is a sufficient condition. His statement is true for his examples but not for my counter example.
     
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