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A From POVM to PVM

  1. Jul 13, 2016 #1

    naima

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    Could you comment this optical device? (I can provide a link)
    A polarized photon a H> + b V> begins to hit a beam splitter. Then In the device components can rotate its phases, mix paths
    and so on. And all this depends on parameters (possibly functions of time). There are two output channels.
    in the first it is in the state with the probability output by the other case it is in the state [tex] a cos \alpha H> +b sin \beta V>[/tex] with the probability [tex] a^2 cos^2 \alpha +b^2 sin^2 \beta [/tex]
    In the second path it is in the state [tex] a sin \alpha H> +,b cos \beta V>[/tex] with the probability [tex] a^2 sin^2 \alpha +b^2 cos^2 \beta [/tex] The POVM is easily seen with two effects summing to Id.
    If alpha and beta are equal to pi / 4 the initial state is not modified. both effects are equal to Id/2 and the probability for each arm is 1/2
    in the opposite case if alpha and beta = pi/2 the measuresure is like in a projective Stern Gerlach.
    As these angles may depend on time, we have continuous way to evolve from a set of equal effects to a set of orthonormal effects which define a PVM with a privileged basis (H> <H and V> <V)

    Is this peculiar to this device or is this a general procedd?
     
    Last edited: Jul 13, 2016
  2. jcsd
  3. Jul 15, 2016 #2

    naima

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    Let us take a density matrix [tex] \rho (t) = \begin {pmatrix}
    aa^*&ab^* exp(-t)\\
    a^*b exp(-t)&bb^*\\
    \end {pmatrix} [/tex]
    It is a relaxing density matrix by decoherence.
    If V is the vector [tex] \begin {pmatrix} a \\
    b \\
    \end {pmatrix} [/tex]
    how can we find the Kraus operators K1 (t) and K2 (t) with
    [tex] \rho (t) = K1 (t) V V ^ \dagger K1 ^ \dagger (t) + K2 (t) V V ^ \dagger K2 ^ \dagger (t) [/tex]
    That gives the same intermediate states for rho (t)?
    In short how to calculate the "effects" from decoherence?
     
    Last edited: Jul 15, 2016
  4. Jul 16, 2016 #3
    For #2, the decoherence doesn't affect the diagonal elements of ## \rho(t) ## while sticking a factor of ## e^{-t} ## to the off-diagonal elements. So a guess is $$ K_1 = \sqrt{k} \hat{1}, \\ K_2 = \sqrt{1-k} Z. $$ Let ##U## be the Pauli ##X## or ##Y## and note that ## ZUZ = -U ##, we want $$ k \hat{1} U \hat{1} - (1-k) ZUZ = (2k-1) U = e^{-t} U. $$ Solving for ##k## gives $$ k = \frac{1 + e^{-t}}{2}. $$ Typically, you probably derive the Kraus operators from an environment model that decoheres the system. A Z rotation with a random angle will also do the job. In that case, we could have an infinite number of Kraus operators, one for each angle of rotation. (Kraus operators are not unique, so that's fine.)
     
  5. Jul 16, 2016 #4

    naima

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    Thank you
    So according to you what are here the two effects [tex] K_1(t) K_1^\dagger (t) and K_2(t) K_2^\dagger(t)[/tex] which sum to Id?
     
  6. Jul 16, 2016 #5
    $$ K_1 K_1^{\dagger} + K_2 K_2^{\dagger} = \frac{1+e^{-t}}{2} \hat{1} + \frac{1-e^{-t}}{2} \hat{1} = \hat{1}. $$ P.S. Doesn't look like a meaningful POVM.
     
  7. Jul 16, 2016 #6

    naima

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    I have not the answer but
    The problem is that your K1(t) and K2(t) do not obey this equation.
     
  8. Jul 16, 2016 #7
    How? Write the initial density matrix (your ## VV^{\dagger} ##) in the Pauli basis. $$ \rho = \frac{1}{2} \left[ \hat{1} + (ab^*+a^*b)X + i(ab^* - a^*b) Y + \left( |a|^2 - |b|^2 \right) Z \right]. $$ The map ## \rho \mapsto \rho(t) = K_1 (t) \rho K_1^{\dagger} (t) + K_2 (t) \rho K_2^{\dagger} (t) ## commutes with the identity and the Z term, so it does nothing to them. For the X term, \begin{align} X & \mapsto \sqrt{\frac{1+e^{-t}}{2}} X \sqrt{\frac{1+e^{-t}}{2}} + \sqrt{\frac{1-e^{-t}}{2}} ZXZ \sqrt{\frac{1-e^{-t}}{2}} \\ & = \left( \frac{1+e^{-t}}{2} - \frac{1-e^{-t}}{2} \right)X = e^{-t} X. \end{align} The Y term works similarly. So $$ \rho(t) = \frac{1}{2} \left[ \hat{1} + (ab^*+a^*b)e^{-t} X + i(ab^* - a^*b) e^{-t} Y + \left( |a|^2 - |b|^2 \right) Z \right], $$ which is your relaxing density matrix.
     
  9. Jul 16, 2016 #8

    naima

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    I am sorry for my first reaction. Your answer is very good. I can explain why i did not believe it.
    As time passes you get 2 effects from K1 and K2 which tend to Id/2 with Id as their sum.
    I was hoping that one of them would give |H><H| and the other |V><V| (with the same sum)!!!
    And that the effects would be equal when t = 0.
    As there are many other possibilities? Can such a solution exist for the kraus operators?
    The advantage would be that we would pass from POVMs to final othonormal PVMs
    (You also wrote that it did not give meaningful effects)
     
  10. Jul 16, 2016 #9
    No worries!

    Now I think I understand the effects that I got better. The decoherence channel that you gave is unital, meaning that it has the maximally mixed state as a fixed point. For a qubit (and only for a qubit), every unital map can be written as an evolution with random unitaries, which was what I did. This means that ##K_j K_J^{\dagger} \propto \hat{1} ## for every Kraus operator, which makes sense because unitary evolutions don't give us any information so the effects are useless.

    From this, we can make up a class of quantum operations that does what you described in the OP. Couple the system to an ancilla with an adjustable coupling strength, then measure the ancilla in an appropriate basis. When the coupling is zero, measuring the ancillia doesn't give us any information about the system so the POVM for the system will just be a bunch of identity operators. When the coupling is turned on, the POVM will change gradually until it hits the point where the system and the ancilla is maximally entangled. Then a strong measurement of the ancilla will also be a strong measurement on the system. (Each effect is a projection operator and all the effects always add up to the identity.) Since every quantum operation (CP map) is equivalent to a measurement model on an ancilla (Nielsen & Chuang), I expect that you can extend the description of any measurement to have this transition that you described.
     
  11. Jul 17, 2016 #10

    naima

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    I think i have it:
    take [tex] K_1(t) = \begin {pmatrix}
    cos \alpha(t)&0\\
    0 & sin \alpha(t) \\
    \end {pmatrix} [/tex]
    and
    [tex] K_2(t) = \begin {pmatrix}
    sin \alpha(t)&0\\
    0 & cos \alpha(t) \\
    \end {pmatrix} [/tex]
    When you compute K1(t) rho(0) K1(t) + K2(t) rho(0) K2(t)
    the diagonal elements of the initial density matrix are not changed and the off diagonal elements are multiplied by [tex]sin 2 \alpha (t) = exp(-t)[/tex]
    so the POVMs evolves with time from equal effects (of a POVM) to orthonormal projectors.
    I think that it is a general process in decoherence.
    the device for all POVMs is here
     
    Last edited: Jul 17, 2016
  12. Jul 17, 2016 #11

    naima

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    Truecrimson showed that with a given evolution of the density matrix (a given interaction hamiltonian) we can associate different sets of events.
    Is it possible that a unique time dependent POVM can be deduced from the Hi = g A P (hamiltonian). for t = 0 they wold be parallel. As each event evolve to a peculiar projector to one of the preferref basis, we should get sub hamiltonians for each evolution.
     
  13. Jul 18, 2016 #12

    naima

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    As Truecrimson and i found two solutions i wondered if one of them is more natural and if measurement devices could help.
    His solution is more natural than mine at t = 0.
    At this moment there still no decoherence. the state is pure. He also start with 2 effects but the first is Id (probability = 1) and the other is null.
    I start with 2 equal effects = Id/2 and we can ask why.
    I win a point at the end : he gets equal effects (Id/2) where i find projectors on the preferred basis.

    In the link i gave we see that we can build devices which can mimick any POVM result. Is there still one hope?
    I think so.
    Some devices are more simple than others.
    Let us take the example of a Stern Gerlach in which we can change the strengh of the magnetic field. The POVM device is made of 2 half plane which can click.
    If t = 0 the electron has a 1/2 probability to hit each half plane detector. if the SG is strong we get the probability ##|a|^2## and ##|b|^2## like in my model. i hope that we could find higher dimensional examples.
     
  14. Jul 18, 2016 #13
    Edit: Never mind. Bad question.

    Different sets of Kraus operators for the same quantum operation are related by a unitary matrix ##U## (could be rectangular). An interesting thing is that in this example, one set of Kraus operators are unitary (reversible), while the other are projection operators (the most irreversible). It makes me curious about what the physical meaning of ##U## is.
     
    Last edited: Jul 18, 2016
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