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Quantum Measurement

  1. Jan 18, 2016 #1
    I am taking a course in quantum information theory. In this theory, the state of a system is given by a density matrix, while a measurement is completely positive, trace preserving map of the form:

    Λ = ∑il i >< i l ⊗ Λi

    , where Λi is completely positive.

    Can anyone tell me how this is equivalent to the usual definition of a measurement on a quantum state, where the state is collapsed into some eigenstate of the operator measured.

    Also it seems crucial to quantum information theory that quantum channels are completely positive and trace preserving. The trace preserving part I get since you want the state of the system to remain and allowed quantum state. But why is the completely positive part needed?
  2. jcsd
  3. Jan 18, 2016 #2


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    The question Intuition on positive-operator valued measures (POVM) on the physics stackexchange might be useful to you.

    The "usual" definition of a measurement that you're thinking about is "Projective measurements". They're defined in terms of an "observable". The observable is a Hermitian matrix, the possible measurement outcomes correspond to the matrix' eigenvalues, and the measurement outcome tells you that the state is now aligned with the eigenvector(s) corresponding to the measured eigenvalue.

    In quantum information land we often use a different abstraction for measurement: POVM measurements. They're a different way of specifying the same thing. Their advantage is mostly that they abstract away some boring details. A POVM is a set of positive operators ##E_m## satisfying ##\sum_m E_m = I##. The probability of outcome ##m## given the density matrix ##\rho## of a state is ##tr(E_m \cdot \rho)##, etc.

    The reason each ##E_m## is a positive operator comes down to the fact that they must sum to identity and that they are secretly, under the hood, the product of a "measurement operator" and its adjoint: ##E_m = M_m^\dagger M_m##. Consider what happens to an orthogonal eigendecomposition of ##M_m## when you multiply by ##M_m^\dagger##. An eigenvalue ##a e^{i b}## will get multiplied by its conjugate ##a e^{-i b}##, causing the phase factors to cancel and leaving a non-negative ##a## as the new eigenvalue.

    The motivating reason for the eigenvalues of the ##E_m##'s to be positive, instead of using some totally different definition and constraints, probability has to do with wanting something that acts like probabilities.

    (Note: You only need the ##M_m## values when working with pure states described by a vector instead of a density matrix.)

    (Note: do not confuse "measurement operator" with "observable".)
    Last edited: Jan 18, 2016
  4. Jan 18, 2016 #3


    Staff: Mentor

    You really need to see an axiomatic development - see post 137:

    To see how its equivalent to usual QM see the first 3 chapters of Ballentine. I wish I could explain the last bit simply - but it requires a long explanation with some tedious math.

  5. Jan 19, 2016 #4

    A. Neumaier

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    It is not equivalent but far more flexible. The latter is a very special case, only realized in measurements of discrete variables and when you obtain perfect information.
    Complete positivity is essential for adequate behavior under composition of subsystems (tensor product formation), and trace preservation is needed to ensure that after a signal passing an instrument, all probabilities still add up to one. When properly modeled, all physical processes satisfy both conditions automatically.

    Relevant Wikipedia articles are
  6. Jan 19, 2016 #5
    Okay so I think I can conclude that I am trying too hard to make sense using ideas from previous courses in quantum mechanics. But then I need to get used to this new way of thinking.

    As stated a measurement is apparently represented by an instrument, which is a CPTP map of the form:

    Λ(X) = Σ li><il ⊗ Λi

    So can anyone tell me exactly what this operator does to the state of a system. As discussed the state of a system is represented by some density matrix ρ, so how should I act with Λ on this?
    My book says without proof that the probability to obtain outcome i is TrΛi and that the post-measurement state is 1/p(i) * Λi. Can anyone clarify why this is? (I think you have already touched upon it, but I am not sure I understand it).
    Also can someone say in a few words what the role of tensor products in quantum information theory is? From my understanding they are used when you divide a larger system into subsystems, but I am not sure about this. For example when a quantum channel is applied to part of larger system its action is Λ⊗id
  7. Jan 19, 2016 #6
    I guess what I want to know is also what is the interpretation of a measurement when we write it as Λ(X) = Σ li><il ⊗ Λi? What does the tensor product do?
  8. Jan 19, 2016 #7


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    Could you include some more context around the equation? It looks like some kind of "measure these qubits then, if the outcome was 0, do measurement A, else if the outcome was 1, do measurement B, else if ..." observable instead of a general definition.
  9. Jan 19, 2016 #8
    Well my book just says that on the attached picture. In general a measurement is represented as an instrument which is a map of the type above. When this acts on the system the probability to obtain outcome i is TrΛi and that the post-measurement state is 1/p(i) * Λi. But I don't know how to see this.

    Attached Files:

  10. Jan 21, 2016 #9


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    It looks like a definition for a specific kind of measurement, defined in terms of a POVM. So you should learn about POVMs in order to understand this definition. I think the general idea is that the ##\left| i \right\rangle##s are the instrument outputs and the ##\Lambda_i##s are the corresponding states of the system given the output. Other than that it looks unfamiliar to me.
  11. Jan 28, 2016 #10


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    Look at the attached picture you gave in the link. a measurement maps a density matrix (it is your X) to two things: the output li><il and Λi(X)
  12. Jan 29, 2016 #11


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    For every state ##\rho## to be measured the initial pointer state of the apparatis is its ground state |0><0|.
    We start from ##|0><0| \otimes \rho##
    then during the premeasurement they interact to become
    ##\Sigma |i><i| \otimes \Lambda_i(\rho)##
    We could write in an abstract way that we have a map from ##H_a \otimes H_b## to itself but as the ground state of the apparatus is a constant we say that the channel sends ##H_a## to ##H_a \otimes H_b##
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