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Sakurai Question regarding density matrix

  1. Jul 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Sakurai Modern Quantum Mechanics Revised Edition. Page 81. density matrix p = 3/4 [1 0; 0 0] + 1/4 [1/2 1/2; 1/2 1/2]. We leave it as an exercise to the reader the task of showing this ensemble can be decomposed in ways other than 3.4.24


    2. Relevant equations 3.4.24 w( sz + ) = 0.75; w(sx +) = 0.25.

    3. The attempt at a solution Various. First I found pure states SX+, SY+, SZ+,SX+, SX+, SY-, and SZ-. The fact that p = [7/8 1/8 ; 1/8 1/8 ] has no i's suggest that the weights for Sy+ = Sy-.
    a SX+ + b Sy+ + c Sz+ ... = [7/8 1/8 ; 1/8 1/8 ] can be but it always seems to lead to some weights which are negative. are negative weights allowed (I do not think so).

    Another attempt I tried was to rotate the basis states themselves to get more pure states. I was able to get the rotated density matrix equal to the linear combination of rotated basis states, but I could not get the density matrix [7/8 1/8 ; 1/8 1/8 ] to be a legitimate linear combination of the rotated basis states. I believe I used ideas that have not been developed by Sakurai at this point in this book. I may be making this problem harder than it is (I hope so ). Otherwise I tend to think the exercise is impossible.

    I tend to think Sakurai throws this out and does not assign it as a problem because it is more easy and straightforward.

    This has not been assigned to me as a problem. I am not taking a course in QM (not in the last decade). I am just interested I would appreciate the solution or any comments.
     
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  3. Jul 20, 2015 #2

    andrewkirk

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    When you say it leads to 'negative weights' are you sure you are not referring to elements of the matrix (which are not weights)? Negative elements of the matrix are fine, as long as the matrix is positive definite.

    If we do an eigen decomposition of the matrix we get eigenvalues 0.895 and 0.105, which are both positive. These are the weights for a decomposition ##\hat{\rho}=\sum_{j=1}^2 \lambda_j |\psi_j\rangle\langle\psi_j|## where the eigenstates are

    $$|\psi_1\rangle=-0.987 |\phi_1\rangle-0.160|\phi_2\rangle$$
    $$|\psi_2\rangle=+0.160 |\phi_1\rangle-0.987|\phi_2\rangle$$

    where ##|\phi_j\rangle## are the basis kets of your original basis.

    I don't have a copy of Sakurai, so it may be that this reply is off topic if there is more context in the question but not in your post, that I have not taken into account.
     
  4. Jul 20, 2015 #3
    Sakurai's text is Modern Quantum Mechanics. The mathematics there addresses spinors. This is different from the eigenvalue problem. Sakurai suggest the matrix [7/8 1/8; 1/8 1/8] can be decomposed in more than one way. The example in the text shows [7/8 1/8; 1/8 1/8] = 3/4 [1 0; 0 0] +1/4 [1/2 1/2; 1/2 1/2]. The matrices [1 0; 0 0] and [1/2 1/2; 1/2 1/2] are "pure" states (pure because the matrices are idempotent.) This is clear. Sakurai leaves as a exercise for the reader that the same matrix [7/8 1/8; 1/8 1/8] can be decomposed into another combination of pure (idempotent) matrices. This does not seem to be obvious. I am not even sure it is true anymore. The weights are 3/4 and 1/4 in the example above. Can another selection of weights be used (that add up to one) multiply pure states to come up with [7/8 1/8; 1/8 1/8].
     
  5. Jul 20, 2015 #4

    andrewkirk

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    The decomposition I gave is a positive (in fact convex) sum of two pure states: ##|\psi_1\rangle\langle \psi_1|## and ##|\psi_2\rangle\langle \psi_2|##.

    That they are idempotent follows immediately from the fact that each is a projection matrix - onto the kets ##|\psi_1\rangle## and ##|\psi_2\rangle## respectively.
     
  6. Jul 20, 2015 #5
    Computer is acting up will follow up on these thoughts tomorrow. I have prepared three threads (replys) and the computer has cut me off tonight. Thank you, Andrewkirk, but I am still not sure this addresses the point of non-uniqueness that Sakurai is looking for. Still the square of the coefficients in your decomposition do square to one. (I.e the coefficients of the projection operators are positive and sum to one). this suggests an approach I have not considered yet. Thanks again. I hope to hear from a reader of Sakurai to see how they confronted this point (context).
     
  7. Jul 24, 2015 #6
    To Andrekirk

    I took your eigenvalues and multiplied them by the density matrices E1 * [ cos A cosA cos A sin A; cos A sin A sin A sin A ]
    + E2 * [sin A sin A -cos A sin A; -cos A sin A sin A sin A]; where E1 = 0.895 and E2 = 1 - E1 and tan A/2 = 1/3). I obtained [7/8 1/8 ; 1/8 1/8] as per the conditions of the problem. Thank you for putting me on the right track. Sincerely Mpresic
     
  8. Mar 23, 2017 #7
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