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Kronecker sum of more than two matrices?

  1. Feb 27, 2015 #1
    1. The problem statement, all variables and given/known data
    The question arises from this quote from wikipedia's article about kronecker product:

    Kronecker sums appear naturally in physics when considering ensembles of non-interacting systems. Let Hi be the Hamiltonian of the i-th such system. Then the total Hamiltonian of the ensemble is
    c2e0b6679eb0d88a8ec2d35a7d1a448f.png

    I have to write this Htot as a ordinary sum over kronecker products of unity matrix and Hi-s.

    2. Relevant equations

    Kronecker sum for two matrices is defined as

    3f1676452a4f1311f5d7a165e319b184.png

    If A is n × n, B is m × m and Ik denotes the k × k identity matrix.

    3. The attempt at a solution

    Well, as I undesratnd, now instead of A and B we have simply Hi and there should be sum kind of sum over i. But the Kronecker sum is defined only for a pair of matrices and it isn't commutative, so the order is important. I tried something like this, for three H-s:

    upload_2015-2-27_12-59-47.png

    But it doesn't look very elegant and I have no idea if this could be true. Any advice?
     

    Attached Files:

  2. jcsd
  3. Mar 1, 2015 #2
    The definition is as follows

    c2e0b6679eb0d88a8ec2d35a7d1a448f.png

    [tex] H_{\text{Tot}} =\left( H_1 \otimes \mathbb{I} \otimes \mathbb{I} \otimes ... \right)+ \left(\mathbb{I} \otimes H_2 \otimes \mathbb{I} \otimes ... \right) + \left(\mathbb{I} \otimes \mathbb{I} \otimes H_3 \otimes ... \right) + ...[/tex]
     
  4. Mar 1, 2015 #3
    Are you sure? Thank you!
     
  5. Mar 1, 2015 #4
    Proof:
    Kronecker sum is associative.

    In other words.
    The Kronecker sum of two matrices is, as you wrote,
    [tex] X=A\oplus B = A\otimes\mathbb{I}_B + \mathbb{I}_A\otimes{B} [/tex]

    Now, since the sum ##A\oplus B## is a matrix, ##X##, the Kronecker sum
    [tex] Y= X\oplus C = X\otimes\mathbb{I}_C + \mathbb{I}_X\otimes C = (A\otimes\mathbb{I}_B + \mathbb{I}_A\otimes{B})\otimes\mathbb{I}_C + \mathbb{I}_X\otimes{C} [/tex]
    Of course ##\mathbb{I}_X=\mathbb{I}_A\otimes\mathbb{I}_B##, which gives
    [tex] Y= A\otimes\mathbb{I}_B\otimes\mathbb{I}_C + \mathbb{I}_A\otimes B\otimes\mathbb{I}_C + \mathbb{I}_A\otimes\mathbb{I}_B\otimes{C} [/tex]

    ##Z= Y\oplus D = ## Keep going... :)
     
  6. Mar 1, 2015 #5
    Thank you very much!
     
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