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Writing a random 2N by 2N matrix in terms of Pauli Matrices

  1. Oct 2, 2013 #1

    Wasn't sure if I should post this to Linear Algebra or here.

    My question is really simple:

    Can a 2N by 2N random, and Hermitian Matrix ( Hamiltonian ) be always written as:

    [itex]H = A \otimes I_{2\times 2} + B \otimes \sigma_x + C \otimes \sigma_y + D \otimes \sigma_z[/itex]

    where A,B,C,D are all N by N matrices, while the sigma's are the Pauli spin matrices.

    My question is, as long as A,B,C,D are random and complex Hermitian matrices of size N by N, do I cover the
    whole 2N by 2N complex Hermitian space with this representation?

    If yes, do you know a reference, a theorem, or a simple proof of this?

    A very simple case is when N = 1 , and I know that any 2 x 2 complex , Hermitian matrix can be written as a linear combination of Pauli Matrices.

    Many thanks,
  2. jcsd
  3. Oct 2, 2013 #2
    Yes, it is. It's just a matter of counting degrees of freedom
  4. Oct 2, 2013 #3


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    Yes, because the Pauli operators form a basis of the set of operators acting on qubit (two-dimensional Hilbert) spaces. If you're not convinced then note that you can always write your operator in the form

    [tex]H = A_{00} \otimes \lvert 0 \rangle \langle 0 \rvert + A_{01} \otimes \lvert 0 \rangle \langle 1 \rvert + A_{10} \otimes \lvert 1 \rangle \langle 0 \rvert + A_{11} \otimes \lvert 1 \rangle \langle 1 \rvert[/tex]​

    just by writing it out explicitly in some basis and collecting the terms in [itex]\lvert 0 \rangle \langle 0 \rvert[/itex], [itex]\lvert 0 \rangle \langle 1 \rvert[/itex], etc., and then substituting

    \lvert 0 \rangle \langle 0 \rvert &=& \tfrac{1}{2} ( \mathbb{I} + \sigma_{z} ) \,, \\
    \lvert 0 \rangle \langle 1 \rvert &=& \tfrac{1}{2} ( \sigma_{x} + i \sigma_{y} ) \,, \\
    \lvert 1 \rangle \langle 0 \rvert &=& \tfrac{1}{2} ( \sigma_{x} - i \sigma_{y} ) \,, \\
    \lvert 1 \rangle \langle 1 \rvert &=& \tfrac{1}{2} ( \mathbb{I} - \sigma_{z} ) \,.

    This works for any operator. If [itex]H[/itex] happens to be Hermitian then this imposes additional constraints. For instance, as you pointed out, the [itex]A[/itex], [itex]B[/itex], [itex]C[/itex], and [itex]D[/itex] from your post must also be Hermitian in that case.
  5. Oct 4, 2013 #4
    Hi , Thank you for the responses ... However, I still don't understand it from a matrix point of view.

    Let's take N = 2 , and have a 4x4 H matrix ... can one prove that my representation will always cover the full space ?

    I didn't follow it from the Dirac notation,

    Many thanks for responses.
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