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A Normalization of 4x4 matrix

  1. Dec 14, 2017 #1
    I am trying to normalize 4x4 matrix (g and f are functions):

    \begin{equation}
    G=\begin{matrix}
    (1-g^2) &0& 0& 0&\\
    0& (1+f^2)& (-g^2-f^2)& 0 \\
    0 &(-g^2-f^2)& (1+f^2)& 0 &\\
    0& 0& 0& (1-g^2)
    \end{matrix}
    \end{equation}

    It's a matrix that's in a research paper (which I dont have) which gives the normalization constant as: N=4-2g^2+2f^2.
    I've been looking up online and found that N can be found with:

    method 1: N=\sqrt{\sum{X^2}} where X represents the elements of the matrix.

    method 2: I also found somewhere which said that I need to find the determinant.

    method 3: The ratio between the integral of excited state matrix and the integral of normal state of the matrix.

    I'm not sure who's right, but I'm not getting what was on paper.

    For method [1] I'm getting as far as: N^2 = 4(1+f^4+f^2g^2+f^2) . So I backtracked to see if their N^2 matches my N^2. But their N^2=16+4g^4+4f^4+16g^2-8g^2f^2+16f^2.

    Note that the normalization must satisfy the following condition G^2=1

    Any comments about all methods mentioned and how to implement it?
     
  2. jcsd
  3. Dec 14, 2017 #2

    Strilanc

    User Avatar
    Science Advisor

    What about the matrix are you trying to normalize?

    I notice that if you sum down the diagonal of the matrix that you get ##4-2g^2-2f^2##, which is the normalization constant you quote. So presumably you're trying to normalize the trace? In that case... just divide the matrix by its trace. The resulting matrix will have a trace of 1, unless the trace was 0 in which case you're not going to be able to normalize.
     
  4. Dec 14, 2017 #3
    Yes, you are correct. The paper normalizes the trace and your answer is correct for that part. I admit that I asked to many question on this one, because when I mentioned normalizing condition $G^2=1$ that is a different story....That is probably normalizing the whole matrix which can be implement with method 3 which I am not familiar with.
     
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