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Geometric and Physical Interpretation of Diagonalization

  1. Oct 28, 2006 #1
    I am a second year student in quantum mechanics. I heard in lecture that simultaneous diagonalization of matrices is important in quantum mechanics. I would like to know why is it significant when two matrices can be simultaneous diagonalized, and what is the geometric and physical interpretation of simultaneous diagonalization of matrices in quantum mechanics and in physics in general.

    Thank you for everyone's help.
     
  2. jcsd
  3. Oct 28, 2006 #2
    The physical interpretation of this result is that if you can simultaneously diagonalise two matrices then the corresponding operators (if the matrices in question are Hermitian) are compatiable. This means that they commute since it is possible to find a commom eigenvector basis for both of the matrices. Ultimately, this means that one can simultaneously measure both of those observables on the quantum system of interest. Thus, in this case there should be no uncertainty relation between the operators and one can be expressed as a function of the other.
     
  4. Oct 28, 2006 #3
    To put it a little more physically, (hermitian) matrices represent observable physical quantities. If you have a state which, when acted upon by a matrix, returns itself multiplied by a constant, that state has a definite value of the observable associated with the matrix. A compete set of such states will diagonalize the matrix.

    If a common set of states can diagonalize two matrices simultaneously, it means that those states have definite values of both observables. So, the statement that there is no uncertainty relation between the observables is quite correct. However, it is not the case that one can be expressed as a function of the other. In general, the observables are independent (like the energy and angular momentum of the electron in a hydrogen atom).
     
  5. Oct 28, 2006 #4

    Hurkyl

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    It's often useful, when studying a matrix, to look at the eigenvectors. Even better if you can choose a basis whose vectors are all eigenvectors.

    So, if two matrices can be simultaneously diagonalized...
     
  6. Sep 20, 2010 #5
    while diagonalizing an operator A with a matrix S (formed from eigenvectors of A), does S need to be unitary (i.e., whether I have to form it with the NORMALIZED eigenvectors)? Even if S is not unitary, A gets diagonalized.
     
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