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Quick question about Hermitian operators

  1. Aug 14, 2012 #1
    If H is a Hermitian operator, then its eigenvalues are real. Is the converse true?
     
  2. jcsd
  3. Aug 14, 2012 #2

    micromass

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    No. Take as counterexample the matrix [tex]\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right).[/tex]
    However, if you know that the operator is normal (that is: if [itex]AA^*=A^*A[/itex]), then it is true that it is Hermitian iff the eigenvalues are real.
     
  4. Aug 17, 2012 #3
    Good counterexample!

    A few more useful rules: a matrix ##M## has "good" eigenvectors if and only if it is normal. "Good" eigenvectors means it is possible to choose eigenvectors such that they form an orthonormal basis for whatever vector space ##M## acts on.

    Equivalently, a matrix is normal if and only if it can be diagonalized by a unitary transformation:
    ##
    M = U D U^{-1}
    ##
    for some unitary matrix ##U## and diagonal matrix ##D## whose diagonal elements are eigenvalues of ##M##.

    Hermitian matrices are always normal, and their eigenvalues are always real. But if ##M## is abnormal, it might have real eigenvalues and "bad" eigenvectors which are not orthogonal. Here's an example:

    ##
    \left[\begin{array}{cc}
    1 & 1 \\
    0 & 2 \\
    \end{array}\right]
    ##

    This thing has eigenvalues 1 and 2, but its eigenvectors

    ##
    \left[\begin{array}{c}
    1 \\ 0 \\
    \end{array}\right]
    \quad
    \left[\begin{array}{c}
    1 \\ 1 \\
    \end{array}\right]
    ##

    are not orthogonal to each other.

    The example micromass gave is even more badly-behaved: it is defective, which means it doesn't have enough linearly-independent eigenvectors to span the vector space. So we can't even form a "bad" basis using its eigenvectors.
     
    Last edited: Aug 17, 2012
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