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I Hermitian operators, matrices and basis

  1. Apr 7, 2017 #1
    Hello, I would just like some help clearing up some pretty basic things about hermitian operators and matricies.

    I am aware that operators can be represented by matricies. And I think I am right in saying that depending on the basis used the matrices will look different, but all our valid representations of the operator.

    As I understand it, there exists a basis where hermitian operators can be represented by a diagonal matrix

    Is the only basis that this can occur in the basis of the eigenvectors of the hermitian operators. And the diagonal matrix elements of the hermitian operators are the eigenvalues of the operator??

    In summary I am asking, is the only basis in which a hermitian operator is represented by a diagonal basis the basis of the eigenvectors? And for the diagonal matrix are the elements the eigenvalues of the operator?


    Thank you
     
  2. jcsd
  3. Apr 7, 2017 #2

    stevendaryl

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    Given any basis [itex]|\psi_j\rangle[/itex], you can use that basis to represent arbitrary states as column matrices. Letting [itex]\mathcal{R}(|\psi\rangle)[/itex] mean the matrix representation of [itex]|\psi\rangle[/itex], we can choose [itex]\mathcal{R}[/itex] so that:

    [itex]\mathcal{R}(|\psi_1\rangle) = \left( \begin{array} \\ 1 \\ 0 \\ . \\ . \\ . \end{array} \right) [/itex]

    [itex]\mathcal{R}(|\psi_2\rangle) = \left( \begin{array} \\ 0 \\ 1 \\ . \\ . \\ . \end{array} \right) [/itex]

    [itex]\mathcal{R}(|\psi_3\rangle) = \left( \begin{array} \\ 0 \\ 0 \\ 1 \\ 0 \\ . \\ . \\ . \end{array} \right) [/itex]

    To say that an operator [itex]\hat{O}[/itex] is diagonal in this basis is to say that its representation is given by:

    [itex]\mathcal{R}(\hat{O}) = \left( \begin{array} \\ \lambda_1 & 0 & ... \\ 0 & \lambda_2 & 0 & ... \\ 0 & 0 & \lambda_3 & ... \\ . \\ . \\ . \end{array} \right) [/itex]

    Yes, [itex]\lambda_1, \lambda_2, ...[/itex] are the eigenvalues of [itex]\hat{O}[/itex] and [itex]|\psi_j\rangle[/itex] are the eigenvectors.
     
  4. Apr 7, 2017 #3

    hilbert2

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    Yes, but note that you can put the eigenvalues to many different orders on the diagonal, in which case different unit vectors of Rn correspond to the values. Also, an operator can be defined on an infinite-dimensional vector space, in which case you can't write it as a matrix.
     
  5. Apr 7, 2017 #4

    PeroK

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    Why not?
     
  6. Apr 7, 2017 #5

    hilbert2

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    Well, you can do it if the basis vectors can be ordered in such a way that there's some obvious pattern in the matrix elements, but think about a vector space that has a basis with the same cardinality as the set of real numbers.
     
  7. Apr 7, 2017 #6
    Got it, thank you!
     
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