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I Quantum Operators

  1. Feb 2, 2017 #1
    My question is, if I understand the question.
    For every "observable" physical corresponds a quantum operator. This operator can be represented as an infinite dimensional matrix in a Hilbert space. Only Hermitian matrices each may be quantum mechanical operators, and at the same time to an observable corresponds to a Hermitian matrix. It's correct?

    Thanks
     
  2. jcsd
  3. Feb 2, 2017 #2

    jfizzix

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    Each observable corresponds to a quantum operator, but not all quantum operators are observables.

    Quantum operators can be represented as matrices in the same dimension Hilbert space as resides the quantum state.
    For example, the spin-state of a spin-1/2 particle has a 2-dimensional Hilbert space, so all operations on that spin can be expressed with 2x2 matrices.
    Alternatively, the position wavefunction of a particle lives in an infinite-dimensional Hilbert space, so operations on the wavefunction can be expressed as infinite-dimensional matrices, or more often, continuous functions over the real numbers.

    Quantum operators, may be hermitian, unitary, orthogonal, or many other things.
    For example, the time evolution operator is unitary.
    Only Hermitian operators have real eigenvalues, which is why only Hermitian matrices can correspond to observables (and vise versa).
     
  4. Feb 3, 2017 #3
    "Only Hermitian operators have real eigenvalues, which is why only Hermitian matrices can correspond to observables (and vise versa)"

    Thank you very much, very clear
    Karolus
     
  5. Feb 3, 2017 #4

    A. Neumaier

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    The matrix ##\pmatrix{0 & 1 \cr 0 & 2}## has all eigenvalues real but is not Hermitian.
     
  6. Feb 3, 2017 #5
    Good point. But then the proposed matrix (for real eigenvalues but not hermitian) can be theoretically a quantum operator?
    I think not, because among other properties the hermitian matrix admits a basis of orthonormal eigenvectors with distinct eigenvalues, in other words it is diagonalizable, and also believe it can be shown, but I have no proof in-hand, which if you calculate the average value of an observable, it follows that the corresponding operator must necessarily be hermitian. It's correct?
     
  7. Feb 3, 2017 #6

    PeroK

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    I wouldn't worry about the mathematical niceties of what is true in general for infinite dimensional operators. Instead, there are essentialy two assumptions:

    1) The operator relating to an observable is Hermitian. Note that the eigenvalues of a Hermitian operator must be real.

    2) Any operator relating to an observable has a complete spectrum of linearly independent eigenvectors.

    There is a proof of 2) for finite dimensional operators (it's called the finite-dimensional spectral theorem). But, for infinite dimensional operators things get more complicated, so best to assume 2) holds for the operators you are interested in.
     
  8. Feb 4, 2017 #7

    jfizzix

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    I stand corrected.


    So, to be more correct, only Hermitian operators correspond to observables because they have real eigenvalues, and because the eigenvectors associated to those eigenvalues form an orthogonal basis (the non-Hermitian matrix previously mentioned does not have this second property). This is necessary because if you measure the spin of a particle to be pointed up, the subsequent probability for up vs down must be (1,0), respectively,
     
    Last edited: Feb 4, 2017
  9. Feb 4, 2017 #8
    The proposed matrix has an eigenvalue, with a value of 2, and infinite eigenvectors that lie on the line y = 2x
    Now, and I return to the question, you can apply to be a quantum operator? The doubt is that a single eigenvalue correspond infinite eigenstates, which is unpleasant from a quantum point of view
     
  10. Feb 7, 2017 #9
    So to recap the conditions to be met by an operator to be a quantum operator they are:
    1. Must possess real eigenvalues (which corresponds to the possible values of a dynamic variable must be real)
    2. For each eigenvalue must match one eigenvector, in other words to different eigenvalues must match different eigenvectors.
    3. In the vector space in which the operator must exist a basis of eigenvectors of the operator, linearly independent, where any carrier (or state) of the vector space in question is expressible as a linear combination of the basis of eigenvectors considered. In other words, the operator in question must be diagonalizable.

    These properties are satisfied if the operator is Hermitian, then we can deduce that only hermitian operators are acceptable physically.
    As a result it can be shown that these properties are equivalent to the fact that the eigenvectors that make up the base are mutually "orthogonal" to each other.

    This is a miraculous result of how a purely mathematical and abstract properties, such as the fact that an operator is hermitian, has a striking physical application.
     
  11. Feb 7, 2017 #10

    PeterDonis

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    This isn't correct, there are cases where multiple eigenvectors can have the same eigenvalue (degeneracy).
     
  12. Feb 7, 2017 #11
    In general, if an eigenvalue ωi is sometimes degenerate mi, the symbol |ωi> not report
    to a single ket but to a generic element of mi- dimensional eigenspace V-mi(ωi)
    In this eigenspace we can choose mi orthogonal vectors with each other and that will
    distinguished by an additional index α that will take mi values
    Sorry, but Idont know LateX, i think the concept is clear.

    I had avoided to consider the degenerate case, not to over complicated the concept.
     
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