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Born rule for degenerate eigenvalue

  1. Sep 18, 2012 #1
    The probability of measuring a value [itex]a[/itex] for an observable [itex]A[/itex] if the system is in the normalized state [itex]|\psi\rangle[/itex] is
    [tex]|\langle a|\psi\rangle|^2[/tex]
    where [itex]\langle a|[/itex] is the normalized eigenbra with eigenvalue [itex]a[/itex].

    This is more-or-less the formulation of the Born rule as it appears in my text. But this seems to only make sense if [itex]\langle a|[/itex] is non-degenerate. So, what's the rule if we have a degeneracy?
  2. jcsd
  3. Sep 18, 2012 #2


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    In the general case one should use the spectral decomposition of the self-adjoint operator which describes the observable A, thus one uses projectors. The projector for a subspace belonging to a degenerate eigenvalue (for simplicity, assume the spectrum to be purely a point spectrum) is the sum of each projector according to the rule

    [tex] P_ n = \sum_{i=1}^{g_n} |ni\rangle \langle ni| [/tex]

    with g_n the dimension of the subspace which corresponds to the degenerate eigenvalue a_n in which [itex] |ni\rangle [/itex] form an orthonormal subbasis. This P_n goes then in the general Born rule (again pure point spectrum):

    [tex] p(a_n)_{|\psi\rangle} = \langle \psi |P_n|\psi\rangle [/tex]
  4. Sep 18, 2012 #3
    The general formulation of the Born rule uses projectors onto subspaces. The probability of finding a eigensubspace of the measurement operator is equal to the expectation of the projector in the state: p = <psi|P|psi> You can easily see that if you project onto a 1-dimensional subspace P can be written as P = |n><n| and the probability becomes p = <psi|n><n|psi> = |<psi|n>|^2
  5. Sep 18, 2012 #4
    The simple version of the above replies is: if there are multiple eigenstates with eigenvalue a, add up the Born probabilities for all those states to get the probability to measure the value a.
  6. Sep 18, 2012 #5
    I have a question Dexter,

    If the projector of a subspace which belongs to a generate eigenvalue, can one say that the projector is a type of generator of those which belong in the subgroup?
  7. Sep 19, 2012 #6


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    I'd interpret Born's rule for a degenerate eigenvalue from the point of view of statistical mechanics. Given a system to be prepared in some state, represented by the statistical operator [itex]\hat{R}[/itex], we ask the question about the probability (density) to find a specific value [itex]a[/itex] of an observable [itex]A[/itex], represented by a self-adjoint operator [itex]\hat{A}[/itex]. If [itex]|a,\beta \rangle[/itex] is a complete set of (generalized) eigenvectors of [itex]\hat{A}[/itex] for the eigenvalue [itex]a[/itex] normalized to unity (or to the [itex]\delta[/itex] distribution), then the probability (density) that the system is found in a specific state given by one of these eigenvalues is
    [tex]P(a,\beta)=\langle a,\beta|\hat{R}|a,\beta \rangle.[/tex]
    This probability can be found experimentally by measuring a complete set of compatible observables (including [itex]A[/itex]).

    If you know only measure [itex]A[/itex] you have to sum (integrate) over all the non-measured observables since, because the basis vectors are orthonormalized, the outcomes are mutually exclusive, i.e., you have
    [tex]P(a)=\sum_{\beta} P(a,\beta) \quad \text{or} \quad \int \mathrm{d} \beta \; P(a,\beta).[/tex]
    So the Born rule for an incomplete measurement in the case of degenerate eigenvalues follows directly from the Born rule for a complete measurement and basic rules of probability theory.
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