Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Born rule for degenerate eigenvalue
  1. Born's rule (Replies: 2)

  2. Born rule derivation? (Replies: 2)

Loading...