Born rule for degenerate eigenvalue

[dEdt]

The probability of measuring a value $a$ for an observable $A$ if the system is in the normalized state $|\psi\rangle$ is
$$|\langle a|\psi\rangle|^2$$
where $\langle a|$ is the normalized eigenbra with eigenvalue $a$.

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 $\langle a|$ is non-degenerate. So, what's the rule if we have a degeneracy?

 

[dextercioby]

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

$$P_ n = \sum_{i=1}^{g_n} |ni\rangle \langle ni|$$

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

$$p(a_n)_{|\psi\rangle} = \langle \psi |P_n|\psi\rangle$$

 

[Jazzdude]

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

 

[The_Duck]

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.

 

[Meselwulf]

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

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?

 

[vanhees71]

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 $\hat{R}$, we ask the question about the probability (density) to find a specific value $a$ of an observable $A$, represented by a self-adjoint operator $\hat{A}$. If $|a,\beta \rangle$ is a complete set of (generalized) eigenvectors of $\hat{A}$ for the eigenvalue $a$ normalized to unity (or to the $\delta$ distribution), then the probability (density) that the system is found in a specific state given by one of these eigenvalues is
$$P(a,\beta)=\langle a,\beta|\hat{R}|a,\beta \rangle.$$
This probability can be found experimentally by measuring a complete set of compatible observables (including $A$).

If you know only measure $A$ 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
$$P(a)=\sum_{\beta} P(a,\beta) \quad \text{or} \quad \int \mathrm{d} \beta \; P(a,\beta).$$
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.