# What is a physical example of an observable with degenerate eigenvalues? [closed]

If eigenvalues of an observable have the physical meaning of a possible result after a measurement, what's the interpretation of degenerate eigenvalues, and what is an example of such an observable?

If the value of the measurement is $$a_2$$, how would we physically be able to tell if the system jumped to the state $$|a_2\rangle$$ or $$|a_3\rangle$$

Measuring $$A$$ alone we wouldn't be able to. Since $$|a_2\rangle$$ and $$|a_3\rangle$$ are degenerate eigenstates of $$A$$, then $$|a_{23}\rangle=b_2|a_2\rangle +b_3|a_3\rangle$$ is also its eigenstate with the same eigenvalue, so $$|\psi\rangle$$ could collapse into it instead.

is the deduction that $$|a_2\rangle$$ and $$|a_3\rangle$$ are two different states that the system can jump into purely mathematical (based on algebra) or was it experimentally observed and verified?

The distinction between these two states will become apparent when you measure some other observable whose eigenvalues for $$|a_2\rangle$$ and $$|a_3\rangle$$ are distinct.

what is an example of such an observable?

Consider kinetic energy and momentum. Kinetic energy $$K$$ of a free 1D particle is twice degenerate for all values (except $$K=0$$): for each momentum of a particle going to the right there exists corresponding momentum of the particle going to the left, while kinetic energy of both is the same. Momentum doesn't have this degeneracy, its whole spectrum is non-degenerate.

If you measure kinetic energy, getting an eigenstate (with possibly superposed momenta, e.g. $$\alpha|{-p}\rangle+\beta|{+p}\rangle$$), you can then measure momentum and finally get non-degenerate momentum eigenstate: either $$|{+p}\rangle$$, or $$|{-p}\rangle$$.

• $|\psi\rangle$ will, not could, collapse to $|a_{23}\rangle$ if the measurement yields $a_2$. Apr 11, 2020 at 6:52
• @nanoman you're right. I was just not sure how to prove this stronger statement. Apr 11, 2020 at 6:54

What is a physical example of an observable with degenerate eigenvalues?

A simple explicit example is the energy of a hydrogen atom. In the Schrodinger approximation, the $$2s$$ state and the $$2p$$ states all have the same energy. These states differ in other quantum numbers involving angular momentum, and they are experimentally verified as distinguishable states.

I'm adding another important example. In quantum mechanics angular momentum is quantised. The operator $$J^2$$ measures the length of the angular momentum vector and $$J_z$$ measures the z-component of the vector. You can label one of these states by $$|j\ m\rangle$$ where $$j$$ corresponds to the length of the angular momentum and $$m$$ to the z-component. $$m$$ takes on values $$-j,\, -j+1,\, ...,\,j-1,\,j$$. Each value of $$m$$ still has the same length so $$|j\ m\rangle$$ is degenerate in $$J^2$$ with degeneracy $$2j+1$$.

Any system of multiple particles (taken as distinguishable for simplicity) illustrates degenerate eigenvalues. The full Hilbert space is the tensor product of each particle's Hilbert space, and an observable of one particle acts as the identity operator on all other particles. Thus, a given eigenvalue of position, momentum, or spin for one particle has an eigenspace consisting of the corresponding eigenstate of that particle tensored with all possible states of the other particles.

Something like this happens generically, because a practical measurement does not tell us a unique quantum state of the whole universe. Particles not involved in the measurement remain in their original quantum states.

In your example, if the measurement of $$A$$ on $$|\psi\rangle$$ yields $$a_2$$ (which has probability $$|b_2|^2 + |b_3|^2$$, not $$|b_1|^2 + |b_2|^2$$), then the new state will be $$(b_2|a_2\rangle + b_3|a_3\rangle)/(|b_2|^2 + |b_3|^2)^{1/2}$$. It will not jump to $$|a_2\rangle$$ unless $$b_3 = 0$$; it will not jump to $$|a_3\rangle$$ unless $$b_2 = 0$$. The collapse upon measuring a degenerate eigenvalue is a projection onto the corresponding eigenspace.