# 4.1: Distinguishing between eigenstates, Quantum numbers as labels

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How can we distinguish between quantum states $$| \alpha_n \rangle$$ which have degenerate values of A? The obvious way is to measure the quantised observables and use them to label the state. We must be sure not to make measurements which change the state. Thus all measurements should correspond to commuting operators (Compatible observations: see QP3). In the non-degenerate case measuring energy is sufficient, but in hydrogen, for example, we used quantum numbers $$n$$ (for energy, operator $$\hat{H}$$), $$l$$ (for total angular momentum, $${\bf \hat{L}}^2$$) and $$m_l$$ (one component of angular momentum, $$\hat{L}_z$$).

Continuing the example of twofold degeneracy (3.6), suppose that some operator $$\hat{B}$$ is compatible with $$\hat{A}$$. This means that $$[A, B] = 0$$ and $$\hat{A}$$ and $$\hat{B}$$ have a common eigenbasis. i.e. some $$\theta$$ and $$\theta + \pi /2$$ give eigenstates of both $$\hat{A}$$ and $$\hat{B}$$ in the form $$| \alpha ( \theta ) \rangle = (\cos \theta | \alpha_1 \rangle + \sin \theta | \alpha_2\rangle$$).

To find the appropriate value of $$\theta$$, we have a similar problem to that encountered in 3.1 and must solve for the eigenvectors of:

$\begin{pmatrix} \langle \alpha_1|\hat{B}| \alpha_1\rangle & \langle \alpha_1|\hat{B}| \alpha_2\rangle \\ \langle \alpha_2|\hat{B}| \alpha_1\rangle & \langle \alpha_2|\hat{B}| \alpha_2\rangle \end{pmatrix}\nonumber$

The eigenvalues of this equation are the quantised measurable values of $$\hat{B}$$. If both of these are equal, there must be another measurable C which will distinguish the two states.

The generalisation to many degenerate levels is straightforward. If there are $$n$$ orthogonal degenerate eigenstates of $$\hat{A}$$, (therefore an $$n$$-dimensional space in which every unit vector is an eigenstate of $$\hat{A}$$), compatibility of eigenbases means there are at least $$n$$ eigenstates of $$\hat{B}$$. It is now possible that all these have different B eigenvalues, or that at least two have the same eigenvalue, in which case if we want a specific set of orthogonal eigenstates, we must look for another compatible operator $$\hat{C}$$.

When the set of operators is sufficiently large that there is a unique set of eigenvalues for each eigenstate, we call it a complete commuting set of operators. An example is $$\hat{H}$$, $${\bf \hat{L}}^2$$, $$\hat{S}_z$$ and $$\hat{L}_z$$ in hydrogen. The complete commuting set is not unique for a given Hamiltonian, for hydrogen we could have used $$\hat{H}$$, $${\bf \hat{L}}^2$$, $$\hat{S}_x$$ and $$\hat{L}_x$$ or $$\hat{H}$$, $${\bf \hat{L}}^2$$, $$\hat{J}$$ and $$\hat{L}_z$$ If one of the quantum numbers can be written in terms of the others then it is redundant. If two of the quantum numbers come from non-commuting operators, then the set does not define a state since the full set of measurements could not be performed without changing the wavefunction.

This page titled 4.1: Distinguishing between eigenstates, Quantum numbers as labels is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to the style and standards of the LibreTexts platform.