Degeneracy in Quantum Mechanics – Meaning and Implications

eigenvaluehilbert-spaceoperatorsquantum mechanicswavefunction

Let $\hat{A}$ be a Hermitian operator that represents the observable $A$. Its eigenstate-equation would be:

$$\hat{A}\psi_n=a_n\psi_n \tag{1}$$

After solving it, we would get a set of the eigenfunctions $\{\psi_n\}$ and eigenvalues $\{a_n\}$ of $\hat{A}$, with $a_i$ the eigenvalue corresponding to the eigenstate $\psi_i$, and vice versa.

Then, what do we mean when we say that there is degeneracy:

  1. That, for the same eigenstate $\psi_i$, there are different eigenvalues $a_i$ that verify $(1)$, or
  2. That, for the same eigenvalue $a_i$, there are different eigenstates $\psi_i$ that verify $(1)$?

Best Answer

  • We mean the second option. For the same eigenvalue $a_i$, there are multiple linearly independent eigenvectors $\psi_{ij}$ where $j$ denotes the degeneracy. When you measure something, if there are multiple linearly independent states giving the same measurement value, those states are degenerate.

  • Also note that the same eigenstate cannot have multiple eigenvalues. It is unique for a given state. Suppose that $A \psi_i = a_i \psi_i$ and $A \psi_i = a_i' \psi_i$. Then $a_i \psi_i − a_i' \psi_i = 0 $, which implies that at least one of $a_i - a_i'$ or $\psi_i$ is equal to zero. Eigenvectors are nonzero by definition, so it must be the case that $a_i=a_i'$.

  • Operators are linear transformations acting on the Hilbert space. The transformations usually involve stretching, squeezing and rotations. The vectors which still remain on their span after the transformations are the eigenvectors for that operation. In other words, the eigenvectors will stay in the same direction even after applying the operator, although they might be squeezed or stretched. The value by which they stretch (a number > $|1|$) or squeeze (a number < $|1|$) is their eigenvalue. It should be clear now, why we can't have multiple eigenvalues for a vector.

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