# Meaning of degeneracy in Quantum Mechanics

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.

We mean (2) : for the same *eigenvalue* $a_i$ there are more than one *eigenfunction* $\psi_i$. In this case one needs an additional index to distinguish different eigenfunctions corresponding to the same eigenvalue:
$$\hat{A}\psi_{n\nu} = a_n\psi_{n\nu}.$$

From a more mathematical point of view, we say there is degeneracy when the *eigenspace* corresponding to a given eigenvalue is bigger than one-dimensional. Suppose we have the eigenvalue equation
$$
\hat A\psi_n = a_n\psi_n\;.
$$
Here $a_n$ is the eigenvalue, and $\psi_n$ is the eigenfunction corresponding to this eigenvalue. But this eigenfunction is of course not uniquely defined: any multiple of $\psi_n$ also satisfies the eigenvalue equation, $\hat A(\lambda\psi_n) = \lambda \hat A\psi_n = \lambda a_n\psi_n = a_n(\lambda\psi_n)$ by linearity. Thus we talk about *eigenspace* belonging to given eigenvalue $a_n$. This eigenspace may be one-dimensional, i.e. every eigenstate must be proportional to $\psi_n$. In this case there is no degeneracy. But is is possible that this eigenspace has dimension higher than 1. For example this eigenspace may be two-dimensional, which means there are two **linearly independent functions**, say $\psi_{n1}$ and $\psi_{n2}$, which are both eigenfunctions of $\hat A$ with eigenvalue $a_n$:
$$
\hat A\psi_{n1} = a_n \psi_{n1}\;,\quad\hat A\psi_{n2} = a_n \psi_{n2}\;.
$$
Then every linear combination of the form $\alpha\psi_{n1} + \beta\psi_{n2}$ is also an eigenstate of $\hat A$ with eigenvalue $a_n$. That is, the eigenspace corresponding to eigenvalue $a_n$ has dimension 2. In this case we say there is (double) degeneracy.