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 $(1)$?
  • $\begingroup$ So, even if the correct option is $(2)$, could happen option $(1)$? Would it make any sense? $\endgroup$
    – Invenietis
    Aug 28, 2020 at 11:55

3 Answers 3

  • 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.

  • $\begingroup$ Ok, thank you. And what do we say that it is degenerate: the eigenfunctions $\psi_{i,1}, \psi_{i,2},...$ sharing the same eigenstate $a_i$, or the eigenstate $a_i$ associated to the different eigenfunctions $\psi_{i,1}, \psi_{i,2},...$? $\endgroup$
    – Invenietis
    Aug 28, 2020 at 12:21
  • $\begingroup$ We say that states are degenerate. Also, $a_i$ is not an eigenstate, it is the eigenvalue, a scalar. Eigenstates are column vectors. $\endgroup$
    – exp ikx
    Aug 28, 2020 at 12:27
  • $\begingroup$ This is all a bit of a long time ago but the $p_x$, $p_y$ and $p_z$ hydrogen orbitals are degenerates because they have the same energy value, amiright? $\endgroup$
    – Gert
    Aug 28, 2020 at 13:22
  • 2
    $\begingroup$ @AbhayHegde I think you must have mis-spoken on your last bullet point here. It is eigenvalues -1 < a < 1 which would correspond to a "squeezing". Negative eigenvalues in general would just point the eigenvector in the opposite direction, not necessary squeeze it. $\endgroup$ Aug 28, 2020 at 21:25
  • 1
    $\begingroup$ When talking about degeneracy I think it is important to stress that it's not just that there are "multiple eigenvectors for the same eigenvalue", but rather that there are "multiple linearly independent eigenvectors for the same eigenvalue". Obviously a multiple of any eigenvector is also an eigenvector. $\endgroup$
    – printf
    Aug 29, 2020 at 3:17

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.


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