Degeneracy and non degeneracy in quantum mechanics

[MatinSAR]

Homework Statement

Why does a non-degenerate operator always lead to a unique orthonormal basis, while a degenerate operator allows infinitely many orthonormal bases to be constructed?

Relevant Equations

$\hat A \left| \psi_n \right\rangle =a_n\left| \psi_n \right\rangle$

This isn’t homework. I’ve been self-studying quantum mechanics from Zettili’s book, and I came across this statement with no explanation. Here’s a screenshot from the book for reference.

Then he proves it:

My problem is with this part:

I cannot understand why there are infinite number of basis sets.

IMO, If there is no degeneracy, then there is only one eigenvector for each eigenvalue. This means there is only one eigenbasis and it's uniquely determined. When there is degeneracy, there are multiple eigenvectors corresponding to the same eigenvalue. Can we say that just one of those eigenvectors is sufficient to form a complete basis? For example, can we construct the basis by using the eigenvectors of the non-degenerate eigenvalues along with just one eigenvector from the degenerate eigenvalue? Is this why there are infinite ways to create the basis when degeneracy occurs?

My idea:
When we talk about a complete basis, we mean that if we express a quantum state, like $\left| \psi_n \right\rangle$, in terms of this basis, we use the components of that state in the basis instead of the $\left| \psi_n \right\rangle$ itself. By saying the basis is complete, we imply that no information about the state is lost during this process.
When there is degeneracy—where more than one eigenvector corresponds to the same eigenvalue so we can choose just one eigenvector from that degenerate subspace while still retaining all the information about the state. This way we form a complete basis when degeneracy exists.

Any insight would be appreciated.

 

[vela]

I cannot understand why there are infinite number of basis sets.

IMO, If there is no degeneracy, then there is only one eigenvector for each eigenvalue. This means there is only one eigenbasis and it's uniquely determined. When there is degeneracy, there are multiple eigenvectors corresponding to the same eigenvalue. Can we say that just one of those eigenvectors is sufficient to form a complete basis?

No. Consider, for example, a vector space of finite dimension $n$. Any basis has to have $n$ vectors, so you can't toss some vectors just because they have degenerate eigenvalues.

Is this why there are infinite ways to create the basis when degeneracy occurs?

Consider a case where the degeneracy is two-fold. Suppose $\lvert \phi_1 \rangle$ and $\lvert \phi_2 \rangle$ are different eigenvectors of $\hat A$ with the same eigenvalue $\lambda$. They'll span a two-dimensional subspace. Any element in this subspace will also be an eigenvector of $\hat A$ with eigenvalue $\lambda$. Choose any two linearly independent vectors in this subspace, and they will be a basis for the subspace. Clearly, there's an infinite number of choices you can make.

 

[MatinSAR]

Consider a case where the degeneracy is two-fold. Suppose $\lvert \phi_1 \rangle$ and $\lvert \phi_2 \rangle$ are different eigenvectors of $\hat A$ with the same eigenvalue $\lambda$. They'll span a two-dimensional subspace. Any element in this subspace will also be an eigenvector of $\hat A$ with eigenvalue $\lambda$. Choose any two linearly independent vectors in this subspace, and they will be a basis for the subspace. Clearly, there's an infinite number of choices you can make.

That makes sense. I appreciate your time.