# Are eigenvalues in quantum mechanics related to eigenfunctions (in the PDE sense) or to linear algebra and eigenvectors?

I'm in 10th grade and a beginner in the amazing world of quantum physics, I want to become a mathematician but I like quantum mechanics as well. The eigenvalues in Schrödinger wave equation used to describe the motion on electrons and other fundamental particles. Are they mathematically related to the solutions of partial differential equations (eigenfunctions) or to the eigenvalues of a square matrix (eigenvectors and eigenvalues)?

## 1 Answer

Both. There's essentially no difference at all between those two settings, when seen from the perspective of abstract linear algebra: in both, you have some vector space $V$ and some linear operator $\mathcal L:V\to V$, and you look for members $v\in V$ of the vector space which obey a property of the form $$\mathcal Lv=\lambda v, \tag 1$$ where $\lambda$ is known as the eigenvalue. The vector $v$ is normally known as an eigenvector, but it may also be appropriate to use the terms eigenstate or eigenfunction ─ which are nevertheless completely synonymous with eigenvector. If $V$ is finite-dimensional, then the eigenvalue equation $(1)$ can be phrased in terms of square matrices and row vectors, but if $\mathcal L$ is some differential operator (as in the PDE form of the time-independent Schrödinger equation) then that is no longer possible, but that fact does not detract in any way from the abstract properties of the eigenvalue relation.

• The vector $v$ in the definition above should be non-zero (if else $0$ would be an eigenvector of any linear operator for any $\lambda\in \mathbb{C}$, thus making the notions of resolvent set and spectrum essentially useless). It is also probably worth to point out that, for (bounded) operators on an infinite dimensional (Banach) space (the ones you are considering), the notions of resolvent and spectrum are the ones that should really be looked at, since the set of eigenvectors may often be empty, or not describing with sufficient accuracy the whole operator. – yuggib Sep 4 '17 at 9:22
• @yuggib Those are boring technicalities that are far, far out of scope on a thread where OP is in 10th grade. – Emilio Pisanty Sep 4 '17 at 9:46
• These are just "boring technicalities" in your opinion, that may however differ from the one of others (like mine). In addition, an answer on this site should be useful not only for the OP, but also to other users. So even if the answer should take in some account the eventual background of the OP, it should also at least contain the least number of inaccuracies. – yuggib Sep 4 '17 at 10:13
• And your answer contains some inaccuracies not unrelated to my "boring" comment above. For example, the "PDE form" of which Schrödinger equation? The one for the evolution of a particle in some potential $V$? or the (horribly) so-called "time-independent" Schrödinger equation? The first is not an eigenvalue equation, and the second fails to describe the physical behavior of a free (or in general non-trapped) particle, since it has no solution in $L^2$ (the Laplacian operator has purely continuous spectrum). – yuggib Sep 4 '17 at 10:17
• @user167920 The one take-home message from all this mathematical arguing is that in a first linear algebra course, you study vector spaces with a finite number of dimensions (often 2 or 3, corresponding to the Euclidean view of "real world" 2D or 3D space) and therefore you can use matrix notation for calculations. However vector spaces involving functions have an infinite number of dimensions (in almost all situations), which brings in some more mathematical complications, and means that matrix notation is not very useful. – alephzero Sep 4 '17 at 12:09