When we study quantum mechanics, teacher often says that $H = -\Delta$ is a Hermitian operator (more precisely, a self-adjoint operator), and then reference linear algebra to note that its eigenvectors can form a complete basis.
The function $e^{i \mathbf{p}\cdot \mathbf{x}}$ is such an eigenvector. This aligns with our intuition, as it represents a Fourier transform:
$$ f(\mathbf{x}) = \int \tilde{f}(\mathbf{p}) e^{i \mathbf{p}\cdot \mathbf{x}} d\mathbf{p} $$
However, I find myself confused because, in principle, harmonic functions are also eigenvectors of the Laplacian operator, such as $u = xy$ or $u = x^2 - y^2$, where:
$$ \Delta u = 0 $$
Why then don't we include harmonic functions in our basis? Some have mentioned to me that harmonic functions are not square-integrable functions belonging to $L^2(\mathbb{R}^n)$, implying that the only eigenvector with eigenvalue zero is the constant function. However, I feel this is not a sufficient reason, as $e^{i \mathbf{p}\cdot \mathbf{x}}$ is also not square-integrable. Others have told me that harmonic functions are not bounded, and that bounded harmonic functions are only constant functions. Yet, there are square-integrable functions that are not bounded either.
I have not studied functional analysis and am unfamiliar with the spectral theorem; some of the terminology is currently beyond my grasp. However, I find the theorems as presented by my physics instructor to be overly simplistic and lacking in rigor. Could someone clarify these misunderstandings?
Let us consider the domain to be the entire $\mathbb{R}^n$. The choice of domain as circle or segments with boundary conditions will circumvent the issues, although the reason behind this remains unclear to me.
PS
In fact, if we draw an analogy to linear algebra, eigenvectors with the same eigenvalue form a subspace, which is not necessarily one-dimensional. Moreover, vectors within this subspace are orthogonal to vectors associated with different eigenvalues. Clearly, harmonic functions cannot be expanded solely by constant functions. Therefore harmonic function is linear independent of constant function.
When studying the Fourier transform, we find that even functions such as $x^n$ can possess a Fourier transform, represented by $\delta^{(n)}(x)$. Consequently, harmonic functions like $x^2 - y^2$, which have an eigenvalue of zero, can be expanded using terms like $e^{ipx}$, which is quite wierd.
One simple answer would be that in QM one generally works only with square integrable functions, so that they have some sort of decay at infinity. So it is not surprising that you don't need a function like $x^2-y^2$ in your basis to represent a square-integrable function $f$.
You "need" all the functions $e^{i\mathbf p\cdot\mathbf x}$ to represent a square-integrable function $f$ essentially because of the fact that the inverse Fourier transform is injective on $L^2(\mathbb R^n)$, so you can express a square-integrable function in a unique way as a superposition of harmonic oscillators $e^{i\mathbf p\cdot\mathbf x}$ (with "square-integrable coefficients"). This is analogous to standard linear algebra, where you can prove that a set of vectors $A$ forms a basis if you can write any vector $v$ as a linear combination of the elements of $A$ in a unique way.
The reason why you don't "need" the harmonic polynomials to represent $f$ is because of the surjectivity of the inverse Fourier transform from $L^2$ to itself, that is, the plane waves are enough to represent any function in $L^2(\mathbb R^n)$.