# Imaginary Eigenvalue Of A Hermitian Operator

The eigenfunctions of a Hermitian operator are real. But consider a function $\psi(x)=e^{-\kappa x}$, $x\in\mathbb{R}$, where $\kappa$ is a real constant. Then, $$\hat p \psi(x)=-i\hbar \frac{d}{dx}e^{-\kappa x}=i\kappa \hbar \psi(x).$$ This gives a pure imaginary eigenvalue. Is it not a contradiction? Or am I missing some crucial point?

• Interesting question. Just pointing out the eigenvalues of a Hermitian operator are real, not the eigenfunctions. – JeffDror Feb 12 '14 at 13:48
• this function is unphysical – Danu Feb 12 '14 at 14:06
• Unphysical doesn't mean that it is mathematically impossible. I believe that the right answer lies on the rigged hilbert space associated is $L^2(\mathbb{R})$, i have to check but I think that $\psi(x)=e^{-\kappa x}$ with real $\kappa$ don't lie there. Anyway, you don't need eigenvectors to define eigenvalues, and the spectral theorem, in this case, rules out non-real eigenvalues – Hydro Guy Jun 6 '14 at 18:39
• Related: physics.stackexchange.com/q/81041/2451 , physics.stackexchange.com/q/221027/2451 and links therein. – Qmechanic Feb 17 '17 at 16:39

What is your Hilbert space? In $L^2(\mathbb R)$ your eigenfunction would have infinite norm. If you dealt instead with a bounded set $L^2([a,b])$, your operator would not be Hermitian unless you impose suitable boundary conditions to discard boundary terms. These boundary conditions, however, would rule out your candidate eigenvector!
• V. Moretti-But even $e^{ikx}$ is not a member of $L^2(-\infty,\infty)$. But this gives real eigenvalue for the same operator. Therefore, what I understood from your answer is that unless the domain is specified it is not guaranteed that the eigenvalues of the hermitian operator are real, it could be be anything, real or imaginary. – SRS Feb 12 '14 at 14:21
• (I rewrite, since it was unclear as I sent it from my mobile phone). Yes, $e^{ikx}$ is a generalized eigenvector like $\delta(x)$ for the position operator...For self-adjoint operators, reality condition holds for generalized eigenvectors, too, but it is more technical to prove. – Valter Moretti Feb 12 '14 at 18:57
The spectral theorem applies to "self-adjoint" operators. Hermitian symmetric operators are not necessarily self-adjoint. One of the equivalent definitions is that $A$ is self-adjoint if there is no $f$ in the Hilbert space such that $(f,(A-\lambda)g)=0$ for all g in the domain of $A$ with non-real $\lambda$. Your example is a proof that the operator is self-adjoint.
An example of a non self-adjoint symmetric operator is $i\frac{d}{dx}$ on the domain of smooth functions on $(0, \infty)$. Consider $f=e^{-x}$ .