# 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. Commented Feb 12, 2014 at 13:48
• this function is unphysical
– Danu
Commented Feb 12, 2014 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 Commented Jun 6, 2014 at 18:39
• Related: physics.stackexchange.com/q/81041/2451 , physics.stackexchange.com/q/221027/2451 and links therein. Commented Feb 17, 2017 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!

• Okay. I was guessing that while specifying an operator we must also specify its domain. Is that right?
– SRS
Commented Feb 12, 2014 at 14:08
• Yes you must also specify the Hilbert space and the domain therein! Commented Feb 12, 2014 at 14:10
• 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
Commented Feb 12, 2014 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. Commented Feb 12, 2014 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 $f=e^{-x}$ .