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?
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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!
Valter Moretti
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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 $f=e^{-x}$ .
Keith McClary
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