# Does there exist a self-adjoint operator whose spectrum consists wholly of prime numbers?

The zeros of the canonical Riemann zeta function have been compared to the prime numbers, and they have a number of special, definite connections. The infamous zeros have also been conjectured to be the spectrum of some Hermitian operator given certain distributional similarities that have been evidenced with large computer calculations. So I muse, is there such an operator with prime numbers as its eigenvalues?

If $$(a_n)_n$$ is any sequence of real numbers, then the operator $$T:(x_n)_n\mapsto(a_nx_n)_n$$ is a Hermitian operator defined on a dense subspace of $$\ell^2$$. Each $$a_n$$ is an eigenvalue of $$T$$, and if $$(a_n)_n$$ is the sequence of primes, then the spectrum of $$T$$ is precisely the set of primes. More generally, the spectrum of $$T$$ is the closure of $$\{a_n\}_n$$.
• Don't we need $(a_n)_{n \in \mathbb N} \in \ell_{\infty}$ in order for $T$ to be well-defined or doesn't that matter as $T$ is only densely defined? – Viktor Glombik Feb 28 at 21:15
• @ViktorGlombik: That's why $T$ is only densely defined. It's an unbounded operator. – Jonas Meyer Mar 5 at 4:33