Consider the quantum anharmonic oscillator, with Hamiltonian $H=p^2/2+q^2/2+gq^4$ for some real $g\geq 0$, with $p$ and $q$ obeying the usual Heisenberg commutation relations. For $g=0$, the ground energy is equal to $1/2$. Suppose $g>0$ and $g$ is algebraic. I would guess that the ground state energy is then not an algebraic number. Are there any results along these lines?
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There is no exact expression for the ground state energy $E_0$ for any nonzero $g$, but there are upper and lower bounds: for $g=1/2$ the upper bound for $2E_0$ is 1.3923516415302918570 and the lower bound for $2E_0$ is 1.3923516415302918502 , see Upper and lower bounds of the ground state energy of anharmonic oscillators using renormalized inner projection. There is no indication that $E_0$ can be expressed as the root of a polynomial, for all we know it's a transcendental number.
answered Jun 7, 2022 at 21:19