quantum mechanics - How can an inverted anharmonic potential $V(x)=-x^4$ have discrete bound states?

I've been watching the lectures on mathematical physics by Carl Bender on youtube where he uses the non-Hermitian Hamiltonian methods to prove that the inverted anharmonic potential $V(x)=-x^4$ has a discrete bound states with positive energy. How can it be?

Answer

More generally, Carl Bender et al. are considering $PT$-symmetric Hamiltonians of the form

$$H~=~ p^2 + x^2 (ix)^{\varepsilon}, \qquad \varepsilon\in\mathbb{R} ,$$

cf. e.g. Refs. 1-3. The Hamiltonian $H$ is not self-adjoint in the usual sense, but self-adjoint in a $PT$-symmetric sense. OP's case corresponds to $\varepsilon=2$. The trick is to analytically continue the wave function $\psi$ with real 1D position $x\in\mathbb{R}$ into the complex position plane $x\in\mathbb{C}$, and prescribe appropriate boundary behaviour in the complex position plane.

See e.g. Refs. 1-3 and references therein for further details and applications. Note that Refs. 1-3 mainly discuss the point spectrum of the operator $H$.

References:

1. 
   

   C.M. Bender, D.C. Brody, and H.F. Jones, Must a Hamiltonian be Hermitian?, arXiv:hep-th/0303005.

   
   


2. 
   

   C.M. Bender, Introduction to $PT$-Symmetric Quantum Theory, arXiv:quant-ph/0501052.

   
   


3. 
   

   C.M. Bender, D.W. Hook, and S.P. Klevansky, Negative-energy $PT$-symmetric Hamiltonians, arXiv:1203.6590.