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?

• Good choice of lecture series; if you want more from the Perimeter institute you'll find them at pirsa.org, they have hundreds there. Some have been imported to YouTube. – JamalS Jan 27 '15 at 16:09
• This seems really interesting! I found some lectures of his here: pirsa.org/index.php?p=speaker&name=Carl_Bender , there appears to be a whole talk on 'non-hermitian hamiltonians', which I'll check out now. It probably answers that question. – Spine Feast Jan 27 '15 at 16:16

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.

• How are non-self-adjoint Hamiltonians justified from a physical point of view? – Phoenix87 Jan 27 '15 at 19:42
• Note that $H$ is self-adjoint in an appropriate PT-symmetric sense. – Qmechanic Jan 27 '15 at 19:45
• Ok but my question still applies. Could you point me towards some literature on the subject? I have never seen the motivations for taking non-self-adjoint operators in place of self-adjoint ones I would be interested in knowing something about this. Thanks! :) – Phoenix87 Jan 27 '15 at 19:49
• There are some applications mentioned in the references. – Qmechanic Jan 27 '15 at 20:16

If you have a reflecting boundary condition at infinity ($\psi_{\infty}=0$), then the particle crosses the distance between this potential and infinity for a finite time (infinite velocity for this potential is reached quickly in a non relativistic case). A finite-time periodic motion is quantized in QM, like in Bohr-Sommerfeld quantization of a quasi-periodical motion.

• How can a wave function vanish if the kinetic energy tends to infinity for large $x$? – user17116 Jan 27 '15 at 19:10
• An infinite reflecting wall will suffice. In fact, the wave function amplitude is the largest where the particle makes its turn, where it is slow, i.e., when $E\approx -x^4$. – Vladimir Kalitvianski Jan 27 '15 at 20:24