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How does Hamiltonian $p^2-x^4$ have a discrete spectrum?

+ 4 like - 0 dislike
1048 views

In Carl Bender's lecture, at around 21:30 onward of the video, he claimed that the Hamiltonian of a repulsive potential $p^2-x^4$ has a discrete spectrum. This is completely counter-intuitive since usually discrete energy levels are associated with bound states, and I did not expect such system to have any bound state. 

The trick that Bender briefly showed is, consider a parametrized family of Hamiltonians 

$$H(x,p,\epsilon)=p^2+x^2(ix)^\epsilon,$$

then this smoothly connects the hamonic oscillator potential to $-x^4$ potential as $\epsilon$ goes from 0 to 2, hence the spectrum of the former smoothly deforms to the latter, rendering the latter also discrete. Frankly speaking this seems like a very dirty trick to me. Could someone explain to me the underlying mathematics? 

asked Apr 20, 2014 in Theoretical Physics by Jia Yiyang (2,465 points) [ no revision ]

2 Answers

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The trick is called PT-symmetric quantum mechanics. http://ptsymmetry.net/?page_id=10

There the inner product is not, as usual, $\psi^*\psi$ but $\psi^*M\psi$ with a suitable positive definite operator $M$, constructed from  nonhermitian Hamiltonian $H$ in such a way that the latter becomes self-adjoint and bounded below in the Hilbert space with the modified inner product. The construction often works when $H$ is PT-symmetric.

Apart from that, it is ordinary quantum mechanics. Indeed, with the tranformation $\Psi=M^{1/2}\psi$ that restores the standard inner product, one gets the transformed Hamiltonian $M^{1/2}HM^{-1/2}$ which is (in the cases of interest) self-adjoint and bounded below in the ordinary way.

answered Apr 20, 2014 by Arnold Neumaier (12,425 points) [ revision history ]

But in the special cases that you restore a real potential, you should restore the ordinary interpretation, the intermediate stages in the interpolation are the ones with a naively non-unitary Hamiltonian. At the end, you get a normal interpretation, except for a different set of allowed wavefunctions (not the usual ones called square-normalizable, but new ones relative to the ground state) plus some recentering of the ground state energy. You should be able to get Bender's spectrum without the full continuation, doing the limit by hand, but the method I tried just now doesn't work (but something works).

It was some years ago that I had looked at the details. It seems to me that the procedure for constructing $M$ wasn't necessarily continuous when the limit of a real potential was taken. 

There is an intriguing link to the Landau pole problem in QFT, as the Hamiltonian of the Lee model, which develops Landau ghosts when a parameter crossing some threshold, becomes complex beyond the threshold, but can still be treated by PT-symmetric methods. Cf. pp.3-5 in http://arxiv.org/pdf/hep-th/0408028.pdf

Possibly something similar happens in QED at the Landau pole. 

+1 for the useful pointers, which lead me to this paper, I'll try to read it when I get some free time.

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The discrete spectrum is because the potential drops fast enough that you classically reach infinity in finite time. The system is bound around infinity, rather than bound at zero. The precise boundary conditions at infinity are defined by the analytic continuation, and make the spectrum well defined.

The finite energy motion has a velocity which is proportional to $\sqrt{E-V}$ ~ $x^2$, and the time it takes to classically go from 0 to infinity is $\int {1\over x^2} dx$ which is convergent. The integral for J is divergent:

$$ J = 2\int_{-\infty}^\infty p dx =2 \int \sqrt{2E+x^4} dx $$

This is infinite, which means that doing the phase-matching to define the bound states for finite energy is ambiguous. This ambiguity is resolved by finding the ground state, setting its energy to zero, and talking about J relative to the ground state. The derivative of J with respect to E, the inverse level spacing, is the classical period, and is finite.

$$ {\partial J \over \partial E} = 2\int {1\over \sqrt{2E+x^4}} dx = T_\mathrm{cl} $$

The Bender prescription picks you exactly which bound wavefunctions to use for this problem, by linking it with the Harmonic oscillator by this particular continuation. The prescription near $V=-x^4$ is $V = -x^4  + x^2(1 + i\mathrm{sgn}(x) \epsilon log |x|)$, which is a particular kind of circularizing boundary condition, where wavefunction is removed at one end (positive imaginary potential) and introduced at the other end of space (negative imaginary potential). Although this is locally non-unitary, you can make it unitary by reinterpreting the Hilbert space inner product.

The same sort of thing happens when you start Bender's deformation at V= x^2(ix)^{i\epsilon), you get a negative and postive imaginary contribution at the two ends of space, and the space of allowed wavefunctions in the Hilbert space deforms. But the imaginary part on the two ends is opposite in sign near $x^2$ and near $x^4$, the two ends of the circuit, so that the situation has smoothly moved the bulk of the Hilbert space wavefunction mass to lie at infinity rather than at zero, in this particular way.

If you ignore Bender, the standard answer for this potential is not really unbound, it is ambiguous, depending on boundary conditions at infinity. If you take a symmetric interval [-L,L] with L large, impose the potential plus a constant which normalizes the ground state energy to be zero, rather than negative infinity. With circular boundary conditions, then you get some silly spectrum which is not Bender's spectrum, and which depends on L strongly.

Benders spectrum comes from this particular analytic continuation, which defines its own Hilbert space. The space of allowed wavefunctions and operators, while still representable as functions on R, are not the usual square-integrable functions, but are a certain collection which are defined relative to the naive non-normalizable ground state of the potential. But the reason these potentials give a discrete spectrum is simply that you oscillate around and come back classically in finite time, the Bender prescription is simply one path which defines precisely how to link up negative infinity and positive infinity. There are other prescriptions which give different answers, as usual for analytic continuation.

answered Apr 20, 2014 by Ron Maimon (7,535 points) [ revision history ]
edited Apr 20, 2014 by Ron Maimon

Thanks Ron, I enjoy reading your answer as always, and don't fully understand it as always. Modulo the technical parts of your answer(I might come back to the technicalities if I get some time finish reading Bender's paper), I'm intrigued by your intuition when you stated "The discrete spectrum is because the potential drops fast enough that you classically reach infinity in finite time. The system is bound around infinity, rather than bound at zero.", I suppose you are saying we can consider the system to be bounded at infinity because infinity can be reached in finite time. This argument makes sense to me on a intuitive level, but also is blur enough for me to cast some doubts. So is there a (slightly?) more rigorous justification for this intuition?  

No, please make one! It is just what I remember working out for myself when I heard Bender speak. I haven't thought about it in years, and working out the precise Hilbert space in the intermediate stages is what Bender did. The problem is that the inner product is not local in the intermediate stages of the continuation, and at the end, when you get to the real inverted potential, it picks out some strange set of wavefunctions. I never found what these are.

The intution, though, is relatively easy to make precise--- the finiteness of the level-spacing integral should be required to have a finite level spacing in a continued unbound potential, simply because when the potential ends up real, the level spacing ends up the WKB level spacing. This is not a theorem, because I don't know the precise WKB approximation for Bender's Hilbert space, just because I didn't work it out, honestly. I don't think this has been said anywhere else, so it requires some work to see if it is even accurate. Let me know if there is a mistake in the reasoning.

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