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  Spectrum of H = cos(x) + cos(p)

+ 2 like - 0 dislike

What is the spectrum for H = cos(x) + cos(p)? Here we choose x, p to have the unit of constant and make h (planck constant) a variable.

What I know now is that in classical physics, the phase space is periodic and is divided into multiple tilted square like regions.

One can also ask the question in classical physics, but I also don't have an answer for that. 

asked Nov 2, 2018 in Mathematics by anonymous [ no revision ]

As $\cos(x)\approx 1-x^2/2$ and $\cos(p)\approx 1-p^2/2$, your Hamiltonian is approximately $2-H_{\text{SHO}}$ with the energy spectrum $E_n$ containing the negative SHO spectrum $2-\left(E_n\right)_{\text{SHO}}$.

For fun, I propose you to consider $x$ and $p$ Grassman variables, so the spectrum is simple: $E=2$. Now you may quantize it, if you like :-)

@VladimirKalitvianski: Your derivation is not good, as the expansions are valid only for small numerical $x$ and $p$. But being operators with unbounded spectrum, the expansion is in fact invalid. This can also be seen from the fact that $H$ is bounded below by $-2$, whereas your spectral formula is unbounded below.

What is the reason you are interested in this Hamiltonian?

@ArnoldNeumaier: I know, you are right. The Hamiltonian is weird, so I underlined this fact. By the way, we must consider the total Hamiltonian rather than separate operators in it.

@VladimirKalitvianski: The Hamiltonian is self-adjoint and bounded below, so it is not weird in a mathematical sense. It may be physically weird but its spectrum is some well-defined subset of $[-2,2]$. 

@ArnoldNeumaier: I am not interested in further discussion.

1 Answer

+ 0 like - 0 dislike

You can certainly shift your phase-space origin into the potential well (by a Galilean boost and translation). Then you know that the energy eigenstates will share the symmetry of the Hamiltonian, so they will be periodic on the tiles.

Now make an approximation of small deviations $p$ and $x$ away from the new origin, and you find that it is approximately the harmonic oscillator. Just take the harmonic oscillator wave-functions, make them periodic and normalized on the tile, and these are your approximate bound states. For the unbound states things will be a bit tricky, but I assume you could use a WKB quantization of the classical solutions to get a basic idea (again, the discrete symmetry on the tiles will help a lot).

answered Nov 2, 2018 by Void (1,620 points) [ no revision ]


I see. Actually what I would prefer is the exact solution, and would like to see the effect on the system when I change the planck constant. (for example how many state we have for a specific h)

The SHO approximation only works when h << 1 and that is the regime of classical physics. I know that get the exact solution may be hard, so asymptopic behavior would also be great assuming that we are in the quantum regime.

You also mentioned the unbounded states, which I believe it doesn't exist for this system by looking at the phase diagram. In some sense, the space is glued between x, x+2pi and p, p+2pi. 

Thank you for your reply!

What I mean by "bound" are states that have $E\lesssim 0$, since then the classical trajectories would always stay confined to their individual valleys. (Perhaps a better term would be "classically non-overlapping".)

There is likely no known exact solution, so you will probably have to work with approximations and/or numerics. The bounds on the applicability of the WKB+harmonic approximation are the following: the phase-space volume of the tile is $V =4\pi^2 \approx 40$, and the approximations I talk about will apply when $h \ll V$. When $h \gtrsim V$, already the ground-state will have significant periodic overlaps with itself. I assume that then one will be able to expand the wave-function in a low-order Fourier expansion in $x,p$ because the details of the Hamiltonian will be sub-dominant compared to the periodicity forced by the wave-packet size.

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