Instanton configurations of an even anharmonic oscillator

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The discussion focuses on the Hamiltonian of an even anharmonic oscillator, defined as H_N(g) = - \frac{1}{2} ∂_q^2 + \frac{1}{2} q^2 + g q^N, where N is even. The use of the Euclidean path integral formalism is essential for determining the eigenenergies of this system, specifically through instanton configurations, which only exist for negative values of g. The scaling of q(t) with the prefactor (-g)^{-1/(N-2)} is necessary to properly define these configurations, as it relates to the double-well potential required for the existence of two vacua.

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Syrius
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Cheers everybody,

the Hamiltonian of an even anharmonic oscillator is defined as
[itex]H_N(g) = - \frac{1}{2} ∂_q^2 + \frac{1}{2} q^2 + g q^N[/itex] (N even).
In a paper (PRl 102, 011601) I found that to determine the eigenenergies of this system the Euclidean path integral formalism is used. They concluded that they have to use instanton configurations.

Since some googling of this term gave me only explanations in terms of Yang-Mills theories or string theory stuff and nothing else, my first question is: How are these instanton configurations defined for a given Hamiltonian.

From the Hamiltonian they further conclude that instanton configurations only exist for negative g. To find then the configurations they scale [itex]q(t) = (-g)^{-1/(N-2)} ζ(t).[/itex]. Maybe it will be obvious once I know the exact definition of instanton configuration, but why do they then scale q(t) with this prefactor?

Many thanks in advance for your answers!

Greetings, Syrius
 
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An instanton, in general, is a solution that interpolates between two vacua of the theory in such a way that it becomes topologically "trapped" at infinity.

You can realize an extremely simple example using a leather belt on a table. If you lay the belt on the table, it wants to lie flat, because of gravity. This is the vacuum configuration. But there are two distinct vacua: it can lie flat on its front, or its back.

Now take the belt and put a half-twist in it, and then hold the ends of it down on the table with your hands. At either end, the belt is in one of its vacuum configurations; lying on its front on one end, and on its back on the other. But somewhere in the middle, the belt will be sticking straight up, in a configuration with positive energy. This part in the middle cannot fall down, because the ends of the belt are trapped; the belt is kept in a positive-energy configuration by topology.

This "belt trick" is actually the sine-Gordon "kink soliton", more or less. The distinction between a soliton and an instanton is mostly semantics; an instanton is topologically trapped in the past and the future, so that the lump of positive energy occurs for an "instant" and then disappears.

As for why g has to be negative in your Hamiltonian, the reason is that there must be a double-well potential of some kind. So the anharmonic term must come with the opposite sign to the harmonic term. This way you will have two vacua to interpolate between.
 

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