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  Why does the Phi-cubed theory have no ground state?

+ 1 like - 0 dislike

In the book of Sredinicki's, he claimed that the $\phi^3$ theory has no ground state, hence this is not a physical theory.
My question is that I can't see why this system has no ground state. And I don't understand either the explaination he gave. For example, what does "roll down the hill" really mean? What's the case for a harmonic oscillator pertuibed by a "q^3" term? Maybe it's better if someone can explain it using the quantum harmonic case.


asked Feb 23, 2019 in Theoretical Physics by andy95220 (5 points) [ no revision ]

1 Answer

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Sredinicki means the exact $\phi^3$ theory. If you understand the exact solutions to the exact "harmonic oscillator stuffed with $x^3$ term", then you can get a gist of a field theory where the occupation numbers replace continuous $x$ of harminic oscillator problem.

If you do not understand the exact solutions to the exact "harmonic oscillator stuffed with $x^3$ term", then you may consider an infinite reflecting wall  $U(x)= 0,\; x<0,\; U(x)=+\infty,\; x\ge0$ as an example where the solutions are not localized. (There is a "ground state" with $E=0$, though.)

A "cubic oscillator" with big $g$ is similar to a slightly inclined wall (still reflecting), but with no "bottom" for negative-valued $x$, so there is no minimal $E_0$ for such a potential, no ground state, no localization.

answered Feb 24, 2019 by Vladimir Kalitvianski (102 points) [ revision history ]

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