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Interpreting Argyres' spectrum of spontaneously broken SUSY QM

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I can't understand the spectrum in the figure on page 19 from Argyres' lecture notes on supersymmetry: http://www.physics.uc.edu/~argyres/661/susy1996.pdf

AryresFigure

Argyres is considering a supersymmetric quantum mechanical system of an anharmonic oscillator, with a superpotential $W\sim x^3$. The plots of $W$ and $V$ make perfect sense. What doesn't make sense is the spectrum on the right.

Why are there both x's and o's over each Hamiltonian $H_1$ and $H_2$?? I thought $H_1$ is exclusively the spin-up Hamiltonian and $H_2$ is exclusively the spin-down Hamiltonian, so therefore the spectrum consist of a column of just x's over $H_1$ and a column just of o's over $H_2$.

Additional request: Would someone please write down the form of $H_1$ and $H_2$ in their answer, so to make sure we're on the same page? A graph of the the respective potentials $V_1$ and $V_2$ would be even better.

Take a look at the much more sensible figure on page 7. This is something that I can comprehend.

enter image description here

This post imported from StackExchange Physics at 2014-05-04 11:36 (UCT), posted by SE-user QuantumDot
asked Apr 26, 2014 in Theoretical Physics by QuantumDot (195 points) [ no revision ]

1 Answer

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The plots you compare are actually not the same. The one on page 7 shows what the spectrum looks like if the ground state is chosen to be spin-up. The spectrum is degenerate and the theory is supersymmetric as we can show the vacuum to correspond to a zero-energy state. In principle, one can also choose it to be the other way around and arrive at the same spectrum. The vacuum is still at zero energy and the spectrum should be degenerate.

It turns out that this is only true classically. Non-perturbative tunneling-effects (instantons) destroy this equivalence: the spectrum in which the vacuum is a spin-down state is no longer degenerate with the spin-up vacuum. This "lifting" of the energy levels is shown in the figure on page 19. $H_1$ and $H_2$ correspond to the aforementioned classically degenerate, but quantum mechanically non-degenerate ground states.

This post imported from StackExchange Physics at 2014-05-04 11:36 (UCT), posted by SE-user Frederic Brünner
answered May 2, 2014 by Frederic Brünner (1,060 points) [ no revision ]

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