• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Interpreting Argyres' spectrum of spontaneously broken SUSY QM

+ 5 like - 0 dislike

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


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

+ 1 like - 0 dislike

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 (1,130 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights