I am puzzled by a statement that is made regarding non-perturbative corrections breaking supersymmetry.

Specifically, consider a one-dimensional system, described by the Hamiltonian

$$H = \frac{1}{2}(P^2 + (W')^2) - \frac{\hbar}{2}W''(x)\sigma_3$$

where $\sigma_3 = \left(\begin{array}{cc}1 & 0\\0 & -1\end{array}\right)$ is the usual Pauli matrix, primes denote differentiation with respect to $x$ (so $W'(x) = \partial_x W(x), W''(x) = \partial_x^2 W(x)$, and $P = -i\hbar\partial_x$ is the momentum operator.

For this Hamiltonian,

$$Q^\dagger = \left(\begin{array}{cc}0 & 1\\0 & 0\end{array}\right)(P + iW')$$ $$Q = \left(\begin{array}{cc}0 & 0\\1 & 0\end{array}\right)(P - iW')$$

so that $H = \frac{1}{2}\{Q,Q^\dagger\}$.

So the vacuum solution, annihilated (by definition) by $Q$ and $Q^\dagger$ can have either one of the following forms:

$$\psi = \left(\begin{array}{c}\psi_+\\0\end{array}\right) \text{ and } (P-iW')\psi_+ = 0 \implies \psi_+ \propto e^{-W(x)/\hbar}$$ or $$\psi = \left(\begin{array}{c}0\\\psi_-\end{array}\right) \text{ and } (P+iW')\psi_- = 0 \implies \psi_- \propto e^{W(x)/\hbar}$$

(The solution with nonzero $\psi_+$ and $\psi_-$ is rejected on grounds of normalizability.)

It is clear that if $W(\pm \infty) = \pm \infty$, then a ground state wavefunction exists (either one of the above two). Also this means $W'(x)$ must vanish somewhere, and so the "potential" term in the Hamiltonian has a minimum value of zero $\implies$ SUSY is unbroken.

So far so good.

But lets consider the possibility $W(\infty) = -W(-\infty)$. In this case, it is said (see http://www.christiansaemann.de/files/LecturesOnSUSY.pdf) that

(A) there can be only an even number of points where $W'(x) = 0$ (counted with multiplicities).

(B) the vacuum is a superposition of wavefunctions localized at the points $x \in \{x_0: W'(x_0) = 0\}$.

This makes sense. But it is also stated that nonperturbative corrections will lift W'(x) to make it strictly positive, thus breaking supersymmetry.

How is it obvious that nonperturbative corrections will necessarily lift W'(x) to make it strictly positive?

  • 1
    $\begingroup$ I think it's important to notice that in this simple case you have the exact form for possible solutions. And it should be pretty obvious that neither of them is normalizable in the second case without appeal to any perturbation theory. Of course in general case it's impossible to get solution exactly. The discussion of (B) ignores the exact solution and tries to get the same result from perturbation theory perspective. $\endgroup$
    – OON
    Mar 22, 2016 at 2:16

1 Answer 1


I believe the answer stems from the arguments in Witten's paper "Constraints on Supersymmetry Breaking" (pgs. 259-260) where he discusses the possibility that $Tr(-1)^{F}$ is affected by either large field strengths (which may cause a coupling not originally present to become significant) or setting a parameter to zero if it changes the asymptotic behavior of the energy in field space (under a continuous deformation of such a parameter, the Witten index would be invariant, as long as the parameter stays nonzero).

So in conclusion, such corrections (which can be mimicked by comparing two cases $g = 0$ and $g \neq 0$ for the potential term $V(\phi) = (m\phi-g\phi^2)^2$ can indeed lift the vacuum energy from its zero value, and this would naturally break supersymmetry.


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