Spectral Pairing, Ground States, and Supersymmetry Breaking

Every isolated positive-energy eigenspace of a supersymmetric Hamiltonian has equal even and odd multiplicity because the normalized maps $A/\sqrt{2E}$ and $A^\dagger/\sqrt{2E}$ are inverse isometries there. Zero energy is exceptional: its even states lie in $\ker A$, its odd states in $\ker A^\dagger$, and normalizability plus the operator domain decide whether either kernel contains a physical state. Supersymmetry is unbroken when at least one normalized zero-energy state exists; a vanishing difference of the two kernel dimensions does not decide the question.

Required background. Supercharges, partner Hamiltonians, and positive energy fixes the grading, normalization, and domains used below. Helpful background. Vacua, states, and representations distinguishes a normalizable vacuum vector from a formal wavefunction or a spectral threshold.

The positive-energy pairing theorem

Let $A:\operatorname{Dom}A\subset\mathcal H_{\bar0}\to\mathcal H_{\bar1}$ be densely defined and closed, and let

$$H_{\bar0}=\frac12A^\dagger A, \qquad H_{\bar1}=\frac12AA^\dagger.$$

Suppose $E>0$ is an eigenvalue and $\psi\in\operatorname{Dom}H_{\bar0}$ obeys $H_{\bar0}\psi=E\psi$. Then $A\psi\ne0$, because otherwise $E\lVert\psi\rVert^2=\langle\psi|H_{\bar0}|\psi\rangle=0$. Moreover,

$$H_{\bar1}(A\psi) =\frac12AA^\dagger A\psi =A(H_{\bar0}\psi) =E(A\psi).$$

This equation is legitimate on the stated domains: $\psi\in\operatorname{Dom}(A^\dagger A)$ implies $A\psi\in\operatorname{Dom}A^\dagger$, and $A^\dagger A\psi=2E\psi\in\operatorname{Dom}A$ implies $A\psi\in\operatorname{Dom}(AA^\dagger)$. Its norm is

$$\lVert A\psi\rVert^2 =\langle\psi|A^\dagger A|\psi\rangle =2E\lVert\psi\rVert^2.$$

Consequently

$$U_E=\frac{A}{\sqrt{2E}}: \ker(H_{\bar0}-E)\longrightarrow\ker(H_{\bar1}-E)$$

is an isometry. The inverse is $A^\dagger/\sqrt{2E}$, because $A^\dagger A=2E$ and $AA^\dagger=2E$ on the respective eigenspaces. Hence the two eigenspaces have the same dimension, including degeneracy. This is the domain-complete version of the familiar partner-potential argument Cooper, Khare, and Sukhatme 1995, §2, pp. 13–16 in the open version.

The spectral-pairing and index-flow figure tracks this isometry to its graded-trace consequence and marks the point where a continuum regulator must replace eigenstate counting.

For continuous spectrum, generalized eigenfunctions can obey analogous intertwining relations, but they are not Hilbert-space vectors. The correct global statement uses the polar decomposition $A=U(A^\dagger A)^{1/2}$: away from zero, the partial isometry $U$ identifies the positive spectral subspaces. Threshold states, scattering normalization, and trace regularization require the separate analysis on the continuum and boundary page.

Zero energy and the supercharge kernels

At zero energy the normalization $1/\sqrt{2E}$ is unavailable. Positivity instead gives

$$\begin{aligned} \ker H_{\bar0}&=\ker A,\\ \ker H_{\bar1}&=\ker A^\dagger,\\ \ker H&=\ker\mathcal Q\cap\ker\mathcal Q^\dagger. \end{aligned}$$

Indeed, $\langle\psi|H_{\bar0}|\psi\rangle=\lVert A\psi\rVert^2/2$, and similarly in the odd sector. A normalized vector in either kernel is annihilated by both Hermitian supercharges and is therefore a supersymmetric ground state.

Three situations must be kept separate:

Spectrum near zero	Normalized zero mode?	Conclusion
Discrete, with $E_0=0$	Yes	Supersymmetry is unbroken.
Discrete, with $E_0>0$	No	Supersymmetry is spontaneously broken; the ground states occur in positive-energy pairs.
$\inf\sigma(H)=0$ but zero is only continuous spectrum	No Hilbert-space vector	There is no normalizable supersymmetric vacuum. An infrared regulator and a specified representation are needed before importing finite-volume breaking language.

The last line is why “the energy can approach zero” is not equivalent to “there is a zero-energy state.”

Solving the first-order equations

For the one-dimensional realization

$$A=\frac{\mathrm d}{\mathrm dx}+w(x), \qquad A^\dagger=-\frac{\mathrm d}{\mathrm dx}+w(x),$$

let $W'(x)=w(x)$. The zero-mode equations integrate exactly:

$$\begin{aligned} A\psi_{\bar0}=0 &\quad\Longrightarrow\quad \psi_{\bar0}(x)=C_{\bar0}e^{-W(x)},\\ A^\dagger\psi_{\bar1}=0 &\quad\Longrightarrow\quad \psi_{\bar1}(x)=C_{\bar1}e^{+W(x)}. \end{aligned}$$

These are candidate states. A physical zero mode must also belong to $L^2$, satisfy all boundary conditions, and lie in the chosen operator domain. On the full line:

• if $W(x)\to+\infty$ at both ends fast enough, the even candidate is normalizable and the odd one is not;
• if $W(x)\to-\infty$ at both ends fast enough, the odd candidate is normalizable and the even one is not;
• if $W$ has opposite signs at the two ends, neither exponential is normalizable.

“Fast enough” is essential. If $W\sim c\log|x|$, square-integrability depends on $c$, not only on the sign. Singular points and finite endpoints likewise require the actual self-adjoint boundary condition. The first-order test and its relation to broken and unbroken supersymmetry are developed explicitly in Cooper, Khare, and Sukhatme 1995, §§2.1–2.2, pp. 18–25 in the open version.

Two contrasting superpotentials

Linear coefficient: one unpaired vacuum

For $w(x)=\omega x$, $W(x)=\omega x^2/2$ tends to $+\infty$ at both ends. The normalized even zero mode is

$$\psi_{\bar0,0}(x) =\left(\frac{\omega}{\pi}\right)^{1/4} e^{-\omega x^2/2},$$

while $e^{+\omega x^2/2}$ is not square-integrable. The positive energies are the paired oscillator levels $E=n\omega$, $n=1,2,\ldots$, and supersymmetry is unbroken.

Even coefficient: no zero mode

Now take

$$w(x)=x^2-a^2, \qquad W(x)=\frac{x^3}{3}-a^2x, \qquad a>0.$$

As $x\to+\infty$, $W\to+\infty$; as $x\to-\infty$, $W\to-\infty$. Therefore $e^{-W}$ diverges at the left end and $e^{+W}$ diverges at the right end. Neither is a state. Yet the partner potentials grow as $x^4/2$, so the spectrum is discrete and has a lowest eigenvalue. Positivity and absence of a zero mode force $E_0>0$: supersymmetry is broken, and even the lowest level is paired. This normalizability criterion is the one-dimensional mechanism used in Witten 1981, §2, pp. 515–523.

The two zeros of $w$ produce semiclassical wells, but local wells do not determine the exact ground-state energy. Tunneling couples their approximate states and removes the would-be zero-energy degeneracy. The Morse and tunneling page explains the geometric version of this mechanism.

What the pairing theorem does not say

It does not say that the total dimensions of $\mathcal H_{\bar0}$ and $\mathcal H_{\bar1}$ are equal. It equates only positive-energy spectral multiplicities under the domain hypotheses.

It does not say that every formal solution of $A\psi=0$ is a vacuum. Normalizability, endpoints, singularities, and self-adjoint extension data are decisive. In singular half-line models, changing the extension can preserve the full two-supercharge algebra, reduce it, or break it Falomir and Pisani 2005, abstract and §§3–4.

It also does not say that zero Witten index means broken supersymmetry. The index is a signed difference of even and odd zero modes; both can be nonzero and cancel. That inference is treated carefully on the Witten-index page.

Exercises

Sector exchange. Classify the supersymmetric ground states for $w(x)=-\omega x$ on $\mathbb R$, with $\omega>0$.

Solution

Here $W(x)=-\omega x^2/2$. The even candidate $e^{-W}=e^{+\omega x^2/2}$ is not normalizable, while the odd candidate $e^{+W}=e^{-\omega x^2/2}$ is. There is one odd supersymmetric vacuum. This is the sector exchange $w\mapsto-w$ of the ordinary supersymmetric oscillator; the positive spectrum is unchanged and the signed index changes from $+1$ to $-1$.

The logarithmic borderline. Let

$$W(x)=\frac c2\log(1+x^2), \qquad w(x)=W'(x)=\frac{cx}{1+x^2}.$$

For which real values of $c$ is the even candidate $e^{-W}$ normalizable? For which values is the odd candidate $e^{W}$ normalizable? Include the borderline values.

Solution

Near the origin both candidates are bounded, so only the two asymptotic ends matter. Their squared magnitudes behave as

$$|e^{-W}|^2=(1+x^2)^{-c}\sim |x|^{-2c}, \qquad |e^{W}|^2=(1+x^2)^c\sim |x|^{2c}.$$

The integral $\int^\infty x^{-2c}\,dx$ converges precisely when $2c>1$, so the even candidate is normalizable for $c>1/2$. The odd candidate is normalizable precisely when $2c<-1$, or $c<-1/2$. At $c=\pm1/2$ the relevant integral diverges logarithmically. Hence neither candidate is normalizable for $|c|\leq1/2$.

Isolated zero mode or continuum threshold? Compare three full-line systems: $w=\omega x$ with $\omega>0$; the logarithmic model above with $c>1/2$; and $w=0$. In each case decide whether zero is an isolated eigenvalue, a normalizable eigenvalue at the continuum threshold, or merely the bottom of continuous spectrum with no normalizable state.

Solution

For $w=\omega x$, the even Gaussian is a normalizable zero mode and the next energy is $\omega$, so zero is isolated by a gap. In the logarithmic model with $c>1/2$, $(1+x^2)^{-c/2}$ is a normalizable even zero mode. However, $w(x)\to0$ and both partner potentials tend to zero, so the essential spectrum begins at zero: the eigenvalue sits at threshold and is not isolated. For $w=0$, the formal zero modes are constants, which are not in $L^2(\mathbb R)$; zero is only the continuum threshold. Thus a zero-energy solution, a normalizable zero mode, and an isolated supersymmetric vacuum are three different statements.

References

• Cooper, Fred, Avinash Khare, and Uday Sukhatme. “Supersymmetry and Quantum Mechanics.” Physics Reports 251, nos. 5–6 (1995): 267–385. doi:10.1016/0370-1573(94)00080-M. Open arXiv version.
• Falomir, H., and P. A. G. Pisani. “Self-Adjoint Extensions and SUSY Breaking in Supersymmetric Quantum Mechanics.” Journal of Physics A: Mathematical and General 38, no. 21 (2005): 4665–4683. doi:10.1088/0305-4470/38/21/011. Open arXiv version.
• Witten, Edward. “Dynamical Breaking of Supersymmetry.” Nuclear Physics B 188, no. 3 (1981): 513–554. doi:10.1016/0550-3213(81)90006-7.