Continuum Spectra, Boundaries, Wall Crossing, and Ground-State Bundles

The algebra of supersymmetric quantum mechanics pairs positive energies, but the global answer also depends on what happens at infinity, at physical boundaries, and as external parameters vary. Continuous spectra can leave a relative spectral-density term; boundary conditions determine the operator domain and hence the cohomology; and a gapped family of vacua forms a vector bundle with Berry–Wilczek–Zee transport. These effects are related, but they must not be conflated.

Required background. The Witten index supplies the protected signed count, and the Hilbert-complex formulation identifies vacua with harmonic representatives. Helpful background. Vector bundles and connections are used for parameter-space transport, while self-adjoint extensions explain why a differential expression alone does not determine a boundary problem.

The continuum contribution is scattering data

Four cases organize the page. The distinction is the hypothesis that fails or the additional datum that must be supplied:

Case	Decisive datum	Valid object	Main warning
Discrete, trace-class spectrum	Common supercharge domains and a convergent heat trace	Integer graded trace equal to the zero-mode difference	A zero value may be cancellation rather than absence.
Open scattering continuum	Matched comparison operator, phase convention, threshold prescription, and order of limits	Relative heat supertrace, possibly $\beta$ dependent	Delta-normalized pairing does not imply pointwise density cancellation.
Physical spatial boundary	A compatible closed domain for $d$ and $d^\dagger$	Absolute, relative, or another explicitly defined complex	Changing the boundary condition changes the operator and can change its index.
Gapped parameter family	Constant kernel rank, smooth projector, and a uniform spectral gap	Ground-state vector bundle with Berry transport	A closing gap destroys the fixed-rank bundle even when the signed index stays fixed.

The spectral-pairing and index-flow figure shows where the first row hands off to the continuum data in the second.

Assume that the even and odd Hamiltonians are self-adjoint scattering operators with the same positive continuum threshold. Neither $e^{-\beta H_{\bar0}}$ nor $e^{-\beta H_{\bar1}}$ need be trace class on an infinite-volume Hilbert space. A relative regulator may nevertheless define

$$\begin{aligned} I_{\mathrm{rel}}(\beta) ={}&n_{\bar0}^{(0)}-n_{\bar1}^{(0)} +\sum_{E_j>0}^{\mathrm{bound}} e^{-\beta E_j} \bigl(m_{\bar0}(E_j)-m_{\bar1}(E_j)\bigr)\\ &+\int_{E_{\mathrm{th}}}^{\infty} e^{-\beta E} \bigl[\rho_{\bar0}(E)-\rho_{\bar1}(E)\bigr]\,dE . \end{aligned}$$

The positive-energy bound multiplicities cancel by the ordinary pairing theorem. The continuum term need not vanish pointwise, because delta-normalized partners can differ by their asymptotic normalization or threshold behavior.

To see where the density difference comes from, put a one-channel problem in a box of length $L$ with the same distant-wall convention in both sectors. If

$$kL+\delta_{\bar a}(k)=\pi n, \qquad \bar a\in\{\bar0,\bar1\},$$

then, away from thresholds,

$$\rho_{\bar0}(k)-\rho_{\bar1}(k) =\frac1\pi\frac{d}{dk} \bigl[\delta_{\bar0}(k)-\delta_{\bar1}(k)\bigr].$$

The extensive $L/\pi$ pieces cancel; the surviving term is encoded by the relative scattering phase. With several open channels, a basis-independent version is

$$\rho_{\bar0}(E)-\rho_{\bar1}(E) =\frac{1}{2\pi i}\frac{d}{dE} \log\det\!\left(S_{\bar0}(E)S_{\bar1}(E)^\dagger\right),$$

for a fixed continuous branch of the logarithm and matched channel normalizations. Bound-state poles, half-bound states, and a channel opening at threshold must be added separately. A different distant-wall regulator can move a finite term between the density integral and the threshold contribution, so a reported continuum index is incomplete unless its prescription and order of limits are stated. Exactly soluble examples and their anomalous $\beta$ dependence are analyzed in Akhoury and Comtet 1984, pp. 253–278; the relation to open-space index theorems is developed in Niemi and Semenoff 1986, §§2–4, pp. 136–155.

Exact continuum benchmark: a tanh superpotential

Take

$$\mathcal H=L^2(\mathbb R)_{\bar0}\oplus L^2(\mathbb R)_{\bar1}, \qquad A=\frac{d}{dx}+m\tanh(mx), \qquad m>0,$$

with $\operatorname{Dom}A=H^1(\mathbb R)$ and the Hilbert-space adjoint on the same Sobolev domain. Since the coefficient is smooth and bounded, the partner Hamiltonians have their standard $H^2(\mathbb R)$ realizations:

$$\begin{aligned} H_{\bar0} &=\frac12\left[-\frac{d^2}{dx^2}+m^2 -2m^2\operatorname{sech}^2(mx)\right],\\ H_{\bar1} &=\frac12\left[-\frac{d^2}{dx^2}+m^2\right]. \end{aligned}$$

The even sector has the normalized zero mode

$$\psi_{\bar0,0}(x)=\sqrt{\frac m2}\operatorname{sech}(mx), \qquad A\psi_{\bar0,0}=0,$$

and the odd sector has none. Both continua begin at $E_{\mathrm{th}}=m^2/2$, with $E(k)=(k^2+m^2)/2$. At $k=0$, the Pöschl–Teller channel has the bounded but non-normalizable solution $\tanh(mx)$, while the free comparison channel has the constant threshold solution. We treat these as half-bound threshold solutions, not as $L^2$ bound states, and fix their relative contribution by the continuous scattering-phase convention below.

Use the same symmetric box and the same distant-wall condition in both sectors. Form the even–odd difference first, take the box size to infinity at fixed $\beta$, and only afterward take a $\beta$ limit. Normalize the phase so the relative scattering matrix tends to the identity as $k\to\infty$. The reflectionless transmission amplitude and the two-channel relative determinant are

$$t(k)=\frac{k+im}{k-im}, \qquad \det\!\left(S_{\bar0}S_{\bar1}^\dagger\right)=t(k)^2.$$

Consequently the full-line density difference in the $k$ variable is

$$\rho_{\bar0}(k)-\rho_{\bar1}(k) =\frac{1}{2\pi i}\frac{d}{dk}\log t(k)^2 =-\frac{2m}{\pi(k^2+m^2)}.$$

The bound zero mode contributes one, while the continuum subtracts a $\beta$-dependent amount:

$$\begin{aligned} I_{\mathrm{rel}}(\beta) &=1-\frac{2m}{\pi}\int_0^\infty \frac{e^{-\beta(k^2+m^2)/2}}{k^2+m^2}\,dk\\ &=1-\operatorname{erfc}\!\left(m\sqrt{\frac\beta2}\right) =\operatorname{erf}\!\left(m\sqrt{\frac\beta2}\right). \end{aligned}$$

Thus $I_{\mathrm{rel}}(\beta)\to1=\operatorname{ind}A$ as $\beta\to\infty$, but it tends to zero as $\beta\to0$. The analytic Fredholm index and this matched relative heat trace agree in the infrared; the latter is not $\beta$ independent because the unbounded supercharges cannot be moved cyclically through the regulated open-space trace. This is the concrete continuum mechanism analyzed in Akhoury and Comtet 1984, pp. 253–278.

Fredholm asymptotics on the line

For

$$A=\frac{d}{dx}+w(x), \qquad w(x)\longrightarrow w_\pm \quad\text{as}\quad x\longrightarrow\pm\infty,$$

suppose $w_\pm$ are nonzero and the approach is sufficiently regular that the essential spectrum is controlled by the limiting constant-coefficient operators. Then the partner Hamiltonians have a positive essential threshold

$$E_{\mathrm{ess}} =\frac12\min\{w_-^2,w_+^2\}>0,$$

and $A$ is Fredholm. The zero-mode test gives

$$\boxed{ \operatorname{ind}A =\frac12\bigl(\operatorname{sgn}w_+ -\operatorname{sgn}w_-\bigr). }$$

Indeed, $\psi_{\bar0}\propto \exp[-\int^x w(y)\,dy]$ is square-integrable at both ends precisely when $w_-<0<w_+$, whereas $\psi_{\bar1}\propto\exp[+\int^x w(y)\,dy]$ is square-integrable precisely when $w_+<0<w_-$. If $w_+$ or $w_-$ crosses zero, the essential gap closes and the Fredholm hypothesis fails exactly where the index can change. This one-dimensional result is the elementary analogue of the mass-at-infinity condition in Callias 1978, Theorems 1–2, pp. 216–224.

This example separates two statements:

• a normalizable state can merge into the continuum when the asymptotic gap closes; and
• while the family remains Fredholm, its signed index cannot jump.

The first is a mechanism for wall crossing. The second is the reason a genuine index is protected.

Boundary conditions define the supersymmetric complex

On a manifold with boundary, writing $A=d$ is not enough. One must choose domains for $d$ and $d^\dagger$ so that integration-by-parts boundary terms vanish and the supercharges map their domains into the appropriate partner domains. The interval $I=[0,L]$ makes this visible without geometric overhead.

For the de Rham complex

$$0\longrightarrow\Omega^0(I) \xrightarrow{\,d\,}\Omega^1(I) \longrightarrow0,$$

the two canonical elliptic choices give different cohomologies:

Boundary problem	Zero-form condition	One-form condition for $g(x)\,dx$	Harmonic representatives	Index
Absolute	$f'(0)=f'(L)=0$	$g(0)=g(L)=0$	Constant zero-form	$+1$
Relative	$f(0)=f(L)=0$	$g'(0)=g'(L)=0$	Constant one-form $dx$	$-1$

Absolute conditions compute $H^\bullet(I)$, so their index is $\chi(I)=1$. Relative conditions compute $H^\bullet(I,\partial I)$, whose only nonzero group is $H^1(I,\partial I)\cong\mathbb C$, so the index is $\chi(I,\partial I)=-1$. Nothing discontinuous has happened to a fixed operator: these are different closed complexes with different domains.

More general Robin or point-interaction conditions require matching partner domains. A self-adjoint Hamiltonian in one sector does not by itself guarantee that the supersymmetric descendant is self-adjoint. Explicit half-line, interval, and punctured-line classifications appear in Al-Hashimi, Salman, Shalaby, and Wiese 2013, §§2–4, pp. 3–17.

Wall crossing versus pair creation at zero

For a continuous Fredholm family $A_s$, the integer

$$\operatorname{ind}A_s =\dim\ker A_s-\dim\ker A_s^\dagger$$

is locally constant. The individual kernel dimensions need not be. An even and an odd positive-energy state may reach zero together, increasing both kernel dimensions by one and leaving the index unchanged. Conversely, a jump of the index signals that the path left the Fredholm family—for example through a continuum threshold, a changed asymptotic mass, or a changed operator domain.

It is therefore useful to distinguish:

Quantity	Behavior in a continuous Fredholm family	Additional condition
Signed index	Locally constant	A jump means Fredholm control or the problem itself changed
Total vacuum degeneracy	May jump through even–odd pairs	The gap above zero closes at the crossing
Vacuum-bundle holonomy	May vary continuously	It is defined on a constant-rank gapped region

The phrase wall crossing should state which row is meant. A jump in a degeneracy is not automatically a jump in the index.

Ground states form a bundle only while rank and gap persist

Let $\lambda$ vary in a parameter manifold $\mathcal M$, and suppose $H(\lambda)$ is a smooth self-adjoint family. On an open set $U\subset\mathcal M$, assume:

1. zero is an isolated eigenvalue of constant finite multiplicity $r$;
2. a uniform gap $\Delta>0$ separates it from the rest of the spectrum; and
3. the spectral projector $P(\lambda)$ depends smoothly on $\lambda$.

Then $\operatorname{im}P\to U$ is a rank-$r$ Hermitian vector bundle. For a local orthonormal frame $\{\lvert a(\lambda)\rangle\}_{a=1}^r$, define

$$\mathcal A_i^{ab} =i\langle a(\lambda)\rvert\partial_i b(\lambda)\rangle .$$

Under a unitary frame change $\lvert a\rangle\mapsto\lvert b\rangle g_{ba}$,

$$\mathcal A\longmapsto g^\dagger\mathcal A g+i\,g^\dagger dg, \qquad \mathcal F=d\mathcal A-i\mathcal A\wedge\mathcal A.$$

The curvature can be written without choosing a frame:

$$\mathcal F =i\,P(dP\wedge dP)P \quad\text{on }\operatorname{im}P.$$

Adiabatic transport around a closed path $C$ is the holonomy

$$U_C =\mathcal P\exp\!\left(i\oint_C\mathcal A\right).$$

The approximation requires motion slow relative to the gap; schematically, matrix elements of $\dot H$ divided by $\Delta^2$ must be small. The geometric phase for a nondegenerate level was identified by Berry 1984, §§2–3, pp. 47–51, and its non-Abelian form for a degenerate subspace by Wilczek and Zee 1984, pp. 2111–2114.

If the grading $\Gamma$ is parameter independent, then $P$ commutes with $\Gamma$ and the connection preserves the even and odd vacuum subbundles. The ordinary Witten index records only the difference of their ranks. It contains no information about their Berry curvature or holonomy.

An exact family with protected index and nontrivial holonomy

Take

$$\mathcal H_{\bar0}=\mathbb C^2, \qquad \mathcal H_{\bar1}=\mathbb C,$$

and, for $z=re^{i\theta}\in\mathbb C$, define the differential

$$D(z) =\frac12 \begin{pmatrix} z&\sqrt3\,\bar z \end{pmatrix} :\mathbb C^2_{\bar0}\longrightarrow\mathbb C_{\bar1}.$$

Set

$$H(z) =\frac12 \begin{pmatrix} D^\dagger D&0\\ 0&DD^\dagger \end{pmatrix}.$$

For $r>0$, $DD^\dagger=r^2$. Hence there is one even zero mode, together with one paired even–odd level at energy $r^2/2$. A normalized periodic zero-mode frame is

$$\lvert u(\theta)\rangle =\frac12 \begin{pmatrix} -\sqrt3\,e^{-i\theta}\\ e^{i\theta} \end{pmatrix}, \qquad D(z)\lvert u(\theta)\rangle=0.$$

Its Berry connection and holonomy are

$$\mathcal A_\theta =i\langle u\rvert\partial_\theta u\rangle =\frac12, \qquad U_{S^1} =\exp\!\left(i\int_0^{2\pi}\mathcal A_\theta\,d\theta\right) =-1.$$

Thus a one-dimensional ground-state bundle can have nontrivial holonomy even though its rank and index are constant.

At $z=0$, the gap $r^2/2$ closes and $H(0)=0$. There are then two even and one odd ground states. The total degeneracy jumps from one to three, so the rank-one spectral subbundle over $\mathbb C\setminus\{0\}$ cannot extend through the origin as the ground eigenspace of $H$. Nevertheless,

$$\operatorname{ind}D =1-0=2-1=1$$

both for $r>0$ and at the origin: the extra zero modes appeared as an even–odd pair. This small matrix model cleanly separates a stable index, a failing spectral gap, a jump in total vacuum count, and nontrivial geometric transport.

The exact comparison is:

Parameter region	$\dim\ker D$	$\dim\ker D^\dagger$	Positive spectrum	Total vacua	$\operatorname{ind}D$	Ground-state geometry
$r>0$	1	0	One paired even–odd level at $E=r^2/2$	1	$+1$	Rank-one bundle on $\mathbb C\setminus\{0\}$; on a circle, $\mathcal A_\theta=1/2$ and $U_{S^1}=-1$
$z=0$	2	1	None; $H(0)=0$	3	$+1$	The gap is closed and the rank-one ground eigenspace has no continuation through the origin

A practical decision sequence

For a new supersymmetric quantum-mechanical problem, ask in this order:

1. What are the closed domains? Specify boundary and asymptotic conditions for $A$ and $A^\dagger$.
2. Is $A$ Fredholm? Check finite-dimensional kernels, closed range, and absence of zero in the essential spectrum.
3. What trace is being taken? Prove trace class or specify a relative regulator and its threshold terms.
4. Is the vacuum multiplicity locally constant and gapped? Only then is there a fixed-rank ground-state bundle.
5. What is being claimed to cross a wall? Separate signed index, total degeneracy, and holonomy.

This sequence prevents a formal cancellation in the bulk from hiding the actual data at a boundary, at infinity, or in parameter space.

Exercises

Berry holonomy under a deformation. For the matrix family above, replace $\sqrt3$ by a real parameter $c>0$:

$$D_c(z) =\frac{1}{\sqrt{1+c^2}} \begin{pmatrix} z&c\,\bar z \end{pmatrix}.$$

Find a normalized even zero mode for $r>0$, its Berry connection around a circle centered at the origin, and the holonomy.

Solution

A periodic normalized kernel vector is

$$\lvert u_c(\theta)\rangle =\frac{1}{\sqrt{1+c^2}} \begin{pmatrix} -c\,e^{-i\theta}\\ e^{i\theta} \end{pmatrix}.$$

Direct differentiation gives

$$\mathcal A_\theta =i\langle u_c\rvert\partial_\theta u_c\rangle =\frac{c^2-1}{c^2+1}.$$

Therefore

$$U_{S^1} =\exp\!\left[ 2\pi i\,\frac{c^2-1}{c^2+1} \right].$$

For $c=\sqrt3$, this reduces to $e^{i\pi}=-1$. The holonomy varies continuously with $c$ even though the ground-state rank and index remain one. This is geometric response, not wall crossing of the index.

Boundary domains determine the index. On $I=[0,L]$, solve the harmonic equations for the absolute and relative de Rham boundary problems in the table above. Compute each index. Why would imposing Neumann conditions on both zero-forms and one-forms fail to define either of these supersymmetric complexes?

Solution

For the absolute problem, a harmonic zero-form obeys $f''=0$ and $f'(0)=f'(L)=0$, so $f$ is constant. A harmonic one-form has coefficient $g''=0$ and $g(0)=g(L)=0$, so it vanishes. Thus

$$n_{\bar0}^{(0)}=1, \qquad n_{\bar1}^{(0)}=0, \qquad I_{\mathrm{abs}}=+1.$$

For the relative problem, Dirichlet conditions remove the harmonic zero-form, whereas $g'(0)=g'(L)=0$ leaves the constant one-form $g\,dx$. Hence

$$n_{\bar0}^{(0)}=0, \qquad n_{\bar1}^{(0)}=1, \qquad I_{\mathrm{rel}}=-1.$$

If both coefficients merely obey Neumann conditions, the boundary form $[g^*f]_0^L$ need not vanish for arbitrary vectors in the two proposed domains. The displayed first-order $d$ and $d^\dagger$ are then not adjoints on those domains, so the choice does not define the absolute or relative supersymmetric complex even though each second-order differential expression can be made self-adjoint separately.

References

• Akhoury, Ratindranath, and Alain Comtet. “Anomalous Behavior of the Witten Index—Exactly Soluble Models.” Nuclear Physics B 246, no. 2 (1984): 253–278. doi:10.1016/0550-3213(84)90296-7.
• Al-Hashimi, M. H., M. Salman, A. Shalaby, and U.-J. Wiese. “Supersymmetric Descendants of Self-Adjointly Extended Quantum Mechanical Hamiltonians.” Annals of Physics 337 (2013): 1–24. doi:10.1016/j.aop.2013.06.002. Open PDF.
• Berry, Michael V. “Quantal Phase Factors Accompanying Adiabatic Changes.” Proceedings of the Royal Society of London A 392, no. 1802 (1984): 45–57. doi:10.1098/rspa.1984.0023.
• Callias, Constantine. “Axial Anomalies and Index Theorems on Open Spaces.” Communications in Mathematical Physics 62, no. 3 (1978): 213–234. doi:10.1007/BF01202525.
• Niemi, Antti J., and Gordon W. Semenoff. “Index Theorems on Open Infinite Manifolds.” Nuclear Physics B 269, no. 1 (1986): 131–169. doi:10.1016/0550-3213(86)90370-6.
• Wilczek, Frank, and A. Zee. “Appearance of Gauge Structure in Simple Dynamical Systems.” Physical Review Letters 52, no. 24 (1984): 2111–2114. doi:10.1103/PhysRevLett.52.2111.