What the Witten index can tell us

The supersymmetric oscillator has one even zero-energy state and an infinite sequence of even–odd pairs. Can we count the first without keeping track of every pair? The Witten index does this, but it keeps signed information. It is neither the total number of supersymmetric states nor a guarantee that an arbitrary deformation leaves the answer unchanged.

We remain in quantum mechanics with two real supercharges, no gauge group, and no spatial field coordinate. Use the same mass-rescaled coordinate and Hamiltonian normalization:

$$\begin{gathered} \mathcal H=L^2(\mathbb R)\otimes\mathbb C^2,\qquad \Psi=(\psi_B,\psi_F)^{\mathsf T},\qquad \Gamma=(-1)^F=\operatorname{diag}(1,-1),\\ A=\partial_x+h',\qquad Q=\begin{pmatrix}0&0\\A&0\end{pmatrix},\qquad H=\frac12\{Q,Q^\dagger\} =\frac12\operatorname{diag}(A^\dagger A,AA^\dagger). \end{gathered} \tag{QM02.1}$$

Here $h$ is real and dimensionless, $x$ has dimension $E^{-1/2}$, and the full-line operators have the closed domains explained in the prerequisite. The grading anticommutes with $Q$ and commutes with $H$. Tong’s upper component is instead called fermionic: in his (1.5)–(1.9), $Q=iQ_{\rm Tong}$ and $\Gamma=-\Gamma_{\rm Tong}$. Thus our index has the opposite sign to his for the same differential Hamiltonian.

Calculate the oscillator trace first

For $\beta>0$, define the graded thermal trace

$$I(\beta)=\operatorname{Tr}_{\mathcal H} \bigl(\Gamma e^{-\beta H}\bigr) =Z_B(\beta)-Z_F(\beta). \tag{QM02.2}$$

The parameter $\beta$ has dimension $E^{-1}$. The notation $Z_B$ means a trace over the even subspace, with no minus sign built into it. Initially require both sector traces to be finite. Equivalently here, $e^{-\beta H}$ is trace class: its positive eigenvalues have a finite sum. This makes the subtraction and rearrangement of the sums legitimate.

For $h=\omega x^2/2$, $\omega>0$, put $q=e^{-\beta\omega}$, so $0<q<1$. The spectra already found give

$$Z_B=\sum_{n=0}^{\infty}q^n=\frac1{1-q}, \qquad Z_F=\sum_{n=0}^{\infty}q^{n+1}=\frac q{1-q}, \qquad \boxed{I=1.} \tag{QM02.3}$$

The ordinary partition function is instead $Z=Z_B+Z_F=(1+q)/(1-q)$; it depends on temperature. The index has discarded the positive-energy pairs and retained the even zero mode. It is dimensionless.

Do not replace this calculation by the unregulated series $1-1+1-1+\cdots$. Setting $\beta=0$ before taking the traces removes their convergence. Taking the limit of the already evaluated difference is a different operation. Tong’s discussion in §1.2.2 makes this regulator distinction explicitly.

Why the cancellation is general

For a discrete eigenvalue $E>0$, the maps $A/\sqrt{2E}$ and $A^\dagger/\sqrt{2E}$ are inverse isometries between the even and odd eigenspaces. Thus their dimensions agree, including any degeneracy. If the heat traces converge, we may group the two contributions at each energy:

$$\begin{aligned} I(\beta) &=n_B^0-n_F^0 +\sum_{E>0}\bigl(n_B(E)-n_F(E)\bigr)e^{-\beta E}\\ &=n_B^0-n_F^0 =\dim\ker A-\dim\ker A^\dagger. \end{aligned} \tag{QM02.4}$$

The superscript zero counts normalizable zero-energy eigenstates in the chosen domains. This proves $\beta$ independence in this setting, without cyclically moving unbounded operators through an undefined trace.

Consequently, a nonzero index guarantees at least $|I|$ supersymmetric ground states. An index of zero permits either no zero states or equal nonzero numbers in the two sectors. Positive-energy pairing gives no way to distinguish those alternatives. The circle exercise below supplies an explicit zero-index theory with two supersymmetric states.

For a continuous family of supersymmetric Hamiltonians, ordinary finite-energy levels can move. A pair can also reach zero and contribute one state of each parity. Neither operation changes their difference. A useful sufficient setting for this argument is a fixed graded Hilbert space, compatible self-adjoint domains, convergent heat traces, and continuous finite-dimensional low-energy spectral subspaces that stay separated from the remaining spectrum during the deformation. No state may be lost at spatial infinity or through a changed boundary condition. These are conditions on the family, not merely on a formula for $h$ at one point.

For polynomial examples, a confining leading term with a coefficient bounded away from zero on a bounded parameter range supplies such control. A pointwise-small change of the potential near the origin need not be small at arbitrarily large $|x|$. This is where a seemingly harmless deformation can invalidate the conclusion.

A second superpotential with zero index

Take

$$h_g(x)=\frac{\omega x^2}{2}+\frac{g x^3}{3}, \qquad \omega>0,\quad g>0. \tag{QM02.5}$$

The parameter $g$ has dimension $E^{3/2}$. The two scalar Hamiltonian potentials are

$$V_{B,F}(x)=\frac12\left[(\omega x+gx^2)^2 \mp(\omega+2gx)\right]. \tag{QM02.6}$$

Both grow as $g^2x^4/2$ at either end. They therefore have discrete, confining spectra, finite heat traces for $\beta>0$, and an attained lowest eigenvalue. But $e^{-h_g}$ grows at negative infinity, while $e^{h_g}$ grows at positive infinity. Neither candidate zero mode is normalizable. Hence

$$n_B^0=n_F^0=0,\qquad I=0,\qquad E_0>0. \tag{QM02.7}$$

The strict inequality uses two facts together: the factorization gives nonnegative energy, and confinement gives an actual lowest eigenstate. Excluding a zero eigenfunction then makes that eigenvalue positive. This is supersymmetry breaking in the quantum-mechanical ground-state sense; the Hamiltonian and its algebra remain supersymmetric. The same conclusion is obtained for the polynomial cubic example in Tong, §1.2.1, with his opposite parity labels.

There is a check stronger than counting classical minima. Set $y=x+\omega/(2g)$. Then

$$h_g'=gy^2-\frac{\omega^2}{4g},\qquad h_g''=2gy, \qquad V_F(y)=V_B(-y). \tag{QM02.8}$$

Reflection is unitary on the full line and exchanges the two Hamiltonians, so their entire spectra agree. For example, at $g/\omega^{3/2}=1$, independent converging spectral calculations give $E_0/\omega\simeq0.16851138$ in both sectors. The proof of positivity and the exact zero index do not depend on this numerical estimate.

At $g=0$ the model becomes the oscillator and has index one. What failed? The two critical points of $h_g$ are $0$ and $-\omega/g$. The second recedes to infinity. Its curvature is $h_g''=-\omega$, so locally it supports a low odd wave packet; near the origin the corresponding low wave packet is even. For nonzero $g$ neither packet is an exact zero mode. As $g$ tends to zero, there is no uniform control of this remote well or its low states. The leading confining power changes. Continuity of the coefficients at each fixed $x$ is insufficient for the index-invariance argument.

We can also bound how low these states lie. Take the normalized oscillator Gaussian centered at zero in the even sector and at $-\omega/g$ in the odd sector. The operators $A$ and $A^\dagger$, respectively, act on them as $g$ times the squared displacement from their centers. Each trial energy is therefore $g^2\langle y^4\rangle/2=3g^2/(8\omega^2)$. The variational principle gives $0<E_0\leq3g^2/(8\omega^2)$. Thus a positive-energy pair really does approach zero as the distant packet escapes; counting only states that remain near the origin loses one member.

Follow escape to infinity exactly

We can make the loss of a state quantitative without approximating a cubic spectrum. Choose

$$h_\epsilon(x)=\frac{\epsilon x^2}{2}+\mu x,\qquad \epsilon>0,\quad \mu>0, \tag{QM02.9}$$

where $[\epsilon]=E$ and $[\mu]=E^{1/2}$. Completing the square translates an oscillator of frequency $\epsilon$. Its even zero mode is

$$\psi_{0,\epsilon}(x) =\left(\frac{\epsilon}{\pi}\right)^{1/4} \exp\left[-\frac{\epsilon}{2} \left(x+\frac{\mu}{\epsilon}\right)^2\right], \qquad \langle x\rangle=-\frac{\mu}{\epsilon},\quad \operatorname{Var}(x)=\frac1{2\epsilon}. \tag{QM02.10}$$

For every $\epsilon>0$ it has norm one and $I=1$. Yet its probability in every fixed bounded interval tends to zero as $\epsilon\to0^+$. A normalized state can converge to zero locally without converging to zero in Hilbert-space norm.

At the endpoint, $H_B=H_F=(-\partial_x^2+\mu^2)/2$. Its spectrum is continuous from $\mu^2/2$ upward, with no square-integrable lowest-energy eigenvector. Each heat trace is infinite because the line has infinite volume. Therefore (QM02.2) is not an ordinary trace at $\epsilon=0$; it is not legitimate to write “infinity minus infinity equals zero” and call that a contradiction.

A local heat-kernel calculation shows the nonuniform limit particularly well. The heat kernel is the position-space matrix element $K(x,z;\beta)=\langle x|e^{-\beta H}|z\rangle$. Write $y=x+\mu/\epsilon$. For the ordinary oscillator,

$$K_{\rm osc}(y,y;\beta) =\sqrt{\frac{\epsilon}{2\pi\sinh(\beta\epsilon)}} \exp\left[-\epsilon y^2 \tanh\left(\frac{\beta\epsilon}{2}\right)\right].$$

For a direct derivation, insert $K=N(\beta)\exp[-a(\beta)(y^2+z^2)+b(\beta)yz]$ into $\partial_\beta K=(\partial_y^2-\epsilon^2y^2)K/2$. Matching powers gives $N'/N=-a$, $a'=\epsilon^2/2-2a^2$, $b'=-2ab$, and $4a^2-b^2=\epsilon^2$. The short-time free kernel fixes $a=\epsilon\coth(\beta\epsilon)/2$, $b=\epsilon/\sinh(\beta\epsilon)$ and the displayed normalization. Setting $z=y$ gives the diagonal formula. The constant energy shifts give $K_B=e^{\beta\epsilon/2}K_{\rm osc}$ and $K_F=e^{-\beta\epsilon/2}K_{\rm osc}$. Their local difference is

$$\rho_{\epsilon,\beta}(x) =\sqrt{\frac{\alpha}{\pi}}\, e^{-\alpha(x+\mu/\epsilon)^2}, \qquad \alpha=\epsilon\tanh\left(\frac{\beta\epsilon}{2}\right). \tag{QM02.11}$$

Its integral over the full line is exactly one. For fixed $\beta>0$ and bounded $x$, however, $\alpha\sim\beta\epsilon^2/2$ and $\rho_{\epsilon,\beta}(x)\to0$. Thus

$$\lim_{\epsilon\to0^+}\int_{\mathbb R}\rho_{\epsilon,\beta}\,dx=1, \qquad \int_{\mathbb R}\lim_{\epsilon\to0^+}\rho_{\epsilon,\beta}\,dx=0. \tag{QM02.12}$$

There is no integrable bound uniform in $\epsilon$ that would justify interchanging these operations. At fixed $\epsilon$, the separate limit $\beta\to\infty$ gives the ground-state probability in (QM02.10). At fixed $\beta$ while $\epsilon\to0$, one must use the full heat kernel because the excited-state gap is also closing.

At the endpoint one could choose to subtract the identical free kernels before integration, obtaining zero. That is an additional relative-trace prescription, not the convergent thermal trace along the family. More general continuum problems can require boundary or spectral-density terms. The elementary pairing argument does not settle them.

The absence of a normalizable vacuum must likewise be separated from an attained positive-energy vacuum. This is the distinction stressed by Intriligator and Seiberg at the start of §2. Their field-theory example in §2.1 also has a vacuum that recedes when a deformation is removed; our Gaussian calculation is a quantum-mechanical realization with its own explicitly stated Hilbert space.

Practice

1. Zero index with and without a vacuum on a circle

Replace the line by a circle of circumference $L$. Let $h(x)$ be smooth, real and periodic, and impose the same boundary condition $\psi(x+L)=e^{i\theta}\psi(x)$ on both components and their operator domains, with $\theta$ understood modulo $2\pi$. Determine the zero modes and the index. For $h=0$, find the spectrum and decide when the attained ground energy is positive.

Hint 1

The first-order solutions are still $e^{-h}$ and $e^{h}$. They are bounded on a compact interval, but must also satisfy the boundary condition.

Hint 2

For $h=0$, use plane waves with $p_n=(2\pi n+\theta)/L$. The two sectors have identical energies.

Solution

For $\theta=0$, both first-order solutions are periodic and normalizable: $\psi_B=e^{-h}/\sqrt{\int_0^L e^{-2h}dx}$ and $\psi_F=e^{h}/\sqrt{\int_0^L e^{2h}dx}$ in their respective components. There is exactly one in each sector, so $I=1-1=0$ despite unbroken supersymmetry.

For $\theta\ne0$ modulo $2\pi$, a nonzero periodic exponential cannot acquire the required phase, so neither solution is allowed. The differential operators preserve the common twisted domain, and their adjoints have the same twist by integration by parts. Positive states still pair. Compactness gives discrete spectra and finite heat traces, so the index remains zero although the zero modes have disappeared in a pair.

At $h=0$ the exact energies are

$$E_{n,B}=E_{n,F}=\frac{(2\pi n+\theta)^2}{2L^2}, \qquad n\in\mathbb Z.$$

The ground energy is zero for the trivial twist and strictly positive otherwise. Each sector trace is a convergent Gaussian sum for $\beta>0$. Thus the same zero index occurs on both sides of a change in the existence of supersymmetric ground states. This changes global boundary data, not the local supercharge formula.

2. Which weighting preserves cancellation?

For the positive-frequency oscillator insert an additional real positive weight $y^F$, with $F=\operatorname{diag}(0,1)$: $I_y(\beta)=\operatorname{Tr}(\Gamma y^F e^{-\beta H})$. Compute it and its $\beta$ derivative. Why is this generally not another protected index? State a sufficient condition on an inserted operator $U$ for positive-energy cancellation.

Hint 1

The weight of an even state is one, while the weight of every odd state is $y$. Keep the shifted oscillator spectra.

Hint 2

Cancellation requires both members of a pair to receive the same inserted weight. Compare $FQ$ with $QF$.

Solution

With $q=e^{-\beta\omega}$,

$$I_y=\frac{1-yq}{1-q},\qquad \frac{dI_y}{d\beta} =-\frac{\omega(1-y)q}{(1-q)^2}.$$

It is independent of $\beta$ only at $y=1$. In our convention $[F,Q]=Q$, so the insertion changes along a supercharge pair. Commuting with $H$ alone is insufficient.

A sufficient condition is that $U$ preserve the grading, commute with both $Q$ and $Q^\dagger$ on their domains, and give a convergent inserted trace. The normalized pairing map then intertwines $U$ on the two positive-energy eigenspaces; their inserted traces cancel even if there are degeneracies. Fugacities used in more elaborate indices must be checked against the actual chosen charges, not merely described as conserved.

3. A zero mode does not make every heat trace finite

On the full line take $h(x)=\tfrac12\log(1+x^2/L^2)$, where $L>0$ has the dimension of $x$. Normalize its even zero mode and find both partner potentials. Explain why its zero-mode count does not by itself justify the ordinary trace (QM02.2), even though the zero mode is normalizable.

Hint 1

Use $\int_{\mathbb R}dx/(1+x^2/L^2)=\pi L$ and differentiate $h$ before expanding the two Hamiltonians.

Hint 2

Inspect both potentials as $|x|\to\infty$. An infinite line with asymptotically free scattering states has a volume-divergent thermal trace.

Solution

The even zero mode is $\psi_B=(\pi L)^{-1/2}(1+x^2/L^2)^{-1/2}$. It and the derivatives needed for the operator domain are square integrable; $A\psi_B=0$. The odd candidate grows, so the difference of zero-mode dimensions is one.

Since $h'=x/(L^2+x^2)$ and $h''=(L^2-x^2)/(L^2+x^2)^2$,

$$V_B=\frac{2x^2-L^2}{2(L^2+x^2)^2}, \qquad V_F=\frac{L^2}{2(L^2+x^2)^2}.$$

Both tend to zero. There are scattering states down to zero energy, as well as the even normalizable state at that threshold. The asymptotically free diagonal heat kernels approach the nonzero constant $(2\pi\beta)^{-1/2}$ far along the line, so their separate spatial integrals diverge. Neither sector heat operator is trace class.

The kernel difference and any regulated continuum contribution require their own definition and analysis; one may not simply invoke (QM02.4) for these divergent thermal traces. This example isolates the missing assumption: normalizability of an individual zero mode does not provide either a discrete spectrum or a finite heat trace.

An index is useful because it forgets much of the spectrum. To use it well, first identify the grading, supercharge, Hilbert space and regulator, then determine what the cancellation leaves behind. A nonzero result is a ground-state guarantee in the stated spectral setting; a zero result calls for further physics.

For a field-theory continuation, choose a three-dimensional superconformal index. Its positive anticommutator differs from the physical cylinder Hamiltonian, so its signed trace can retain excited states.