Supercharges, Partner Hamiltonians, and Positive Energy

The Hamiltonian of supersymmetric quantum mechanics is nonnegative because it is built from a closed operator and its Hilbert-space adjoint, not merely because a differential expression can be written as a formal square. The same construction produces two partner Hamiltonians, fixes their domains, and makes every positive oscillator level appear once in each grading sector while allowing an unpaired zero mode.

Required background. Bilinear and Hermitian forms, adjoints, and isometries supplies the adjoint operation used below, and self-adjointness, extensions, and unitary evolution supplies the domain criteria. Helpful background. Vacua, states, and representations gives the state-space interpretation, while graded spacetime symmetry and the four-dimensional N=1 algebra explain how this one-dimensional algebra sits inside relativistic supersymmetry.

A graded Hilbert space and its domains

Let

$$\mathcal H=\mathcal H_{\bar 0}\oplus\mathcal H_{\bar 1}, \qquad \Gamma=(-1)^F= \begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$

Take a densely defined closed operator

$$A:\operatorname{Dom}A\subset\mathcal H_{\bar0} \longrightarrow\mathcal H_{\bar1}.$$

Its adjoint $A^\dagger$ is defined by the inner product and by its domain: $\chi\in\operatorname{Dom}A^\dagger$ precisely when the functional $\psi\mapsto\langle\chi|A\psi\rangle$ is bounded in the $\mathcal H_{\bar0}$ norm. This definition includes boundary conditions and behavior at infinity. A formal integration-by-parts symbol $A^\sharp$ is not yet $A^\dagger$.

Define the complex supercharges

$$\mathcal Q= \begin{pmatrix}0&0\\A&0\end{pmatrix}, \qquad \mathcal Q^\dagger= \begin{pmatrix}0&A^\dagger\\0&0\end{pmatrix}.$$

Their domains are $\operatorname{Dom}A\oplus\mathcal H_{\bar1}$ and $\mathcal H_{\bar0}\oplus\operatorname{Dom}A^\dagger$, respectively. They are odd, $\{\Gamma,\mathcal Q\}=0$, and nilpotent in the domain sense: $\mathcal Q$ maps its domain into itself and $\mathcal Q^2=0$. The two Hermitian supercharges are

$$Q_1=\frac{\mathcal Q+\mathcal Q^\dagger}{\sqrt2}, \qquad Q_2=\frac{i(\mathcal Q^\dagger-\mathcal Q)}{\sqrt2},$$

The two descriptions serve different purposes. The complex operator $\mathcal Q$ raises the grading and is nilpotent, so it is the differential used to define cohomology; $\mathcal Q^\dagger$ lowers the grading. By contrast, $Q_1$ and $Q_2$ are self-adjoint symmetry generators. Each combines the raising and lowering maps and therefore sends an even–odd pair into both directions. The factors of $\sqrt2$ are chosen so that each Hermitian supercharge squares to the same Hamiltonian. Both are defined on $\operatorname{Dom}A\oplus\operatorname{Dom}A^\dagger$; closedness of $A$ makes these block operators self-adjoint. Their square is the Hamiltonian

$$H=Q_1^2=Q_2^2 =\frac12\{\mathcal Q,\mathcal Q^\dagger\} =\frac12 \begin{pmatrix} A^\dagger A&0\\ 0&AA^\dagger \end{pmatrix}.$$

The partner domains are part of this statement:

$$\begin{aligned} \operatorname{Dom}(A^\dagger A) &=\{\psi\in\operatorname{Dom}A:A\psi\in\operatorname{Dom}A^\dagger\},\\ \operatorname{Dom}(AA^\dagger) &=\{\chi\in\operatorname{Dom}A^\dagger:A^\dagger\chi\in\operatorname{Dom}A\}. \end{aligned}$$

Both $A^\dagger A$ and $AA^\dagger$ are self-adjoint and nonnegative. Their closed quadratic form on $\operatorname{Dom}A\oplus\operatorname{Dom}A^\dagger$ is

$$h[\Psi] =\frac12\left( \lVert A\psi_{\bar0}\rVert^2 +\lVert A^\dagger\psi_{\bar1}\rVert^2 \right)\ge 0.$$

For $\Psi\in\operatorname{Dom}H$, this form equals $\langle\Psi|H\Psi\rangle$. A general vector in the form domain need not lie in $\operatorname{Dom}H$, so writing the operator expectation there would be incorrect. Positivity nevertheless follows before solving a Schrödinger equation. Witten’s original construction and systematic partner-Hamiltonian treatments use this algebraic square as the starting point Witten 1981, §2, pp. 515–523, de Crombrugghe and Rittenberg 1983, §§2–3, pp. 101–109, and Cooper, Khare, and Sukhatme 1995, §2, arXiv PDF pp. 13–21.

One-dimensional partner potentials

Work on $L^2(\mathbb R,\mathrm dx)$ in units with particle mass $m=1$. Let $w:\mathbb R\to\mathbb R$ be locally absolutely continuous and choose a closed realization of

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

On the full line, standard Sobolev domains and suitable growth of $w$ give the displayed adjoint pair. On an interval or for singular $w$, this line must be replaced by an explicit self-adjoint extension.

Acting on a smooth core gives

$$\begin{aligned} H_{\bar0}=\frac12A^\dagger A &=\frac12\left[-\frac{\mathrm d^2}{\mathrm dx^2}+w(x)^2-w'(x)\right],\\ H_{\bar1}=\frac12AA^\dagger &=\frac12\left[-\frac{\mathrm d^2}{\mathrm dx^2}+w(x)^2+w'(x)\right]. \end{aligned}$$

The $w'$ term is not an arbitrary correction: it is the commutator between differentiation and multiplication. The two scalar potentials are therefore

$$V_{\bar0}=\frac12(w^2-w'), \qquad V_{\bar1}=\frac12(w^2+w').$$

Changing $w\mapsto-w$ interchanges the two sectors. Calling the primitive $W(x)=\int^x w(y)\,\mathrm dy$ avoids confusing the coefficient $w$ with the function that appears in the zero-mode exponent.

The supersymmetric oscillator

Set $w(x)=\omega x$ with $\omega>0$. If

$$a=\frac{1}{\sqrt{2\omega}} \left(\frac{\mathrm d}{\mathrm dx}+\omega x\right), \qquad [a,a^\dagger]=1,$$

then $A=\sqrt{2\omega}\,a$ and

$$H_{\bar0}=\omega a^\dagger a, \qquad H_{\bar1}=\omega aa^\dagger =\omega(a^\dagger a+1).$$

The complete spectrum is transparent:

Sector	Normalized states	Energies
Even	$\lvert n,\bar0\rangle$, $n=0,1,\ldots$	$E_{n,\bar0}=n\omega$
Odd	$\lvert n,\bar1\rangle$, $n=0,1,\ldots$	$E_{n,\bar1}=(n+1)\omega$

For every $n\ge0$,

$$\begin{aligned} \mathcal Q|n+1,\bar0\rangle &=\sqrt{2\omega(n+1)}\,|n,\bar1\rangle,\\ \mathcal Q^\dagger|n,\bar1\rangle &=\sqrt{2\omega(n+1)}\,|n+1,\bar0\rangle. \end{aligned}$$

Thus every $E>0$ state is visibly paired. The Gaussian $|0,\bar0\rangle$, annihilated by $A$, has $E=0$ and no odd partner. This is the smallest exact model of an unpaired supersymmetric vacuum; the next page proves that the same pairing holds for an arbitrary closed $A$.

The cohomology–Hodge correspondence then shows how the same positive-energy pairing leaves precisely the unpaired kernel vectors as cohomology representatives.

Boundaries change the operator

On an interval $[a,b]$, integration by parts gives the boundary form

$$\langle\chi|A\psi\rangle -\langle A^\sharp\chi|\psi\rangle =\left[\chi(x)^*\psi(x)\right]_{a}^{b}, \qquad A^\sharp=-\frac{\mathrm d}{\mathrm dx}+w.$$

A domain for $A$ and a domain for its adjoint must make this form vanish for all allowed pairs. It is not enough to choose self-adjoint boundary conditions for the two second-order expressions independently: $A$ must map the even domain into the odd one, $A^\dagger$ must map back, and the block supercharge must be self-adjoint. On a half-line, for example, a Robin condition for one Hamiltonian can induce a Dirichlet condition for its supersymmetric descendant; more general self-adjoint extensions do not automatically have self-adjoint supersymmetric partners Al-Hashimi et al. 2013, §§2.2–2.3.

The same warning applies at singularities. A differential expression can obey the supersymmetry algebra on compactly supported test functions while the chosen physical extension preserves only one real supercharge—or none. The domain, not the local formula, decides the symmetry.

Common pitfalls

A formal adjoint is not the adjoint. The sign flip on $\mathrm d/\mathrm dx$ follows from integration by parts, but endpoint terms and integrability determine $\operatorname{Dom}A^\dagger$.

Factorization does not guarantee a zero mode. It guarantees $H\ge0$. A solution of $A\psi=0$ must still be normalizable and satisfy the boundary conditions.

An additive energy shift is physical here. Replacing $H$ by $H+c$ generally destroys $H=\{\mathcal Q,\mathcal Q^\dagger\}/2$ unless the algebra is changed. Zero energy is fixed by the superalgebra, not by an arbitrary choice of origin.

Exercises

Oscillator normalization and pairing. Starting from $w(x)=\omega x$, use $V_{\bar0,\bar1}=(w^2\mp w')/2$ to recover the two oscillator Hamiltonians. Then verify the energy and normalization of $\mathcal Q|n+1,\bar0\rangle$.

Solution

Because $w'=\omega$,

$$H_{\bar0}=\frac12\left(-\frac{d^2}{dx^2}+\omega^2x^2-\omega\right) =\omega a^\dagger a,$$

whereas

$$H_{\bar1}=\frac12\left(-\frac{d^2}{dx^2}+\omega^2x^2+\omega\right) =\omega(a^\dagger a+1).$$

Thus $|n+1,\bar0\rangle$ and $|n,\bar1\rangle$ both have energy $E=(n+1)\omega$. Since $A=\sqrt{2\omega}\,a$,

$$\mathcal Q|n+1,\bar0\rangle =\sqrt{2\omega(n+1)}\,|n,\bar1\rangle =\sqrt{2E}\,|n,\bar1\rangle.$$

The coefficient also follows from $\lVert A\psi\rVert^2=2E\lVert\psi\rVert^2$.

A kernel vector and the Hamiltonian domain. Suppose $A$ is closed and $A\psi=0$ for a normalized $\psi\in\operatorname{Dom}A$. Show that $(\psi,0)$ lies in $\operatorname{Dom}H$ and has zero energy.

Solution

Because $A\psi=0$ and $0$ belongs to $\operatorname{Dom}A^\dagger$, the definition of $\operatorname{Dom}(A^\dagger A)$ is satisfied. Hence $H(\psi,0)=\tfrac12(A^\dagger A\psi,0)=0$. Equivalently, both $\mathcal Q$ and $\mathcal Q^\dagger$ annihilate $(\psi,0)$.

A self-adjoint pair that is not supersymmetric. On $I=[0,L]$, suppose one chooses

$$H_{\bar0}=H_{\bar1}=-\frac12\frac{d^2}{dx^2}$$

with Neumann conditions in both sectors, and then claims that $A=d/dx$ and $A^\dagger=-d/dx$ factorize the pair. Use the boundary form and the action of $A$ on a Neumann eigenfunction to find the failure. Give one compatible repair.

Solution

Take $\psi(x)=\cos(\pi x/L)$ and $\chi(x)=1$. Both obey Neumann conditions, but

$$\langle\chi,A\psi\rangle- \langle-\chi',\psi\rangle =\left[\chi^*\psi\right]_0^L=-2\ne0.$$

Thus $-d/dx$ on the same Neumann domain is not the Hilbert-space adjoint of $d/dx$. The domain is also not preserved by the proposed pairing: $A\psi=-(\pi/L)\sin(\pi x/L)$ obeys Dirichlet conditions, but its derivative does not vanish at the endpoints, so it is not in the odd Neumann Hamiltonian domain.

A compatible choice is $\operatorname{Dom}A=H^1(I)$ and $\operatorname{Dom}A^\dagger=H_0^1(I)$. The composites then give Neumann conditions for $A^\dagger A$ and Dirichlet conditions for $AA^\dagger$. Reversing the minimal and maximal extensions gives the relative complex. The two second-order operators must descend from this adjoint first-order pair; choosing them independently loses the supersymmetry algebra.

References

• 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 arXiv version.
• 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.
• de Crombrugghe, M., and V. Rittenberg. “Supersymmetric Quantum Mechanics.” Annals of Physics 151, no. 1 (1983): 99–126. doi:10.1016/0003-4916(83)90316-0.
• Witten, Edward. “Dynamical Breaking of Supersymmetry.” Nuclear Physics B 188, no. 3 (1981): 513–554. doi:10.1016/0550-3213(81)90006-7.