A supersymmetric oscillator you can solve

The two-state example in What a supercharge does fixed the energy and constructed its partner states. Can we instead specify a Hamiltonian, find its entire spectrum, and see where that pairing stops? A harmonic oscillator with two internal states lets us do all three calculations.

This is quantum mechanics: there is one time dimension, no spatial field coordinate, no gauge group and no curved background. The system has two real supercharges, both displayed below. Its internal even and odd states implement a grading; their names do not assert a spacetime spin for a particle moving on the line. No three-dimensional spinor or superspace preparation is needed.

Give a wavefunction two components

Use $\hbar=1$ and a coordinate $x$ normalized so that the kinetic energy is $p^2/2$, with $p=-i\partial_x$. This is a mass-rescaled particle coordinate. With energy as the unit, $x$ has dimension $E^{-1/2}$, $p$ has dimension $E^{1/2}$, and time has dimension $E^{-1}$. The Hilbert space and its norm are

$$\mathcal H=L^2(\mathbb R)\otimes\mathbb C^2, \qquad \Psi=\begin{pmatrix}\psi_B\\\psi_F\end{pmatrix}, \qquad \|\Psi\|^2=\int_{\mathbb R}dx\,(|\psi_B|^2+|\psi_F|^2).$$

Here $B$ is even and $F$ is odd, following the earlier lesson. Thus $\Gamma=(-1)^F=\operatorname{diag}(1,-1)$; in this model we can also use the integer fermion-number operator $F=\operatorname{diag}(0,1)$. The matrices act on internal components, while differential operators act on $x$.

Choose a smooth real, dimensionless function $h(x)$ and introduce

$$A=\partial_x+h'(x),\qquad A^\dagger=-\partial_x+h'(x), \qquad Q=\begin{pmatrix}0&0\\A&0\end{pmatrix}, \quad Q^\dagger=\begin{pmatrix}0&A^\dagger\\0&0\end{pmatrix}. \tag{QM01.1}$$

The prime denotes an $x$ derivative. Integration by parts gives the adjoint: the boundary term must vanish. We first compute on smooth rapidly decreasing functions. For the polynomial examples below, use the closed differential operator $A$ on the full line and its Hilbert-space adjoint; $A^\dagger A$ and $AA^\dagger$ define the nonnegative self-adjoint Hamiltonians. Introducing a finite interval would require a new boundary-condition analysis.

The function $h$ is often called the quantum-mechanical superpotential. It is a real function of a particle coordinate, not a holomorphic field-theory superpotential. Its derivative has the same dimension as $\partial_x$, so $Q$ has dimension $E^{1/2}$.

Matrix multiplication immediately gives $Q^2=(Q^\dagger)^2=0$ and $\Gamma Q=-Q\Gamma$. Define the Hamiltonian through the same normalization as before:

$$H=\frac12\{Q,Q^\dagger\} =\begin{pmatrix}H_B&0\\0&H_F\end{pmatrix}, \qquad H_B=\frac12A^\dagger A,\quad H_F=\frac12AA^\dagger. \tag{QM01.2}$$

To expand the products, apply them to a test function $f$. In $A^\dagger Af$, the derivative of $h'f$ supplies $-h''f-h'f'$, and the last term $h'\partial_xf$ cancels the second contribution. Reversing the order changes the sign of $h''$. Consequently,

$$H_B=\frac12[-\partial_x^2+(h')^2-h''], \qquad H_F=\frac12[-\partial_x^2+(h')^2+h'']. \tag{QM01.3}$$

These are partner Hamiltonians. Their potential terms differ, but their positive-energy spectra will agree. The two Hermitian charges are $S_1=(Q+Q^\dagger)/\sqrt2$ and $S_2=(Q-Q^\dagger)/(i\sqrt2)$, with $\{S_a,S_b\}=2\delta_{ab}H$. Thus $S_a^2=H$, whereas the complex charge is nilpotent.

Tong’s construction, (1.5)–(1.9), uses the same differential Hamiltonian but calls the upper component fermionic. Our translation is $Q=iQ_{\rm Tong}$ and $\Gamma=-\Gamma_{\rm Tong}$. The phase leaves the anticommutator unchanged; the grading translation exchanges the names of the ground-state sectors. Carry that distinction into any signed state count.

Pair states with their correct norm

Associativity gives $AH_B=H_FA$ and $A^\dagger H_F=H_BA^\dagger$. Suppose $f_B$ is a normalized eigenfunction of $H_B$ with energy $E>0$. Then

$$f_F=\frac{Af_B}{\sqrt{2E}},\qquad H_Ff_F=Ef_F,\qquad \|f_F\|^2=\frac{\langle f_B,A^\dagger Af_B\rangle}{2E}=1. \tag{QM01.4}$$

Applying $A^\dagger/\sqrt{2E}$ returns $f_B$. This constructs an invertible map between the two eigenspaces; it does more than identify equal numbers of formal solutions. The eigenfunctions must belong to the operator domains. In the confining examples here, the decaying eigenfunctions meet that requirement.

For any normalized state in the Hamiltonian domain, $2\langle H\rangle=\|Q\Psi\|^2+\|Q^\dagger\Psi\|^2$. At zero energy both norms vanish. We therefore solve first-order equations, separately in the two sectors:

$$A\psi_B=0\ \Longrightarrow\ \psi_B=C_Be^{-h(x)}, \qquad A^\dagger\psi_F=0\ \Longrightarrow\ \psi_F=C_Fe^{h(x)}. \tag{QM01.5}$$

They are candidate wavefunctions. The actual tests are $\int_{\mathbb R}e^{-2h}dx<\infty$ and $\int_{\mathbb R}e^{2h}dx<\infty$, together with the operator-domain condition. Growth to $+\infty$ by itself is not a general integrability theorem: the rate matters. For example, $h=c\log(1+x^2)/2$ gives $e^{-2h}=(1+x^2)^{-c}$, whose integral converges only for $c>1/2$. For a polynomial with positive even leading coefficient, the required decay is sufficient. The first-order solution in each sector is unique up to its constant, so there is at most one zero mode per sector in this full-line scalar problem.

Solve the oscillator completely

Take $h=\omega x^2/2$ with $\omega>0$. The parameter $\omega$ has dimension energy. The partner potentials are $V_B=\omega^2x^2/2-\omega/2$ and $V_F=\omega^2x^2/2+\omega/2$. Define the ordinary oscillator lowering operator $a=(\partial_x+\omega x)/\sqrt{2\omega}$, with $[a,a^\dagger]=1$. Then $A=\sqrt{2\omega}\,a$ and

$$H_B=\omega a^\dagger a,\qquad H_F=\omega(a^\dagger a+1),\qquad H=\omega(a^\dagger a+F). \tag{QM01.6}$$

The spatial oscillator functions are $\phi_n=(a^\dagger)^n\phi_0/\sqrt{n!}$, where $\phi_0=(\omega/\pi)^{1/4}e^{-\omega x^2/2}$. Let $|n,B\rangle=(\phi_n,0)^T$ and $|n,F\rangle=(0,\phi_n)^T$. Their energies are

$$E_{n,B}=\omega n,\qquad E_{n,F}=\omega(n+1), \qquad n=0,1,2,\ldots. \tag{QM01.7}$$

The label $n$ counts spatial excitations; it is not the full energy in both sectors. The pairing is shifted by one spatial level:

Energy	Even state	Odd state
$0$	$\lvert0,B\rangle$	None
$\omega$	$\lvert1,B\rangle$	$\lvert0,F\rangle$
$2\omega$	$\lvert2,B\rangle$	$\lvert1,F\rangle$
$j\omega$, $j\geq1$	$\lvert j,B\rangle$	$\lvert j-1,F\rangle$

Indeed, $Q|n,B\rangle=\sqrt{2\omega n}\,|n-1,F\rangle$ for $n\geq1$, and $Q$ annihilates every odd state and $|0,B\rangle$. Its adjoint reverses each positive-energy arrow. These are precisely the two-state matrices of the prerequisite, now embedded in an entire Hilbert space.

The unique zero-energy state is

$$\Psi_0(x)=\left(\frac{\omega}{\pi}\right)^{1/4} e^{-\omega x^2/2}\begin{pmatrix}1\\0\end{pmatrix}, \qquad Q\Psi_0=Q^\dagger\Psi_0=0. \tag{QM01.8}$$

The Gaussian integral fixes its norm. The formal odd solution $e^{+\omega x^2/2}$ is not square integrable and supplies no extra state. The ordinary oscillator zero-point energy $\omega/2$ is canceled by the fixed $-\omega/2$ in $H_B$. This is the Hamiltonian required by the supercharges. Adding an arbitrary constant to $H$ while retaining those charges would spoil (QM01.2).

There is no conflict between $V_B(0)=-\omega/2$ and nonnegative energy: positivity is a statement about the full differential operator, including kinetic energy. It does not require its potential to be pointwise positive. Here $\langle p^2/2\rangle=\omega/4$ and $\langle\omega^2x^2/2\rangle=\omega/4$ in the ground state, which cancel the constant exactly.

What happens when confinement disappears?

At $\omega=0$, both partners become the free Hamiltonian $-\partial_x^2/2$. The formal zero modes are constants, neither of which is in $L^2(\mathbb R)$. The spectrum approaches zero continuously, but there is no normalized lowest-energy eigenstate.

This differs from a system whose attained ground energy is strictly positive. The latter has a ground state that is not annihilated by both charges; the free full-line example has no ground-state vector to test. The same distinction matters in field theory when a potential runs away: absence of a supersymmetric vacuum does not by itself establish a positive-energy static vacuum. Intriligator and Seiberg, (1.2)–(1.4) and the opening of section 2.

For $\omega>0$, the zero-mode probability has variance $1/(2\omega)$. Its spreading as $\omega\to0^+$ explains how a normalizable state can be lost at an endpoint. It also warns against applying a discrete-spectrum argument without checking its assumptions.

The Witten index turns this pairing into a signed trace. Its usefulness depends on knowing when that trace exists and when states can escape to infinity under a deformation.

Practice

1. Move the oscillator and reverse its sign

Set $h=\omega(x-x_0)^2/2$, with real $x_0$ and nonzero real $\omega$. Find both energy spectra, the normalized zero mode and its parity, and the mean and variance of $x$. Does moving $x_0$ break supersymmetry?

Hint 1

The ordinary oscillator frequency is $|\omega|$, while the sector-dependent constant is still $\mp\omega/2$.

Hint 2

Test $e^{-h}$ and $e^{h}$ on the entire line before selecting a sector.

Solution

Translation to $y=x-x_0$ is unitary. Therefore

$$E_{n,B}=|\omega|(n+\tfrac12)-\frac{\omega}{2},\qquad E_{n,F}=|\omega|(n+\tfrac12)+\frac{\omega}{2}.$$

For positive $\omega$, the zero mode is even; for negative $\omega$, it is odd. In its nonzero component it equals $(|\omega|/\pi)^{1/4}e^{-|\omega|(x-x_0)^2/2}$. The distribution gives $\langle x\rangle=x_0$ and $\langle(x-x_0)^2\rangle=1/(2|\omega|)$. The spectra remain nonnegative and positive levels remain paired, checking the sign exchange. Translation changes the state’s location, not the ground energy or its annihilation by the charges. The singular limit $\omega=0$ is excluded from this solution.

2. Deform the potential without solving every level

Take $h=\omega x^2/2+g x^4/4$, with $\omega>0$ and $g>0$. Find the partner potentials and determine whether a supersymmetric ground state survives. Check the first-order correction in $g$ to its energy using ordinary perturbation theory around the oscillator. Does the result determine the excited spectrum?

Hint 1

First compute $h'$ and $h''$. The normalizable candidate follows directly from (QM01.5).

Hint 2

In the undeformed Gaussian, $\langle x^2\rangle=1/(2\omega)$ and $\langle x^4\rangle=3/(4\omega^2)$.

Solution

Here $g$ has dimension $E^2$. Differentiation gives

$$V_{B,F}=\frac12\bigl[(\omega x+gx^3)^2 \mp(\omega+3gx^2)\bigr].$$

The even zero mode is $C\exp(-\omega x^2/2-gx^4/4)$, with $C^{-2}=\int_{\mathbb R}e^{-\omega x^2-gx^4/2}dx$. This integral is finite and positive, and the wavefunction and its derivatives have sufficient decay for the operator domain. The odd candidate grows. The even zero mode is therefore unique and its energy is exactly zero, without an expansion in $g$.

For a perturbative check, write $H_B=H_B^{(0)}+g(\omega x^4-3x^2/2)+g^2x^6/2$. The first-order correction is

$$\delta E_0^{(1)} =g\left[\omega\frac{3}{4\omega^2} -\frac32\frac1{2\omega}\right]=0.$$

Both terms are needed. The check tests the coefficient of $h''$ independently of merely guessing a decaying wavefunction. Excited positive levels remain paired by the intertwining argument, but their actual energies are not fixed by this ground-state solution. Perturbation theory for them requires $g/\omega^2$ small; the exact zero-mode argument does not.

3. An extra zero mode from a cutoff

For the positive-frequency oscillator, retain spatial states $n=0,\ldots,K$ in each parity sector, with $K\geq1$. Define $a_K$ by the usual lowering rule and let its adjoint vanish on the top state. Build $Q_K$ using $A_K=\sqrt{2\omega}\,a_K$ and define $H_K=\{Q_K,Q_K^\dagger\}/2$. Find its spectrum. Why does exact matrix supersymmetry in this truncation not guarantee the correct continuum ground states? Give a finite cutoff that retains complete positive-energy pairs.

Hint 1

Compare $a_K^\dagger a_K$ with $a_Ka_K^\dagger$ on the top spatial state.

Hint 2

The two sectors need not have the same dimension. Cut off by total energy using the spectrum table.

Solution

The even block has energies $0,\omega,\ldots,K\omega$. In the odd block, $a_Ka_K^\dagger$ has diagonal entries $1,2,\ldots,K,0$. The artificial state $|K,F\rangle$ therefore has zero energy instead of the physical $(K+1)\omega$. It sits at the boundary of the retained basis, even though its assigned energy is low.

The truncated commutator is

$$[a_K,a_K^\dagger]=I-(K+1)|K\rangle\langle K|.$$

Its trace vanishes, as every finite matrix commutator must; it cannot equal the identity. Simply compressing the true odd Hamiltonian would give diagonal entries $\omega,\ldots,(K+1)\omega$, but that compression is not the anticommutator of these square truncated charges. A choice between these operations matters at the boundary.

Instead retain $|0,B\rangle$ and the pairs $(|j,B\rangle,|j-1,F\rangle)$ for $j=1,\ldots,K$. There are $K+1$ even states and $K$ odd states. The restricted $A$ is a rectangular map, with matrix element $\sqrt{2\omega j}$ from the $j$th even state to the $(j-1)$st odd state. Both charges preserve this subspace; their anticommutator gives the exact retained energies and just one zero mode. This finite example checks a cutoff prescription. It does not by itself establish convergence for a different interacting Hamiltonian.

The oscillator has turned an algebraic pairing statement into a calculation of actual wavefunctions, their norms, and a complete spectrum. A next question is whether a signed count can retain the unpaired information while all positive-energy pairs cancel. Answering it will require specifying which trace exists and what happens when states spread to infinity.