What a supercharge does

An ordinary internal symmetry can rotate two scalar fields into one another. Supersymmetry relates a scalar field to a spinor field. How can that be consistent with the different statistics of bosons and fermions?

The answer requires distinguishing three objects: a supercharge, which is a symmetry generator; a transformation parameter, which specifies an infinitesimal variation; and a field, on which the generator acts. We will first separate them in a scalar–fermion example, then construct a pair of quantum states whose supersymmetry can be checked by multiplying two small matrices.

This lesson assumes the familiar distinction between quantum fields and states, together with commutators and Hermitian conjugation. The first argument works in both three and four spacetime dimensions. The matrix example is quantum mechanics, with one time dimension, and requires no spinor conventions.

Begin with a scalar and a fermion

Work in flat Lorentzian spacetime with gauge-neutral fields. Let $\phi(x)$ be a bosonic scalar field and $\psi_\alpha(x)$ a fermionic spinor field. The label $\alpha$ identifies a spinor component. A supercharge $Q_\alpha$ also carries a spinor index. Schematically, one part of its action can take the form

$$[Q_\alpha,\phi(x)]=\psi_\alpha(x).$$

Here $[A,B]=AB-BA$. We have chosen the normalization of this schematic relation to be one. An actual component theory may have a factor such as $\sqrt{2}$, and also has transformation rules for the fermion and any other fields. This single relation is the beginning of a multiplet, not a complete definition of a supersymmetric theory.

Introduce a constant spinor parameter $\epsilon^\alpha$ and define the corresponding part of the variation by

$$\delta_\epsilon\phi(x) =\epsilon^\alpha[Q_\alpha,\phi(x)] =\epsilon^\alpha\psi_\alpha(x).$$

Repeated $\alpha$ is summed. The contraction removes the spinor index, so the right-hand side is a Lorentz scalar. In this convention any overall phase in the definition of an infinitesimal transformation is included in the parameter; other sources may write an explicit $i$.

The parameter is Grassmann odd: exchanging it with a fermionic field introduces a minus sign,

$$\epsilon^\alpha\psi_\beta=-\psi_\beta\epsilon^\alpha.$$

The product of two odd objects is even. Thus $\delta_\epsilon\phi$ has the same parity as $\phi$, and the transformed field

$$\phi'=\phi+\delta_\epsilon\phi$$

remains bosonic. It is the action of $Q_\alpha$ alone that changes parity. Multiplying by an odd parameter makes the complete variation preserve it. A Grassmann parameter is a formal algebraic variable, not an ordinary real-valued angle that continuously rotates a physical boson state into a fermion state.

Object	Role	Parity
$Q_\alpha$	Operator generating the symmetry	Odd
$\epsilon^\alpha$	External transformation parameter	Odd
$\phi$	Scalar field	Even
$\psi_\alpha$	Spinor field	Odd
$\epsilon^\alpha Q_\alpha$	Generator with its parameter attached	Even
$\delta_\epsilon\phi$	Change of the scalar field	Even

The word spinor describes transformation under spacetime rotations and boosts; the word odd describes grading. They are distinct pieces of information. For example, an ordinary commuting spinor can be used to label an odd supercharge. This convention is useful on curved spaces. Attaching a further odd coefficient then produces an even variation of the kind used here.

What makes a generator fermionic?

Let $\Gamma=(-1)^F$ denote fermion parity on the Hilbert space. The notation only needs to distinguish an even and an odd number of fermions; a separately conserved integer fermion number is unnecessary. A state $|b\rangle$ of even parity satisfies $\Gamma|b\rangle=|b\rangle$, whereas an odd state $|f\rangle$ satisfies $\Gamma|f\rangle=-|f\rangle$.

An odd operator $Q$ anticommutes with $\Gamma$:

$$\Gamma Q=-Q\Gamma.$$

Apply this equation to an even state:

$$\Gamma(Q|b\rangle) =-Q\Gamma|b\rangle =-Q|b\rangle.$$

If $Q|b\rangle$ is nonzero, it is therefore odd. The same calculation starting from an odd state gives an even result. A particular supercharge may annihilate a state; “changes bosons into fermions” always allows the answer zero.

On operators, the appropriate action is the graded commutator. If $p(\mathcal O)$ is zero for an even operator and one for an odd operator, define

$$[Q,\mathcal O\} =Q\mathcal O-(-1)^{p(\mathcal O)}\mathcal OQ.$$

Consequently,

$$[Q,\phi\}=Q\phi-\phi Q, \qquad [Q,\psi\}=Q\psi+\psi Q.$$

The second expression is an anticommutator, denoted $\{Q,\psi\}$. With an odd parameter $\epsilon$ placed on the left, our variation convention is

$$\delta_\epsilon\mathcal O=\epsilon[Q,\mathcal O\}.$$

The bracket reverses parity, and the parameter reverses it again. This explains why a scalar variation is computed from a commutator while a fermion variation is computed from an anticommutator.

Relativistic supersymmetry adds such odd, spinorial generators to spacetime symmetry. Their algebra includes translations. For example, the four-dimensional algebra relates an anticommutator of conjugate supercharges to momentum, and the charges commute with momentum. These are algebraic assumptions with physical consequences, rather than consequences of merely putting scalars and fermions in the same Lagrangian. Bilal, sections 3.1–3.2.

From a field transformation to a state transformation

A field is an operator, not a normalized one-particle state. Acting with a suitably smeared field on the vacuum produces a state, which may contain several particle contributions. Nevertheless, the relation between field transformations and state transformations is direct.

Suppose the vacuum is invariant under the supercharge under discussion:

$$Q_\alpha|0\rangle=0.$$

Insert and subtract $\phi Q_\alpha$:

$$\begin{aligned} Q_\alpha\bigl(\phi|0\rangle\bigr) &=(Q_\alpha\phi-\phi Q_\alpha)|0\rangle +\phi Q_\alpha|0\rangle\\ &=[Q_\alpha,\phi]|0\rangle\\ &=\psi_\alpha|0\rangle. \end{aligned}$$

Thus the field relation tells us how the states created by those operators are related. To identify definite particles, one must additionally isolate the relevant energy and momentum states. A supermultiplet is a collection of states or fields closed under the supersymmetry transformations; specify which meaning is intended.

The assumption about the vacuum matters. If $Q_\alpha|0\rangle\ne0$, the extra term $\phi Q_\alpha|0\rangle$ remains. A theory can have a supersymmetry of its equations and action while its chosen vacuum fails to preserve that symmetry. The simple particle-pair interpretation above then cannot be inferred by dropping that term.

A pair you can calculate completely

To see the algebra without spinor indices, consider a quantum-mechanical system with an even state $|b\rangle$ and an odd state $|f\rangle$, both of energy $E>0$. This is one energy block, not a complete oscillator or a model of spacetime spin. There is no gauge group or curved background.

Take an orthonormal ordered basis $(|b\rangle,|f\rangle)$ and set

$$H=E\begin{pmatrix}1&0\\0&1\end{pmatrix}, \qquad \Gamma=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad Q=\sqrt{2E}\begin{pmatrix}0&0\\1&0\end{pmatrix}.$$

Here $H$ is the Hamiltonian and the dagger denotes the Hilbert-space adjoint. These definitions give

$$Q|b\rangle=\sqrt{2E}|f\rangle,\qquad Q|f\rangle=0,$$

while $Q^\dagger$ takes $|f\rangle$ back to $\sqrt{2E}|b\rangle$ and annihilates $|b\rangle$. Multiplication yields

$$Q^\dagger Q=2E\begin{pmatrix}1&0\\0&0\end{pmatrix}, \qquad QQ^\dagger=2E\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$

Adding and comparing with $H$ gives the supersymmetry algebra in our normalization:

$$Q^2=(Q^\dagger)^2=0,\qquad \{Q,Q^\dagger\}=2H,\qquad [H,Q]=0.$$

This one complex supercharge contains two real, Hermitian generators,

$$S_1=\frac{Q+Q^\dagger}{\sqrt2}, \qquad S_2=\frac{Q-Q^\dagger}{i\sqrt2}.$$

Both are manifest here; direct substitution gives $\{S_a,S_b\}=2\delta_{ab}H$. In particular, $S_1^2=S_2^2=H$, whereas $Q^2=0$. The slogan “a supercharge squares to a translation” depends on which real or complex combination is meant.

Why must a nonzero partner have the same energy? For any energy eigenstate $|u\rangle$ in a system with $[H,Q]=0$,

$$H(Q|u\rangle)=QH|u\rangle=E\,Q|u\rangle.$$

If $Q|u\rangle\ne0$, it is another energy-$E$ state. In a relativistic theory, commuting with every momentum component similarly preserves momentum and therefore the mass of particle states. The transformation can change spin and parity while preserving these quantities.

There is also a useful norm check. For a normalized state in the quantum-mechanical algebra above,

$$2\langle H\rangle =\langle QQ^\dagger+Q^\dagger Q\rangle =\|Q^\dagger|u\rangle\|^2+\|Q|u\rangle\|^2 \ge0.$$

At positive energy at least one of the two actions is nonzero. At zero energy both must vanish, assuming a positive-definite inner product and a normalizable state in the operators’ domain. A zero-energy state therefore need not have a partner generated by these charges. The zero of energy here is the one fixed by the algebra; shifting $H$ by a constant while retaining the same $Q$ changes the stated relation.

Check the dimensions and the number of charges

Use $\hbar=c=1$ and let $[X]$ mean the mass dimension of $X$. Since $[H]=1$, the algebra $\{Q,Q^\dagger\}=2H$ requires

$$[Q]=\frac12.$$

An infinitesimal variation has the same dimension as the field being varied, so $\epsilon Q$ is dimensionless and $[\epsilon]=-1/2$. For canonically normalized free relativistic fields in dimension $d$, the scalar kinetic term has two derivatives and the fermion kinetic term has one. Requiring each Lagrangian density to have dimension $d$ gives

$$[\phi]=\frac{d-2}{2}, \qquad [\psi]=\frac{d-1}{2}.$$

Therefore $[\epsilon\psi]=[\phi]$ in both $d=3$ and $d=4$. These are engineering dimensions from the free kinetic terms, not a claim about anomalous dimensions at an interacting fixed point.

The symbol $\mathcal N$ counts different spinor copies in a dimension-dependent convention. It is not itself the number of real supercharges:

Setting	Real Lorentzian supercharges
The quantum-mechanical algebra just constructed	2
Three-dimensional $\mathcal N=2$	4
Four-dimensional $\mathcal N=1$	4

In four dimensions one two-component complex spinor charge has four real components; its adjoint does not double the independent count. In three dimensions $\mathcal N=2$ also has four real components, organized using three-dimensional spinors. The relevant flat-space algebras are displayed in Bilal, equation (3.2) and Closset and collaborators, equation (1.2). These are full flat-space counts. A chosen background or vacuum can preserve fewer, and a calculation may display only a subalgebra. Gauge rank is a separate quantity, conventionally written $N$ rather than $\mathcal N$.

Practice

1. Diagnose a proposed transformation

A student writes $\delta\phi=a\psi$, with $a$ an ordinary commuting number. Explain the parity problem. Repair it using an odd parameter, and check the dimensions in three dimensions.

Hint 1

An allowed variation must have the same parity as the field to which it is added.

Hint 2

For free fields in three dimensions, use $[\phi]=1/2$ and $[\psi]=1$.

Solution

Because $a$ is even, $a\psi$ is odd. It cannot be the even variation of $\phi$. Use an odd spinor parameter and contract its index: $\delta\phi=\epsilon^\alpha\psi_\alpha$. This product is even and a Lorentz scalar. Its dimension must be $1/2$, giving $[\epsilon]+1=1/2$, hence $[\epsilon]=-1/2$. The parity and dimension checks are independent: fixing one does not automatically fix the other.

2. Can the partners have different energies?

Replace the matrix example by

$$H=\begin{pmatrix}E_b&0\\0&E_f\end{pmatrix}, \qquad Q=\begin{pmatrix}0&0\\c&0\end{pmatrix},$$

where $c$ is complex. Determine when $[H,Q]=0$ and $\{Q,Q^\dagger\}=2H$ both hold. Include the case $c=0$.

Hint 1

Only the lower-left entry of $[H,Q]$ can be nonzero.

Hint 2

Compute $Q^\dagger Q$ and $QQ^\dagger$ separately before adding.

Solution

The commutator is

$$[H,Q]=\begin{pmatrix}0&0\\c(E_f-E_b)&0\end{pmatrix}.$$

If $c\ne0$, commutation requires $E_f=E_b$. The anticommutator is $|c|^2I$, so it imposes the stronger normalization condition

$$E_f=E_b=\frac{|c|^2}{2}.$$

If $c=0$, the commutator alone puts no restriction on the energies, but the anticommutator requires $E_b=E_f=0$. This distinguishes a conserved odd operator from an operator satisfying the full stated supersymmetry algebra. As a check, $c=\sqrt{2E}$ reproduces the worked example.

3. Which vacuum assumption did we use?

Keep $[Q,\phi]=\psi$, but suppose $Q|\Omega\rangle=|\chi\rangle\ne0$. Compute $Q(\phi|\Omega\rangle)$. Does a supersymmetric transformation law of the fields alone justify the simple pairing argument based on this vacuum?

Hint 1

Move $Q$ past $\phi$ without discarding its final action on the vacuum.

Hint 2

Compare the answer with the derivation using $Q|0\rangle=0$.

Solution

The operator identity gives

$$Q(\phi|\Omega\rangle) =\psi|\Omega\rangle+\phi|\chi\rangle.$$

The additional term is not controlled by the scalar–fermion relation alone. The action’s symmetry and the vacuum’s invariance are separate assertions. Thus this field law by itself does not justify identifying the two operator-created states as the simple pair discussed above. Setting $|\chi\rangle=0$ recovers the invariant-vacuum result. This does not deny the operator identity $[H,Q]=0$ when it holds; it identifies the missing assumption in translating fields into particle states around a chosen vacuum.

A first line of the three-manifold paper

In equation (6.8) of Supersymmetric Field Theories on Three-Manifolds, v3, the scalar transformation starts with $\delta\phi=\sqrt2\,\zeta\psi$. You can already recognize the scalar, the spinor field, the spinor contraction, and a normalization factor. The paper’s operations use commuting Killing-spinor labels and are odd; attaching an odd coefficient recovers an even variation as above. Its Euclidean spinors also require different conjugation bookkeeping from a Lorentzian Hilbert-space adjoint. Neither convention changes the distinction between a generator and its parameter.

The next calculation is Grassmann variables and their calculus. It makes the sign rule systematic and introduces differentiation and integration of odd variables. Before proceeding, answer this exit question in words: why does $[Q,\phi]$ have odd parity while $\phi+\epsilon[Q,\phi]$ remains a bosonic field?

You can also explore a supersymmetric harmonic oscillator. That optional branch starts with a Hamiltonian and derives its spectrum, making the distinction between paired positive-energy states and an unpaired normalizable zero mode explicit.