Weyl spinors and the N=1 supersymmetry algebra

Two supersymmetry operations are supposed to produce a translation. In four dimensions each operation carries a two-component spinor index, while a translation carries a vector index. The matrices that connect these indices determine both the algebra and the signs in its first field-theory realization.

We will construct those matrices, check their identities, and count the independent generators. The only site preparation is Grassmann variables and their calculus, including its distinction between odd variables and operators. Four-dimensional Lorentz transformations and elementary QFT are assumed; three-dimensional spinors and curved backgrounds are not.

Our setting is flat four-dimensional Lorentzian spacetime, with the full $\mathcal N=1$ Poincaré supersymmetry algebra manifest. There is no gauge group, background source, or chosen interacting action in this calculation. We use $\hbar=c=1$ and

$$\eta_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1), \qquad \mu,\nu=0,1,2,3.$$

Why two kinds of spinor index?

Write the Pauli matrices as

$$\tau_1=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad \tau_2=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\qquad \tau_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$

They obey $\tau_i\tau_j=\delta_{ij}I+i\epsilon_{ijk}\tau_k$, with $\epsilon_{123}=1$. Package a real four-vector into a Hermitian matrix,

$$X=x^0I+x^i\tau_i =\begin{pmatrix} x^0+x^3&x^1-ix^2\\ x^1+ix^2&x^0-x^3 \end{pmatrix}.$$

Its determinant is $(x^0)^2-\boldsymbol x^2$. If $S$ is a complex two-by-two matrix with $\det S=1$, then $X'=SXS^\dagger$ is Hermitian and has the same determinant. This constructs a proper, time-orientation-preserving Lorentz transformation. The matrices $S$ and $-S$ give the same vector transformation: the group $SL(2,\mathbb C)$ is the double cover used to act on spinors.

An undotted Weyl spinor is a column $\psi_\alpha$, $\alpha=1,2$, transforming as $\psi'=S\psi$. A lower dotted spinor transforms with $S^*$, the entrywise complex conjugate matrix. In Lorentzian field theory the conjugate of $\psi$ is therefore written

$$\bar\psi_{\dot\alpha}=(\psi_\alpha)^\dagger.$$

For classical Grassmann fields the dagger denotes conjugation with reversal of product order; for quantum operators it is the adjoint. A dot labels the conjugate Lorentz representation. It is neither another spacetime index nor a second independent Lorentzian Weyl field. Bilal, equations (2.6)–(2.7)

For example, $S=\operatorname{diag}(e^{\rho/2},e^{-\rho/2})$ produces

$$x'^0=\cosh\rho\,x^0+\sinh\rho\,x^3,\qquad x'^3=\sinh\rho\,x^0+\cosh\rho\,x^3.$$

The two spinor components acquire the half-rapidity factors $e^{\pm\rho/2}$. This is a concrete transformation law, not a statement about two particle polarizations.

Raise indices before suppressing them

Use the following antisymmetric tensors, also for dotted indices:

$$(\epsilon^{\alpha\beta})= \begin{pmatrix}0&1\\-1&0\end{pmatrix}, \qquad (\epsilon_{\alpha\beta})= \begin{pmatrix}0&-1\\1&0\end{pmatrix}, \qquad \psi^\alpha=\epsilon^{\alpha\beta}\psi_\beta.$$

Thus $(\psi^1,\psi^2)=(\psi_2,-\psi_1)$; lowering with $\epsilon_{\alpha\beta}$ reverses this operation. The identity $S^{\mathsf T}\epsilon S=\epsilon$ follows from $\det S=1$, so epsilon contraction is invariant.

Our scalar contractions are

$$\psi\chi=\psi^\alpha\chi_\alpha =\psi_2\chi_1-\psi_1\chi_2, \qquad \bar\psi\bar\chi=\bar\psi_{\dot\alpha}\bar\chi^{\dot\alpha}.$$

For odd fields, $\psi\chi=\chi\psi$: the antisymmetric epsilon and the exchange of two odd components contribute two minus signs. In particular,

$$\psi\psi=-2\psi_1\psi_2$$

need not vanish. For commuting spinor columns the same contraction is antisymmetric and a column contracted with itself vanishes. Spinor transformation and Grassmann parity remain different properties.

Conjugation reverses the order:

$$(\psi^\alpha\chi_\alpha)^\dagger =\bar\chi_{\dot\alpha}\bar\psi^{\dot\alpha} =\bar\psi\bar\chi$$

for odd fields. These are Bilal’s epsilon and contraction conventions, including the opposite displayed contraction order for dotted indices. Equations (2.9)–(2.15)

The matrices that turn spinors into vectors

Define the Lorentzian sigma pair by

$$\sigma^\mu_{\alpha\dot\alpha}=(I,-\tau_1,-\tau_2,-\tau_3), \qquad \bar\sigma^{\mu\dot\alpha\alpha}=(I,+\tau_1,+\tau_2,+\tau_3).$$

The bar on $\bar\sigma$ names the matrix with the other index placement; it does not instruct us to conjugate $\sigma^\mu$ entry by entry. More precisely,

$$\bar\sigma^{\mu\dot\alpha\alpha} =\epsilon^{\dot\alpha\dot\beta}\epsilon^{\alpha\beta} \sigma^\mu_{\beta\dot\beta}.$$

Raising both indices and putting the dotted one first produces the stated matrices. Lowering the spacetime index instead uses $\eta_{\mu\nu}$. Consequently the earlier vector matrix is $X=x_\mu\sigma^\mu=x^\mu\sigma_\mu$. Keeping these operations separate prevents a common spatial-sign error.

The first identity is

$$\boxed{\sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu =2\eta^{\mu\nu}I.}$$

The dotted index is summed in this matrix product, leaving one lower and one upper undotted index. With $\sigma$ and $\bar\sigma$ exchanged, the analogous identity acts on dotted indices.

Check every type of pair. For $\mu=\nu=0$, the left side is $2I$. For a time-space pair it is $\tau_i-\tau_i=0$. For two spatial indices it is

$$-\tau_i\tau_j-\tau_j\tau_i=-2\delta_{ij}I.$$

For instance, $\sigma^1\bar\sigma^2=-i\tau_3$ and $\sigma^2\bar\sigma^1=+i\tau_3$ cancel. Squaring a single spatial matrix product instead gives $\sigma^1\bar\sigma^1=-I$, checking the metric sign.

Why is this identity useful? On a smooth field in flat space, the two derivatives commute. Therefore

$$(\sigma^\mu\partial_\mu)(\bar\sigma^\nu\partial_\nu) =\frac12(\sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu) \partial_\mu\partial_\nu =I\Box,\qquad \Box=\partial_0^2-\sum_i\partial_i^2.$$

The antisymmetric matrix part disappears against a symmetric second derivative. This is the same operation that removes unwanted second-derivative terms in a closure calculation.

A second identity changes how spinor indices are paired:

$$\boxed{\sigma^\mu_{\alpha\dot\alpha} \bar\sigma_{\mu}^{\dot\beta\beta} =2\delta_\alpha{}^\beta\delta_{\dot\alpha}{}^{\dot\beta}.}$$

Here the spacetime index is contracted, while all four spinor indices remain free. To derive it, expand an arbitrary two-by-two matrix $M$ in the basis $I,\tau_i$:

$$M=\frac12\operatorname{tr}(M)I +\frac12\sum_i\operatorname{tr}(\tau_iM)\tau_i.$$

The coefficients follow from $\operatorname{tr}\tau_i=0$ and $\operatorname{tr}(\tau_i\tau_j)=2\delta_{ij}$. Comparing the coefficient of each entry of $M$ gives

$$\delta_{ab}\delta_{cd}+\sum_i(\tau_i)_{ab}(\tau_i)_{cd} =2\delta_{ad}\delta_{bc}.$$

Take $(a,b,c,d)=(\alpha,\dot\alpha,\dot\beta,\beta)$. The two spatial minus signs in $\sigma^i\bar\sigma_i$ cancel, giving the boxed identity. Contracting it with four odd spinors requires an additional Grassmann reordering; Exercise 1 keeps that sign visible. Both identities follow directly from the source’s matrices, rather than from a three-dimensional gamma identity. Bilal, equations (2.11)–(2.18)

The algebra and its four real generators

A supercharge $Q_\alpha$ is an odd operator transforming as an undotted Weyl spinor. Its Lorentzian adjoint is $\bar Q_{\dot\alpha}=(Q_\alpha)^\dagger$. The ordinary $\mathcal N=1$ super-Poincaré algebra is

$$\begin{aligned} \{Q_\alpha,\bar Q_{\dot\beta}\} &=2\sigma^\mu_{\alpha\dot\beta}P_\mu,\\ \{Q_\alpha,Q_\beta\}&=0,\qquad \{\bar Q_{\dot\alpha},\bar Q_{\dot\beta}\}=0,\\ [P_\mu,Q_\alpha]&=[P_\mu,\bar Q_{\dot\alpha}]=0. \end{aligned}$$

The Lorentz transformation laws are the ones just constructed; the translations commute with one another. The mixed anticommutator transforms as a vector, matching its momentum-valued right side. This is an algebraic definition, not a consequence of naming a scalar and a fermion.

On a positive-frequency c-number mode $e^{-ip_\nu x^\nu}$, the wave-function momentum operator $i\partial_\mu$ has eigenvalue $p_\mu$. Distinguish this from the adjoint action on a quantum field, $[P_\mu,\Phi(x)]=-i\partial_\mu\Phi(x)$: a field’s annihilation term removes momentum from a state. Bilal’s superspace realization also uses $P_\mu=-i\partial_\mu$, as stated in equation (4.10). These are different actions of translations, not a rephasing of mutually adjoint supercharges. When comparing differential formulas, specify which action is being represented. Bilal, equation (4.10)

There is no Lorentz-scalar central term in the same-chirality anticommutator for this $\mathcal N=1$ algebra: a candidate proportional to $\epsilon_{\alpha\beta}$ is antisymmetric, whereas $\{Q_\alpha,Q_\beta\}$ is symmetric. We are discussing the ordinary particle algebra, without additional charges carried by extended defects. Bilal, equations (3.1)–(3.3)

To count real generators, for each $\alpha=1,2$ form

$$S_{\alpha,1}=\frac{Q_\alpha+Q_\alpha^\dagger}{\sqrt2}, \qquad S_{\alpha,2}=\frac{Q_\alpha-Q_\alpha^\dagger}{i\sqrt2}.$$

These are four Hermitian generators. Conversely $Q_\alpha=(S_{\alpha,1}+iS_{\alpha,2})/\sqrt2$. Writing $Q$ and $\bar Q$ therefore does not give eight independent real generators. It displays one complex two-component charge and its adjoint. All four belong to the algebra being used; $\mathcal N=1$ counts its single Weyl copy, not a single real generator. This count is also distinct from the physical polarizations of a matter field.

The normalization can be tested without a model. Taking the two diagonal components and using $\operatorname{tr}\sigma^\mu=2\delta^\mu_0$ gives

$$P_0=\frac14\sum_{\alpha=1}^2\{Q_\alpha,Q_\alpha^\dagger\}.$$

On a normalized state in a positive-norm Hilbert space, in the common domain of these operators,

$$\langle P_0\rangle =\frac14\sum_\alpha \left(\|Q_\alpha|u\rangle\|^2+ \|Q_\alpha^\dagger|u\rangle\|^2\right)\ge0.$$

This is a flat-space unitary statement. It is not a positivity proof for an arbitrary Euclidean functional integral. Since $[P_\mu]=1$, the algebra fixes $[Q]=1/2$. An odd parameter $\epsilon^\alpha$ in an even variation has dimension $-1/2$; for a free four-dimensional scalar and Weyl field, $[\phi]=1$ and $[\psi]=3/2$, so $\epsilon\psi$ has the scalar’s dimension. Source norm argument, equation (3.4)

Worked momentum check

Take a momentum eigensector with contravariant momentum $p^\mu=(E,0,0,p)$, hence $p_\mu=(E,0,0,-p)$. The supercharge anticommutator matrix is

$$2\sigma^\mu p_\mu =2(EI+p\tau_3) =\begin{pmatrix}2(E+p)&0\\0&2(E-p)\end{pmatrix}.$$

At rest, $p=0$ and $E=m>0$, both diagonal entries are $2m$. The operators $a_\alpha=Q_\alpha/\sqrt{2m}$ obey $\{a_\alpha,a_\beta^\dagger\}=\delta_{\alpha\beta}$ and $\{a_\alpha,a_\beta\}=0$. This supplies a direct normalization check using familiar fermionic operators.

At $E=p>0$ the second diagonal entry vanishes. Positivity implies that $Q_2$ and its adjoint annihilate this momentum sector. The first pair remains nonzero, with anticommutator $4E$. This is a restriction of the representation at null momentum, not a reduction in the number of generators defining the full theory. Reversing the momentum exchanges which component vanishes.

Euclidean continuation changes conjugation

For later Euclidean calculations, first complexify the field variables and then continue $x^0=-ix^4$, so $\partial_0=i\partial_4$. This substitution alone does not specify a path-integral contour.

Define four-by-four matrices in chiral blocks,

$$\gamma_L^\mu= \begin{pmatrix}0&\sigma^\mu\\\bar\sigma^\mu&0\end{pmatrix}, \qquad \gamma_E^4=\gamma_L^0,\qquad \gamma_E^i=-i\gamma_L^i.$$

The checked Lorentzian identity gives $\{\gamma_L^\mu,\gamma_L^\nu\}=2\eta^{\mu\nu}I_4$. The factors of $-i$ then give $\{\gamma_E^a,\gamma_E^b\}=2\delta^{ab}I_4$, $a,b=1,2,3,4$, and

$$\sigma_E^a=(i\tau_1,i\tau_2,i\tau_3,I),\qquad \bar\sigma_E^a=(-i\tau_1,-i\tau_2,-i\tau_3,I),$$ $$\gamma_L^\mu\partial_\mu =i\gamma_E^a\partial_a.$$

This last identity checks the spatial signs against the chosen Wick rotation. The Euclidean sigma pair satisfies the corresponding identity with $\delta^{ab}$. The Lorentzian chiral blocks are Bilal’s equation (2.16); the Euclidean pair above is our explicitly derived continuation.

Why treat the two chiralities independently? The Euclidean spin group is $\operatorname{Spin}(4)=SU(2)_L\times SU(2)_R$. Its vector action can be exhibited on

$$Y=x^4I+ix^i\tau_i,\qquad \det Y=(x^4)^2+\boldsymbol x^2,\qquad Y'=U_LYU_R^\dagger.$$

The two matrices $U_L,U_R$ are independently chosen $SU(2)$ matrices. The two Weyl blocks transform under the separate factors. In Lorentzian signature a raised conjugate column transforms as $(S^\dagger)^{-1}\bar\psi^{\dot\alpha}$; the same $S$ determined both chiralities. No relation using ordinary complex conjugation can tie together two independently chosen Euclidean rotation factors. Exercise 3 demonstrates the failure with one finite transformation.

Thus a barred Euclidean field or parameter is an independent complex variable unless a separate, compatible condition is explicitly supplied. Complexifying four real Lorentzian supercharges gives a four-complex-dimensional algebra; it does not redefine the parent theory as having eight real Lorentzian supercharges. State the signature before applying an adjoint or a counting rule.

Practice

1. Keep the sign in a four-spinor rearrangement

Let $\xi,\chi,\bar\eta,\bar\lambda$ be independent odd spinors. Use the completeness identity to reduce

$$(\xi\sigma^\mu\bar\eta)(\bar\lambda\bar\sigma_\mu\chi)$$

to scalar contractions. What changes if all four spinors are commuting?

Hint 1

Write every spinor index and apply the identity before reordering the spinors.

Hint 2

Bring $\chi_\alpha$ next to $\xi^\alpha$, then put the dotted contraction in lower-left, upper-right order.

Complete solution

The matrix identity gives

$$2\xi^\alpha\bar\eta^{\dot\alpha} \bar\lambda_{\dot\alpha}\chi_\alpha.$$

Moving $\chi$ left through the two dotted odd factors contributes two minus signs. Interchanging the dotted factors contributes one more:

$$(\xi\sigma^\mu\bar\eta)(\bar\lambda\bar\sigma_\mu\chi) =-2(\xi\chi)(\bar\lambda\bar\eta).$$

For commuting spinors no exchange signs arise, and the coefficient is $+2$. The matrix identity itself has not changed.

For an explicit odd check, use four exterior generators $\theta_1,\ldots,\theta_4$ and lower columns

$$\xi=(\theta_1,0)^{\mathsf T},\quad \chi=(0,\theta_2)^{\mathsf T},\quad \bar\eta=(\theta_3,0)^{\mathsf T},\quad \bar\lambda=(0,\theta_4)^{\mathsf T}.$$

Then $\xi\chi=-\theta_1\theta_2$ and $\bar\lambda\bar\eta=\theta_3\theta_4$. Both the scalar expression and explicit sigma multiplication give $2\theta_1\theta_2\theta_3\theta_4$.

2. Can the algebra act at spacelike momentum?

For a general real $p^\mu=(E,\boldsymbol p)$, find the eigenvalues of $2\sigma^\mu p_\mu$. Determine the condition for positivity. Explain what a zero eigenvalue implies for a supercharge combination, and why this does not change the theory’s value of $\mathcal N$.

Hint 1

Use $(\boldsymbol p\cdot\boldsymbol\tau)^2=|\boldsymbol p|^2I$, or rotate the spatial momentum to the third axis.

Hint 2

For an eigenvector $v$, form the corresponding linear combination of $Q_\alpha$ and use the sum of its two squared norms.

Complete solution

The matrix is $2(EI+\boldsymbol p\cdot\boldsymbol\tau)$, with eigenvalues $2(E+|\boldsymbol p|)$ and $2(E-|\boldsymbol p|)$. Both are nonnegative precisely when $E\ge|\boldsymbol p|$. In particular, a spacelike eigensector cannot carry this algebra on a positive-norm Hilbert space with the stated adjoint.

Let $v$ be a normalized eigenvector and set $q_v=\sum_\alpha v_\alpha^*Q_\alpha$. In this momentum sector,

$$\|q_v|u\rangle\|^2+\|q_v^\dagger|u\rangle\|^2 =v^\dagger(2\sigma^\mu p_\mu)v\,\|u\|^2.$$

A zero eigenvalue forces both actions to vanish. For $E=|\boldsymbol p|>0$ there is one such complex combination. For $E=|\boldsymbol p|=0$, all charges annihilate the state, subject to the same domain and positivity assumptions. These are statements about a representation at fixed momentum. The defining algebra still has four real generators. At $\boldsymbol p=0,E>0$, both eigenvalues become $2E$, checking the rest-frame calculation.

3. Test a proposed Euclidean conjugation rule

Let $\psi_L$ and $\chi_R$ denote the two Euclidean Weyl columns, transforming by $U_L$ and $U_R$. Someone imposes the Lorentzian-looking rule $\chi_R=\epsilon\,\psi_L^*$, with $\epsilon=(\epsilon^{\alpha\beta})$. Test it under $U_L=-I$, $U_R=I$. Does treating the Euclidean columns independently mean that the Lorentzian theory had twice as many supercharges?

Hint 1

The proposed transformation is allowed because the two $SU(2)$ factors are independent.

Hint 2

Compare the transformed right column with the conjugate of the transformed left column before imposing any equality.

Complete solution

The actual transformation gives $\psi_L'=-\psi_L$ and $\chi_R'=\chi_R$. The proposed rule would instead require

$$\chi_R'=\epsilon(\psi_L')^* =-\epsilon\psi_L^*=-\chi_R.$$

It fails for a nonzero column. The allowed Euclidean vector transformation is $Y'=-Y$; it preserves the positive quadratic norm and orientation in four dimensions. Thus this is a valid rotation test, not an inadmissible reflection of a single coordinate.

Even within one Euclidean Weyl factor, the invariant antilinear map $J\psi=\epsilon\psi^*$ has $J^2=\epsilon\epsilon^*=-I$. It is not an ordinary reality condition $J\psi=\psi$ for a single nonzero Weyl column.

Independent complex variables describe the complexified continuation. Returning to a Lorentzian theory requires its Lorentzian adjoint structure; the original four-real-supercharge count is unchanged. A Euclidean functional integral also requires its own contour and reality assumptions, which this rotation test does not choose.

The next calculation

You can now specify each index and sign in $\{Q_\alpha,\bar Q_{\dot\beta}\}=2\sigma^\mu_{\alpha\dot\beta}P_\mu$, verify the sigma identities, and explain why a Weyl charge and its adjoint contain four real generators. A useful exit test is to reproduce the null-momentum matrix and identify which norm argument makes one combination act trivially.

The next question is how scalar, Weyl, and auxiliary fields realize this algebra while preserving a free action. Chiral and vector multiplets performs that component calculation using these four-dimensional conventions directly. The sigma identities prepare it; they do not by themselves prove closure of a particular multiplet or invariance of its action.