research32 min studyLesson 706 in Physics

BRST transformations and nilpotent charge

Use a graded differential to encode gauge algebra and define physical equivalence within its perturbative domain.

Before you begin

On this page

The question for today

What you’ll understand

Apply the graded product rule with anticommuting ghosts.
Check nilpotency in a concrete SU(2) ghost component.
Explain BRST-exact gauge fixing and the conditional cohomological physical-state construction.

Gauge fixing changes the manifest symmetry of a path-integral description, but a larger algebra involving ghosts can retain a nilpotent invariance. BRST symmetry organizes that algebra. Its central operation squares to zero, allowing both invariant quantities and physically redundant quantities to be identified systematically. The construction is especially effective in anomaly-free perturbation theory with a compatible regulator and gauge patch. Extending the same formulas to every nonperturbative field configuration requires additional global analysis.

Promote the gauge parameter to a ghost

Use the component adjoint derivative inherited from the Hermitian-generator convention. Replace the infinitesimal gauge parameter by a Grassmann ghost c, and let the odd BRST differential s raise ghost number by one. The ghost itself must transform quadratically so that two successive transformations cancel. Treating its coefficients as ordinary commuting numbers would erase this term and destroy the representation of the non-Abelian algebra.

Add an antighost and a commuting auxiliary field b with transformations that form a simple doublet. This auxiliary field allows the differential to square to zero without imposing the equations of motion. Eliminating it can make nilpotency appear to hold only after using a field equation, obscuring the clean off-shell algebra. Overall factors of i in the auxiliary-field definition vary by convention; here they are absorbed into b and kept fixed.

sA_\mu^a=(D_\mu c)^a,\qquad sc^a=-\frac g2 f^{abc}c^bc^c,\qquad s\bar c^a=b^a,\qquad sb^a=0

The ghosts and antighosts are odd, while A and b are even. These transformations use the module convention D=partial-minus-i-g-A; auxiliary-field factors have been absorbed into b.↔ Scroll wide equations horizontally.

Use a graded rule rather than ordinary differentiation

For a product, s passes an odd factor with a minus sign. That rule combines with anticommutation of ghost components and the Lie-algebra Jacobi identity to produce nilpotency. Checking one field is not enough in a general theory, but a component calculation exposes the signs concretely. The identity for s squared on the connection additionally uses the covariant derivative and the quadratic ghost transformation, rather than assuming gauge transformations simply commute.

The same structure can be represented by a conserved BRST charge Q in a suitable gauge-fixed quantum state space. Q has ghost number one and is nilpotent when the quantum symmetry is maintained. The larger state space includes indefinite-metric and ghost sectors, so its vectors are not automatically observable states. The operator statement requires attention to its domain, regularization, and anomalies; it is not established merely by manipulating classical transformation symbols.

s(XY)=(sX)Y+(-1)^{|X|}X(sY),\qquad s^2=0,\qquad Q^2=0

The parity |X| is zero or one. Classical nilpotency follows from closure and Jacobi; the quantum charge statement assumes a well-defined anomaly-free realization with the stated gauge-fixing domain.↔ Scroll wide equations horizontally.

Build exact gauge fixing and physical equivalence

An odd gauge-fixing fermion Psi produces a BRST-invariant addition s Psi because applying s again vanishes. This packages the gauge-fixing and ghost terms together and helps explain why changing a gauge prescription need not change properly defined physical quantities. Such reasoning assumes the measure is invariant and no boundary of the integration region contributes. A restriction introduced to handle gauge copies can make those assumptions nontrivial.

In a controlled perturbative construction, physical states are BRST closed, while states differing by a BRST-exact vector represent the same physical class. Exact vectors lie inside the closed space because Q squares to zero. Additional positivity and asymptotic-state arguments establish a physical Hilbert space in the intended setting. This cohomological quotient does not by itself prove the spectrum of a confining gauge theory, and gauge-fixed elementary field states need not survive as physical particles there.

S_{\rm gf+gh}=s\Psi,\qquad \mathcal H_{\rm phys}=\frac{\ker Q|_{{\rm gh}=0}}{\operatorname{im}Q|_{{\rm gh}=-1}}

The quotient identifies ghost-number-zero closed states modulo exact states coming from ghost number minus one. It describes a BRST construction under its anomaly, domain, and physical-positivity assumptions, not an unconditional nonperturbative solution.↔ Scroll wide equations horizontally.

Work it through

An SU(2) nilpotency check

In a dimensionless SU(2) ghost algebra take $g=2$ . Compute $sc^1$ explicitly and verify $s^2c^1=0$ , retaining graded signs and the independent Grassmann generators.

Antisymmetry gives two contributions to the first component. Since $c^3c^2=-c^2c^3$ , they combine to $sc^1=-2c^2c^3$ , rather than cancelling as they would for commuting coefficients.
Use the graded product rule: $s^2c^1=-2[(sc^2)c^3-c^2(sc^3)]$ . The remaining transformations are $sc^2=-2c^3c^1$ and $sc^3=-2c^1c^2$ .
Substitution gives products proportional to $c^3c^1c^3$ and $c^2c^1c^2$ . Each vanishes after anticommuting repeated generators together, since the square of a Grassmann generator is zero. Thus the component is nilpotent.

The result

The first ghost transforms as minus twice c-two c-three and has zero second BRST variation; graded multiplication is essential to the calculation.

The living notebook

Explore the model

Interactive lab

Interactive lab · quantum

Matching probabilities and physical equivalence

Reset and set mixing angle ninety degrees and phase zero; predict equal basis probabilities and Bloch x one. Compare a fresh Reset run at ninety degrees and phase 180 degrees; predict Bloch x minus one. These states have different relative phase, so they are not the same physical ray. The pure-qubit lab has no ghosts, graded differential, BRST quotient, or gauge-state space.

Bloch vector / P(0)P(1)Phase latitude

Probability |0⟩: 71.1%
Probability |1⟩: 28.9%
Bloch (x, y, z): 0.78, 0.45, 0.42

Mixing angle θ65°Relative phase φ30°

t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forPhysical equivalence needs a precise rule; matching one measurement distribution is insufficient and does not illustrate BRST cohomology by itself.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Nilpotency gives Q(Q chi)=0. This permits a quotient of closed states by exact ones, while positivity, operator domains, and a physical interpretation require additional assumptions.

Your turn

Try it yourself

In a separate dimensionless toy algebra let $sx=c,sc=0,s\bar c=b,sb=0$ . For $\Psi=\bar c(x+\xi b/2)$ compute s-Psi. Set $x=3,b=1,\xi=2$ to give its bosonic part, and verify that a second variation vanishes.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The antighost is odd, so passing s through it contributes a minus sign. Keep the ghost term until the second variation is checked.

Compare with the solution

The first variation is $s\Psi=bx+\xi b^2/2-\bar c c$ . Its bosonic part at the supplied values is $3+1=4$ , while the Grassmann term remains minus antighost times ghost. Applying s gives $bc-bc=0$ because sb and sc vanish. Dropping the ghost term would leave bc and spoil the cancellation. This is an algebraic finite model, not a gauge-field path integral.

Keep the idea

Take it with you

BRST nilpotency depends on ghost anticommutation and graded differentiation.
An auxiliary field keeps the basic algebra off shell.
The physical-state quotient needs quantum consistency and global-domain assumptions.

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