Null States and BPZ Differential Equations

The previous lesson translated conformal Ward identities into radial quantization. A primary field creates a highest-weight state; negative Virasoro modes create descendants. The next question is subtle and extremely powerful: are all descendants independent?

The Poincaré–Birkhoff–Witt monomials are linearly independent in the Verma module for every $(c,h)$. What changes at special values is that this module becomes reducible: a nonzero positive-level descendant can itself be a highest-weight state. Such a state is called a singular vector; the submodule it generates is null with respect to the standard contravariant form and is quotiented out in the irreducible module. Only in that quotient do its descendants become relations. In correlation functions, the same quotient becomes a differential equation.

The simplest nontrivial example occurs at level two. A relation of the form

$$\left(L_{-2}-\kappa L_{-1}^{2}\right)|\psi\rangle=0$$

holds in the irreducible quotient and turns an insertion of $L_{-2}\psi$ into $\kappa\partial_z^2\psi$. But $L_{-2}\psi$ can also be computed by inserting the stress tensor and using the Ward identity. Equating the two gives a second-order differential equation for every correlator containing $\psi$. This is the Belavin–Polyakov–Zamolodchikov, or BPZ, mechanism.

Verma modules and null descendants

Holomorphic conventions

We work in the holomorphic sector of a two-dimensional CFT. The Virasoro algebra is

$$[L_n,L_m]=(n-m)L_{n+m}+{c\over 12}n(n^2-1)\delta_{n+m,0}.$$

A primary state $|h\rangle$ obeys

$$L_0|h\rangle=h|h\rangle, \qquad L_n|h\rangle=0\quad(n>0).$$

The corresponding primary field has OPE

$$T(z)\phi(w)\sim {h\phi(w)\over (z-w)^2}+{\partial\phi(w)\over z-w}.$$

Many CFT books write the holomorphic conformal weight as $\Delta$. On this page $h$ denotes the holomorphic weight, while $\Delta=h+\bar h$ is reserved for the full scaling dimension when needed. The antiholomorphic sector has the same story with barred modes.

Given a highest-weight state $|h\rangle$, the Virasoro lowering operators $L_{-n}$ with $n>0$ generate descendants:

$$L_{-n_1}L_{-n_2}\cdots L_{-n_k}|h\rangle, \qquad n_i>0.$$

The level of this descendant is

$$N=n_1+n_2+\cdots+n_k,$$

and the $L_0$ eigenvalue is $h+N$. For example, level one has one state,

$$L_{-1}|h\rangle,$$

while level two has two natural basis states,

$$L_{-2}|h\rangle, \qquad L_{-1}^{2}|h\rangle.$$

A Verma module is the vector space spanned by all such descendants before imposing any additional null relations. A positive-level descendant $|\chi\rangle$ is a singular vector if it obeys

$$L_n|\chi\rangle=0\quad(n>0), \qquad L_0|\chi\rangle=(h+N)|\chi\rangle,$$

For the standard contravariant form, this highest-weight condition makes $|\chi\rangle$ orthogonal to the whole Verma module and hence zero norm. The converse is not true in an indefinite inner-product space: a zero-norm vector need not be singular. In a unitary theory positivity makes every null vector physically invisible; in a general highest-weight representation the safe algebraic statement is that $|\chi\rangle$ generates a proper submodule and the irreducible representation is the corresponding quotient.

A null descendant is both a descendant of $|h\rangle$ and a highest-weight state in its own right. All of its descendants form a null submodule, which is removed in the irreducible representation.

The quotient language is not cosmetic. It is what turns representation theory into differential equations. A null descendant can be represented locally as a differential operator acting on a field, and the statement that it vanishes in the quotient becomes an identity among correlation functions.

The level-one warm-up

At level one the only descendant is $L_{-1}|h\rangle$. It is primary if it is killed by all $L_n$ with $n>0$. It is enough to check $L_1$, since higher positive modes are even easier:

$$L_1L_{-1}|h\rangle=[L_1,L_{-1}]|h\rangle=2L_0|h\rangle=2h|h\rangle.$$

Therefore

$$L_{-1}|h\rangle\quad\text{is null only if}\quad h=0.$$

Thus every highest-weight module with $h=0$ has a level-one singular vector. In the identity module the quotient identifies it with zero:

$$L_{-1}|0\rangle=0.$$

In local language, this is just

$$\partial \mathbf 1=0.$$

The identity module does not begin with a genuine level-one state. Its first descendant not forced to vanish by global invariance is

$$L_{-2}|0\rangle,$$

which corresponds to the stress tensor $T(0)|0\rangle$. This is why the stress tensor lives in the identity module but is not itself the identity.

The level-one example is almost too simple, but it shows the pattern: a descendant becomes primary only when a coefficient forced by the Virasoro algebra vanishes. At level two the condition is no longer just $h=0$; it becomes a relation between $h$ and the central charge $c$.

The level-two null vector

At level two, take the most general descendant modulo normalization:

$$|\chi\rangle=(L_{-2}+aL_{-1}^{2})|h\rangle.$$

We want $|\chi\rangle$ to be primary:

$$L_n|\chi\rangle=0\qquad(n>0).$$

It is enough to impose $L_1|\chi\rangle=0$ and $L_2|\chi\rangle=0$. The higher conditions follow from commutators, because for example $L_3$ is proportional to $[L_1,L_2]$.

First compute the $L_1$ condition. Since $L_1|h\rangle=0$,

$$L_1L_{-2}|h\rangle=[L_1,L_{-2}]|h\rangle=3L_{-1}|h\rangle.$$

Also

$$\begin{aligned} L_1L_{-1}^{2}|h\rangle &=[L_1,L_{-1}^{2}]|h\rangle \\ &=\left(2L_0L_{-1}+2L_{-1}L_0\right)|h\rangle \\ &=2(2h+1)L_{-1}|h\rangle. \end{aligned}$$

Therefore

$$L_1|\chi\rangle=\bigl[3+2a(2h+1)\bigr]L_{-1}|h\rangle.$$

The condition $L_1|\chi\rangle=0$ gives

$$\boxed{ a=-{3\over 2(2h+1)}. }$$

Now impose the $L_2$ condition. The Virasoro algebra gives

$$L_2L_{-2}|h\rangle=[L_2,L_{-2}]|h\rangle =\left(4L_0+{c\over2}\right)|h\rangle =\left(4h+{c\over2}\right)|h\rangle,$$

and

$$L_2L_{-1}^{2}|h\rangle=6h|h\rangle.$$

Thus

$$L_2|\chi\rangle=\left(4h+{c\over2}+6ah\right)|h\rangle.$$

Substituting the value of $a$ gives the level-two null-vector condition

$$4h+{c\over2}-{9h\over 2h+1}=0.$$

Equivalently,

$$\boxed{ 16h^2+(2c-10)h+c=0. }$$

When this quadratic relation holds, the null state is

$$\boxed{ |\chi\rangle=\left(L_{-2}-{3\over 2(2h+1)}L_{-1}^{2}\right)|h\rangle. }$$

A level-two vector $|\chi\rangle=(L_{-2}+aL_{-1}^2)|h\rangle$ is primary only if its images under $L_1$ and $L_2$ vanish. These two requirements fix $a$ and impose one quadratic relation between $h$ and $c$.

The two roots are

$$h={5-c\pm\sqrt{(c-1)(c-25)}\over16}.$$

For the Ising value $c=1/2$, these are

$$h={1\over16}, \qquad h={1\over2}.$$

These are precisely the holomorphic weights of the spin field $\sigma$ and the energy field $\varepsilon$ in the Ising CFT. Their antiholomorphic weights are the same, so the corresponding full scaling dimensions are $1/8$ and $1$. This is the first hint that null-vector constraints know the operator algebra of the model.

Kac labels and the Coulomb-gas parametrization

It is useful to package the special weights by two integers. For $c\le1$, choose the real branches

$$\alpha_\pm =\sqrt{{1-c\over24}}\pm\sqrt{{25-c\over24}}, \qquad \alpha_+\alpha_-=-1.$$

For other values of $c$, the same formulas are understood by analytic continuation with a consistent branch choice.

The Kac weights are

$$\boxed{ h_{r,s} ={c-1\over24}+{1\over4}\left(r\alpha_+ + s\alpha_-\right)^2, \qquad r,s\in\mathbb Z_{>0}. }$$

The two level-two solutions are

$$h_{1,2}, \qquad h_{2,1}.$$

The notation $\psi_{1,2}$ means “the degenerate primary with Kac labels $(1,2)$,” not “$\psi$ divided by $2$.” This tiny notation trap is responsible for a surprising amount of blackboard chaos.

For $c=1/2$ one finds

$$\alpha_+={2\over\sqrt3}, \qquad \alpha_-=-{\sqrt3\over2},$$

and hence

$$h_{1,1}=0, \qquad h_{1,2}={1\over16}, \qquad h_{2,1}={1\over2}.$$

The next page will develop the Kac table more systematically. Here we only need the fact that $h_{1,2}$ and $h_{2,1}$ are exactly the two weights for which the level-two null vector exists.

Null-field decoupling

Let $\psi(z)$ be a primary field whose state obeys the level-two null relation

$$\left(L_{-2}-\kappa L_{-1}^{2}\right)|\psi\rangle=0, \qquad \kappa={3\over2(2h_\psi+1)}.$$

In local operator language this becomes

$$\left(L_{-2}\psi\right)(z)-\kappa\partial_z^2\psi(z)=0$$

inside all correlators, after passing to the irreducible module. This statement is often called null-vector decoupling:

$$\left\langle \left[\bigl(L_{-2}\psi\bigr)(z)-\kappa\partial_z^2\psi(z)\right] \prod_{i=1}^N\phi_i(z_i) \right\rangle=0.$$

The power of this identity is that the two terms can be evaluated in very different ways. The $L_{-1}^2$ term is just a second derivative with respect to the insertion point $z$. The $L_{-2}$ term is computed by inserting the stress tensor and using the Ward identity.

The mode $L_{-2}$ acting on a field at $z$ is represented by the contour integral

$$\bigl(L_{-2}\psi\bigr)(z) ={1\over2\pi i}\oint_z {dw\over w-z}\,T(w)\psi(z).$$

Therefore, for

$$G(z;z_i)=\left\langle\psi(z)\prod_{i=1}^N\phi_i(z_i)\right\rangle,$$

we have

$$\left\langle\bigl(L_{-2}\psi\bigr)(z)\prod_i\phi_i(z_i)\right\rangle ={1\over2\pi i}\oint_z {dw\over w-z}\, \left\langle T(w)\psi(z)\prod_i\phi_i(z_i)\right\rangle.$$

Now deform the contour away from $z$. The global Ward identities make the stress-tensor correlator fall sufficiently rapidly at infinity, so the contour there gives no contribution for this kernel. The small contour around $z$ is therefore equal to minus the sum of small contours around the other insertions. The OPE

$$T(w)\phi_i(z_i)\sim {h_i\phi_i(z_i)\over(w-z_i)^2}+{\partial_{z_i}\phi_i(z_i)\over w-z_i}$$

gives the residues. The result is

$$\boxed{ \left\langle\bigl(L_{-2}\psi\bigr)(z)\prod_i\phi_i(z_i)\right\rangle = \sum_{i=1}^N \left( {h_i\over(z-z_i)^2}+{1\over z-z_i}\partial_{z_i} \right)G(z;z_i). }$$

The operator $L_{-2}\psi(z)$ is represented by a stress-tensor contour with kernel $(w-z)^{-1}$. Deforming the contour to the other insertions converts it into a differential operator acting on their positions.

Combining this with null-vector decoupling gives the BPZ equation

$$\boxed{ \left[ \kappa\partial_z^2 - \sum_{i=1}^N \left( {h_i\over(z-z_i)^2}+{1\over z-z_i}\partial_{z_i} \right) \right] G(z;z_i)=0, \qquad \kappa={3\over2(2h_\psi+1)}. }$$

This is not merely a global Ward identity. Global conformal invariance constrains correlators by first-order equations. A null vector adds a higher-order equation because the representation itself is reducible.

Four-point functions and the cross ratio

A particularly important case is a four-point function with one degenerate insertion. Place three primary fields at $0$, $1$, and $\infty$:

$$G(z)=\lim_{R\to\infty}R^{2h_3} \left\langle\psi(z)\phi_1(0)\phi_2(1)\phi_3(R)\right\rangle.$$

The remaining coordinate $z$ is the cross ratio. The BPZ equation becomes an ordinary differential equation with regular singular points at

$$z=0, \qquad z=1, \qquad z=\infty.$$

This is why hypergeometric functions appear so naturally in minimal-model four-point functions.

After using global conformal invariance to put three insertions at $0$, $1$, and $\infty$, a correlator with a level-two degenerate field satisfies a second-order differential equation in the cross ratio $z$.

For the raw four-point function above, let $\partial_0$ and $\partial_1$ mean derivatives with respect to those insertion points before setting them to $0$ and $1$. After taking the normalized $R\to\infty$ limit, the global Ward identities give

$$\partial_0G=(z-1){dG\over dz}+A G, \qquad \partial_1G=-z{dG\over dz}-A G,$$

where

$$A=h_\psi+h_1+h_2-h_3.$$

Substitution into the BPZ equation gives

$$\boxed{ \left[ \kappa {d^2\over dz^2} +{2z-1\over z(z-1)}{d\over dz} -{h_1\over z^2} -{h_2\over(z-1)^2} +{h_\psi+h_1+h_2-h_3\over z(z-1)} \right]G(z)=0. }$$

Different choices of prefactor for the reduced conformal block shift the first-derivative and potential terms, but the content is the same: its holomorphic dependence lies in the two-dimensional local solution space of a second-order Fuchsian equation. The differential equation alone does not fix the physical correlator; OPE coefficients, antiholomorphic pairing, single-valuedness, and crossing symmetry select the allowed linear combination.

Near $z=0$, let

$$G(z)\sim z^p.$$

The leading terms give the indicial equation

$$\kappa p(p-1)+p-h_1=0.$$

In OPE language the exponent is

$$p=h_{\text{int}}-h_\psi-h_1,$$

where $h_{\text{int}}$ is the weight of the intermediate primary appearing in the OPE $\psi\times\phi_1$.

For the Ising spin field, $c=1/2$ and

$$h_\psi=h_1={1\over16}, \qquad \kappa={4\over3}.$$

The indicial equation becomes

$${4\over3}p(p-1)+p-{1\over16}=0,$$

with solutions

$$p=-{1\over8}, \qquad p={3\over8}.$$

These correspond to

$$h_{\text{int}}=0, \qquad h_{\text{int}}={1\over2},$$

so the BPZ equation recovers the Ising fusion rule

$$\sigma\times\sigma=\mathbf 1+\varepsilon.$$

A second-order BPZ equation has two local solutions near an OPE limit. Algebraically, the level-two null state restricts the fusion of a degenerate field to two possible channels.

For the standard Kac family, this statement takes the form

$$\phi_{1,2}\times\phi_{r,s} =\phi_{r,s-1}\oplus\phi_{r,s+1},$$

before applying field identifications or boundary truncations of a particular minimal model. At an edge of the Kac table one nominal channel may therefore be absent.

This is the key conceptual point. The singular vector is an algebraic relation in the state space. The BPZ equation is the same relation written in position space. The allowed OPE channels are encoded in the local exponents of that differential equation.

Null-state quotients are not gauge fixing

The null-state quotient is already an intrinsic statement about a stand-alone CFT representation. Generic Virasoro descendants are genuine states and operators; one does not remove them merely because Virasoro generators implement local conformal transformations. Only the proper submodule generated by a singular vector is set to zero in the irreducible representation.

When a CFT is coupled to two-dimensional gravity, or used as worldsheet matter in string theory, diffeomorphism and Weyl gauge fixing introduces constraints and a ghost sector. Physical states are then defined by a BRST cohomology problem involving the combined matter-plus-ghost system. BRST-exact states are gauge redundancies, but that construction is not identical to quotienting a matter Verma module by a BPZ null submodule.

Both procedures use quotients, which explains the useful analogy in the manuscript, but their origins must remain distinct: one is reducibility of a Virasoro representation, while the other is gauge redundancy of a gravitational or string theory.

Summary

A Virasoro Verma module is freely generated by the lowering modes $L_{-n}$. At special values of $(c,h)$ it becomes reducible because a nonzero descendant is itself highest weight. Such a singular vector generates a null submodule, and the irreducible representation is obtained by quotienting it out.

At level one, $L_{-1}|h\rangle$ is singular precisely when $h=0$; in the identity module its vanishing reflects $\partial\mathbf 1=0$. At level two, the null vector is

$$\left(L_{-2}-{3\over2(2h+1)}L_{-1}^{2}\right)|h\rangle,$$

provided

$$16h^2+(2c-10)h+c=0.$$

The two solutions are the Kac weights $h_{1,2}$ and $h_{2,1}$. If a correlator contains such a degenerate field $\psi$, null-vector decoupling and the stress-tensor Ward identity imply the BPZ equation

$$\left[ {3\over2(2h_\psi+1)}\partial_z^2 - \sum_i \left( {h_i\over(z-z_i)^2}+{1\over z-z_i}\partial_{z_i} \right) \right] \left\langle\psi(z)\prod_i\phi_i(z_i)\right\rangle=0.$$

This is one of the great miracles of two-dimensional conformal field theory: representation theory turns correlation functions into solvable differential equations.

Common pitfalls

A null state is not just any descendant with zero expectation value. It is a descendant that lies in a null submodule and is set to zero in the irreducible representation. In a unitary theory this agrees with zero norm and orthogonality to all states; in nonunitary models the quotient language is safer.

Conversely, zero norm alone is not enough in an indefinite inner-product space. The decisive algebraic property here is that a positive-level descendant is also highest weight and therefore generates a proper submodule.

The operator $L_{-2}\psi$ is not generally equal to a second derivative. It becomes proportional to $\partial^2\psi$ only for a degenerate field whose state obeys a level-two null relation.

The Kac label $\psi_{1,2}$ is a pair of integers. It is not a fraction. The comma matters.

The BPZ equation for a reduced conformal block depends on the prefactor convention. The coordinate-invariant statement is the unreduced equation with derivatives with respect to all other insertion points.

Two local BPZ solutions mean at most two candidate holomorphic channels, not two automatically nonzero OPE coefficients. The spectrum, Kac-table identifications, and crossing-consistent OPE data decide which allowed channels are actually present.

Finally, BPZ null states should not be identified wholesale with gauge states. Matter Virasoro descendants are generally physical; worldsheet gauge reduction requires the separate BRST construction with ghosts.

Exercises

Exercise 1: The level-one singular vector

Show that $L_{-1}|h\rangle$ is a null primary only when $h=0$.

Solution

A primary state satisfies $L_n|h\rangle=0$ for $n>0$ and $L_0|h\rangle=h|h\rangle$. At level one the only descendant is $L_{-1}|h\rangle$. It is primary if $L_nL_{-1}|h\rangle=0$ for all $n>0$.

The nontrivial condition is $n=1$:

$$L_1L_{-1}|h\rangle=[L_1,L_{-1}]|h\rangle=2L_0|h\rangle=2h|h\rangle.$$

Therefore $L_1L_{-1}|h\rangle=0$ only if $h=0$. For $n\ge2$,

$$[L_n,L_{-1}]=(n+1)L_{n-1},$$

and $L_{n-1}|h\rangle=0$ because $n-1>0$. Hence $h=0$ is the only condition.

Exercise 2: The level-two degeneracy condition

Derive the level-two null-vector condition

$$16h^2+(2c-10)h+c=0.$$

Solution

Start with

$$|\chi\rangle=(L_{-2}+aL_{-1}^2)|h\rangle.$$

The $L_1$ condition gives

$$L_1L_{-2}|h\rangle=3L_{-1}|h\rangle,$$

and

$$L_1L_{-1}^2|h\rangle=2(2h+1)L_{-1}|h\rangle.$$

Thus

$$3+2a(2h+1)=0,$$

so

$$a=-{3\over2(2h+1)}.$$

The $L_2$ condition gives

$$L_2L_{-2}|h\rangle=\left(4h+{c\over2}\right)|h\rangle,$$

and

$$L_2L_{-1}^2|h\rangle=6h|h\rangle.$$

Therefore

$$4h+{c\over2}+6ah=0.$$

Substituting $a$ gives

$$4h+{c\over2}-{9h\over2h+1}=0.$$

Multiplying by $2(2h+1)$ yields

$$16h^2+(2c-10)h+c=0.$$

Exercise 3: Null-field decoupling

Let $\psi$ be a level-two degenerate primary with

$$\left(L_{-2}-\kappa L_{-1}^2\right)|\psi\rangle=0.$$

Use the stress-tensor Ward identity to prove

$$\left[ \kappa\partial_z^2 - \sum_{i=1}^N \left({h_i\over(z-z_i)^2}+{1\over z-z_i}\partial_{z_i}\right) \right] \left\langle\psi(z)\prod_i\phi_i(z_i)\right\rangle=0.$$

Solution

Let

$$G(z;z_i)=\left\langle\psi(z)\prod_i\phi_i(z_i)\right\rangle.$$

The null relation gives

$$\left\langle(L_{-2}\psi)(z)\prod_i\phi_i(z_i)\right\rangle = \kappa\partial_z^2G.$$

Represent $L_{-2}$ by a contour integral around $z$:

$$\left\langle(L_{-2}\psi)(z)\prod_i\phi_i(z_i)\right\rangle ={1\over2\pi i}\oint_z{dw\over w-z} \left\langle T(w)\psi(z)\prod_i\phi_i(z_i)\right\rangle.$$

Deform this contour to contours around the other insertions. The OPE

$$T(w)\phi_i(z_i)\sim {h_i\phi_i(z_i)\over(w-z_i)^2}+{\partial_{z_i}\phi_i(z_i)\over w-z_i}$$

gives the residue

$${h_i\over(z-z_i)^2}G+{1\over z-z_i}\partial_{z_i}G.$$

Thus

$$\left\langle(L_{-2}\psi)(z)\prod_i\phi_i(z_i)\right\rangle =\sum_i\left({h_i\over(z-z_i)^2}+{1\over z-z_i}\partial_{z_i}\right)G.$$

Equating this expression with $\kappa\partial_z^2G$ proves the BPZ equation.

Exercise 4: Ising fusion exponents

For the Ising spin field, take $c=1/2$ and $h_\psi=h_1=1/16$. Use the BPZ equation near $z=0$ to find the two possible OPE exponents in $\psi(z)\phi_1(0)$.

Solution

For $h_\psi=1/16$,

$$\kappa={3\over2(2h_\psi+1)}={3\over2(9/8)}={4\over3}.$$

Near $z=0$, write $G(z)\sim z^p$. The leading terms in the BPZ equation give

$$\kappa p(p-1)+p-h_1=0.$$

Substituting $h_1=1/16$ and $\kappa=4/3$ gives

$${4\over3}p(p-1)+p-{1\over16}=0.$$

Multiplying by $48$,

$$64p^2-16p-3=0.$$

The roots are

$$p={3\over8}, \qquad p=-{1\over8}.$$

Since an OPE exponent has the form

$$p=h_{\text{int}}-h_\psi-h_1,$$

and $h_\psi+h_1=1/8$, these correspond to

$$h_{\text{int}}={1\over2}, \qquad h_{\text{int}}=0.$$

Thus the Ising spin OPE has the two channels

$$\sigma\times\sigma=\mathbf 1+\varepsilon.$$

References

• A. A. Belavin, A. M. Polyakov, and A. B. Zamolodchikov, “Infinite conformal symmetry in two-dimensional quantum field theory,” Nuclear Physics B 241 (1984), 333–380.
• P. Di Francesco, P. Mathieu, and D. Sénéchal, Conformal Field Theory, Springer (1997), Chapters 6–8.
• P. Ginsparg, “Applied Conformal Field Theory,” in Fields, Strings and Critical Phenomena, Les Houches Session XLIX, Elsevier (1989), pp. 1–168.