I think that gauge invariance of a Lagrangian is not a sufficient condition for the Ward identity to be valid. So why does the Ward identity happen to hold in Yang-Mills theory, and maybe in many other gauge theories which I'm not familiar with? Or why do physical states with equivalent polarizations have the following relation: $$ |e^\prime,p\rangle=|e,p\rangle+Q|\mathrm{some\,state}\rangle $$ where $e,e^\prime$ are polarizations that can be related by a gauge transformation with the coupling constant set to be zero, and $Q$ is the BRST charge. Is it a coincidence or is there a way to determine whether the Ward identity holds in a general gauge theory other than to explicitly carry out the calculations?


1 Answer 1


The Ward identities are the statement that if we write the scattering amplitude for an external photon with polarization $\zeta$ and momentum $k$ as $M = \zeta^\mu M_\mu$, then we have $k^\mu M_\mu = 0$. This equation is important because it shows that the spurious longitudinal polarizations proportional to the photon momentum decouple from all physical processes because their scattering amplitudes are generically zero.

The Ward identity holds for every Yang-Mills gauge theory in which the gauge field couples to a conserved current, and it also holds for massive vector field theories that don't have gauge symmetries, provided they still couple to a current in the form of the interaction Lagrangian being $A^\mu j_\mu$ where $j_\mu$ is some function of the matter fields the gauge field couples to. Gauge invariance of the Lagrangian and this form of couplied lead directly to the Ward identities through the general Ward-Takahashi identity:

For $j^\mu$ the conserved current of a global continuous symmetry (which is part of every gauge symmetry), we have that $$ \partial_\mu \langle T j^\mu(x) \phi(x_1)\dots \phi(x_n)\rangle = - \mathrm{i}\sum_{j=1}^n \langle \phi(x_1)\dots\phi(x_{j-1}) \delta\phi(x_j)\phi(x_{j+1})\dots \phi(x_n)\tag{1}$$ holds for all fields $\phi$, where $\delta \phi$ is the variation of $\phi$ under the symmetry. A scattering amplitude involving an external photon is schematically $$ M(k) = \zeta^\mu \mathrm{i} \int \mathrm{e}^{-\mathrm{i}kx} \partial_x^2 \langle T A_\mu(x)\cdot\text{other stuff}\rangle$$ and since $\frac{\delta S}{\delta A^\mu} = \partial_x^2 A_\mu(x) - j^\mu$ holds classically, we get $$ \partial_x^2 \langle TA_\mu(x) \dots\rangle = \langle T j_\mu(x) \dots \rangle + \text{contact terms}$$ upon use of the Dyson-Schwinger equation. The contact terms are $n-1$-point functions and don't contribute to the connected $n$-point function we're trying to compute, so we may neglect them.

Finally, for $\zeta^\mu = k^\mu$, we may use the Fourier relationship between $k$ and $\partial$ to pull the $\zeta^\mu$ into the integral, giving us $\partial_\mu\langle j^\mu\dots\rangle$ inside the integral. By Ward-Takahashi (eq. (1)), this also only consists of contact terms that don't contribute to the scattering amplitude, and therefore $\zeta^\mu M_\mu = 0$.

The only assumptions that went into this argument are that we have a global continuous symmetry with conserved current $j_\mu$, and that the equation of motion for the gauge field is $\partial_x^2 A = j_\mu$. This is for the abelian case.

For the non-Abelian Yang-Mills case, the corresponding identities get more complicated, although they still follow from the general logic of the Ward-Takahashi identities. The non-Abelian versions are called Slavnov-Taylor identities and also involve the Faddeev-Popov ghost fields, but effectively also mean that the longitudinal polarization decouple from all physical processes.

Finally, we can address the "equality" $$ \lvert e,p\rangle = \lvert e',p'\rangle + Q\lvert \psi\rangle.$$ This should be thought of as an equality we impose on the Hilbert space of states for get rid of negative/zero norm states. A priori, $\lvert e,p\rangle$ and $\lvert e',p'\rangle$ are different states, but we force this equation in the usual manner of a quotient space onto the physical space of states. Two elements that differ by an image of $Q$ are declared to be equal, more precisely, the physical Hilbert space is hte cohomology of $Q$, that is, $\ker(Q)/\mathrm{im}(Q)$. The Ward identity ensures that this quotient is physically harmless - only because the longitudinal polarizations (which correspond to $\mathrm{im}(Q)$ in the non-Abelian BRST formulation) decouple we are allowed to say that two states differing only by such a polarization are equal, since this guarantees that the scattering amplitudes fo all states we just declared equal actually are equal. Without the Ward identity, taking the quotient by the zero norm states is physically inconsistent.

  • $\begingroup$ 1. Do the Slavnov-Taylor identities hold for any gauge theory or do they hold only for Yang-Mills theories? 2. The states $A(x)$ and $A^\prime (x)$ are classically equivalent if they can be related by a gauge transformation and satisfy the same boundary condition, because they both describe the evolution of the system from a particular initial state. But how to understand that the quantum mechanical states $|e,p\rangle$ and $|e^\prime,p\rangle$ are equivalent? $\endgroup$
    – Xavier
    Commented Nov 14, 2016 at 15:34
  • $\begingroup$ 3. Why can the equation $|e,p\rangle=|e^\prime,p\rangle+Q|\psi\rangle$ be used to get rid of negative-norm states? Isn't the nonexistence of negative-norm states be guaranteed by the fact that they do not exist in another gauge like the axial gauge in Yang-Mills theory? 4. Most importantly, I'd like to know whether there's an easy way to check the invariance of an amplitude under the gauge transformation of a polarization (if the Slavnov -Taylor identities do not hold for a general gauge theory). $\endgroup$
    – Xavier
    Commented Nov 14, 2016 at 15:36
  • $\begingroup$ @Xavier 1. Only for Yang-Mills. However, the unphysical states in the naive state space (i.e. the states in $\mathrm{im}(Q)$) must decouple by the general BRST procedure, if that's what you're concerned about (this answers also 4.). 2. The BRST operator encodes the gauge symmetry, so the quantum mechanical states are equivalent because they are related by a gauge transformation, nothing unclassical here. 3. The niave state space fo a gauge theory has neg. norm and zero norm states. I don't have enough space here to rehash the entire BRSt procedure to get rid of these. $\endgroup$
    – ACuriousMind
    Commented Nov 14, 2016 at 15:48

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