In a gauge theory (non-abelian for this question), I am told that two states $|\psi\rangle$ and $|\phi\rangle$ are to be identified if they are related by a gauge transformation $U(x)$


But means that two states are to be identified when the gauge transformation is homogenous, i.e. independent of $x$.

Does this formally mean that for the one-particle quark state, the red quark $|1_\text{Red}\rangle$, green quark $|1_\text{Green}\rangle$, and blue quark $|1_\text{Blue}\rangle$ are all identified as a single state, as they are all related by gauge transformations? I would say yes...

So then I ask, does this mean we are overcounting the number of color states when we calculate the color-averaged scattering amplitude? I want to say, no because there are many quarks in this world, and the color orientation of the incoming and outgoing quarks with respect to some other colored quark elsewhere is gauge-invariant....

Is this correct reasoning or poor thinking?


2 Answers 2


No, in a gauge theory, the two states are identified even if $U$ is the operator of the most general gauge transformation, i.e. one whose gauge parameter depends on $\vec x$. That's also a fact that is needed to eliminate unphysical (i.e. longitudinal and time-like) polarizations of the photon (or other gauge bosons).

Yes, a one-particle state with a blue quark is physically identified with the red quark state or the green quark state. (In QCD, we have confinement, so only color-neutral bound states may exist in isolation, anyway. That means that the ability to recolor particles doesn't give us any new freedom when we consider objects that may exist in isolation: they are color-neutral i.e. invariant under $SU(3)$, anyway.)

However, if there are at least two colored particles, e.g. two quarks, only the "uniform transformations" of their colors – not independent changes of the two quarks' colors, separately – are a gauge transformation. In the first paragraph, I said that the gauge transformation may depend on $\vec x$ so we actually can find an equivalent state in which e.g. all the quarks (at various positions) are changed to red quarks. But the gauge transformations needed to achieve this recoloring has $\lambda(x,y,z)$ that nontrivially depends on the location in space. And because it does, $\partial_\mu \lambda(x,y,z)$ is nonzero, and therefore the value of the gauge field (e.g. gluon field in this example) in between the quarks is also changing.

So every state with quarks at various positions may be changed, by a gauge transformation, to an equivalent state in which all the quarks are red; however, the value of the gluon field in between has to be changed, too. So if it was zero at the beginning, it won't stay zero after the gauge transformation. If you only consider gauge-equivalent states that don't change the gluon field, you have to consider gauge transformations whose $\lambda$ is location-independent, and those gauge transformations can never change the "relative color between two quarks" (or other colored particles).

We are not multiply counting anything when we're averaging because the average $(A+A+A)/3$ is nothing else than $A$ again, so it doesn't matter whether you average three equal things or not. And if you compute the inclusive cross section of a process in which colorful particles are produced, you have to sum up over all different colors of the final quarks because they're determined in an "absolute way": their color may be fixed relatively to the colors of quarks in the initial state etc. and relative colors are gauge-invariant (if we don't allow the gauge field to change), as I have mentioned.

So be sure that textbooks of QFT don't make any mistake if they ignore the fact that the differently colored one-quark states are physically equivalent.

Let me mention one more thing. At the beginning, you said that only states related by gauge transformations $U$ where $U$ is given by a location-constant $\lambda$ are identified. I said it wasn't true: $U$ may come from a location-dependent $\lambda(x,y,z)$ and they're still identified.

Correction: when careful, red quark and blue quark one-particle states are not identified

In fact, the truth was even "further in my direction" than what I said. It's true that states related by gauge transformations with location-dependent gauge parameters are identified. But if we are more rigorous, there is an extra constraint: we should only consider the transformations in which $\lambda$ goes to the identity ($1$ or $0$, depending on whether we use the multiplicative or additive notation: additive is usual for $U(1)$, multiplicative is needed in the non-Abelian case) when $|\vec x|\to\infty$. The physical states don't have to be invariant under gauge transformations that are nontrivial even at infinity. That's why the "global $SU(3)$ transformations" are actually not required to keep the physical states invariant, and we shouldn't consider one-blue-quark and one-red-quark states to be identified. That's a good thing because if we made this identification, the $U(1)$ counterpart of this assertion would be that a one-electron state and its multiple by $\exp(i\alpha)$ are identified – implying that the state is identified with the zero vector. That would be too bad because charged particles would be prohibited (unphysical).

  • $\begingroup$ Two questions: (1) Where does the constraint that states are identified only when related by a gauge transformation satisfying $\lambda\rightarrow 1$ as $|\vec x|\rightarrow\infty$ come from? (2) In the $U(1)$ counterpart, why does the identification of the electron state with its phase multiple $e^{i\alpha}$ imply that a state is identified with the zero vector? That doesn't sound right. $\endgroup$
    – QuantumDot
    Nov 24, 2012 at 16:15
  • $\begingroup$ Dear QuantumDot, (1) the actual physical reason why gauge symmetry is a part of physics is that it's needed to eliminate the negative-norm excitations of gauge bosons anywhere in the bulk of the space, and this job is already achieved by the gauge transformations that are trivial at infinity. It's consistent to require the invariance under these "at infinity trivial" gauge transformations only and it's really desirable because of (2) - we don't want charged states to be banned. $\endgroup$ Nov 26, 2012 at 12:55
  • $\begingroup$ (2) If $\psi\sim e^{i\alpha}\psi$ where $\sim$ is an equivalence and the equivalence acts linearly on the Hilbert space, then one may simply subtract $\psi$ from both sides to get that also $0\sim (e^{i\alpha}-1)\psi$, so the state has to be equivalent to zero because the numerical prefactor isn't zero. I am not saying this is true for the single-electron state. On the contrary, I am saying that we surely can't afford this to be the case, so the assumption can't hold, either. We can't identify states that differ by the overall phase only because that would mean that we identify them with zero. $\endgroup$ Nov 26, 2012 at 12:58
  • $\begingroup$ Many thanks for your response! So, to summarize if $\psi\sim e^{i\alpha}\psi$ then therefore $\psi\sim 0$. And because of this, we can't afford $\psi\sim e^{i\alpha}\psi$ to be true. Do I interpret your response correctly? $\endgroup$
    – QuantumDot
    Nov 26, 2012 at 21:52

Nope. Two states related by a (local) gauge transformation: $$ \psi(x)\mapsto e^{i\alpha(x)}\psi(x), \quad A_\mu(x) \mapsto A_\mu(x) - \partial_\mu \alpha(x), $$ where $$ \quad \alpha(x)\to 0,\quad x\to \infty $$ are the same state.

But a global gauge transformation $\alpha(x)=const\ne 0$ is not a gauge transformation! The electric charge $Q$ for example, is a conserved quantity given by the global U(1) transformation: $$ \psi(x) \mapsto e^{i\alpha} \psi(x) $$ and thus global transformation is physically observable. Or writing down its action on state space, a state $\vec{a}$ is called having charge $q$ if: $$ Q\vec{a} = q \vec{a} $$ If ‘two states differ by a global gauge transformation are considered the same’, then any nonzero state can NOT have nonzero charge!!! This is NOT what we want!!!

P.s. \vec should be understood as \ket but I found PSE can’t recognize \ket……


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