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I have come across the following statement:

The anomalies of QCD cannot be reproduced by a collection of free fermions carrying $U(1)_V$, $SU(N_f)_L$ and $SU(N_f)_R$ quantum numbers. That is why one must either have a CFT or a spontaneous symmetry breaking with massless bosons and Wess-Zumino-Witten terms.

I do not understand this statement. Can somebody elaborate this statement and recommend a good reference about it? I appreciate the help.

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    $\begingroup$ where did you read this? $\endgroup$
    – AccidentalFourierTransform
    Commented Oct 7, 2017 at 16:25
  • $\begingroup$ I haven't read it somewhere, I heard it from a person. $\endgroup$
    – QGravity
    Commented Oct 10, 2017 at 7:43

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The statement is relatively simple. The pure QCD has the so-called internal anomalies - the non-zero coefficients $$ \tag 1 D_{abc} = \text{tr}\left[ [t_{a},t_{b}]_{+}t_{c}\right] $$ for the triangle diagram with running inside currents of the QCD global group $$ G\simeq SU_{L}(3)\times SU_{R}(3)\times U_{B}(1), $$ with $a,b,c$ corresponding to the generators of $G$. Although there are no gauge fields associated to these currents, these coefficients must be reproduced at any scale of the theory. This statement is called the anomaly matching, and firstly it was discovered by 't Hooft. In particular it means that if below some scale a degree of freedom making the contribution into $(1)$ die out, then below this scale there must be new degrees of freedom making exactly the same contribution as the died one. For the QCD with its confinement, this means that $$ \tag 2 D_{abc}\bigg|_{quarks} = D_{abc}\bigg|_{\text{quark bound states}}, $$ where the quark bound states belong to some representation of $G$.

In general, it is hard to calculate the right hand-side of $(2)$. Fortunately, there is non-trivial statement that the contribution into $D_{abc}$ can come only from massless degrees of freedom (or from particles whose mass directly violates the given generator symmetry). Spin-3/2 and higher degrees of freedom are forbidden because of the Lorentz covariance (the statement is known as Weinberg-Witten no-go theorem), the similar statement is true about the spin-1 degrees of freedom, so the only possible candidates are massless fermions and massless spin-zero particles. The latter are typically associated with the spontaneous symmetry breaking and then are called the Goldstone bosons. Therefore, the anomaly matching tells us that

either in the QCD there are massless spin $1/2$ bound states reproducing $(1)$, or there is the SSB with the Goldstone bosons reproducing $(1)$.

For the QCD it was found that, assuming first existence of massless bound fermions, it is impossible to construct their representations which match $(2)$. Contrary, there is possible to construct an effective action composed from the Goldstone bosons which matches $(2)$. This action is called the Wess-Zumino-Witten action.

As for the resource I would recommend you Weinberg's QFT, Volume 2, 22.5.

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  • $\begingroup$ I'm curious how these results cohabit with the proposed existence of the mass gap (which is essential for QCD to be experimentally successful, and backed up by numerical lattice computations). Nambu-Goldstone bosons are always massless unless the underlying $SU(3)\times U(1)$ symmetry is not exact. $\endgroup$
    – Prof. Legolasov
    Commented Oct 7, 2017 at 23:12
  • $\begingroup$ Thank you very much for the nice answer. There are two issues in that the section you mentioned that I do not fully understand. On page 391, the second paragraph, it is mentioned that 1) $N$ for the $SU(N)$ color group is taken to be odd so there could be untrapped fermionic bound state. How could that happen?; 2) The anomaly constant for $SU_L(n)−SU_R(n)−U(1)_Y$ is non-zero. What does $Y$ stand for in $U(1)_Y$? Is it hypercharge?; 3) Why does $k$ in (22.5.2) should be odd? $\endgroup$
    – QGravity
    Commented Oct 10, 2017 at 7:41
  • $\begingroup$ @QGravity : the $k$ number is odd because you need to construct the representations corresponding to the fermionic quarks bound states. This can be done by taking odd total number of quarks and antiquarks. The $U_{Y}(1)$ symmetry corresponds to the baryon number global symmetry, which is $U_{B}(1)$. Actual QCD classical approximate symmetry is $G_{\text{full}} \simeq U_{L}(3)\times U_{R}(3) \simeq SU_{L}(3)\times SU_{R}(3)\times U_{B}(1)\times U_{A}(1)$, but its $U_{A}(1)$ part is broken by the anomaly. $\endgroup$
    – Name YYY
    Commented Oct 10, 2017 at 12:37
  • $\begingroup$ @SolenodonParadoxus : are you talking about the non-zero mesons mass? If not, what mass gap do you mean? $\endgroup$
    – Name YYY
    Commented Oct 10, 2017 at 17:31
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    $\begingroup$ @NameYYY to prove the existence of the mass gap in a (nonperturbative) QCD theory, one has to demonstrate that there exists a real number $\Delta > 0$ such that any fluctuation of the vacuum (meaning any physical state not proportional to the Poincare vacuum state of the full (interacting) theory) has energy of at least $\Delta$. $\endgroup$
    – Prof. Legolasov
    Commented Oct 11, 2017 at 6:54

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