# A Question about a $U(1)_{B-L}$

I know I can write the QCD lagrangian like this:

$$\mathcal{L} = (i\bar{q}_{R} \gamma_{\mu}\partial_{\mu} {q}_{R} + i\bar{q}_{L}\gamma_{\mu}\partial_{\mu} {q}_{L}) + \text{other terms}$$

When written this way we say it is invariant under $SU(3)_{R}\times SU(3)_{L} \times U(1) \times U(1)$.

But in a book, "Stefan Scherer & Matthias Schindler - A Primer for Chiral Pertubation Theory", it says: $SU(3)_{R}\times SU(3)_{L} \times U(1)_{B-L}$.

Why that?

• "In a book" is a useless reference. If you are citing something, tell us what. Also, whoever writes down $\mathrm{U}(1)_{B-L}$ should explain what is meant by that. May 24, 2016 at 20:41
• @ACuriousMind: The book is: "Stefan Scherer & Matthias Schindler - A Primer for Chiral Pertubation Theory" May 24, 2016 at 21:33
• You should edit such information into the question. May 24, 2016 at 21:54

I can't find any reference to $U(1)_{B-L}$ in the book mentioned. The $L$ here refers to lepton number, and it's suspicious that this should arise in a book on QCD.

This aside, what symmetries does the QCD Lagrangian have? In the case that we have $N$ flavours of massless quark, we are free to rotate the left-handed quarks amongst themselves (in flavour space) and the right-handed quarks amongst themselves independently. This leads to the global symmetry:

$$U(N) \times U(N)$$

At the level of the algebra, this symmetry can also be written

$$SU(N)_L \times SU(N)_R \times U(1)_L \times U(1)_R$$

where the subscripts indicate whether the symmetry acts on the left- or right-handed quarks. Instead of independently rotating left or right, sometimes it can be useful to consider rephasing left and right by the same amount, or rephasing them by opposite amounts. These are referred to as vector and axial symmetries, respectively. One can show that any rephasing of just left, or just right, can be achieved by a combination of axial and vector rephasings. So $$U(1)_L \times U(1)_R = U(1)_V \times U(1)_A$$

Now for the punchline. In the quantum theory, the axial symmetry $U(1)_A$ is anomalous. This means that it is not a symmetry quantum mechanically, despite being a symmetry classically. So the symmetry group of massless QCD as a quantum theory is:

$$SU(N)_L \times SU(N)_R \times U(1)_V$$

One can check that a simultaneous rephasing of all quark flavours and chiralities by the same amount is precisely the symmetry corresponding to baryon number. So we can also write $U(1)_V = U(1)_B$. This is almost the end of the story, but we haven't yet explained why $U(1)_{B-L}$ should appear. It turns out that when we embed QCD into the standard model, and couple the quarks to the weak and hypercharge gauge bosons as well as the gluon, $U(1)_B$ is also anomalous. The same is true of $U(1)_L$, the symmetry that corresponds to rephasing all leptons simultaneously by the same amount. The only $U(1)$ global symmetry that survives in the quantum theory of the standard model is $B - L$, the symmetry corresponding to rotating all quarks by the same amount, and all leptons by the opposite amount.

I searched through the whole book and didn't find a single instance of "$U(1)_{B-L}$", so a page number reference would be helpful. But for the purposes of pure QCD, $U(1)_V$ is the same as $U(1)_B$, because all quarks carry the same baryon number, and that's the same as $U(1)_{B-L}$ since nothing carries lepton number.

One might prefer to write $U(1)_{B-L}$ rather than $U(1)_V$ if one has the rest of the Standard Model in mind. If you include the other Standard Model gauge fields, it turns out that $U(1)_V$ is anomalous, and $U(1)_L$ is anomalous as well; the unique non-anomalous combination is $U(1)_{B-L}$. This is nice because it gives us an absolutely conserved quantity, as well as a symmetry that may be gauged.

That still doesn't fully explain it, because if they were only going to write down symmetries that were non-anomalous in the SM, they shouldn't have written down $SU(3)_L \times SU(3)_R$ either, since this symmetry is anomalous as well. I suspect they were just writing out of habit without looking too closely at what they were doing.