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Chiral current VEV below the QCD scale

+ 3 like - 0 dislike
186 views

Let's have pure QCD. I know that after spontaneous symmetry breaking quark bilinear form are replaced by their averaged values:
$$
\bar{q}_{i}q_{j} \to  \langle \bar{q}_{i}q_{j}\rangle \approx \Lambda_{QCD}^3, \quad \bar{q}_{i}\gamma_{5}q_{j} \to  \langle \bar{q}_{i}\gamma_{5}q_{j}\rangle \approx 0
$$

What can be said about VEVs of $\partial_{\mu}\bar{q}_{i}\gamma^{\mu}\gamma_{5}q_{i}$, $(\partial_{\mu}(\bar{q}_{i}\gamma^{\mu}\gamma_{5}q_{i}))^2$ (for example, in terms of the path integral approach)?

More precisely, I've asked about correlator

$$
\int d^4x d^4t \delta (t)\langle 0| T(\partial_{\mu}^{x}J^{\mu}_{5}(x)\partial_{\nu}^{t}J^{\nu}_{5}(t))|0\rangle ,
$$

which appears when we want to calculate axion or axion-like particle mass which is generated during QCD crossover.

It seems that it is zero since no massless states couples to correlator
$$
\int d^{4}x e^{ikx}\langle 0 |T(J_{\mu}^{5}(x)J_{\nu}^{5}(0)) |0\rangle
$$

asked Dec 7, 2015 in Theoretical Physics by NAME_XXX (1,010 points) [ revision history ]
edited Dec 10, 2015 by NAME_XXX

For the first operator, it seems $\langle\partial_{\mu}\bar{q}_{i}\gamma^{\mu}\gamma_{5}q_{i}\rangle=\partial_{\mu}\langle\bar{q}_{i}\gamma^{\mu}\gamma_{5}q_{i}\rangle=\partial_\mu 0=0$. Or is there any reason you have in mind that the partial derivative doesn't commute with taking VEV?

EDIT: Or did you mean $\langle(\partial_{\mu}\bar{q}_{i})\gamma^{\mu}\gamma_{5}q_{i}\rangle$?

@JiaYiyang: It is customary that the derivative operator only applies to the shortest term afterwards for which it makes sense. Thus your second interpretation is the default meaning.

The second example by the OP is syntactically faulty, hence not clearly interpretable. I guess that the first opening parenthesis is spurious; the remaining parentheses then make sense and force the derivative operator to apply to the whole expression. Thus the first half of your remark applies that one can get its expectation as the derivative of the expectation of the square.

@ArnoldNeumaier, in that case I suspect there isn't much that can be easily said about that VEV, after all even the seemingly simpler $\langle\bar{q}q\rangle$ value comes from either phenomenology or numerical lattice study.

@JiaYiyang: It looks like one might be able to give nonrigorous order of magnitude estimates for some of the polynomial expressions, based on dimensional arguments. This would be in the spirit of the question. But I have currently no time to figure this out.

@dimension10 @Dilaton can we import the answer linked by tmshaefer?

@ArnoldNeumaier, given that the SE answer is taken as the right answer by OP, OP's notation of  $\partial_{\mu}\bar{q}_{i}\gamma^{\mu}\gamma_{5}q_{i}$ probably simply means $\partial_{\mu}(\bar{q}_{i}\gamma^{\mu}\gamma_{5}q_{i})$.....

1 Answer

+ 3 like - 0 dislike

By the chiral anomaly equation $$ \partial^\mu \bar{q}_f\gamma_\mu\gamma_5 q_f = \frac{N_f}{16\pi^2} \tilde{G}^a_{\alpha\beta}G^{a\,\alpha\beta} $$ this correlator is proportional to the topological susceptibility $$ \chi_{top}= \frac{1}{V}\frac{1}{(16\pi^2)^2}\int d^4x \int d^4 y \; \langle T\, \tilde{G}^a_{\alpha\beta}G^{a\,\alpha\beta}(x) \tilde{G}^b_{\gamma\delta}G^{b\,\gamma\delta}(y) \rangle $$ The topological susceptibility is zero if one of the quarks is massless, but it is non-zero in general, and of $O(\Lambda^4_{QCD})$ in pure gauge theory or the large $N_c$ limit.

This post imported from StackExchange Physics at 2015-12-14 11:01 (UTC), posted by SE-user Thomas
answered Dec 10, 2015 by tmschaefer (705 points) [ no revision ]

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