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In Quantum Field Theories (QFT) there is a well known phenomenon of anomalies, where a classical symmetry is broken in the quantum theory due to a so called anomaly. This symmetry breaking can be understood in the path intergral formulation to be a result of the non-invariance of the functional measure under the classical symmetry transformation. Although the action itself is invariant, the functional measure might not be, and therefore, if that is the case the path integral wouldn't be invariant either. For further reading you can look in wikipedia.

My question is: is it possible that the action wouldn't be invariant, and the measure neither, but the path integral would? That is to say, the lack of invarince of the action and the measure would cancel each other, to form an invariant path integral, in such a way that would give have a symmetry in the quantum theory, but not in the classical one.

I would guess that it isn't possible, but is there a proof?

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    $\begingroup$ I think I have a suggestion for a proof - in order for the theory to be invariant under the suggested transformation, all correlation functions must obey the symmetry. Therefore, by relating this correlation functions to the derivative of the partition function, you get many derivatives of the action. These derivatives must be invariant under the symmetry, and therefore, if all derivatives are invariant, you would expect the action itself to be invariant too. $\endgroup$ – itamarhason Jan 25 '16 at 8:19
  • $\begingroup$ Unless the action is not analytic, then there is a contraction. Maybe you could write your proof as an answer, and we could see if it makes sense or not. $\endgroup$ – Adam Jan 25 '16 at 12:16
  • $\begingroup$ @Adam - will do. $\endgroup$ – itamarhason Jan 25 '16 at 15:09
  • $\begingroup$ Rethinking about it, I think my "proof" is flawed because in getting the correlation functions you take the functional derivatives with respect to an auxiliary function. The fact that these derivative should vanish isn't in itself sufficient because these derivative don't encapsulate the behavior of the derivative with respect to the fields themselves, which is what I should be interested in. $\endgroup$ – itamarhason Jan 25 '16 at 15:22
  • $\begingroup$ In some situations, one can have quantum anomaly cancelled by a classical anomaly (i.e. coming from the action). An example is a Chern-Simons theory on a manifold with a boundary, with chiral fermions living on the boundary. There are also situations, if I recall correctly, when a quantum anomaly in one description of the system, say, in the UV theory, is reproduced by a classical anomaly in the IR description. $\endgroup$ – Peter Kravchuk Jan 28 '16 at 9:34
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Comments to the question (v2):

  1. Traditionally, the classical action $S$ sits in the Boltzmann factor $\exp\left[\frac{i}{\hbar} S\right]$ behind an inverse power of $\hbar$ in the path integral, while the path integral measure is independent of $\hbar$. In the conventional way of counting, we say that the Jacobian $J$ from the path integral measure is a one-loop effect proportional to $\hbar$, while the variation of the classical action $S$ is tree-level, i.e. independent of $\hbar$. Anyway, the upshot is, that in a usual setting, the two variations carry different $\hbar$-orders, and cannot cancel.

  2. However, in principle one may introduce a quantum action $$\tag{A} W(\hbar)~=~S+ \sum_{n=1}^{\infty}\hbar^n M_n~=~ S+\hbar M_1+{\cal O}(\hbar^2)$$ with quantum terms. Then the Boltzmann factor becomes $$\tag{B} \exp\left[\frac{i}{\hbar} W\right]~=~\exp\left[\frac{i}{\hbar} S\right]e^{iM_1}(1+{\cal O}(\hbar)),$$ so that a cancellation may formally take place between the $M_1$-action factor and the path integral measure.

  3. It seems appropriate to mention that such cancellation is the main idea behind the quantum master equation (QME) $$ \frac{1}{2}(W,W)~=~i\hbar\Delta W\tag{QME}$$ in the Batalin-Vilkovisky formalism. The lhs. and rhs. of the above QME are associated with the action and Jacobian, respectively, leading to a (generalized) BRST symmetry of the path integral.

  4. Nevertheless, in practice in a local QFT, the BV operator (aka. the odd Laplacian) $\Delta$ is singular object. The QME is typically only satisfied if both sides of the QME are zero separately, i.e. the action and the measure parts cancel separately in practical applications.

References:

  1. I.A. Batalin & G.A. Vilkovisky, Gauge Algebra and Quantization, Phys. Lett. B 102 (1981) 27–31.

  2. W. Troost, P. van Nieuwenhuizen & A. Van Proeyen, Anomalies and the Batalin-Vilkovisky lagrangian formalism, Nucl. Phys. B333 (1990) 727.

  3. nLab.

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  • $\begingroup$ 1. I know that the variation of the measure gives a one-loop effect, but the action's variation contributes to all loops, not only to the tree-level. I mean, a variation in the action would contribute to all loops via propagators which are induced from the action. The higher-loops contributions are induced from the action. The classical picture is induce from the action via the tree-level diagrams, but the loop diagrams are also induced from the action. Maybe you mean there is a tree level variation that cannot be removed by the measure variation - this I can accept. $\endgroup$ – itamarhason Jan 25 '16 at 15:28
  • $\begingroup$ I think we can reformulate your first comment in a simpler way: The measure variation is independet of $\hbar$ while the action has an $\hbar$ coefficient, so they just cannot cancel each other. $\endgroup$ – itamarhason Jan 25 '16 at 15:34
  • $\begingroup$ I think comments 2, 3 and 4 are just confusing, and 1 is a good answer by itself. $\endgroup$ – itamarhason Jan 25 '16 at 15:36
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jan 25 '16 at 16:07
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There is a simple proof that the cancellation is impossible (at least unless you are willing to add to the classical a term proportional to $\hbar$), I am reformulating an answer by @Qmechanic in a simpler language:

The anomaly, or the measure variation which is the typical source of the anomaly, contributes a term which is independent of $\hbar$, while any variation of the action comes with a $\frac{1}{\hbar}$ coefficient. Therefore, they just cannot cancel each other.

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