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  Vector field self-energy amplitude and imaginary part

+ 0 like - 0 dislike

Suppose we have some diagram $M(p)$ of self-energy of massless vector field (i.e., indeed the gauge field). Let's compute its imaginary part. Let's assume that we obtain
$$\text{Im}M(p) = c \times p^{4}\theta(p^{2}), \quad \text{where c is finite}$$

What does this mean?

asked Jan 10, 2017 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Jan 10, 2017 by NAME_XXX

Source of your formula?

@ArnoldNeumaier : I've found this result by computing the 6-th order self-energy diagram of the axial gauge field in the anomalous abelian gauge field theory. Precisely, its longitudinal part contains the imaginary part I'vr obtained. This result is connected with the anomaly, so I want to know what is the physical sense of it.

@ArnoldNeumaier : sorry, there is the mistake in the expression above. Instead of $M$ there must be $\text{Im}M$.

In general, an imaginary self-energy indicates (as in resonances) the inverse lifetime of an unstable state. One has it for example in the Lamb shift. I am not sure how to interpret it in a field theory.

@ArnoldNeumaier : indeed, this property stays in the quantum field theory due to optical theorem. However, this is connected to non-zero mass of decaying particle, while I don't understand whether the imaginary part above signals about the non-zero mass.

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