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# Tag Info

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Feynman parametrisation is NOT permitted for an integral which is more than logarithmically divergent. More specifically, the change of integration order after the Feynman parametrisation (or proper-time regularisation) is problematic for a quadratically divergent integral in your case. A hint: never attempt performing the quadratically divergent ...

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To fill in the details of mike stone's post, this does come from a dispersion relation. Since the Hilbert space is spanned by a complete set of physical states, we can introduce a cut-off and write the propagator as $$J_p(q^2) = \int^{\Lambda^2}_0d\kappa^2\frac{\rho(\kappa^2)}{q^2+\kappa^2-i\epsilon},$$ where $\rho$ is real. If we use the Poisson kernel ...

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This look like a dispersion relation integral. If I'm correct (I have not done any of the algebra so this is just a guess based on first appearances) you compute the discontinuity ${\rm Im}J(q^2)$ in across the normal threshold cut that starts from $s=q^2=4m^2$ and us it in a "slot" integral. Some subtraction may be necessary. The discontinuity comes from ...

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I am writing up the work done by OP with the help of some minor comments from me so that it is a record of their progress towards arriving at the answer to the question. OP's expression was missing the "Dirac indices" associated to the gamma matrices that enter the definition of the vertices and the propagators. OP introduces these and found that a product ...

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Consider one of the terms coming from the expansion of the numerator, $$\text{tr }\gamma^\mu\not k\gamma^\nu\not k=k_\alpha k_\beta\,\text{tr }\gamma^\mu\gamma^\alpha\gamma^\nu\gamma^\beta.$$ If you really wanted to write out the explicit matrix-element indices, this would be k_\alpha k_\beta\,(\gamma^\mu)_{ab}(\gamma^\alpha)_{bc}(\gamma^\nu)_{cd}(\...

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What you are writing as red is actually a quantum mechanical state, which is a vector in a Hilbert space. What you are writing as a dot is something called a tensor product; it produces a vector in another Hilbert space. You can add and subtract the vectors in Hilbert spaces, and you can multiply them by scalars like $2$ or $1/\sqrt 3$, just like in any ...

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First of all physics is not about proofs. It is about mathematical models that are fitted to data, and the same models make predictions for new experiments. If the predictions come true, the model is validated. To start with there was the parton model, before quarks as such were accepted as a standard model. When deep inelastic scattering data on protons ...

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