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Cornell University, Department of Physics Question 1: Potentials From May 12, 2017 Cornell University, Department of Physics PHYS 4444, Particle physics, HW # 8, due: 3/30/2017, 11:40 AM Question 1: Potentials from QFT As we talked in class, in the limit where particle production and annihilation can be neglected, one can use the idea of potentials. In QFT, this can be used to find how a potential is generated from particle exchange. In particular, the statement that \the photon is the carrier of the electric force" can be understood. 1. Review scattering in non-relativistic quantum mechanics, and write down the relation between the transition amplitude and the potential. Then, invert it and write the potential in terms of the amplitude. It is this later relation that we will exploit to get potentials from QFT. Note that we use a relativistic normalization such that the we write the non-relativistic scattering amplitude as Z f(~k;~k0) = d3xei(~k−~k0)xV (x) (1) This is different by a fator of 2π=m compared to the usuall normalization. Answer: We invert the potential and get (we omit the vector sign) 3 Z d k 0 V (x) = e−i(k−k )xf(k − k0) (2) (2π)3 First we will look at the Yukawa potential, and afterwards we will do the Coulomb potential. We consider the process AB ! AB, where A and B are charged massive scalars, and there is one more real massive scalar, C. The interaction Lagrangian is: 2 2 −Lint = eC(jAj + jBj ); (3) where e is a coupling constant. 2. Explain why AB ! AB can be used to extract the potential while AA ! BB is not suitable. Answer: We can only use the concept of potentials when particle creation and an- nihilation can be neglected. This is the case in AB ! AB but not in AA ! BB. 1 3. Draw the Feynman diagram for the AB ! AB process and calculate the amplitude. Answer: 2 −e in out 2 2 ; q ≡ pA − pA (4) mc − q 4. Working in the center of mass frame, show that: −e2 A = 2 2 (5) j~qj + mc where ~q is the spatial component of qµ. In other reference frames, this will be true in a non-relativistic approximation. in out Answer: In the CM we have EA = EA and thus qµ = (0; ~q). Remembering the minus sign from the metric we get what we should. 5. Using the relation between the amplitude and the potential show that e−mcr V (r) = a (6) r and calculate a. Recall that Z Z Z 1 3 2 −iqr q −mcr d ~q = q dqdφd(cos θ); dqe 2 2 = −iπe (7) −∞ q + mc Answer: For notational simplicity, we will re-use q for j~qj. Z 3 −i~q·~x 2 d ~q e V (x) = −e 3 2 2 (2π) q + mc 2 Z 2π Z 1 2 Z 1 e q −iqr cos θ = − 3 dφ dq 2 2 d(cos θ)e 8π 0 0 q + mc −1 e2 Z 1 q2 e−iqr − eiqr = − 2 dq 2 2 4π 0 q + mc −iqr 2 Z 1 2 Z −∞ e q −iqr e q −iqr = 2 dq 2 2 e − 2 dq 2 2 e 4π ir 0 q + mc 4π ir 0 q + mc 2 Z 1 e q −iqr = 2 dq 2 2 e 4π ir −∞ q + mc Z 1 1 −iqr q = 2 dqe 2 2 4π ir −∞ q + mc e2 e−mcr = − (8) 4π r 2 where in the fourth line, we perform a change of variables q ! −q for the second set of terms, and in the last line, we used Z 1 −iqr q −mcr dqe 2 2 = −iπe (9) −∞ q + mc Note that even if you had chosen a different sign convention for your Fourier transform, i.e. e+i~q·~x, everything still becomes the same third line onwards. The sign of the potential is robust and independent of the convention. 6. What can you say about the nature of the potential, is it attractive or repulsive? @V Answer: Attractive as − @r is negative. The above potential was used by Yukawa as a rough explanation of the strong force. One assumes that there exists a scalar particle, called the pion (denoted by π), that is responsible for keeping the proton and neutron together. It is known that of the naturally occuring elements, Uranium has the largest nucleon number, with a nuclear size of about R ≈ 15 fm. 7. We are now assuming that the potential of eq. (6) with mc replaced by mπ is the one that describes the strong interaction. Under this assumption estimate the pion mass, mπ. We also assume that the coupling constant for the strong interaction is 1. Answer: Roughly we know that the things break down when the electric repulsion wins over the pion attraction, that is when α e−mπr ∼ (10) r r that leads to mπ ∼ − log α=r ∼ 70 MeV, where we used the fact that 200 MeV fm ≈ 1 and that log(137) ≈ 5. 8. The pion was discovered with a mass of mπ ∼ 140 MeV. What can you say about this pion, is it the one that was predicted by Yukawa? Answer: While we are off by a factor of 2, it still seems like a good model. We have to understand that the model is very crude so a factor of a few is still very good. We now move on to discuss the Coulomb force. The idea is very similar to the Yukawa potential discussed above but with two main differences. First, is the fact that the photon is massless, and second, there is a minus sign in the photon propagator vs the scalar one (we did not discuss the origin of this sign). 9. Derive the Coulomb potential from the amplitude in the non-relativistic limit. Answer: We take the Yukawa case, remove the sign and set m = 0. Note that if you 3 had set m = 0 in the amplitude before performing the intgeral, you might find that the integral is divergent. In reality, the propagator is always assumed to have a small imaginary part in the denominator, i.e. q2 − m2 + i instead of q2 − m2, and you only take to zero after you have done all your calculations and integrals. This avoids the divergence. 10. While theoretically the photon is massless, it is interesting to know it experimentally. Based on what you did so far, can you estimate an upper bound on the photon mass? Answer: Just the fact that we see electric and magnetic forces at macroscopic distance < −1 −7 tells us that mγ ∼ (1 m) ∼ 10 eV where 1 m stand as a typical distance where we still do not see deviation. 4.
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