If the vertex factor for the decay of a vector boson X into two spin- (1)/(2) fermions f1 and f2

If the vertex factor for the decay of a vector boson X into...

Key Concepts

Feynman Vertex Factor

The vertex factor in quantum field theory encapsulates the interaction strength and structure between particles at a given vertex. In the context of decays or scatterings, it contributes to the matrix element calculation. Here, the vertex factor involves gamma matrices and coupling constants, describing how the vector boson couples to the fermions with both vector and axial-vector components.

Phase Space Integration in the Rest Frame

For a two-body decay, the phase space integration simplifies significantly when performed in the rest frame of the decaying particle. In this frame, energy and momentum conservation lead to analytical forms for the final-state momenta which, when integrated, yield the decay rate. This concept is important for deriving analytic expressions for decay widths, as illustrated in the problem where the boson rest frame is used.

Gamma Matrix Trace Techniques

Gamma matrix trace techniques are instrumental in evaluating squared matrix elements involving Dirac spinors. By taking traces over products of gamma matrices, one can simplify expressions that result from summing over fermion spins. This technique uses identities and properties of gamma matrices, including the anticommutation relations and trace formulas, to reduce the product of operators to manageable expressions.

Decay Width Calculation

This concept refers to determining the probability per unit time that a particle decays into a given final state. The decay width is calculated by integrating the squared matrix element over the appropriate final-state phase space. In this problem, the two-body decay formula in the rest frame of the decaying particle is used, which involves both the interaction vertex and the available phase space for the decay products.

Spin Averaging and Summing

In calculating physical observables such as decay widths, one must average over initial state spins and sum over final state spins. This process ensures that the computed quantity is independent of the initial polarization states and correctly accounts for all possible final configurations. The example shows an averaging over boson spins and a summation over fermionic degrees of freedom using appropriate projection operators.

Transcript

00:01 Before we get started on the actual problem, we're going to go ahead and introduce the conversion from kilogram meters per second to mep for speed of light, because that's the unit that we're going to use for momentum in this problem.

00:13 So to do that, a simple way to convert is to actually divide the speed of light by itself and introduce the conversion from jules to mev.

00:22 So what we do is we write the speed of light right here, and the value is 2 .998 times 10 of the 8 meters per second, and we multiply by kilograms right here and meters per second right here, divide by the speed of light c right here.

00:39 So now we multiply meters per second by meters per second again to get meters squared per second squared.

00:46 We have a unit of kilograms right here.

00:49 So now we have meter square per second square times kilogram, which is a jewel.

00:55 We introduced the conversion between mev and jewel, and that is 1 .602 times 10 of the minus 13 joules per mev.

01:05 So now we come to this line and we see that the jewel units will cancel.

01:13 So we have jewel canceling jewel because one mev again equals 1 .602 times 10 of the minus 13 joules.

01:21 And we arrive at this final line right here.

01:26 So one kilogram meters per second is equal to 1 .871 times 10 of the 21 m .ev for the speed of light.

01:35 So our particle reaction that we're looking at is the decay of a positively charged sigma, decaying to a positively charged pion in a neutron.

01:46 And we're given the following information.

01:48 We have a magnetic field of 1 .15.

01:51 Tesla pointing out of the page, we have the sigma moving along a circular path in the magnetic field with a radius of 1 .99 meters, a positively charged pion that results with a positive, or i'm sorry, a circular path of 0 .580 meters.

02:12 A neutron has a rest mass of 939 .6 m .ev per c squared, and a positively charged pion has a mass of 139 .6 mev per c squared.

02:26 Now, the angle between the momentum of the pion and the sigma is going to be 64 .5 degrees at the time of decay.

02:34 So we're going to go ahead and answer some questions based on this information.

02:39 So let's go ahead and proceed here.

02:42 So we go ahead and let me get this going here.

02:50 Kind of skip through some things.

02:55 Once again, the sigma and the pion move on circular paths.

02:58 What this actually means is that there's a centripetal force.

03:02 So q, which is the charge times the velocity of either particle times the magnetic field will equal mv squared over r.

03:09 That's the centripetal force.

03:14 And this will allow us to get the momentum.

03:17 We can determine the momentum of either particle since they move on circular paths.

03:23 And it's going to be equal to mass times velocity, which is actually equal to q the electric charge of either particle times the magnetic field times the radius of curvature of that particle.

03:36 So if we start with the sigma particle, that'll equal the electric charge times the magnetic field times its radius at curvature.

03:44 So we can start out working in the si units and then just use the conversion.

03:49 Factor that we obtained, we just multiply by it.

03:56 So remember, that's the conversion factor we got right here.

03:59 We got 1 .871 times 10 to the 21 meter, or rather, m .ev per c.

04:06 We multiply by that right here, and we obtain a momentum of 686 mev per c.

04:16 We do the same for the pion.

04:18 We have electric charge times magnetic field times the radius of its path.

04:27 And i'm going to move down a little bit here.

04:34 And we get, with the conversion factor, we get 200 mev per c.

04:40 So now i'm going to draw a little diagram here that shows what's going on when the decay occurs.

04:48 So the arrow on the left, we're about to have some labels here, is the sigma particle.

04:54 And let's just emphasize just so we can here.

04:57 These are, this is a positively charged sigma.

04:59 This is a positively charged pion.

05:01 And the neutron is the little in on the bottom.

05:07 We have some relations here that are trigonometric based on directions of these arrows, these momentum.

05:14 That's what these arrows are.

05:17 These are some relations we get.

05:21 So remember that 64 .5 degrees was the angle between the incident momentum, which is the momentum of the sigma particle on the left there.

05:31 If it had continued to keep moving, it would have moved along the dash line.

05:36 So that's 64 .5 degrees from the direction that the pion is moving.

05:41 So that's why you see that angle of 64 .5 degrees drawn that way.

05:46 So we have that the momentum of the sigma particle equals the momentum of the neutron, pn, times cosine of phi.

05:55 That's an angle that we don't know.

05:56 Plus the momentum of the pion times cosine 64 .5 degrees.

06:03 So we can rewrite this and get pn cosine phi equals p, momentum of the sigma minus p, which is the momentum of the pion times cosine 64 .5 degrees.

06:19 I've simply rewritten it.

06:20 Now perpendicular to the incident momentum in this direction, going this way, 90, there is no momentum for the sigma particle...

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