Calculating Cross Section with 4-Fermi Interaction

[indigojoker]

Hi all, I'm trying to teach myself high energy physics and was wondering if anyone knew a good reference site on calculating cross sections. I'm specifically trying to calculate the cross section for \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e using the 4-fermi interaction.

Are there any good sites that talk about the four-fermi interaction and how to calculate the cross section with it? Or could someone here explain it to me?

Thanks!

 

[kdv]

Hi all, I'm trying to teach myself high energy physics and was wondering if anyone knew a good reference site on calculating cross sections. I'm specifically trying to calculate the cross section for \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e using the 4-fermi interaction.

Are there any good sites that talk about the four-fermi interaction and how to calculate the cross section with it? Or could someone here explain it to me?

Thanks!

I don't know of a reference but that would be a very easy one to do. And it's almost identical to the process e^+ e^- \rightarrow \mu^- \mu^+. You simply have to drop the photon propagator, replace e^2 by Fermi's constant and be careful about distinguishing the four-momenta of the incoming and outgoing particles.

Have you done any cross section calculation before? If not, maybe you could try e^+ e^- \rightarrow \mu^- \mu^+ or, in the exchange channel, e^- \mu^- \rightarrow e^- \mu^- first as a practice run. You can find the answers for those in textbooks.

 

[BenTheMan]

Hmm, there ARE a few textbooks out there for free.

Here's on by Georgi:
http://www.people.fas.harvard.edu/~hgeorgi/283.html
The calculation that kdv is talking about is found on page 45ish.

the calculation you are looking for is somewhere in chapter 2 (check section 2).

Good luck!

 

[indigojoker]

is the four-fermi interaction the same as V-A interaction?

 

[Haelfix]

Not necessarily, Fermi's old theory is pure vector, whereas the new (SM) gauge mediated interaction is V-A.

 

[Meir Achuz]

I would recommend you buy a good textbook.
I recommend one by Halzen and Martin for the type of cross section you mention.
Free is worth what you pay.

 

[kdv]

I would recommend you buy a good textbook.
I recommend one by Halzen and Martin for the type of cross section you mention.
Free is worth what you pay.

If you are implying that free stuff is not woth much, I have to say that this is not true when it concerns physics. The book by Georgi cited earlier in this thread looks excellent. Also, Srednicki has a good fraction of his book available for free on the internet. There are sevreal other excellent introductions to QFT available for free on the Internet (some of which are also complete books). I like especially David Tong's one.

 

[kdv]

Hmm, there ARE a few textbooks out there for free.

Here's on by Georgi:
http://www.people.fas.harvard.edu/~hgeorgi/283.html
The calculation that kdv is talking about is found on page 45ish.

the calculation you are looking for is somewhere in chapter 2 (check section 2).

Good luck!

Thanks for this link, Ben! I did not know about that one. You are the Man!

 

[BenTheMan]

I would recommend you buy a good textbook.
I recommend one by Halzen and Martin for the type of cross section you mention.
Free is worth what you pay.

Actually I have heard some people (physics professors) claim outright that Halzen and Martin is wrong in some places, so I'd rather use Georgi's textbook (which used to not be free).

 

[BenTheMan]

Srednicki's QFT book can be found here: http://www.physics.ucsb.edu/~mark/qft.html

 

[indigojoker]

I don't know of a reference but that would be a very easy one to do. And it's almost identical to the process e^+ e^- \rightarrow \mu^- \mu^+. You simply have to drop the photon propagator, replace e^2 by Fermi's constant and be careful about distinguishing the four-momenta of the incoming and outgoing particles.

Have you done any cross section calculation before? If not, maybe you could try e^+ e^- \rightarrow \mu^- \mu^+ or, in the exchange channel, e^- \mu^- \rightarrow e^- \mu^- first as a practice run. You can find the answers for those in textbooks.

I have not done any cross section calculations before. I have found that Perkins (4th) calculates the cross section for e^+ e^- \rightarrow \mu^- \mu^+ on pg. 142

I'm confused about something. What is the difference between calculating the cross section for \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e using the four-fermion interaction different from using intermediate vector boson theory?

 

[kdv]

I have not done any cross section calculations before. I have found that Perkins (4th) calculates the cross section for e^+ e^- \rightarrow \mu^- \mu^+ on pg. 142

Then maybe you should start with that one. You probably need to spend some time learning the tricks of cross section calculations including trace techniques. If you have access to the book by Halzen and Martin, it is very good to show you just what you need. If you use QFT books, you will have to dug out the parts that you need from a lot of other stuff. But I would recommend the book by Srednicki, available for free.

I'm confused about something. What is the difference between calculating the cross section for \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e using the four-fermion interaction different from using intermediate vector boson theory?

Second Edit One important point: in the IVB model the boson is massive. So as long as the center of mass energy of the reaction is much less than the mass of the boson, the propagator of the boson can be approximated by one over the mass squared of the boson. In that limit, the IVB calculation becomes identical to the four-Fermi calculation (at the condition of using the appropriate four-Fermi coupling).
So the two apporaches give the same result at low energies. At high energies, however, the results for the cross sections are completely different.

There is also another issue: renormalizability, but I don't know if you are interested in that issue.

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The four-Fermi interaction is a contact interaction between the four fermions. So the Feynman diagram is simply 4 fermion lines meeting at a point. In the Lagrangian, you have an interaction term with the four fermions fields and nothing else.
In the IVB approach, there is a boson exchanged between the fermions so the Feynman diagram has two fermion lines connecting to a boson line which propagates and connects to the other two fermion lines. In the Lagrangian, you have two interactio terms: a boson field coupled to the first two fermions and a second interaction with the boson interacting with the other two fermions.

In terms of the actual calculation, it is not very different in the two cases but the underlying physics is very different.

EDIT: just to make things clear: in the actual calculation you would have a term for the boson progator which you don't have in the four-fermi calculation but my point is that this does nto change very much the calculation in terms of difficulty. But it does affect the energy dependence of the cross section, obviously

 

[kdv]

I have not done any cross section calculations before. I have found that Perkins (4th) calculates the cross section for e^+ e^- \rightarrow \mu^- \mu^+ on pg. 142

I'm confused about something. What is the difference between calculating the cross section for \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e using the four-fermion interaction different from using intermediate vector boson theory?

Do you mean the book by Donald Perkins? I only have the third edition but in my edition he does not show the calculation at all. He simply gives the answer and draw the Feynman diagram. halzen and Martin show the steps. So does Srednicki

EDIT: I just realized that Perkins does the calculation explicitly in an appendix.

 

[indigojoker]

Halzen and Martin, they calculate the cross section for e^+ e^- \rightarrow \mu^- \mu^+ and it is given in (6.33). However, for \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e wouldn't I need to use charge currents?

It looks to me that the weak interaction amplitude for 4-fermi interaction is given on 12.13 and the amplitude using IVB theory is given in 12.14...?

So essentially, to calculate the cross section, one needs to calculate 12.13 or 12.14 and plug it into eq 4.35?

 

[nrqed]

Halzen and Martin, they calculate the cross section for e^+ e^- \rightarrow \mu^- \mu^+ and it is given in (6.33).[/tex]

yes except that this assumes that the masses are negligible so it's only vali at high energies.

However, for \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e wouldn't I need to use charge currents?

Right.

It looks to me that the weak interaction amplitude for 4-fermi interaction is given on 12.13

Right (although I think that historically, the Fermi theory involved only vector coupling so no chiral projector in J^\mu. I think it's only later that it was realized the interaction was V-A (vector minus axial) ).

and the amplitude using IVB theory is given in 12.14...?

Right (there seems to be a typo in 12.14 in my edition because they wrote for the eelctron neutrino u_{\nu_e} when it should be v_{\nu_e} because it's an antineutrino being produced)

But if you want to do the process \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e
your diagram is different than figure 12.5 so your expression won't be identical to 12.14 (i.e. be careful about using particle and antiparticle states u and v correctly)

So essentially, to calculate the cross section, one needs to calculate 12.13 or 12.14 and plug it into eq 4.35?

Yes.

 

[indigojoker]

Right (there seems to be a typo in 12.14 in my edition because they wrote for the eelctron neutrino u_{\nu_e} when it should be v_{\nu_e} because it's an antineutrino being produced)

The process \mu^- \rightarrow e^- \bar{\nu_e} \nu_\mu is equivalent to \mu^- \nu_e \rightarrow e^- \nu_\mu

In 12.14, H+M treats it as an ingoing electron neutrino, which is the same as an outgoing antineutrino right? I believe we have the same editions but i do not understand the typo.

 

[indigojoker]

Using 12.13, here's what I have for the weak interaction amplitude:

M=\frac{4G}{\sqrt{2}} J^{\mu}J^{\dagger}_{\mu}
M=\frac{4G}{\sqrt{2}} \left( \bar{u_{\nu_e}} \gamma^{\mu} \frac{1}{2} (1-\gamma^5) u_{\mu} \right) \left( \bar{u_e} \gamma_{\mu} \frac{1}{2} (1-\gamma^5) u_{\nu_{\mu}} \right)

Ok, from here, I think that I would follow eq. 12.65 (which refers me back to section 6.3). After chugging though all the math, I believe the result is 12.57 and eventually 12.60?

 

[nrqed]

The process \mu^- \rightarrow e^- \bar{\nu_e} \nu_\mu is equivalent to \mu^- \nu_e \rightarrow e^- \nu_\mu

In 12.14, H+M treats it as an ingoing electron neutrino, which is the same as an outgoing antineutrino right? I believe we have the same editions but i do not understand the typo.

I am not sure what you mean by "equivalent". There is certainly a connection (they are related by crossing symmetry) but the processes are not identical.
If you have an antiparticle you must use a "v" spinor instead of a "u" spinor. And when squaring the amplitude you have to use a different relation see 5.47). Of course if one treats the neutrino as massless then it makes no difference.

 

[nrqed]

Using 12.13, here's what I have for the weak interaction amplitude:

M=\frac{4G}{\sqrt{2}} J^{\mu}J^{\dagger}_{\mu}
M=\frac{4G}{\sqrt{2}} \left( \bar{u_{\nu_e}} \gamma^{\mu} \frac{1}{2} (1-\gamma^5) u_{\mu} \right) \left( \bar{u_e} \gamma_{\mu} \frac{1}{2} (1-\gamma^5) u_{\nu_{\mu}} \right)

Ok, from here, I think that I would follow eq. 12.65 (which refers me back to section 6.3). After chugging though all the math, I believe the result is 12.57 and eventually 12.60?

I am a bit confused about what process you want to do.
First, the expression above is incorrect because you are connecting an electron neutrino to a muon on the left side.

which process are you doing now? Process 12.65 is not the process you mentioned in your initial post! and 12.57 refers to figue 12.8 a.


By the way: if you look at equation 12.31 for muon decay , which should be the same as eq 12.14 (they both refer to figure 12.5) you see that they use a v spinor for the anineutrino. Sothat shows that there is a typo in 12.14.

 

[indigojoker]

whoops, it was late when i posted that, here's what I meant to write:

<br /> M=\frac{4G}{\sqrt{2}} \left( \bar{u_{\nu_\mu}} \gamma^{\mu} \frac{1}{2} (1-\gamma^5) u_{\mu} \right) \left( \bar{u_e} \gamma_{\mu} \frac{1}{2} (1-\gamma^5) u_{\nu_e} \right)<br />

I am trying to calculate the cross section using the 4-fermi interaction (and then later try it with IVB theory)

Sorry about the confusion. And yes, I do agree that there is a typo in the 12.14

Would it be reasonable to follow the calculation of M for the muon decay (12.31)? but instead of v, i would use u since the interaction I'm considering does not contain an anineutrino?

 

[nrqed]

whoops, it was late when i posted that, here's what I meant to write:

<br /> M=\frac{4G}{\sqrt{2}} \left( \bar{u_{\nu_\mu}} \gamma^{\mu} \frac{1}{2} (1-\gamma^5) u_{\mu} \right) \left( \bar{u_e} \gamma_{\mu} \frac{1}{2} (1-\gamma^5) u_{\nu_e} \right)<br />

I am trying to calculate the cross section using the 4-fermi interaction (and then later try it with IVB theory)

Sorry about the confusion. And yes, I do agree that there is a typo in the 12.14

Would it be reasonable to follow the calculation of M for the muon decay (12.31)? but instead of v, i would use u since the interaction I'm considering does not contain an anineutrino?

The expression you wrote corresponds to \mu + \nu_e \Rightarrow e^- + \nu_\mu

 

[indigojoker]

M=\frac{G}{\sqrt{2}}\left(\bar{u_{\nu_e}}(k,t)\gamma^{\lambda}(1-\gamma_5)u_{e}(p,s)<br /> \right)\left(\bar{u_\mu}(p&#039;,s&#039;)\gamma_{\lambda}(1-\gamma_5)u_{\nu_\mu}(k&#039;,t&#039;)\right)

M^{\dagger}=\frac{G}{\sqrt{2}}\left(\bar{u_{e}}(p,s)\gamma^{\sigma}<br /> (1-\gamma_5)u_{\nu_e}(k,t)\right)\left(\bar{u_\nu_\mu}(k&#039;,t&#039;)\gamma_{\sigma}(1-\gamma_5)u_{\mu}(p&#039;,s&#039;)<br /> \right)

MM^{\dagger}=\frac{G^2}{2} \left{ \left(\bar{u_{\nu_e}}(k,t)\gamma^{\lambda}(1-\gamma_5)u_{e}(p,s)<br /> \right) <br /> \left(\bar{u_{e}}(p,s)\gamma^{\sigma}(1-\gamma_5)u_{\nu_e}(k,t)\right)<br /> \right} \left{<br /> \left(\bar{u_\mu}(p&#039;,s&#039;)\gamma_{\lambda}(1-\gamma_5)u_{\nu_\mu}(k&#039;,t&#039;)\right) \left(\bar{u_\nu_\mu}<br /> (k&#039;,t&#039;)\gamma_{\sigma}(1-\gamma_5)u_{\mu}(p&#039;,s&#039;)\right) \right}

= \frac{1}{2} \frac{G^2}{2} \left{ \left( \gamma^\lambda (1-\gamma_5)(\displaystyle{\not}p + m_e) \gamma^\sigma<br /> (1-\gamma_5) \displaystyle{\not}K \right) \left(\gamma_\lambda (1-\gamma_5) \displaystyle{\not}K\prime \gamma_\sigma (1-\gamma_5)<br /> (\displaystyle{\not}p\prime+m_\mu)\right) \right}

= \frac{G^2}{4} \left{ Tr\left( \gamma^\lambda (1-\gamma_5)(\displaystyle{\not}p + m_e) \gamma^\sigma<br /> (1-\gamma_5) \displaystyle{\not}K \right) Tr\left(\gamma_\lambda (1-\gamma_5) \displaystyle{\not}K\prime \gamma_\sigma (1-\gamma_5)<br /> (\displaystyle{\not}p\prime+m_\mu)\right) \right}

since \gamma^\sigma(1-\gamma_5)=(1+\gamma_5)\gamma^\sigma and (1-\gamma^5)(1+\gamma^5)=0, then m_\mu and m_e terms vanish

\bar{M^2}=\frac{G^2}{4} \left{ Tr\left( \gamma^\lambda (1-\gamma_5) \displaystyle{\not}p \gamma^\sigma<br /> (1-\gamma_5) \displaystyle{\not}K \right) Tr\left(\gamma_\lambda (1-\gamma_5) \displaystyle{\not}K\prime \gamma_\sigma (1-\gamma_5)<br /> \displaystyle{\not}p\prime \right) \right}

And then with a little bit of calculations, I got the results of:

\bar{M^2}= 64G^2 (p \cdot k&#039;)(p&#039; \cdot k)

I skipped several steps however I can write them out if you would like to see steps.

How does this look for the weak interaction amplitude for \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e

 

[indigojoker]

is it a coincidence that the amplitude for muon decay \mu \rightarrow e^- \bar{\nu_e} \nu_\mu is the same as the process for \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e?

 

[nrqed]

is it a coincidence that the cross section for muon decay \mu \rightarrow e^- \bar{\nu_e} \nu_\mu is the same as the process for \nu _{\mu} + e^- \rightarrow \mu ^- + \nu_e

Sorry, I did not have the time to go through your derivation in details; it looks good at first sight.

By the way, we don't use the expression "cross section" for a decay. It's then called decay rate. But I don't think you mean decay rate either, I think you mean the square of the amplitude. In any case, it's not surprising that they are related because of "crossing symmetry". When moving the external legs of a Feynman diagram, the amplitudes of the corresponding processes are related by simple relations (in general, for two bodies to two bodies, the Mandelstam variables get exchanged)

 

[indigojoker]

Sorry, I did not have the time to go through your derivation in details; it looks good at first sight.

By the way, we don't use the expression "cross section" for a decay. It's then called decay rate. But I don't think you mean decay rate either, I think you mean the square of the amplitude. In any case, it's not surprising that they are related because of "crossing symmetry". When moving the external legs of a Feynman diagram, the amplitudes of the corresponding processes are related by simple relations (in general, for two bodies to two bodies, the Mandelstam variables get exchanged)

thanks for the response! yes, I realized I miss-used the expression cross section and replaced it with amplitude. I thought the two processes were related due to their similarity when drawing Feynman diagrams but wasnt sure, thanks for clarifying.

Just for further understanding... can we say that the cross section is proportional to G^2 E_{cm}^2 since \bar{M^2}= 64G^2 (p \cdot k&#039;)(p&#039; \cdot k)? If so, then we can say that \frac{d \sigma}{dE_{cm}^2} =const and then could we say that it is safe to assume that the total cross section approaches infinity as E_{cm} increases?

 

[nrqed]

thanks for the response! yes, I realized I miss-used the expression cross section and replaced it with amplitude. I thought the two processes were related due to their similarity when drawing Feynman diagrams but wasnt sure, thanks for clarifying.

Just for further understanding... can we say that the cross section is proportional to G^2 E_{cm}^2 since \bar{M^2}= 64G^2 (p \cdot k&#039;)(p&#039; \cdot k)? If so, then we can say that \frac{d \sigma}{dE_{cm}^2} =const and then could we say that it is safe to assume that the total cross section approaches infinity as E_{cm} increases?

I don't have time to check now but how did you get that the cross section is proportional to G^2 E_{cm}^2 (it might be obvious but I don't see). Did you include the phase space factors (which tend to suppress the cross sections)

 

[indigojoker]

I don't have time to check now but how did you get that the cross section is proportional to G^2 E_{cm}^2 (it might be obvious but I don't see). Did you include the phase space factors (which tend to suppress the cross sections)

I thought the phase space factor comes into play when calculating the differential decay rate:

d \Gamma = \frac{1}{2M} |M|^2 dQ

where as the differential cross section is given by:

\frac{d \sigma}{d \Omega} = \frac{1}{64 \pi^2 E_{cm}^2 } \frac{p_f}{p_i} |M|^2

 

[nrqed]

I thought the phase space factor comes into play when calculating the differential decay rate:

d \Gamma = \frac{1}{2M} |M|^2 dQ

where as the differential cross section is given by:

\frac{d \sigma}{d \Omega} = \frac{1}{64 \pi^2 E_{cm}^2 } \frac{p_f}{p_i} |M|^2

That's what I meant: there is a suppressing 1/E_cm^2 factor from phase space (phase space is everything that multiplies the square of the amplitude).
What is your M^2 expressed in terms of CM quantities?

 

[indigojoker]

\frac{d \sigma}{d \Omega} = \frac{1}{64 \pi^2 s } \frac{p_f}{p_i} |M|^2

<br /> |M|^2= 64G^2 (p \cdot k&#039;)(p&#039; \cdot k) <br />

u=-2 k \cdot p&#039; = -2 p \cdot k&#039;

Seems to me that the cross section is proportional to G^2 \frac{u^2}{s}

I had thought s=p \cdot k&#039;=p&#039; \cdot k

 

[indigojoker]

by the way, how does the calculation for the amplitude look?

 

[indigojoker]

also, would the calculation for the amplitude |M|^2 look the same for IVB theory except there would be a \frac{g^2}{M_W^2 - q^2} term in front of the amplitude? I'm not sure how this "changes" anything compared to the 4 fermi result since they are only constants out in front.