Deriving Ji's Sum Rule - Help with Steps from Diehl Paper

[cylon]

Dear all,

I am new here and hope that I write this in the right place. ;-)

I am seeking help in deriving Ji's Sum Rule, which tells you that the second moment of the nucleon's GPDs equals the total angular momentum of the quarks. E.g. in Diehl, hep-ph/0307382, Sect. 3.6.

The step where I'm stuck is in the previously mentioned paper, from Eq. (68) to Eq. (69). It seems so easy, though: for $\mu$ different from $\nu$, the terms in $g^{\mu\nu}$ in (68) disappear, while the other term in C disappears in the forward limit p'->p. One thus keeps only the terms in A and B. For the B-term, I immediately use the Gordon identity. This gives me:
\[<p|T^{\mu\nu}|p> = (A(0)+B(0))/2 \bar{u} \frac{p^{\mu}\gamma^{\nu} + p^{\nu}\gamma{\mu}}{2} u - B(0) \bar{u} \frac{p^{\mu}p^{\nu}}{m} u\]

If I then fill in some values for the spinors, I always arrive at (A-B)/2 instead of (A+B)/2. I don't see where it goes wrong, which is so frustrating... Is it due to my spinors? Or to something else? I don't seem to need the fact that the proton is at rest (which is written below Eq. (66)). In fact, I always end up with a term ~ Lz, which turns zero in the limit of a proton in its rest frame.

I hope that someone here can help me... I've read so many papers about it, and still don't find the mistake.

Kind regards,
C

 

[AndyW]

atherine

Dear Catherine,

First of all, welcome to the forum! It's great to have new members who are interested in GPDs and their applications. I will try my best to help you with your problem.

From your description, it seems like you are on the right track. The step from Eq. (68) to Eq. (69) is indeed straightforward, as you have mentioned. The terms in $g^{\mu\nu}$ disappear because they are only relevant for the case where $\mu = \nu$, and the term in C disappears in the forward limit. So, you are correct in keeping only the terms in A and B.

As for your calculation, I believe the issue lies in your spinors. The correct expression for $<p|T^{\mu\nu}|p>$ should be:
\[<p|T^{\mu\nu}|p> = \frac{1}{2} (A(0) + B(0)) \bar{u}(p) \gamma^{\mu} u(p) \bar{u}(p) \gamma^{\nu} u(p) - B(0) \bar{u}(p) \frac{p^{\mu}p^{\nu}}{m^2} u(p)\]

Note that the spinors in the first term are squared, while in your expression they are not. This is likely the reason why you are getting a term ~Lz, which should not be present in the final expression.

I hope this helps in solving your problem. If you have any further questions, please don't hesitate to ask. Good luck with your research!

 

[sir-pinski]

.

Dear C.,

Thank you for reaching out for help with deriving Ji's Sum Rule. I can understand your frustration, as these calculations can often be tricky and it's easy to make mistakes along the way. I will try my best to guide you through the steps and hopefully help you find the mistake in your calculation.

Firstly, let's start with Eq. (68) in Diehl's paper:
\[<p'|T^{\mu\nu}|p> = \bar{u}(p') \left[ A(t)\frac{p^{\mu}p^{\nu}}{m}+ B(t) \frac{p^{\mu}\gamma^{\nu} - p^{\nu}\gamma^{\mu}}{2m} + C(t)g^{\mu\nu} \right] u(p)\]

In order to simplify this expression, we need to use the Gordon identity, which tells us that:
\[\bar{u}(p')\gamma^{\mu}u(p) = \bar{u}(p')\frac{p^{\mu}+p'^{\mu}}{2m}u(p)\]

Using this identity, we can rewrite Eq. (68) as:
\[<p'|T^{\mu\nu}|p> = \bar{u}(p') \left[ \frac{A(t)+B(t)}{2} \frac{p^{\mu}\gamma^{\nu} + p^{\nu}\gamma^{\mu}}{2m} + \frac{A(t)-B(t)}{2} \frac{p^{\mu}p^{\nu}}{m} + C(t)g^{\mu\nu} \right] u(p)\]

Now, in the forward limit where $p'=p$, we know that $t=0$ and $A(0)=B(0)$ (from Eq. (66) in the paper). Therefore, the terms in $p^{\mu}\gamma^{\nu}$ and $p^{\nu}\gamma^{\mu}$ will cancel out, leaving us with:
\[<p|T^{\mu\nu}|p> = \bar{u}(p) \left[ \frac{A(0)-B(0)}{2m} p^{\mu}p^{\nu} + C(0)g