Calculation of neutral Kaon Box Diagrams

[johnpatitucci]

Hi everybody,i'm worried about some calculation. I have to write down the amplitude for the neutral Kaon oscillation. There are two Box diagrams which are depicted here:

http://upload.wikimedia.org/wikipedia/de/9/92/FeynmanKaon.png

The two diagrams come along with a combinatorical factor of 1/2 and they have a relative sign according to the possible contractions.

Now I get for diagram on the left with W-exchange in (\xi =1) gauge up to some constants the following dirac structure (external momenta neglected)

[\bar(s) \gamma_{\mu} (\slashed{k} + m_1) \gamma_{\nu} d] \otimes
[\bar(s) \gamma_{\nu} (\slashed{k} + m_2) \gamma_{\mu} d]

Now, wenn calculating the right diagram, i get basically the same, which together with the relative sign would give a vanishing full amplitude.
Does anybody know, what i am doing wrong ?

 

[Hepth]

Just to fix TeX:
$$[\bar{(s)} \gamma_{\mu} (\not k + m_1) \gamma_{\nu} d] \otimes [\bar{(s)} \gamma_{\nu} (\not k + m_2) \gamma_{\mu} d]$$

 

[blechman]

without looking closely, there IS going to be a partial cancellation. This is called the GIM mechanism. But the cancellation will not be complete. You have to be very careful with gamma matrix algebra, etc. That "basically" should translate to: "almost, but NOT QUITE!"

BTW: there are other diagrams that matter, including the ones with a "c" and a "t" on the internal lines. You're going to need those.

 

[johnpatitucci]

Well, thanks for your replies. But just think of my expression with masses replaced according to
m_1 -> m_i , m_2 -> m_j and sum over i,j = u,c,t. I've done that. The masses vanish by moving the helicity projectors to the right and therefore you can write the sum behind this expression, so that it only acts on the masses in the denominator [= (k^2 - m_i^2)(k^2-m_j^2)* (k^2 - M_W^2)].

There has to be something else.

(Sorry, how do i implement TeX here ?)

 

[Hepth]

"[ t e x ]" to open and "[ / t e x ]" to close

 

[blechman]

The masses do NOT vanish! The final answer should be of the form:

$$G_F^2(m_1^2 - m_2^2)$$

That's the GIM mechanism. So the term that is independent of masses DOES vanish, but the term with the masses only vanishes in the particles have the same mass. Check your gamma matrix algebra again! That's probably where you went wrong.

Also, if you want to include tex within the paragraph (like $g_{\mu\nu}$), use "[ i t e x ]" and "[ \ i t e x ]" (without the spaces between the letters, obviously).

 

[Safinaz Salem]

Hi,

I started to calculate the neutral Kaon Box Diagram, can anyone tell me what is the first step to solve this integarl

I = \int \frac{d^4 k^2}{(k^2-M^2_W)^2 (k^2 - m^2_i) (k^2 - m^2_j)}

which appears in the dominator.

I think we need first to make feynman parametrization then wick rotation. Is it right ?
Because I can't do the right parametrization.

thx

 

[jkp]

Hi

I realize that this thread is very old, but it's new to me! Anyway, it seems to me that there are more questions here than answers (and some answers seemed to be misleading). So please allow me to first summarize the calculation in order to clarify the question, and then make a suggestion for the answer.

The Question:

Basically the question is how do the two box diagrams, which are equal and sum with a relative minus sign, not sum to zero? (The sum should actually give a function $A(x_i,x_j)$ rather than zero.)

The calculation:

I've attached the relevant Feynman diagrams for the W boson box diagram which I'm going to discuss (see bottom of post). The reason I include it is because the original post linked to some diagrams which had a few mistakes.

Starting from the diagram on the left, and using the Feynman Gauge, it's easy to find,
$$\left(\frac{g}{\sqrt{2}}\right)^4 \sum_{i,j=u,c,t}\lambda_i \lambda_j \int \frac{d^4 q}{(4\pi)^4} \,(\bar{d}\gamma^\mu P_L \frac{i(\not\! q+m_i)}{q^2-m_i^2}\gamma^\nu P_L s)\, (\bar{d}\gamma_\nu P_L \frac{i(\not\! q+m_j)}{q^2-m_j^2}\gamma_\mu P_L s) \left(\frac{-i}{q^2-M_W^2}\right)^2$$
As the projection operators must match (i.e. $P_L P_R =0$), and recalling that $P_L$ changes to $P_R$ whenever it passes a $\gamma^\mu$, it is clear that the $m_i$ and $m_j$ terms are zero, leaving only the $\not\! q$ terms;
$$\frac{g^4}{4} \sum_{i,j=u,c,t}\lambda_i \lambda_j \int \frac{d^4 q}{(4\pi)^4} \,\frac{q_\alpha q_\beta}{D} \,(\bar{d}\gamma^\mu \gamma^\alpha\gamma^\nu P_L s) (\bar{d}\gamma_\nu \gamma^\beta\gamma_\mu P_L s)$$
Here we have defined, $\lambda_i=V^{}_{is}V^{*}_{id}$, where $V$ is the CKM matrix, and
$$\frac{1}{D}= \frac{1}{(q^2-M_W^2)^2}\frac{1}{q^2-m_i^2}\frac{1}{q^2-m_j^2} \,.$$
The next step is to do the integration using,
$$I_{\alpha\beta}(i,j)\equiv \int d^4 q \, \frac{q_\alpha q_\beta}{(q^2-M_W^2)^2(q^2-m_i^2)(q^2-m_j^2)} = \frac{-i\pi^2}{4 M_W^2}A(x_i,\,x_j)\,g_{\alpha\beta}$$
with $x_i=m_i^2/M_W^2$. The function $A$ is defined by,
$$A(x_i,\,x_j) = \frac{J(x_i)-J(x_j)}{x_i-x_j}$$
and
$$J(x_i)=\frac{1}{1-x_i} + \frac{x_i^2 \,\ln x_i}{(1-x_i)^2}$$
This is actually easy to prove, but I'll leave it out for now. Using this we now get
$$\frac{-i\,g^4\,\pi^2}{16 M_W^2(2\pi)^4} \sum_{i,j=u,c,t}\lambda_i \lambda_j \,A(x_i,\,x_j) \,(\bar{d}\gamma^\mu \gamma^\alpha\gamma^\nu P_L s) \,(\bar{d}\gamma_\nu \gamma_\alpha\gamma_\mu P_L s) \,.$$
Next we simplify the combination, $(\bar{d}\gamma^\mu \gamma^\alpha\gamma^\nu P_L s) \,(\bar{d}\gamma_\nu \gamma_\alpha\gamma_\mu P_L s)$. All you need to do here is make use of the $\gamma$ matrix
identity,
$$\gamma^\mu \gamma^\alpha \gamma^\nu = g^{\mu\alpha}\gamma^\nu + g^{\nu\alpha}\gamma^\mu - g^{\mu\nu}\gamma^\alpha - i\epsilon^{\mu\alpha\nu\beta}\gamma_5 \gamma_\beta$$
to show that,
$$(\bar{d}\gamma^\mu \gamma^\alpha\gamma^\nu P_L s) \,(\bar{d}\gamma_\nu \gamma_\alpha\gamma_\mu P_L s) = 4\, (\bar{d}\gamma^\mu P_L s) \,(\bar{d}\gamma_\mu P_L s)$$
and then the result for the left diagram becomes,
$$-\frac{i\,g^4\,\pi^2}{4 M_W^2\,(2\pi)^4} \sum_{i,j=u,c,t}\lambda_i \lambda_j \,A(x_i,\,x_j) \, (\bar{d}\gamma^\mu P_L s) \,(\bar{d}\gamma_\mu P_L s) \,.$$
Now if we do the same calculation for the box diagram on the right we actually get the same result. So how is it that we don't get zero when we add up the two diagrams?

My suggested answer:

The 4-quark operators which appear in the two diagrams aren't actually the same.
They connect different s-quarks to different d-quarks. In the left diagram it is the two initial(final) state quarks which are connected, while in the right diagram it is one from the initial state and one from the final state. In order to become the same 4-quark operator we need to use a Fierz rearrangement,
$$(\bar{d}_{1L} \gamma^\mu s_{2L}) \,(\bar{d}_{3L} \gamma_\mu s_{4L}) = - (\bar{d}_{1L} \gamma^\mu s_{4L}) \,(\bar{d}_{3L} \gamma_\mu s_{2L})$$
and hence introduce an extra minus sign, so that the total is just twice that for the first diagram given above.

Does anyone know if this is the correct solution to the problem?

 

[Trifis]

Does anybody can give me a hint on how to reach

$$S_0(x_t) = \frac{3}{2}\frac{x_t^3}{(1-x_t)^3} \ln{x_t} - [\frac{1}{4}+\frac{9}{4}\frac{1}{1-x_t}-\frac{3}{2}\frac{1}{(1-x_t)^2}] x_t$$

from

$$A(x_i,\,x_j) = \frac{J(x_i)-J(x_j)}{x_i-x_j}$$

in the limit of large m_t?

 

[mfb]

Please open a new thread, this one is from 2010.