Register to reply

[itex]K^{0}-\bar{K}^{0}[/itex] mixing SQCD

by ChrisVer
Tags: k0bark0, mixing, sqcd
Share this thread:
Jun1-14, 01:25 PM
P: 872
I am trying to understand what the author means by the attachment (the underlined phrase).

In my understanding if the masses were equal I should have:

[itex] \frac{1}{(p^{2}-m^{2}+i \epsilon)^{2}} \sum_{ij} U^{d_{L}}_{di} U^{d_{L} \dagger}_{is}U^{d_{L}}_{dj} U^{d_{L} \dagger}_{js}[/itex]

why is the unitary supposed to make it vanish?
(the attachment is from page 202 of Theory and Phenomenology of Sparticles (2004)- M.Drees, R.Godbole,P.Roy)
Attached Thumbnails
Phys.Org News Partner Physics news on
Scientists uncover clues to role of magnetism in iron-based superconductors
Researchers find first direct evidence of 'spin symmetry' in atoms
X-ray laser probes tiny quantum tornadoes in superfluid droplets
Jun1-14, 02:59 PM
Sci Advisor
Bill_K's Avatar
P: 4,160
I think because then the numerators become UU = I, and I is diagonal and would not couple d to s, which is off-diagonal.
Jun2-14, 02:33 AM
P: 872
Thanks, I just realized that [itex]d,s[/itex] where not free indices, but they denoted the corresponding quarks coming in and out...

Jun2-14, 03:27 AM
P: 872
[itex]K^{0}-\bar{K}^{0}[/itex] mixing SQCD

Also the last equation there is in the attachment should be some kind of Taylor expansion of the above factor, around [itex] m_{i}^{2} = m_{d}^{2} + Δm_{i}^{2}[/itex]
I am also having 1 question...
[itex] \sum_{ij} U^{d_{L}}_{di} U^{d_{L} \dagger}_{is}U^{d_{L}}_{dj} U^{d_{L} \dagger}_{js}\frac{1}{(p^{2}-m_{d}^{2}-Δm_{i}^{2}+i \epsilon)(p^{2}-m_{d}^{2}-Δm_{j}^{2}+i \epsilon)}[/itex]

[itex]\frac{1}{p^{2}-m_{d}^{2}-Δm_{i}^{2}+i \epsilon}= \frac{1}{p^{2}-m_{d}^{2}+i \epsilon} + \frac{Δm_{i}^{2}}{(p^{2}-m_{d}^{2}+i \epsilon)^{2}}+O(\frac{1}{(p^{2}-m_{d}^{2})^{3}})[/itex]

So for j...
[itex]\frac{1}{p^{2}-m_{d}^{2}-Δm_{j}^{2}+i \epsilon}= \frac{1}{p^{2}-m_{d}^{2}+i \epsilon} + \frac{Δm_{j}^{2}}{(p^{2}-m_{d}^{2}+i \epsilon)^{2}}+O(\frac{1}{(p^{2}-m_{d}^{2})^{3}})[/itex]

And these I multiply... the 1st terms will give zero because of the unitarity of [itex]U[/itex]'s as before... the 1-2 and 2-1 terms will also give zero because of the unitarity of one of the [itex]U[/itex] each time ... So the only remaining term is the 2-2:

[itex] \frac{1}{(p^{2}-m_{d}^{2}+i \epsilon)^{4}}\sum_{ij} U^{d_{L}}_{di} U^{d_{L} \dagger}_{is}U^{d_{L}}_{dj} U^{d_{L} \dagger}_{js} Δm_{i}^{2}Δm_{j}^{2}+O(\frac{1}{(p^{2}-m_{d}^{2})^{5}})[/itex]

Which I think is equivalent to the last expression...
However isn't it also zero? because "i" can't be "s" and "d" at the same time... Shouldn't [itex]Δm_{i}[/itex] be between the [itex]U[/itex]'s? if it could be regarded as a matrix to save the day...
Jun2-14, 03:42 AM
P: 872
Oh I'm stupid... I just saw what's going on....
[itex]Δm_{i}[/itex] is going to change the summation way, so it won't give the δds anymore...

Register to reply

Related Discussions
What does [itex]\zeta(s)/s[/itex] converge to as [itex]\Im(s)\rightarrow\infty[/itex] Calculus 10
Show [itex]\phi[/itex][itex]\circ[/itex]f is Riemann integrable Calculus & Beyond Homework 5
Prove [itex]C[a,b][/itex] a closed linear subspace of [itex]L^{\infty}[a,b][/itex] Calculus & Beyond Homework 1
Proofs, [itex]\exists[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]) such that... Calculus & Beyond Homework 6
Show seq. [itex] x_n [/itex] with [itex] |x_{n+1} - x_n| < \epsilon [/itex] is Cauchy Calculus & Beyond Homework 2