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wilson-eft / wilson

A Python package for the running and matching of Wilson coefficients above and below the electroweak scale
https://wilson-eft.github.io
MIT License
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Question: Matching SMEFT and WET at one loop #68

Closed Stefan-Bissmann closed 3 years ago

Stefan-Bissmann commented 3 years ago

Dear all, we currently want to use wilson to test our results from our implementation of the matching between SMEFT and WET (especially the flavio basis). However, I ran into some issues and would like to ask whether it is just me not understanding how to use wilson correctly. So, the example is the following: Say that we want to match 'CuG_33' to 'C8_bs' and as a cross check to 'CdG_23' (in the basis JMS) The commands I use are: 


w = wilson.Wilson({'uG_33': 1e-6}, 80.3, 'SMEFT', 'Warsaw')
w.set_option('smeft_matchingscale', 80.3)
w.set_option('smeft_matching_order', 1)
wc = w.match_run(80.3, 'WET', 'flavio')
print(wc['C8_bs'])
wc2 = w.match_run(80.3, 'WET', 'JMS')
print(wc2['dG_23'])

now, taking conversion factors (vT2 8 pi2 /(g3 mb) 1 / ( Vtb * Vts)) between flavio and JMS basis into account, the results differ by a factor of 10:

C8_bs = -0.028... CdG_23 = 8.725e-10 (= 0.0029... after conversion)

Furthermore, they are in conflict with results obtained when I directly use the corresponding code in 'smeft_loop.py', which agrees with our implementation of the matching in https://arxiv.org/pdf/1512.02830.pdf (after taking the conversion factor into account). This result reads -0.041 Maybe I am just mixing different conventions by mistake or do not use wilson correctly?

Thank you in advance for your help.

DavidMStraub commented 3 years ago

Hint: you can enclose inline code in ` and display code blocks in ```python ... ``` to make it more readable.

DavidMStraub commented 3 years ago

I agree with your conversion factor but disagree with there being a discrepancy. Can you check your numerics?

Stefan-Bissmann commented 3 years ago

Thanks a lot for the fast reply! So, for you everything is just fine when comparing both bases (the results from wilson)?

DavidMStraub commented 3 years ago

I'm too lazy to look up the right values, but using rough values I get:

v=246.22
Vtb=1
Vts=-0.04
mb=3
g3 = 1.2
print(wc2['dG_23'] / (1/v**2*Vtb*Vts/(8*pi**2)*mb*g3))

The result is (-0.029003550946260098+0.000543143668753945j).

Stefan-Bissmann commented 3 years ago

Ok, then it might just be some issue with my numerics. Just wanted to check that I did not mess up anything. Thank you so much for the fast help! It is much appreciated.