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Question: Matching SMEFT and WET at one loop #68

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Stefan-Bissmann opened this issue Feb 24, 2021 · 5 comments
Closed

Question: Matching SMEFT and WET at one loop #68

Stefan-Bissmann opened this issue Feb 24, 2021 · 5 comments

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@Stefan-Bissmann
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Stefan-Bissmann commented Feb 24, 2021

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

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

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

@DavidMStraub
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DavidMStraub commented Feb 24, 2021

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
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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.

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