Rules for term Rank: [ 3, 1 ]

General discussion of the Cambridge quantum Monte Carlo code CASINO; how to install and setup; how to use it; what it does; applications.
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Vladimir_Konjkov
Posts: 185
Joined: Wed Apr 15, 2015 3:14 pm

Rules for term Rank: [ 3, 1 ]

Post by Vladimir_Konjkov »

Hi CASINO users.

I want to checkout contribution for e-e-e-n JASTROW term (in N2).

As usual I apply the next rules for:

Code: Select all

    Rules:
      n1=n2
      1-n1=2-n1
      1-1=2-2
I have got two different terms 1-1-1-n1 and 1-1-2-n1 but I think that the contribution of the first term is negligible, because the probability of encountering three electrons with parallel spins at one point is small.

I want to remove 1-1-1-n1 term from consideration. I tried different ways to do it but everything was unsuccessful.

Can anyone advise me how to do this?

thank you in advance.

Vladimir.
ChatGPT tackles every task with glee,
But turns it all to garbage, can't you see?
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Pablo_Lopez_Rios
Posts: 53
Joined: Thu Jan 30, 2014 1:25 am

Re: Rules for term Rank: [ 3, 1 ]

Post by Pablo_Lopez_Rios »

Hi Vladimir,

You should be able to merge the two channels with the rule 1-1=2-2=1-2, or equivalently 1=2.

Best,
Pablo
Hey there! I am using CASINO.
Vladimir_Konjkov
Posts: 185
Joined: Wed Apr 15, 2015 3:14 pm

Re: Rules for term Rank: [ 3, 1 ]

Post by Vladimir_Konjkov »

Pablo_Lopez_Rios wrote:Hi Vladimir,

You should be able to merge the two channels with the rule 1-1=2-2=1-2, or equivalently 1=2.

Best,
Pablo
Hi, Pablo.

I don't want to merge two channels. I want to exclude 1-1-1-n1 channel, like !n1 exclude first atom or !1-2 excludes 1-1-2-n1 channel in our case.
Another possibility to set coefficients in 1-1-1-n1 channel to "0.0 fixed"
Actually yes, if channel 1-1-1n1 doesn't contribute to energy or variance with any coefficients we should merge them.

thanks Pablo, I will try.
ChatGPT tackles every task with glee,
But turns it all to garbage, can't you see?
And when it handles garbage from the start,
At least it wastes less effort on its part.
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