# A Gaussian 09 Output file_DFT Calcuation

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1. Aug 3, 2016

### darwined

Hi,

I am trying to understand the functional form of B3LYP from the Gaussian output file. I have tried to relate the details in the output file with the functional form of B3LYP. But I am not sure what certain terms correspond to. I have mentioned below the details. Can you pleaese help me.

In the output of TD-DFT calculation files, I see the following (this is an e.g. of B3LYP)

IExCor= 402 DFT=T Ex+Corr=B3LYP ExCW=0 ScaHFX= 0.200000
ScaDFX= 0.800000 0.720000 1.000000 0.810000 ScalE2= 1.000000 1.000000

And in this case, it is well known that the funcitonal B3LYP has 20% Hartree Fock Exchange and 80% Density Exchange, the terms ScaHFX= 0.200000
ScaDFX= 0.800000 which says the scaling factor for HFX is 20% and DFX is 80% respectively. And when summed up, they give the value '1'.

0.720000 corresponds to the coefficient for gradient of Ex(Becke88).

Can someone please tell me what do the last two terms (1.000000 & 0.810000) correspond to.

Please let me know if I am not clear.

2. Aug 3, 2016

### TeethWhitener

From the Gaussian user manual: http://www.gaussian.com/g_tech/g_ur/k_dft.htm [Broken]

Gaussian 09 can use any model of the general form: P2EXHF + P1(P4EXSlater + P3ΔExnon-local) + P6EClocal + P5ΔECnon-local

...and a little later...

Here is a route section specifying the functional corresponding to the B3LYP keyword:

#P BLYP IOp(3/76=1000002000) IOp(3/77=0720008000) IOp(3/78=0810010000)

The output file displays the values that are in use:

IExCor= 402 DFT=T Ex=B+HF Corr=LYP ExCW=0 ScaHFX= 0.200000

where the value of ScaHFX is P2, and the sequence of values given for ScaDFX are P4, P3, P6 and P5.

Last edited by a moderator: May 8, 2017
3. Aug 3, 2016

### darwined

Thank you for your response TeethWhitener.

I almost got all the values from Gaussian 09 website (the link that you have provided).

Comparing the information printed in the output file and Gaussian standard format, I can related that P1=1, P2=0.2, P3=0.72, P4=0.8.
But as far as P5 and P6 are concerned, I am not sure what is the value of P5 and P6 are and what is/are the correlation functional(s) used. Is it VWN or LYP or Both.

Can you kindly clarify.

Last edited by a moderator: May 8, 2017
4. Aug 3, 2016

### TeethWhitener

Again from the Gaussian manual:

B3LYP uses the non-local correlation provided by the LYP expression, and VWN functional III for local correlation (not functional V).

So P5 is LYP and P6
is VWN functional III.

5. Aug 4, 2016

### darwined

Thank you for the responseTeethWhitener.

Now comparing the Gaussian output file and the coeffecients, can you please tell me if the functional form I have arrived at is correct?

Gaussian output

IExCor= 402 DFT=T Ex+Corr=B3LYP ExCW=0 ScaHFX= 0.200000
ScaDFX= 0.800000 0.720000 1.000000 0.810000 ScalE2= 1.000000 1.000000

Functional form after comparison.

P2*HFX + P1*(P3*DFX + P4*Delta DFX) + P5*LYP + P6*VWN

where P1=1, P2=0.2, P3=0.8, P4=0.72, P6=1 and P5= 0.81.

If it is correct, does this not mean there is 1.81*Ec (P5+P6) in the above equation. Kindly clarify.

6. Aug 4, 2016

### TeethWhitener

Again from the Gaussian manual:

Note that since LYP includes both local and non-local terms, the correlation functional used is actually:

C*ECLYP+(1-C)*ECVWN

In other words, VWN is used to provide the excess local correlation required, since LYP contains a local term essentially equivalent to VWN.

That should clear everything up. LYP is a special case, because it's got local and nonlocal contributions.

7. Aug 4, 2016

### darwined

Thank you for your swift response. That is exactly where I am confused now. We now have P5 and P6 left. Also from the Gaussian output file , we have 1.000000 0.810000 (the last two terms).

Now according to the manual, if I assign P5(0.81) to be scaling factor for Ec(LYP), I get a question why 0.19 (scaling factor for Ec(VWN)) is not printed in the output. Also then in that case what does P6=1 correspond to?

I am sorry for the trouble caused, but I am feeling very confused in the correlation terms involved.
Please let me know if I am not clear.

8. Aug 4, 2016

### TeethWhitener

Ok let's put it all in one place:

Becke Three Parameter Hybrid Functionals. These functionals have the form devised by Becke in 1993 [http://www.gaussian.com/g_tech/g_ur/refs.htm#Becke93a [Broken]]:

A*EXSlater+(1-A)*EXHF+B*ΔEXBecke+ECVWN+C*ΔECnon-local

where A, B, and C are the constants determined by Becke via fitting to the G1 molecule set.

...
There are several variations of this hybrid functional. B3LYP uses the non-local correlation provided by the LYP expression, and VWN functional III for local correlation (not functional V). Note that since LYP includes both local and non-local terms, the correlation functional used is actually:

C*ECLYP+(1-C)*ECVWN

In other words, VWN is used to provide the excess local correlation required, since LYP contains a local term essentially equivalent to VWN.

...
Gaussian 09 can use any model of the general form: P2EXHF + P1(P4EXSlater + P3ΔExnon-local) + P6EClocal + P5ΔECnon-local

...

Here is a route section specifying the functional corresponding to the B3LYP keyword:

#P BLYP IOp(3/76=1000002000) IOp(3/77=0720008000) IOp(3/78=0810010000)

The output file displays the values that are in use:

IExCor= 402 DFT=T Ex=B+HF Corr=LYP ExCW=0 ScaHFX= 0.200000

where the value of ScaHFX is P2, and the sequence of values given for ScaDFX are P4, P3, P6 and P5.

So P4 = A = 0.8, P2 = 1-A = 0.2, P3 = B = 0.72, and the correlation functional is ECVWN+CΔECnon-local. Here ECVWN is the local correlation functional, so P6 = 1. Since LYP has a local and a non-local part, we have to subtract the local fraction from VWN to compensate. So the non-local correlation functional is actually ECLYP-ECVWN, making P5 = C = 0.81. But since part of ECVWN was already included in P6, the total amount of ECVWN in the final functional ends up being 1-0.81 = 0.19.

Last edited by a moderator: May 8, 2017
9. Aug 7, 2016

### darwined

Dear TeethWhitener,

Sorry for the delay in response.

Thanks for your help throughout. Now things are clear to me.