# Question about calculate transition electric field

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 Ylniu Member Profile Send PM
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 9:21:00 AM PST - Wed, Jan 18th 2017 Dear all,  I want to calculate transition electric field:  $E_{ba,x}(\mathbf{R}) = \int \rho_{ba}(r) {(x-R_x)/|\mathbf{r}-\mathbf{R}|^3} d^3 x$ where b and a represent two electronic states, ρba(r) is the transition density.  I modify the nwdft/lr_tddft/tddft_analysis.F, and I calculate H2O as an example.  But I read the code of calculating transition dipole moment as an example. I find in source file tddft_analysis.F, when ipol = 1 for closed shell molecule, the quantity should times dsqrt(2.0d0), not 2.0, for spin up and spin down. 1294 if (ipol.eq.1) then 1295 do i=1,20 1296 tmom(i)=tmom(i)*dsqrt(2.0d0) 1297 enddo I think it should be an simple problem, but I still can not understand why should it be like: 1296 tmom(i)=tmom(i)*2.0d0 for two spin statistics. Can any one give me an simple hint of this question?

 Niri Forum:Admin, Forum:Mod, NWChemDeveloper, bureaucrat, sysop Profile Send PM
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 6:38:50 PM PST - Thu, Jan 19th 2017 Hi, In the tddft_analysis, the multipole_density call is being used to calculate the transition moments and this is in turn used to calculate the strengths which is squared.   Hope this helps. Best, -Niri

 Ylniu Member Profile Send PM
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 6:29:02 PM PST - Fri, Jan 20th 2017 Dear Niri, Thanks a lot for your reply! I find that multipole_density calculate the transition multipole moment, then the moment times 2.0 when ipol==1(closed shell, spin-restricted case), then this moment is used to calculate oscillator strength and written in the output file. 1271 call multipole_density(ao_bas_han,cntr,3, 1272 1 g_td(i),tmom,20)  ! transition moments 1294 if (ipol.eq.1) then 1295 do i=1,20 1296 tmom(i)=tmom(i)*dsqrt(2.0d0) 1297 enddo 1328 osc_str = 1329 & 2.0d0/3.0d0*(tmom(2)**2+tmom(3)**2+tmom(4)**2)*apbval(n) 1330 c 1331 write(LuOut,9200) tmom(2),tmom(3),tmom(4), 1332 1 tmom(5),tmom(6),tmom(7), 1333 2 tmom(8),tmom(9),tmom(10), 1334 2 osc_str But in my mind, I think it should times with 2.0, for both spin up and spin down, in the line 1296 and then output in the line 1331. Is there anything wrong for my understanding? Edited On 4:16:38 PM PST - Sat, Jan 21st 2017 by Ylniu

 Ylniu Member Profile Send PM
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 5:07:07 PM PST - Sat, Jan 21st 2017 Dear Niri, I get the answer why I am wrong to understand the coeffient sqrt(2.0d0). First, the normalizing condition of TDDFT coeffient is: $\sum\limits_{ia}\left(X_{ia}^2-Y_{ia}^2\right) = \sum\limits_{ia}\left(X_{ia,\uparrow}^2-Y_{ia,\uparrow}^2\right) + \sum\limits_{ia}\left(X_{ia,\downarrow}^2-Y_{ia,\downarrow}^2\right)=1$ This is true for UDFT(ipol=2). But for the case of RDFT(ipol=1), in the source file nwdft/lr_tddft/tddft_analysis.F, $\sum\limits_{ia}\left(X_{ia,\uparrow}^2-Y_{ia,\uparrow}^2\right)=1$ But the TDDFT coeffient should satisfy the condition: $\sum\limits_{ia}\left(X_{ia,\uparrow}^2-Y_{ia,\uparrow}^2\right)=0.5$ Then $X_{ia,\uparrow}$ should times by $1/ \sqrt{2}$. Consider the contribution from both spin-up and spin-down, the moment should times by 2. This is reason why tmom(i) should times by $1/\sqrt{2} \times 2 = \sqrt{2}$: 1294 if (ipol.eq.1) then 1295 do i=1,20 1296 tmom(i)=tmom(i)*dsqrt(2.0d0) 1297 enddo Thanks again for your helpful reply! And now I can add the code to calculate the transition electric field.

 Niri Forum:Admin, Forum:Mod, NWChemDeveloper, bureaucrat, sysop Profile Send PM
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 12:47:19 AM PST - Sun, Jan 22nd 2017 Dear Yiniu, Indeed. Looks good. Best, -Niri

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