Help-Forum: use of .SECMOM
Dear Dalton Community, I would like some help regarding Dalton functionalities. Since the forum is down, I write in this gitlab environment. If I am writing at the wrong place please let me know.
I would like to compute the properties of molecules (let's say water in vacuum) under a non-spacially homogenous field. Extending the Taylor development of the electromagnetic potential, one can obtain such interacting Hamoltinan:
H = H_0 - \mu \cdot E - 1/2 Q^{[2]} : \nabla E (see Eq 3.186 of Principles and Practices of Molecular Properties, Norman, Wiley) Where \mu is the dipole moment of the molecule (<\psi|\hat{r}|\psi>), Q^{[2]} its quadrupolar moment (<\psi|\hat{r} \hat{r}^T|\psi>), E the value of the (external) electric field and \nabla E its spatial derivative.
Using such formalism, one can compute the polarizability (\alpha) of the molecule by the so-called Finite Field (FF) method. To do so, one should apply a static electric field, solve the Hamiltonian and plot the dipole moment of the molecule with respect to the applied electric field. The evolution should be linear, and the slop is the polarizability.
Practically within Dalton (DFT), I tried:
########################
**DALTON
.RUN WAVE FUNCTIONS
.RUN PROPERTIES
**INTEGRALS
.DIPLEN
**WAVE FUNCTIONS
.DFT
Camb3lyp
*SCF IN
.MAX DIIS
500
*HAMILTONIAN
.FIELD
0.0001
ZDIPLEN
**END OF DALTON
########################
Which should correspond to the Hamiltonian: H = H_0 - \mu_z * 0.0001
Indeed, as I increase the field, the energy decreases: \mu_z is positive, the field is positive: - \mu_z * E_z is negative. The obtained polarizability is in agreement with a response calculation, i.e.: \alpha = <<\mu, -\mu>> = - <<\mu, \mu>> (Kubo relation).
Now, I want to compute the first hyperpolarizability at the quadrupolar level. For instance, I would like to know the evolution of the dipole moment with respect to E and \nabla E. For instance, I can write the induced dipole moment \mu_{NL} to be:
\mu_{NL} = 1/2 \beta^q : E \nabla E + 1/2 \beta^q : \nabla E E = \beta^q : E \nabla E (here the field are time-independent so we even in the T convention (the same as the Kubo relation) we should take into account E \nabla E and \nabla E E)
In other words, if I assume that such development is correct, the induced dipole moment is linear with respect to \nabla E. Therefore, let's do again the Finite Field approach: varying \nabla E (for instance \delta E_y / \delta y) and let's plot the induced dipole moment:
########################
**DALTON
.RUN WAVE FUNCTIONS
.RUN PROPERTIES
**INTEGRALS
.DIPLEN
.SECMOM
**WAVE FUNCTIONS
.DFT
Camb3lyp
*SCF IN
.MAX DIIS
500
*HAMILTONIAN
.FIELD
0.00010
YDIPLEN
.FIELD
0.00010
YYSECMOM
**END OF DALTON
########################
finite_field_beta_q_bbbb_linear_scale.pdf
Hourra: it is linear. According to the previous equation, the \beta^q is the slop (here -300 a.u.)
However, if I check the energy there is a 'problem': as the \nabla E increases, the energy increases.
The YYSECMOM is negative, and \delta E_y / \delta y is positive. Moreover, the difference in energy from \delta E_y / \delta y=10**(-4) and \delta E_y / \delta y=2 10**(-4) correspond exactly to the YYSECMOM*10**(-4).
Thus, if I am not mistaking, the Hamiltonian I have built using the latter input is:
H = H_0 - \mu_y * 0.0001 - YYSECMOM * 0.00010
The 1/2 obtained by the initial equation is gone. Also, if I compute the quadrupolar hyperpolarizability \beta^q (in the T convention!) using:
########################
**DALTON
.RUN WAVE FUNCTIONS
.RUN PROPERTIES
.RUN RESPONSE
**INTEGRALS
.DIPLEN
.SECMOM
**WAVE FUNCTIONS
.DFT
Camb3lyp
*SCF IN
.MAX DIIS
500
**RESPONSE
*QUADRA
.APROP
YDIPLEN
.BPROP
YDIPLEN
.CPROP
YYSECMOM
.SHG
.FREQUE
1
0.0
**END OF DALTON
########################
In the response scheme, \beta^q = + <<\mu, -1/2 Q, -\mu>> = 1/2 <<\mu, Q, \mu>>. The value of <<\mu, Q, \mu>> is about +300 a.u. Therefore, I think that this confirm that the Hamiltonian built in the previous Finite Field case is indeed:
H = H_0 - \mu \cdot E - Q : \nabla E (and not H = H_0 - \mu \cdot E - 1/2 Q : \nabla E )
To obtained the same hyperpolarizability using the response scheme and the Finite Field theory, we should apply 1/2. However, there is still a - sign between the response scheme and the Finite Field one. Which I am struggling to explain.
Questions:
- When I ask:
*HAMILTONIAN
.FIELD
0.00010
YYSECMOM
Does Dalton add + YYSECMOM * 0.00010, - YYSECMOM * 0.00010 or -1/2 YYSECMOM * 0.00010 in the energy
- The output of the response module:
########################
@ Quadratic response function value in a.u. for
@ A operator, symmetry, spin: YDIPLEN 1 0
@ B operator, symmetry, spin: YDIPLEN 1 0
@ C operator, symmetry, spin: YYSECMOM 1 0
@ omega B, omega C, QR value : 0.00000000 0.00000000 307.89403346
########################
Means that: << + YDIPLEN, + YDIPLEN, + YYSECMOM>> = 307.89403346 There is no ''hidden'' - sign?
Thank you a lot for your help, Best regards, Guillaume Le Breton (PhD student)