In this section, we describe the possible ways to solve the TDSE, see Eq. (1.3), focusing on vibrational one-dimensional systems (i.e., we use Rinstead ofR), where the kinetic operator can be described as a diagonal matrix within the BO approximation with elements:65
Tˆ = − ¯h2 2
∂
∂Rg ∂
∂R = pgˆ pˆ
2 , (1.6)
where g represents the inverse of the moved mass in the coordinate R.
This coordinate can be an internal coordinate (e.g., a bond distance) or a collective coordinate (e.g., a normal mode). Depending on the definition ofR,gcan be a function (e.g., the bending angle), or a constant (e.g., the reduced mass belonging to some normal vibrational mode of a polyatomic molecule),65in which case the kinetic operator is justTˆ =gpˆ2/2. Applying this definition of the kinetic operator to the TDSE, we obtain a series of
equations coupled by the electric field, ih¯ ∂
∂tψm(R, t)=(Vm+pgˆ pˆ
2 )ψm(R, t)−
n
ànmε(t)ψm(R, t). (1.7) This set of equations can be solved by applying the Hamiltonian to the wavefunction, where the procedure mainly depends on the chosen basis.
On the one hand, it is possible to choose theivibrational eigenfunctions of the field-free Hamiltonian for themelectronic potentials,Vm, as a basis,
ψm(R, t)=
i
ci,m(t)φi,m(R), (1.8) where(Vm+ ˆT )φi,m(R)= Ei,mφi,m(R)andci,m(t)are the amplitudes at every time. By inserting Eq. (1.8) in Eq. (1.7) and projecting onφj,n, we obtain
ih¯ ∂
∂tcj,n(t)=
i
Ej,ni,m−àj,ni,mε(t)
ci,m(t), (1.9) where the integrals of the time-independent Hamiltonian elements are Ej,ni,m=
φj,n∗ (Vˆ + ˆT )φi,mdR, which in case of orthogonal eigenfunctions are justEj,ni,m =Ei,mδijδmn. Similarly,àj,ni,m are the matrix elements of the dipole moment, which can be related to the electronic dipole momentànm
asàj,ni,m =
φj,n∗ ànmφi,mdR. If there is no electric field, this equation can be analytically solved and the time evolution of the coefficients is just
ci,m(t)=ci,m(0)exp
−i
¯ hEi,mt
. (1.10)
However, neither the calculation of the eigenfunctions nor the solution including an electric field are analytic beyond the harmonic model and the numerical approximation requires many basis functions to solve the above equation.21
On the other hand, we can work directly in the one-dimensional grid R, whereVmandàmnare directly defined. In this case, the problem is the definition of the kinetic operator, which is readily applied in the momentum space but not in the coordinate space. However, the nuclear wavefunction can be easily transformed to the momentum space by a Fourier transform.
In the momentum representation, the kinetic operator is diagonal, ψm(R, t)=
i
ci,m(t)|ri−→FT ψm(p, t)=
i
ci,m(t)|pi, (1.11) where|riand|piare basis sets with zeros in all the grid points and 1 when R= ri andp = pi, respectively. Using this definition, the application of the kinetic operator to the wavefunction is simple. For example,
−1
2gpˆ2
i
ci,m(t)|pi = −
i
ci,m(t)g
2p2i |pi (1.12) and similarly in the potential part, where Vm|ri = Vm(ri)|ri and àmn|ri =àmn(ri)|ri.
1.2.3.1. Second-order differentiator
One of the simplest methods to solve the TDSE on a grid is the second-order differentiator (SOD) method.66In this approach, the wavefunction at time t+tis expanded in a second-order Taylor expansion:
ψ(t+t)=ψ(t)+t∂
∂tψ(t)+t2 2
∂2
∂t2ψ(t), (1.13) where we can obtain the temporal derivative of ψ using Eq. (1.7). To avoid the application of the Hamiltonian twice, it is possible to modify the propagator, so that21
ψ(t+t)=ψ(t−t)+2t∂
∂tψ(t). (1.14) The main problem of this propagator is the numerical instability. Since the propagator is not unitary, the time-steptshould be very small to assure the conservation of the norm.
1.2.3.2. Split-operator method
A more elaborated propagator is the so-called split-operator (SO) tech- nique.67–69 In this method, we integrate the TDSE from Eq. (1.10) and
arrive at a solution as:
ψ(t+t)=exp
−i
¯ hH tˆ
ψ(t)=exp
−i
¯
h(Wˆ + ˆT )t
ψ(t), (1.15) where Wˆ = ˆV −àE represents the potential part of the Hamiltonian, including the field interaction, which is represented in coordinate space, andTˆ is the kinetic part that should be applied in momentum space. As they are not represented on the same grid, it is not possible to apply the exponential including both potential and kinetic parts at the same time and they have to be split into two terms. The problem is that Wˆ and Tˆ do not commute, and it is not exact to describe the exponential term as the multiplication of two noncommuting ones, i.e., exp(H )ˆ =exp(W )ˆ exp(T ).ˆ This problem is solved in the SO by splitting one of the parts in two. For example, splitting the potential part,
ψ(t+t)≈exp
−i
¯ hWˆ t
2
exp
−i
¯ hT tˆ
exp
−i
¯ hWˆ t
2
ψ(t).
(1.16) The application of the potential part is very simple even if it is not diagonal, e.g., when the electric field couples two electronic states. In that case, the operators are 2×2 matrices, where every matrix element depends onRrepresented by the grid pointsri. The potential part of the propagation can be easily carried out after a diagonalization of the W matrix at everyri,
exp
−i
¯ hWt
2
ψ=Zexp
−i
¯
hZ†WZt 2
Z†ψ
=Zexp
−i
¯ hDt
2
Z†ψ, (1.17) whereDis a diagonal matrix containing the eigenvalues ofWandZis the unitary transformation matrix containing the corresponding eigenvectors.
In contrast to the SOD, the SO is unitary and very stable. However, the kinetic operator in the exponent cannot be applied using Fourier transform whengdepends on the coordinate.