In the multi-configurational time-dependent Hartree (MCTDH) theory, one essential step that reduces the memory and time cost is to form a single particle function (SPF) basis ansatz as shown in equations in (1) and (2) in the linked review literature1.

Equation (1):

(f: degrees of freedom, $q_1, \ldots, q_f$: coordinates) \begin{align} \Psi \left( {{q_1}, \ldots {q_f},t} \right) &= \sum\limits_{{j_1}}^{{n_1}} { \ldots \sum\limits_{{j_p}}^{{n_p}} {{A_{{j_1} \ldots {j_p}}}\left( t \right)\varphi _{{j_1}}^{\left( 1 \right)}\left( {{Q_1},t} \right) \ldots } } \varphi _{{j_p}}^{\left( p \right)}\left( {{Q_p},t} \right)\tag{1}\\ &= \sum\limits_J {{A_J}{\Phi _J}}\tag{2} \end{align}

Equation (1) is a direct product expansion of p sets of orthonormal time-dependent basisfunctions $\left\{ {\varphi {}^{\left( \kappa \right)}} \right\}$, known as single-particle functions (SPFs). The coordinate for each set of ${n_\kappa }$ functions is a composite coordinate of one or more system coordinates (equation 2):

\begin{array}{c} {Q_\kappa } = \left( {{q_a},{q_b}, \ldots } \right).\tag{2} \end{array}

However, the reference fails to explain in detail how to construct time-dependent, orthonormal SPFs from a primitive basis set. How is this done?

  1. G.A. Worth, et al. Using the MCTDH wavepacket propagation method to describe multimode non-adiabatic dynamics. Int. Rev. Phys. Chem. 27, 569–606 (2008). DOI: 10.1080/01442350802137656.
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    $\begingroup$ That would be nice if you could add the referred equations as part of your question. $\endgroup$ Jul 7, 2020 at 22:57
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    $\begingroup$ Paulie, I did not vote to close here, but you will have to show Eqs. 1 and 2 in order for the rest of the community to take back their close votes I think. You have 4 close votes and with 1 more the question will become closed. $\endgroup$ Jul 13, 2020 at 23:12
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    $\begingroup$ All right. I will try to add equations when I have time. I am really busy these days. $\endgroup$
    – Paulie Bao
    Jul 14, 2020 at 1:23

1 Answer 1


I believe that Eq.13 of the same reference is the construction you are asking about. Since MCTDH is an initial value problem, you have to construct the SPFs at the initial time by choosing the initial value of the time-dependent coefficients $a^{(\kappa)}_{kj}$ and then MCTDH will propagate these coefficients. Then your question becomes "how to construct the SPFs at initial time?", which is up to you when you choose your initial condition.

The only constraint is that you need to choose the coefficients so that the SPFs are initially orthonormalized (the MCTDH propagation retains the orthonormalization). This amounts to choose the coefficients such that \begin{equation}\tag{1} \sum_{lm}(a^{(\kappa)}_{lj})^*\langle \chi^{(\kappa)}_l | \chi^{(\kappa)}_m \rangle a^{(\kappa)}_{mk}=\delta_{jk}, \end{equation} where $\delta_{jk}$ is the Kronecker delta. If the primitive basis is already orthonormal, $\langle \chi^{(\kappa)}_l | \chi^{(\kappa)}_m \rangle=\delta_{lm}$, the coefficients just represent an ensemble of orthonormalized vectors, but if the primitive basis is not orthonormal, you will have to use your favorite orthonormalization scheme (e.g. Gram–Schmidt or using a singular value decomposition among others).

The initial wavefunction is very often chosen as a single Hartree product of physically "meaningful" SPFs, where the "meaningful" part comes from what you want to achieve with your propagation (e.g. a product of 1-dimensional Gaussian functions). Another possibility is to use the eigenstate of some other operator (e.g. a zeroth order Hamiltonian), which could be obtained from a previous MCTDH calculation (e.g. a relaxation).

So, the answer is that the construction is up to your initial conditions as long as you make sure that the SPFs are initially orthonormalized. I would recommend reading chapter 3 and 5 of the following book:

H.-D. Meyer, F. Gatti, G. A. Worth: Multidimensional Quantum Dynamics (2009)

or the following review, which can be found at the website of the MCTDH developers (pci.uni-heidelberg.de/cms/mctdh.html):

M. H. Beck, A. Jäckle, G. A. Worth, H.-D. Meyer: Phys. Rep. 324, 1 (2000)

I hope this message answers your question.

  • $\begingroup$ +10 beautiful! This was one of our oldest unanswered questions, so I'm glad to have an answer here! Welcome to our community, and thank you for your contributions! We hope to see much more of you here in the future!!! $\endgroup$ Feb 22, 2021 at 19:41

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