In the multi-configurational time-dependent Hartree (MCTDH) method, the equation of motion is derived from the time-dependent Schrödinger equation by substituting the wavefunction ansatz expanded in a single particle function (SPF) basis. The wavepacket dynamics is solved by solving a series of decoupled equations of motion in the SPF basis as follows:

$$ i\dot\phi^{(k)}=(1-\hat{P}^{(k)})(\boldsymbol \rho^{(k)})^{-1}\boldsymbol{H}^{(k)}\phi^{(k)}. $$

For the definition of the quantities, one could refer to the linked article:


I wonder if someone knows how to derive this equation above from the time dependent Schrödinger equation?

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    $\begingroup$ Generally, a review paper won't include derivations, but it will likely reference a paper that did. Even if the whole derivation isn't there, it will likely be easier to answer "what is the missing step in this derivation" than "derive this equation" $\endgroup$
    – Tyberius
    Jul 6, 2020 at 14:24
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    $\begingroup$ I answered the question. I basically followed Tyberius's advice which was to look at the original MCTDH paper from 1990. $\endgroup$ Jul 6, 2020 at 18:43

2 Answers 2


I will outline the way it was derived in the original 1990 paper. We start with an ansatz for the time-dependent wavefunction:

\begin{equation} \tag{1} \psi(x_1,\ldots,x_n;t) = \sum_{j_1=1}^{m_1}\cdots \sum_{j_n=1}^{m_n}a_{j_1\cdots j_n}\phi_{j_1}^{(1)}(x_1,t)\cdots \phi_{j_n}^{(n)}(x_n,t), \end{equation}

with single-particle functions (SPFs) satisfying (the second constraint is to make MCTDH simpler):

\begin{equation} \tag{2}\label{ortho} \langle \phi_i^{(k)} | \phi_j^{(k)}\rangle =\delta_{ij} ~,~ \langle \phi_i^{(k)} | \dot\phi_j^{(k)}\rangle =0. \end{equation}

Now we will use the Dirac-Frenkel variational principle (DFVP) to optimize the parameters:

\begin{equation} \tag{3}\label{DiracFrenkel} \langle \delta \psi |(H-\rm{i}\frac{\partial}{\partial t})|\psi\rangle =0. \end{equation}

Making use of all 4 equations so far, leads to this (you may need some practice with using DFVP):

\begin{equation} \tag{4}\label{} \textrm{i}\dot a_{j_1\ldots j_n}=\langle \phi_{j_1}^{(1)}\cdots\phi_{j_n}^{(n)}|H|\psi\rangle . \end{equation}

If we define the following:

\begin{align} J &\equiv (j_1,j_2,\ldots ,j_{k-1},j_{k+1},\ldots ,j_n)\tag{5}\\ \mathbf{A}^{(k)} &\equiv a_{j_1\ldots j_{k-1},j,j_{k+1}}^{(k)} \equiv A_{Jj}^{(k)} \tag{6}\\ \mathbf{B}^{(k)} &\equiv \left(\mathbf{A}^{(k)\dagger}\mathbf{A}^{(k)\dagger} \right)^{-1}\mathbf{A}^{(k)\dagger}\tag{7}\\ \hat{H}^{(k)}_{IJ} &\equiv \langle \phi_I^{(k)} |H|\phi_J^{(k)}\rangle \tag{8}\\ \hat{P}^{(k)}&\equiv \sum_{j=1}^{m_k}|\phi_j^{(k)}\rangle\langle \phi_k^{(k)}|\tag{9}, \end{align}

we can instead write:

\begin{equation} \tag{10} \textrm{i}|\dot\phi_i^{(k)}\rangle = (1 - \hat{P}^{(k)})\sum_{IJj}B_{iI}^{(k)}\hat{H}_{IJ}^{(k)}A_{Jj}^{(k)}|\phi_j^{(k)}\rangle. \end{equation}

These are the original working equations for MCTDH, and they are also almost exactly written the way you've written, except with $B$ instead of $\rho$: This is enough to get you started. A full derivation of the working MCTDH equation typically takes more than 60 lines, assuming you already have available some handy DFVP expressions.

  • $\begingroup$ Thanks for answering my question. I have briefly read the original paper and corrected some typo in your answer. I agree that your derivation follows the original paper which claim that DFVP is applied to derive equation 4. However, as for me, equation 4 is derived simply by projection of time dependent Schrödinger equation on SPF basis. $\endgroup$
    – Paulie Bao
    Jul 7, 2020 at 7:41
  • $\begingroup$ You're right. My version would have had an $H|\dot \psi \rangle$ which would have been wrong. $\endgroup$ Jul 7, 2020 at 14:21

This response comes late, but I hope you or other readers will find it useful. Regarding your question on how to derive MCTDH equations (directly) from the time-dependent Schrödinger equation (SE), I would add the following comments.

The SE is exact in the full Hilbert space, but for numerical simulations, you need to choose a finite (usually small) working subspace, which is imposed by the form of your ansatz. Since the solution of the SE lies in the full Hilbert space, you cannot generally solve the SE in this working subspace. This means that the coupled MCTDH equations are not equivalent to the SE, and cannot be derived unambiguously from the SE.

Alternatively, you can minimize the error of your approximate solution with respect to the parameters used in your ansatz. The time-dependent variational principle (TDVP) is the tool that does that by projecting the SE error on the correct tangent space. Therefore, one must distinguish between the exact SE and its approximate version coming out of the TDVP, and I believe the answer to your question is in the application of the TDVP.

The projection done by the TDVP is otherwise equivalently achieved using a linear projection of the SE only when the parameterization is "linear", as it is the case for the time-dependent parameters $a_{j_1⋯j_n}$ (the expansion coefficients in the basis of the configurations). Hence, your comment on eq.4 being "derived simply by projection". However, the projection is more complicated for a "non-linear" parameterization, and MCTDH employs such a non-linear parameterization in the expression of the time-dependent SPFs. Hence, the resulting equation of motion for the SPFs are not simple projections of the SE but are obtained using the TDVP.

Thus, the answer to your question would be that MCTDH equations cannot be derived directly from the SE. You need to use the SE in combination with the TDVP to correctly project on the tangent space corresponding to your ansatz. I hope this answer helps you, and if you are interested in more details, I suggest you the following documents to read:

P. Kramer, M. Saraceno: Geometry of the Time-Dependent Variational Principle in Quantum Mechanics (1981)

A. Raab: Chem. Phys. Lett. 319, 674 (2000)

L. Hackl, T. Guaita, T. Shi, J. Haegeman, E. Demler et al.: arXiv:2004.01015 (2020)


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