Slater Determinants for Large numbers of electrons

In summary: The problem of course is that the number of one-particle states (i.e., the size of the basis) is very large in realistic calculations, which means that the number of matrix elements is very large as well. The trick is then to use density fitting techniques or other methods to truncate the one-particle space that is actually used to evaluate the 1RDM or the matrix elements of the operators. This way, the computational cost can be reduced significantly while still obtaining accurate results.
  • #1
t_evans
8
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Disclaimer: If this is the wrong place for this, I apologise, this probably comes somewhere between QM, Atomic, Linear algebra and a spoonful of Quantum chemistry for good measure.

Anyway, for a group of non interacting (mean field) electrons, moving in a potential generated by nuclei and the average interactions with other electrons, the single particle, non relativistic Hamiltonian can be solved numerically quite easily. Then the ground state of the system is first estimated by the Slater determinant of the single electron wavefunctions, building up from the lowest energy wavefunctions to the highest required by the number of electrons.

[itex]\psi(x_1,\ldots,x_N) = \left|\begin{array}{ccc}
\phi_{1}(x_{1}) & \ldots & \phi_{N}(x_{1})\\
\vdots & \ddots\\
\phi_{1}(x_{N}) & & \phi_{N}(x_{N})
\end{array}\right|[/itex]

Where the variables include position and spin. Now, for small numbers of electrons, It is reasonable to just take the determinant, mod square it get the probability density function and then integrate over the required region, to generate the potential, then essentially go through the Hartree-Foch self consistency method. The problem with this is that for large N (say, 40 or 50), the determinant has enormous numbers of terms depending on 40 or 50 variables, which obviously isn't particularly practical for actually computing anything of any interest.

For the issue at hand (i.e. atomic systems), on spherical symmetry and non relativistic (ignoring the LS couping), the states are degenerate in all generality in the projected spin [itex]s_z[/itex] and the projected orbital angular momentum [itex]l_z[/itex]. Furthermore, looking just for a spherical potential means that all the angular integrals can be done ahead of time with the orthogonality relations of the spherical harmonics, so just looking the radial cumulative probability (which is the thing of interest for generating the potential), will look something like;

[itex] P(R) = \int dx_1 ... \int dx_N \psi^*\psi [/itex]

Is there some general rule for splitting up this integral into the non orthogonal bits? For Helium and Lithium it seems that the states separate straightforwardly to (within normalisation, which has been played fast and loose with) the products of the determinants of the bits which are non-orthogonal, or for example, in Lithium, the effective state looks something like

[itex] \left|1s\downarrow;\, x_{1}\right\rangle \otimes\left|\begin{array}{cc}
\left|1s\uparrow;\, x_{2}\right\rangle & \left|2s\uparrow;\, x_{2}\right\rangle \\
\left|1s\uparrow;\, x_{3}\right\rangle & \left|2s\uparrow;\, x_{3}\right\rangle
\end{array}\right| [/itex]

And then further states added in the same manner? (i.e. in separate determinants for the different angular momentum and spin quantum numbers)

I've probably not been particularly clear for which I apologise, but any help would be much appreciated!
(n.b. sorry for mixing wavefunction and Dirac notation)
 
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  • #2
t_evans said:
Now, for small numbers of electrons, It is reasonable to just take the determinant, mod square it get the probability density function and then integrate over the required region, to generate the potential, then essentially go through the Hartree-Foch self consistency method. The problem with this is that for large N (say, 40 or 50), the determinant has enormous numbers of terms depending on 40 or 50 variables, which obviously isn't particularly practical for actually computing anything of any interest.
This is the main use of second quantization algebra for finite particle systems: Getting rid of the determinants (and permanents). Expectation values of one- and two-particle operators can be evaluated without ever touching any determinant. Let's assume that we expand our one-particle states within a orthonormal finite basis |r>. Then a one-particle operator (e.g., the the electron density δ(r-g) (with position operator r and for the density at g)) can be evaluated by contracting the matrix elements of the operator [tex]o_{rs} = \langle r|op|s\rangle[/tex] with the ``one-particle reduced density matrix'' (1-RDM) [tex]\gamma_{rs} = \langle
\Phi| a^r a_s |\Phi\rangle[/tex]. The latter can be evaluated trivially for a determinant [tex]|\Phi\rangle[/tex]: It's one if the spin-orbitals r and s are both occupied in the determinant, and zero otherwise. More complicated expectation values can be evaluated using so-called Wick theorems and normal ordered substitution operators. Those are used to re-express expectation values for more complicated wave functions in terms of Slater derminants or configuration state functions (CSFs).

Using those techniques the actual determinants can usually be avoided.
 

1. What is a Slater Determinant?

A Slater Determinant is a mathematical expression used in quantum mechanics to describe the wave function of a multi-electron system. It is a product of single-electron wave functions that accounts for the antisymmetry of the electrons' wave function.

2. How does a Slater Determinant differ from a Hartree-Fock wave function?

A Slater Determinant is a more general form of a Hartree-Fock wave function, which only considers the ground state configuration of an electron system. The Slater Determinant can be used to describe excited states and systems with more than one electron.

3. What are the challenges in using Slater Determinants for large numbers of electrons?

One of the major challenges is the computational complexity involved in calculating and manipulating large numbers of single-electron wave functions. Additionally, the number of possible Slater Determinants increases exponentially with the number of electrons, making it difficult to accurately describe the wave function of a system with a large number of electrons.

4. How can Slater Determinants be used in electronic structure calculations?

Slater Determinants can be used as a basis for electronic structure calculations, such as density functional theory (DFT). They can also be used to generate trial wave functions in variational Monte Carlo simulations to calculate the ground state energy of a system.

5. Are there any alternative methods to describe large numbers of electrons in quantum systems?

Yes, there are several alternative methods such as the Configuration Interaction (CI) method, which uses a linear combination of Slater Determinants to approximate the wave function of a system. Other methods include Coupled Cluster (CC) theory and Many-Body Perturbation Theory (MBPT).

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