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I've noticed while reading about Hartree-Fock in Modern Quantum Chemistry (Szabo and Ostlund) that most basis sets use only real basis functions. What I'm wondering is why aren't complex basis functions used? I might be mistaken, but from my experience it seems as though integration on the complex plane would have a lot of convenient properties that would make computations easier. Put another way, are complex basis functions not used because of some difficultly in implementation or is it because of difficulty in interpretation of properties like the molecular orbitals?

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    $\begingroup$ I have found a recent article that discusses using complex basis functions that might offer some insight, but I currently don't have access to it. Even if there are some cases where it is used, I'm interested in why it wouldn't be the standard approach. $\endgroup$ – Tyberius Mar 19 '17 at 2:17
  • $\begingroup$ "(...) integration on the complex plane would have a lot of convenient properties that would make computations easier." Could you give us some examples? $\endgroup$ – pentavalentcarbon Mar 19 '17 at 17:53
  • $\begingroup$ @pentavalent I guess I was thinking about it the wrong way in my explanation. It in most cases wouldn't make things easier to compute, but it does open up a wider array of functions to describe an orbital, which might allow fewer basis functions to be used to obtain an accurate description. I'm not certain this would necessarily be useful, but one could also potentially take advantage of path independence of complex integration to solve certain integral. But I admit, its purely speculation on my part whether these properties would be of any use in practice. $\endgroup$ – Tyberius Mar 19 '17 at 20:02
  • $\begingroup$ Also, I read a bit of that article and found this book which discusses using complex basis sets for a specific type of problem. It seems in both cases that complex wavefunctions are used because they allow one to obtain eigenvalues of an analytically extended Hamiltonian and that these eigenvalues are useful in the context of resonance scattering. $\endgroup$ – Tyberius Mar 19 '17 at 20:09
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The basic answer is rather simple: In the non-relativistic limit it is just not necessary. There is no complex operator and hence nothing which could change the complex component of your basis functions. Using complex basis functions would only double the required work (i.e., computer time). There are very efficient schemes in order to calculate the standard integrals (McMurchie-Davidson, Obara-Saika, ...). I don't think that the representability of MOs is a reason for using real basis functions, since, strictly speaking, you can interprete only the square of the MOs and not their amplitude.

However, there are situations, where your MOs can be complex. This can happen when you have a complex operator in your equation. An example are the Pauli spin matrices of which $\sigma_y$ has only an imaginary compound. They occur, e.g., the Dirac equation (see here one formulation where they explicitly state the incorporation of the Pauli matrices). Complex MOs can be obtained either by using complex basis functions and real MO coefficients or by using real basis functions with complex coefficients. Starting from a non-relativistic implementation it might be simpler just to keep your basis functions real and to choose the MO coefficients complex.

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  • $\begingroup$ I haven't had much experience with the particular algorithms used to compute the integrals in these calculations so thank you for introducing them. I'm curious about one point in your answer: why would using complex basis functions "double" the work required? Does it eliminate certain symmetries that allow less integrals to be computed or do these efficient algorithms either not work or work less effectively for complex functions? $\endgroup$ – Tyberius Mar 19 '17 at 20:20
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    $\begingroup$ @Tyberius : If you multiply two real numbers, you basically have one operation: $A*B$. If you instead multiply two complex numbers $(A+iC)*(B+iD)$ you end up with $A*B - C*D + i ( A*D + B*C)$. In this case you would increase thenumber of operations by a factor of 6. So, the "double" was maybe not correct. Indeed, for GHF calculations, as they pop up in relativistic QC, one has to drop the spin symmetry. I.e., the orbitals are spinors which contain $\alpha$, as well as $\beta$ components. Hence you loose the blockdiagonal nature of many operators' matrix representation. $\endgroup$ – TheFox Mar 20 '17 at 20:32

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