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Say we have a system of 2 electrons, an ansatz for the wavefunction is then a Slater determinant consisting of two orbitals. $$ \left|\Psi_{Slater}\right>= \frac{1}{\sqrt{2!}}\left(\left|\phi_1\phi_2\right>-\left|\phi_2\phi_1\right>\right) $$

How is this related to the basis expansion ansatz: $$ \left|\Psi\right>= \sum_i c_i \left|\phi_i\right> $$

The basis expansion formula would give me a function of 1 coordinate x_1 if i projected it on the position space, while the slater determinant ansatz would give me a function of two coordinates x_1, x_2. x_i stands for a set of x,y,z coordinates. Spin is neglected in this example. I understand that for the basis set expansion in an orthonormal basis, one can obtain the coefficients by the inner product $$ \left<\phi_i|\Psi \right>=c_i $$

but how can i use this on a slater determinant ? I only understand how this works if everything relies on one coordinate x_i, then i can calculate this: $$ \left<\phi_i|\Psi \right>=\int\phi_i(x)\Psi(x)dx=c_i $$

But how does this work if the wavefunction is multidimensional like the slater determinant wavefunction ? $$ \left<\phi_i|\Psi_{Slater} \right>=\int\phi_i(x_1)\Psi_{Slater}(x_1,x_2)dx= c_i(x_2)? $$ This expression makes no sense, since the remaining object would still be a function of $$x_2$$

To my knowledge only basis functions of one coordinate are used in calculations, like 3D Gaussians. Assuming that i have a wavefunction given in slater determinant form, how would i find the coefficients of the basis functions ?

I am afraid i am mixing up things that don't belong together, clarifications would be welcome.

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    $\begingroup$ I think I see where you are getting tripped up. LCAO is a way of constructing MOs, not multielectron wavefunctions. The Slater determinant is a way of forming a wavefunction from MOs, which may have themselves been formed using LCAO. @HansWurst $\endgroup$
    – Tyberius ♦
    Aug 30, 2018 at 14:38
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    $\begingroup$ To put @Tyberius comment into equations: your second equation should be: $\phi_i=\sum_\nu c_{\nu,i} \varphi_\nu$, where $\phi_i$ are the molecular orbitals from your first equations, and $\varphi_\nu$ are atomic orbital basis functions of the LCAO ansatz. $\endgroup$
    – Feodoran
    Aug 30, 2018 at 14:49
  • $\begingroup$ That makes sense. But i am still not clear on some other equations then.It is often assumed that you can write $$\left|\Psi \right> = \sum_i c_i \left|\phi_i\right>$$ The small phi's are then not molecular orbitals but some eigenfunctions of an operator for the given space ? I.e. i could use the Hamilton Eigenfunctions as the basis functions and they would dependent for the example above, each on two coordinates x_1 and x_2 ? $\endgroup$
    – Hans Wurst
    Aug 30, 2018 at 17:09
  • $\begingroup$ It is hard to tell what you are asking about. It is unclear where this equation is coming from. What is its physical context, other than being some linear combination? Alternatively we would need to know what exactly $|\phi\rangle$ is here. But without knowing either, I can only make wild guesses: it could be the Hartree-Product, CI, or something completely different. Please clarify your question with some more context. $\endgroup$
    – Feodoran
    Aug 30, 2018 at 20:42

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I found the question somewhat confusing. I think it is because mixing up some concepts.

I understand that for the basis set expansion in an orthonormal basis, one can obtain the coefficients by the inner product $$ \left<\phi_i|\Psi \right>=c_i $$

Please, think about what you mean by the coefficients and where the functions live, before continue reading.

As you thought this expression as an inner product (which is right), both functions must live in the same space so they have the same number of coordinates. What you are doing with that inner product is projecting $\Psi$ over the $\phi_i$. If $\{\phi_i\}$ is an orthogonal set, then you can make the expansion you wrote. But, if the are not, then such expansion is meaning less. You still can make expansions that recover $\Psi$ (in fact, an infinite number of them). So, I get confused about which coefficients are you referring to. As for example, $\phi_1$, appears in both: many Slater's det. and in many terms in one (or more) Slater's det., to ask which is its coefficient in the Slater det. makes not sense, except if you want to expand the Slater det. into a LCAO which is not possible (but you still can project it).

See below.

But how does this work if the wavefunction is multidimensional like the slater determinant wavefunction ?

First, note that even for 1-electron problems the problem is not mono-dimensional. So, the case of Slater's determinants is exactly the same. You only need to integrate over every coordinate.

To my knowledge only basis functions of one coordinate are used in calculations, like 3D Gaussians.

3D-Gaussians have three coordinates. It is obvious using Cartesian coordinates. Under spherical coordinates they only depends on $r$, but they still are functions of $\theta$ and $\phi$. Although you can take only one coordinate ($r$) for its evaluation, it is not true for computing an arbitrary inner product.

Assuming that i have a wavefunction given in slater determinant form, how would i find the coefficients of the basis functions ?

At first, take in mind that those coefficients are for expanding the Slater det. into the subspace spanned by your $\{\phi_i\}$. That is, it they are not an complete set, you only will get an approximation. You just have to perform those integrations. In your example, the $c_i$ will not be a function of $x_2$, because you have to integrate over every coordinate.

I am not sure if I understand correctly your doubts. Please, do not hesitate in ask for clarification. Maybe this answer could be useful.

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  • $\begingroup$ I was initially confused by the fact that the explicit variables are suppressed and was unsure about the dimensionality of the involved objects, but that has been cleared up by now. $\endgroup$
    – Hans Wurst
    Sep 17, 2018 at 11:38
  • $\begingroup$ @Hans Wurst , books often do not characterize rigorously the mathematical objects that they are using. That leads to an extra effort for the reader to realize exactly what those objects are. Normally it is skipped leaving some holes. $\endgroup$
    – user1420303
    Sep 22, 2018 at 14:35

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