I am an outsider to this field, so I am not sure about the validity of my work below.

Let us define the following Hamiltonian from DFT:

$$ \tag{1}H_{ij} \psi_{ij} \equiv (-\frac{\hbar^2 \nabla_i^2}{2m} -\frac{\hbar^2 \nabla_j^2}{2m} + \frac{1}{4 \pi \epsilon_0}\frac{k_2}{\vec r_i ^2} + \frac{1}{4 \pi \epsilon_0}\frac{k_2}{\vec r_j ^2} + \frac{\lambda}{4 \pi \epsilon_0} \frac{k_1}{(\vec r_i - \vec r_j)^2} ) \psi_{ij}$$

where $k_1$ and $k_2$ are constants. Let $\psi_{12}$ be the lowest energy state of the system:

$$\tag{2} \min_\psi H_{12} \psi = E_0(\lambda) $$

We add a constant energy to the system such that the minimum eigenvalue obeys:

$$\tag{3} \min (H_{12} + c)\psi_{12} = (E_0(\lambda) + c)$$

Lets say I want to solve the minimum energy of a $3$ electron material. I play the following trick

$$\tag{4} \min ((H_{12}+ c) \psi_{12})^3 = (E_0(\lambda) + c )^3 $$

Let us use:

$$\tag{5} \!\!\!\!\!\!\!\!\!\!\!\!\! \min ((H_{12}+ c) \psi_{12})^3 = \min ((H_{12}+ c) \psi_{12}) \min ((H_{12}+ c) \psi_{12}) \min ((H_{12}+ c) \psi_{12}) \!\!\!\!\!\!\!$$


$$\tag{6}\min ((H_{12}+ c) \psi_{12})^3 =\min ((H_{12}+ c) \psi_{12}) ((H_{23}+ c) \psi_{23})((H_{31}+ c) \psi_{31}) $$

Focusing on the constant $c^2$ term:

$$\tag{7} \min c^2(H_{12} + H_{23} + H_{31})\psi_{12} \psi_{23} \psi_{31} = 3c^2 E_0 (\lambda) $$

However there will be $n-2$ excess kinetic energy terms and static external potential terms. Thus we choose $\lambda =n-1 $ (where $n=3$ over here)

And get:

$$\tag{8} \min \frac{c^{2}}{2}((H_{12} + H_{23} + H_{31})\psi_{12} \psi_{23} \psi_{31} ) =\frac{3}{2}c^2 E_0( 2) $$

In general for $n$ electrons:

$$\tag{9} H \equiv \min \frac{c^{N-1}}{n-1}(\sum_{i\neq j}^n H_{ij} ) =c^{N-1} \frac{N}{n-1}E_0( n-1) $$

where $N = n(n-1)/2$


Is this trick correct? Can someone cross check if this yields correct answers?


1 Answer 1


There are a few issues with the derivation:

  1. (not necessarily a mistake, but) please note that if you mean the Coulomb Hamiltonian, the RHS of (1) should involve the reciprocals of $|r_i|$ etc., not $r_i^2$ etc. The Coulomb interaction enters the Hamiltonian in the form of potential energy, rather than in the form of force. Please ignore this comment if you are deliberately using the inverse square potential as an example.
  2. In (2), note that the quantity to be minimized is $\langle \psi|H|\psi \rangle$ (in other words, $\int \psi^*(\vec{r})H\psi(\vec{r}) d\vec{r}$), not $H|\psi\rangle$. The latter is not even a number; rather, it's a function, and you cannot minimize a function - you can only minimize a number.
  3. By using $\langle \psi|H|\psi \rangle$ instead of $H|\psi\rangle$, you will be able to correct some of the equations that follow, but you won't arrive at anything that resembles your conclusion. You'll end up at something like $\langle \psi_{12}|H_{12}+c|\psi_{12} \rangle\langle \psi_{23}|H_{23}+c|\psi_{23} \rangle\langle \psi_{31}|H_{31}+c|\psi_{31} \rangle$, but you can't proceed further because the multiplications of bras, kets and operators are in general not commutative. You can't rearrange this into a form that resembles (7). Whenever in doubt, write all bras and kets into the equivalent integral form, write out the differential operators in H, and apply basic calculus rules to see which operations you can do and which you can't.

And finally, a general advice: there is almost certainly no free lunch with the Coulombic many-body problem. It's worthwhile to consider how to numerically approximate it in the best way, but attempts to simplify it analytically would almost certainly be fruitless. I remember that in my first year as a graduate student I expanded the $e^{-iHt/\hbar}$ operator using the BCH approximation, and attempted to simplify and resum the series by using the fact $\nabla^2 V = 0$. This took quite a lot of my spare time and I got nothing, except for the precious lesson that any efforts to analytically simplify the Coulombic many-body problem are likely bound to make little sense. Indeed if you can analytically simplify the problem to a significant degree, you will prove P=NP.

  • $\begingroup$ Ah! I am sure you must be correct if my calculations imply $P=NP$ but do mind responding to the edit? (which is a response to $3$) because if I remain in the position basis I think I can get away? $\endgroup$ Commented Aug 24, 2021 at 4:05
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    $\begingroup$ If you accepted this answer already, and have follow-up questions I'd recommend asking a new question since follow-up back-and-forth questions can go on ad infinitum and that's not always fair for the answerer. $\endgroup$ Commented Aug 24, 2021 at 4:09
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    $\begingroup$ @NikeDattani Fair enough I'll undo the edit in that case $\endgroup$ Commented Aug 24, 2021 at 4:10

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