# What is the difference between product wave function and true wave function for many electron system

I know expression for product wave function, and I know true wave function for many electron system can't be expressed. But can someone tell me what is missing in the product wave function? What is the approximation?

• What is missing is the correlation between the different electrons making up the system. – Lewis Miller Dec 14 '16 at 16:43
• I don't know if this is the case and if I have fully understand the question, but I think that as a product wave function you call the wave function that comes out from separating the variables of the hamiltonian operator of the whole system. Since not all the hamiltonians for many particles systems can be separated, you can use an approximation (for instance, the Born-Oppenheimer Approximation, that separate the nuclei space from the electrons space) and say that you solve your system as if it were separable (getting the product wave function) even if it is not. – JackI Dec 14 '16 at 16:49
• thats what i thought at first, but it is a approximation. if you separate the variables of hamiltonian, thats the exact wavefunction. here product wavefunction is not equal to exact wavefuction. – user133174 Dec 14 '16 at 17:52

The electronic Hamiltonian for a multi-electron atom (or molecule) generally contains three components. (When I write "electronic Hamiltonian", the Born-Oppenheimer approximation - which allows for separation of electronic and nuclear motion - is already implicitly invoked.)

$$\hat{H} = \sum_i{\hat{T}_i} + \sum_i\sum_a{\hat{V}^\mathrm{en}_{ia}} + \sum_i\sum_{j>i}{\hat{V}^\mathrm{ee}_{ij}}$$

where the first summation represents the kinetic energy ($T_i$ being the kinetic energy of the $i$-th electron), the second representing electron-nucleus attractions ($a$ is the summation index for nuclei - if the system is an atom then you can drop this summation), and the last representing electron-electron repulsions.

If we ignore the electron-electron repulsions entirely (or equivalently, treat them as a constant, which has no impact on the eigenfunctions but simply shifts the energy eigenvalues by a constant amount), then you can rearrange the terms and write

$$\hat{H} = \hat{H}_1 + \hat{H_2} + \cdots + \hat{H}_n$$

where $n$ is the total number of electrons and $\hat{H}_i$ is an operator that only acts on one electron at a time:

$$\hat{H}_i = \hat{T}_i + \sum_a \hat{V}^\mathrm{en}_{ia}$$

and it's relatively straightforward to show (using separation of variables) that the product function $\psi = \psi_1\psi_2\cdots\psi_n$ satisfies the Schrodinger equation

$$\hat{H}\psi = E\psi$$

with $\hat{H}_i \psi_i = E_i \psi_i$ and $E = E_1 + E_2 + \cdots + E_n$.

So, the thing that is "missing" is simply the proper treatment of electron-electron repulsions. Sometimes this is also referred to as "correlation", but this has the potential to be ambiguous, as the word correlation can be used to refer to different aspects of electron-electron repulsions.