# SWE for Lithium

1. Nov 10, 2009

### pzona

1. The problem statement, all variables and given/known data
Set up the Schroedinger Wave Equation for Li. Hint: Don’t forget to consider electron-electron
interactions.

2. Relevant equations
H^$$\Psi$$=E$$\Psi$$, etc.

3. The attempt at a solution
I understand how to set up the equation, i.e. how to use the H^ operator, obtain solutions for $$\Psi$$, and I'm fairly sure I can solve it (for H anyway). I'm more unsure of how to solve it for multiple electron atoms than anything else. I'm guessing that I'll have to do it three times, once for each electron. Is this correct? Also, where do I begin to account for electron-electron interactions? I'm not sure what this means, as we never explicitly went over it in class. Does this pertain to the charges, or just positions? I'm not asking anyone to do the problem for me (I'd much rather do it myself actually), but if someone could give me a place to start, I'd appreciate it; I'm completely lost.

2. Nov 11, 2009

### Staff: Mentor

I suppose question asks just for correct H operator, one that takes all interactions into account. But I can be wrong, my Quantum Chemistry long ago fall apart leaving just a pinch of rust.

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methods

3. Nov 11, 2009

### pzona

Thanks, Borek, I actually hadn't considered that. I'm planning on going to a review session tomorrow night so hopefully I'll be able to get some help with this then. Any idea on what factors to consider (i.e. charge? electron energy?) when calculating the Hamiltonian operator for multiple electrons? I don't want to seem like I'm asking the professor to just give me answers.

4. Nov 11, 2009

### Staff: Mentor

e2/ri,j for each pair of electrons (where ri,j is a distance)?

Don't treat me too seriously, I am sneezing. I am sneezing because I had to dust my quantum chemistry book to check if I am right. I don't remember seeing book with that thick layer of dust. I have to take a shower.

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methods

5. Nov 11, 2009

### pzona

Lol, I understand

Well that seems to be a good place for me to start. Thanks a lot for your help.