Born-Oppenheimer approximation confusion


by VortexLattice
Tags: approximation, bornoppenheimer, confusion
VortexLattice
VortexLattice is offline
#1
Jan21-13, 12:35 AM
P: 146
Hi, I'm reading about the Born-Oppenheimer approximation for a solid and they're doing the formalism of it. They say that we can basically consider the ions stationary with respect to the electrons because they move so little and so slowly in comparison to them.

They say that ##R_i## are the positions of the ions and ##r_j## are the positions of the electrons, ##P_i## are the momenta of the ions, ##p_j## are the momenta of the electrons (all vectors but I'm just writing them like this here). Then they say that we'll look at "core" electrons separately from "valence" electrons, because core ones just hang out by the nuclei while valence ones move around. Given all this, the hamiltonian is:

##H = \sum\limits_i \frac{P_i^2}{2M} + \sum\limits_{j = cond. elecs} \frac{p_j^2}{2m} + \sum\limits_{i,i'} V_{i,i'}(|R_i - R_{i'}|) + (e^2/2) \sum\limits_{j,j'=cond. elecs} \frac{1}{|r_j - r_{j'}|} + \sum\limits_{i,j} V_{ei}(|r_j - R_i|) + E_{core}##

(where ##E_{core}## is the energy of the "core" electrons that are "attached" to the nuclei.)

Then they rewrite this as:

##H = T_i + T_e + V_{ii} + V_{ee} + V_{ei} + E_{core}##

Then they say that we can write the full wavefunction as a combination of two functions (here, ##r## and ##R## are the sets of the positions of all the electrons/ions, not single ones):

##\Psi(r,R) = \sum\limits_n \Phi_n(R) \Psi_{e,n}(r,R)##

Then, they just do the eigenvalue equation, ##H\Psi = E\Psi##:

##(T_i + V_{ii} + E_{core})\Psi + \sum\limits_n \Phi_n (T_e + V_{ee} + V_{ei})\Psi_{e,n}(r,R) = E\Psi##

In the second term, the part with the explicit sum, they put ##\Phi_n## out in front because the operators directly following it "only operate on the electron part of the product wavefunction", according to my book. But here's my confusion: doesn't ##V_{ei}## act on the ion part of the wave function? It was defined as ##\sum\limits_{i,j} V_{ei}(|r_j - R_i|)##, which has that ##R_i## in it. What am I missing?

Thank you!
Phys.Org News Partner Physics news on Phys.org
Researchers develop scalable methods for manufacturing metamaterials
Researchers find tin selenide shows promise for efficiently converting waste heat into electrical energy
After 13 years, progress in pitch-drop experiment (w/ video)
Cthugha
Cthugha is offline
#2
Jan21-13, 01:30 AM
Sci Advisor
P: 1,563
Quote Quote by VortexLattice View Post
What am I missing?
You are more or less missing the Born-Oppenheimer approximation itself. Recall what you said in the beginning:

Quote Quote by VortexLattice View Post
They say that we can basically consider the ions stationary with respect to the electrons because they move so little and so slowly in comparison to them.
So your assumption is basically that your electron distribution does not drag your ions around. The distribution of the ions will stay as it is. Therefore you can treat the position of the ions as a parameter instead of a variable. So the potential will only depend on these positions, but not act on them.

If you want a more intuitive explanation, what you do is not solving the coupled system of ions and electrons, but getting a solution for the electron system for a fixed set of ion positions.
DrDu
DrDu is online now
#3
Jan21-13, 01:42 AM
Sci Advisor
P: 3,371
No, up to what VortexLattice has described, the whole wavefunction is still completely general and the BO approximation has not been invoked, yet.
V_ei clearly acts also on the nuclear wavefunction, but as it is a multiplicative operator, it does not matter whether it appears in front or after ##\Phi_n##.

Cthugha
Cthugha is offline
#4
Jan21-13, 01:52 AM
Sci Advisor
P: 1,563

Born-Oppenheimer approximation confusion


Oh sorry. Maybe I misread and should stop posting after midnight.
VortexLattice
VortexLattice is offline
#5
Jan21-13, 12:27 PM
P: 146
Quote Quote by DrDu View Post
No, up to what VortexLattice has described, the whole wavefunction is still completely general and the BO approximation has not been invoked, yet.
V_ei clearly acts also on the nuclear wavefunction, but as it is a multiplicative operator, it does not matter whether it appears in front or after ##\Phi_n##.
Hmmm, this doesn't seem consistent though. If ##V_{ei} = \sum\limits_{i,j} V_{ei}(|r_j - R_i|)##, I see what you're saying about the order of the operator ##R_i## not mattering (that's what you're saying, right?), but then what about ##V_{ii}##? That also just has multiplicative factors of ##R##: ##V_{ii} = \sum\limits_{i,i'} V_{i,i'}(|R_i - R_{i'}|)##.

So why isn't that in the second term as well?

Thank you!
DrDu
DrDu is online now
#6
Jan22-13, 01:33 AM
Sci Advisor
P: 3,371
Quote Quote by VortexLattice View Post
So why isn't that in the second term as well?
Feel free to put it there!
But the ordering chosen is not a matter of mathematical necessity but depends on the approximations which will probably introduced in the continuation of the argument.
VortexLattice
VortexLattice is offline
#7
Jan22-13, 09:42 AM
P: 146
Quote Quote by DrDu View Post
Feel free to put it there!
But the ordering chosen is not a matter of mathematical necessity but depends on the approximations which will probably introduced in the continuation of the argument.
Hmmm, but that changes the final result! The final result they reach for this is separating ##\Phi_n## and ##\Psi_{e,n}## by separating this equation:

##(T_i + V_{ii} + E_{core})\Psi + \sum\limits_n \Phi_n (T_e + V_{ee} + V_{ei})\Psi_{e,n}(r,R) = E\Psi##

and after doing many manipulations, get:

##(T_i + V_{ii} + E_{core} + E_{e,n})\Phi_n = E_n\Phi_n##

and

##(T_e + V_{ee} + V_{ei})\Psi_{e,n}(r,R) = E_{e,n}\Psi_{e,n}(r,R)##

So, the placement of ##V_{ei}## makes it end up in either the nuclear or electronic equation.
DrDu
DrDu is online now
#8
Jan22-13, 10:27 AM
Sci Advisor
P: 3,371
V_ei depends on both nuclear and electronic coordinates. You want an equation for ##\Phi_n## which does not contain electronic coordinates. Likewise you want an equation for ##\Psi_{e,n}## which does not contain derivatives with respect to R.


Register to reply

Related Discussions
born oppenheimer approximation vs adiabatic approximation Quantum Physics 4
Born Oppenheimer Approximation and Product Asnatz Atomic, Solid State, Comp. Physics 34
Born Oppenheimer?? Atomic, Solid State, Comp. Physics 3
Adibatic Approx.(i.e., Born-Oppenheimer) Quantum Physics 12
Born-Oppenheimer approximation Advanced Physics Homework 1