Separating Hamiltonian functions. Helium atom.

[scorpion990]

I'm using McQuarrie's "Quantum Chemistry" book for a little bit of light reading. He included a proof of a theorem that states that if a Hamiltonian function is separable, then the eigenfunctions of Schrodinger's equation are the products of the eigenfunctions of the simpler "separated" equations... Anyway...

I just started studying the helium atom. Schrodinger's equation is written with two different laplacian operators, one that acts "only on the first electron", and one that acts "only on the second electron". However, the laplacian for both electrons is just a differential operator which contains a mixture of partial derivatives with respect to all three coordinates. I don't understand why you can assume that the first laplacian only acts "on the first electron". After all, a differential operator (D) acting on two functions (F and G):
D(F*G) is not F*D(G) if D contains derivatives which pertain to F.

Help?

 

[theUndergrad]

That Hamiltonian operator is looking at each interaction separately and making a statement that all of the interactions taken together describe the system completely. Omitting the inter-nueclon terms (which have little effect on the motion of the electrons), and assuming a stationary nucleus (which we can do since the mass of the nucleus is much larger than the mass of the electrons) there are 5 terms: a laplacian term that deals with the momentum of one of the electrons relative to the nucleus, another laplacian term that deals with the momentum of the other electron relative to the nucleus, and three columb terms: one for each of the electrons interacting with the nucleus and one for the interaction between the two electrons (this one is sometimes omitted to simply the problem).

So there are two momentum terms because each of the electron's momentum is independent of the others, so you need one term that deals with each.

---------------
theUndergrad

http://www.theUndergraduateJournal.com/

 

[denian]

I can offer some clarification on this topic. The concept of separable Hamiltonian functions is an important one in quantum mechanics, especially in the study of multi-electron systems like the helium atom. This theorem states that if a Hamiltonian function can be separated into simpler, independent functions, then the eigenfunctions of the overall system are just the products of the eigenfunctions of these simpler functions.

In the case of the helium atom, the Hamiltonian function can be separated into two independent functions, one for each electron. This is possible because the two electrons do not interact with each other, but only with the nucleus. Therefore, the Hamiltonian function can be written as the sum of two simpler functions, each representing the energy of one electron.

Now, regarding your question about the laplacian operator acting on only one electron. In quantum mechanics, the position and momentum of a particle are represented by operators, not just numbers. These operators act on the wavefunction of the system, which contains information about all particles in the system. So, when we say the laplacian operator acts "only on the first electron", we mean that it acts on the part of the wavefunction that describes the first electron, while the other part of the wavefunction remains unchanged.

In mathematical terms, this means that the laplacian operator is diagonal with respect to each individual electron's coordinates. This allows us to separate the Hamiltonian function and treat the two electrons as independent particles, each with their own set of coordinates and operators.

I hope this explanation helps clarify the concept of separable Hamiltonian functions and the role of operators in quantum mechanics. Keep up the good work in your studies!