Helium Hamiltonian: Derive Ground State & Quantum Numbers

[kramleigh]

Hi,

I have a question in a past exam paper which I can't quite understand how to prove. It reads:

Give an expression for the Hamiltonian of the Helium atom. Neglecting the interaction between the electrons, derive the state function for the Helium ground state in terms of hydrogen-like spatial state functions and spin eigenstates. What are the quantum numbers of this state?

I have attched a word document which shows the progress that I have made on the question so far.

I understand that the spin eigenstates are the "1/2" terms after each chi and that the interaction term is the last in the hamiltonian but am unsure on how to prove that this is its ground state, other than the fact that psi100, is the lowest energy state since n = 1.

Any help would be much appreciated. Thanks

 

[Pietjuh]

Hi,

I have a question in a past exam paper which I can't quite understand how to prove. It reads:

Give an expression for the Hamiltonian of the Helium atom. Neglecting the interaction between the electrons, derive the state function for the Helium ground state in terms of hydrogen-like spatial state functions and spin eigenstates. What are the quantum numbers of this state?

I have attched a word document which shows the progress that I have made on the question so far.

I understand that the spin eigenstates are the "1/2" terms after each chi and that the interaction term is the last in the hamiltonian but am unsure on how to prove that this is its ground state, other than the fact that psi100, is the lowest energy state since n = 1.

Any help would be much appreciated. Thanks

Ok, in the hamiltonian for the helium atom you have terms for the individual electrons and the interaction term. But in this problem you can neglect it, which means that the ground state of the helium atom is just the product of the two ground states of the hydrogen atom, so the ground state for the helium atom is given by
$$\psi_0 (r_1 , r_2 ) = \frac{8}{\pi a^3}e^{-2(r_1 + r_2) / a}$$
If you plug this in your hamiltonian you will obtain an energy of E = -109 eV. This energy does not depend on the spin quantum numbers, because you don't take spin-orbit coupling and all other perturbative effects into account i presume

Now you are right that the total wavefunction must be antisymmetric because the electrons are fermions. This means that the spin part must be antisymmetric. The only possible antisymmetric spin state for 2 electrons, is the spin singlet, which has the quantum numbers s = 0 and m_s = 0. This spin part of the wavefunction is just 1/sqrt(2) (up(1)down(2) - down(1)up(2)), which you also wrote down in your doc file.

Now the final part of the question is easy, your quantum numbers are n = 1, l = 0, m_l = 0, s = 0, m_s = 0. So you basically already solved the problem :)