The ground state of helium - showing that parhelium is ground state

[Emspak]

Homework Statement

The Hamiltonian of helium can be expressed as the sum of two hydrogen Hamiltonians and that of the Coulomb interaction of two electrons.

$\hat H = \hat H_1 + \hat H_2 + \hat H_{1,2}$

The wave function for parahelium (spin = 0) is

$\psi(1,2) = \psi_S(r_1, r_2)\dot \xi_A(s_1, s_2)$ with the first being a symmetric spatial function and the second being an antisymmetric one.

We can separate this into the normalized function

$\psi_S(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1)(\psi_2(r_2)+ \psi_1(r_2)(\psi_2(r_1)]=\psi_S(r_2,r_1)$

For orthohelium the functions look like this:

$\psi(1,2) = \psi_A(r_1, r_2)\dot \xi_S(s_1, s_2)$
$\psi_A(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1)(\psi_2(r_2) - \psi_1(r_2)(\psi_2(r_1)]=\psi_A(r_2,r_1)$

Show the ground state of helium is parahelium. The hint is what happens to the wavefunction.

Homework Equations

OK, so I start with that the Hamiltonian of the given wave function(s)

$$H_1 = \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial^2 r_1^2} = E_1 \psi, H_2 = \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial^2 r_1^2} = E_2 \psi,
H_{1,2} = -\frac{e^2}{4\pi\epsilon_0 r_{1,2}}$$

The Attempt at a Solution

OK, I was trying to get a handle on how to get started with this. My first thought was to treat each Hamiltonian separately and just get a wavefunction the way I would solving any Schrodinger, though I am not sure of the boundary conditions. Then I can get a value for energy. But my sense is there is a simpler way to do it. I don't need a whole walk-through I don't think, I am just trying to understand some of the slutions I do see out there.

 

[Emspak]

and looking into it some more I wanted to check if this was an OK (mathematically speaking) thing to do:
One thing I thought of doing was this (for $H_1$):

$$H_1 = -\frac{\hbar^2}{2m}(\frac{\partial^2 \psi}{\partial r_1^2}+\frac{\partial^2 \psi}{\partial r_2^2})=E_1(\psi_1(r_2)+\psi_1(r_1))$$

but again I don't know if that's kosher

Thanks.

 

[DrClaude]

For orthohelium the functions look like this:

$\psi(1,2) = \psi_A(r_1, r_2)\dot \xi_S(s_1, s_2)$
$\psi_A(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1)(\psi_2(r_2) - \psi_1(r_2)(\psi_2(r_1)]=\psi_A(r_2,r_1)$

You mean
$$
\psi_A(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1) \psi_2(r_2) - \psi_1(r_2)\psi_2(r_1)]= - \psi_A(r_2,r_1)
$$

OK, so I start with that the Hamiltonian of the given wave function(s)

$$H_1 = \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial^2 r_1^2} = E_1 \psi, H_2 = \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial^2 r_1^2} = E_2 \psi,
H_{1,2} = -\frac{e^2}{4\pi\epsilon_0 r_{1,2}}$$

You're missing the Coulomb interaction between the electrons and the nucleus. Both $H_1$ and $H_2$ look like the single-electron (hydrogenic atom) Hamiltonian.

OK, I was trying to get a handle on how to get started with this. My first thought was to treat each Hamiltonian separately and just get a wavefunction the way I would solving any Schrodinger, though I am not sure of the boundary conditions.

That part should be trivial: it is easy to find the solution for a hydrogenic atom.

You have the right approach. Look at the energy of the two-electron system by first neglecting the electron-electron interaction (treat it as a perturbation, even though it is not really one).

 

[Emspak]

OK, adding in the potential terms I wanted to make sure this was o to do:
For the $\psi(r_1,r_2)$ function I should have (With the potential terms)
$$H_1 = -\frac{\hbar^2}{2m}(\frac{\partial^2 \psi_1}{\partial r_1^2}+\frac{\partial^2 \psi}{\partial r_2^2}-\frac{2e^2}{4\pi \epsilon_0 r}-\frac{2e^2}{4\pi \epsilon_0 r})=E_1(\psi_1(r_2)+\psi_1(r_1))$$.

To solve that I can further break that up as follows:

$$-\frac{\hbar^2}{2m}\frac{\partial^2 \psi_1}{\partial r_1^2}-\frac{2e^2}{4\pi \epsilon_0 r_1}=E\psi_1(r_1)$$
$$-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial r_2^2}-\frac{2e^2}{4\pi \epsilon_0 r}=E_1\psi_1(r_2)$$.

Right? Add these energies up and I should have a different result for the assymetric and symmetric cases. (One of the functions is negative, for one thing).

 

[unscientific]

OK, adding in the potential terms I wanted to make sure this was o to do:
For the $\psi(r_1,r_2)$ function I should have (With the potential terms)
$$H_1 = -\frac{\hbar^2}{2m}(\frac{\partial^2 \psi_1}{\partial r_1^2}+\frac{\partial^2 \psi}{\partial r_2^2}-\frac{2e^2}{4\pi \epsilon_0 r}-\frac{2e^2}{4\pi \epsilon_0 r})=E_1(\psi_1(r_2)+\psi_1(r_1))$$.

To solve that I can further break that up as follows:

$$-\frac{\hbar^2}{2m}\frac{\partial^2 \psi_1}{\partial r_1^2}-\frac{2e^2}{4\pi \epsilon_0 r_1}=E\psi_1(r_1)$$
$$-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial r_2^2}-\frac{2e^2}{4\pi \epsilon_0 r}=E_1\psi_1(r_2)$$.

Right? Add these energies up and I should have a different result for the assymetric and symmetric cases. (One of the functions is negative, for one thing).

Just consider what happens to the spatial part of the wavefunction in the ground state. Is it spatially symmetric or anti-symmetric?

Now, what must the spin part be then?