What is the solution for two electrons on a sphere?

[PeterDonis]

If I have to be an expert on solving this problem before I can ask about it here,

Not an expert in solving this specific problem, but it's not clear that you have a general "A" level background in quantum mechanics. Without that you might not be able to get far on this specific problem no matter how much help we give you; it would amount to us just solving the problem and telling you the solution.

 

[PeterDonis]

You will have to give me time to put up my math

That's fine, nobody is under time pressure here. But if you are going to make claims like "the antisymmetric part of the two-particle spatial wave function is zero", you need to back them up with math.

 

[bob012345]

Not an expert in solving this specific problem, but it's not clear that you have a general "A" level background in quantum mechanics. Without that you might not be able to get far on this specific problem no matter how much help we give you; it would amount to us just solving the problem and telling you the solution.

Please define exactly what an "A" level is. I've had a lot of QM at the University of Illinois.

 

[PeterDonis]

lease define exactly what an "A" level is.

Graduate level knowledge of the subject matter. See here:

https://www.physicsforums.com/help/threadprefixes/

 

[HomogenousCow]

https://arxiv.org/abs/1004.3641If anyone’s interested in the state of the problem.

So as one would expect, the groundstate does correspond to the classical antipodal configuration.

 

[hilbert2]

Could it be expected that deforming the sphere to an ellipsoid of revolution would cause splitting of energy levels or only shift the existing ones?

Edit: it would wreck the same symmetry that's with the $p_x ,p_y ,p_z$ orbitals of a hydrogen atom, but the answer is not immediately clear to me.

 

[bob012345]

That's fine, nobody is under time pressure here. But if you are going to make claims like "the antisymmetric part of the two-particle spatial wave function is zero", you need to back them up with math.

That's fine. I'm working on the math now. Then I need to figure out how to show you since LaTeX doesn't seem to work from my iPad. Maybe I can attach a file with my work. That claim was merely my thoughts at the time and I hope to either defend it of refute it as work on the problem develops. Thanks to all participating in this thread!

 

[PeterDonis]

LaTeX doesn't seem to work from my iPad

You can just type the LaTeX directly into a post; it's just text characters. I don't see how it would work any differently from typing any other characters.

Maybe I can attach a file with my work

That's not in line with PF policy; we need to see the math entered directly into the post so we can quote it when we reply.

 

[bob012345]

So as one would expect, the groundstate does correspond to the classical antipodal configuration.

Thank you very much for finding this!

 

[bob012345]

You can just type the LaTeX directly into a post; it's just text characters. I don't see how it would work any differently from typing any other characters.
That's not in line with PF policy; we need to see the math entered directly into the post so we can quote it when we reply.

"Note: the PF apps for iOS (iPhone, iPad) and Android can only display raw LaTeX code as plain text. You must use PF via a web browser in order to see properly-rendered equations."

Thanks. That makes it a little harder for me but not impossible.

 

[Nugatory]

Thanks. That makes it a little harder for me but not impossible.

So many things work better on the iPad in Safari than in the app that I've just never used the app there.

 

[PeterDonis]

the PF apps for iOS (iPhone, iPad) and Android can only display raw LaTeX code as plain text

Aren't the apps outdated now that PF has upgraded to a new version of the forum software? The Android app, at least, no longer works for me, it tells me to use the website and then shuts down.

 

[Vanadium 50]

the groundstate does correspond to the classical antipodal configuration.

Well, kind of. It's not exact. I didn't really expect it to be, but I don't think anything precludes it.

 

[Vanadium 50]

Could it be expected that deforming the sphere to an ellipsoid of revolution would cause splitting of energy levels or only shift the existing ones?

Edit: it would wreck the same symmetry that's with the $p_x ,p_y ,p_z$ orbitals of a hydrogen atom, but the answer is not immediately clear to me.

Do you mean like the n-l degeneracies in the hydrogen atom? That would definitely be broken if it existed, but this system doesn't have any. It does have the usual magnetic degeneracies.

 

[HomogenousCow]

Well, kind of. It's not exact. I didn't really expect it to be, but I don't think anything precludes it.

How could it be exact?

 

[Vanadium 50]

How could it be exact?

Nothing in QM prohibits this. Think about the case where you have a stick holding the two charges, still both constrained to move on the surface of the sphere, a diameter apart. QM can certainly calculate this motion, so it's not a prohibited solution. As I said earlier, the requirement is that the distance between the second charge and the antipode commute with the Hamiltonian, which at the time seemed unlikely, and downthread we know that it doesn't. But the fact that the relationship isn't exact doesn't mean it can't be.

 

[Vanadium 50]

You could probably exploit the rotational symmetry of the problem and "put" one of the spherical harmonics at the north pole

You probably could, but this also means you don't have a well-defined z-axis to quantize around. I suspect that it makes things worse instead of better: I would certain want to keep the Ylm's if I could.

 

[bob012345]

I'm having a problem normalizing a two electron wavefunction on a unit sphere. The normalized solutions for the one electron problem are the $Y_\ell^m (\theta, \varphi)$. This is a prelude to the problem of this thread.

We construct the anti-symmetric wavefunction from the $\Phi_a(\vec r_i)$ representing the $Y_\ell^m (\theta_i, \varphi_i)$ for a set of quantum numbers ${\ell, m}$ labeled $a$ or $b$ and coordinate set $i$ labeled 1 or 2;
$$\Psi_A(\vec r_1,\vec r_2)= \frac{1}{\sqrt{2}}[\Phi_a(\vec r_1) \Phi_b(\vec r_2) + \Phi_b(\vec r_1) \Phi_a(\vec r_2)][singlet>$$
$Y_{0}^{0} (\theta_1, \varphi_1) = {1\over 2\sqrt{\pi} }$ for all coordinate sets and quantum number sets we can write the ground state wavefunction.;
$$\Psi_A(\vec r_1,\vec r_2)= \frac{1}{\sqrt{2}}[ {1\over 2\sqrt{\pi} } {1\over 2\sqrt{\pi} }+ {1\over 2\sqrt{\pi}} {1\over 2\sqrt{\pi} }][singlet>$$

Which gives us;

$$\Psi_A= \frac{1}{\sqrt{2}}[ {1\over 2{\pi}}][singlet>$$

If properly normalized, the probability over all space which is the probability of finding two particles on the sphere, should be 1 but I get 2.

$$<\Psi^{\ast}|\Psi>=\iint( \frac{1}{\sqrt{2}}[ {1\over 2{\pi}}])^2dA_1dA_2= (\frac{1}{\sqrt{2}}[ {1\over 2{\pi}}])^2(4\pi)^2=2$$

I thought for any properly constructed wavefunction this should be 1. So I wonder if it's the product wavefunction $\Phi_a(\vec r_1) \Phi_b(\vec r_2)$ that should be normalized when constructing the anti-symmetric wavefunction? What else could be the issue? I ask because all my texts discuss the full symmetric spatial part is necessary for the ground state of Helium but don't actually use it but do use the product wavefunction $\Phi_a(\vec r_1) \Phi_b(\vec r_2)$. I worked through the Helium atom normalization with the full symmetric wavefunction and got 2 also. The full symmetric spatial wavefunction does normalize with a factor of ${1\over 2}$ rather than ${1\over \sqrt 2}$.Thanks.

 

[bob012345]

Ok, now I see what's happening and why... The formal structure of symmetrization for the spatial part goes as this;$$\Psi(\vec r_1,\vec r_2)= \frac{1}{\sqrt{2}}[\Phi_a(\vec r_1) \Phi_b(\vec r_2) \pm \Phi_b(\vec r_1) \Phi_a(\vec r_2)]$$

Where ${a,b}$ represent a set of quantum numbers and coordinate sets are labeled ${1, 2}$ and the Hamiltonian was separable.;

In the general case, the $\Phi_j(\vec{r}_i)$ represents not just normalized one electron wavefunctions but $orthogonal$ ones. When the quantum numbers are the same, the states are not orthogonal and thus are special cases. This is why all the texts treat the ground state of Helium as a special case because both electrons are in the same state and thus are not orthogonal. Thus, the normalization factor for the full symmetricized wavefunction is $1\over2$ which gives the same result as the product of two single electron functions $\Phi_a(\vec r_1) \Phi_b(\vec r_2)$ but doubles the work. Now I can proceed to the main problem.

 

[bob012345]

Now, on to the main problem of two electrons on a sphere. I start basic and then will build on and explore the solutions in follow up posts.

We start the problem with the Schrödinger equation;$$\hat H \Psi = E \Psi $$

The Hamiltonian for a particle on a sphere with no potential is;

$$\hat H = - \frac{\hbar^2}{2m} \nabla^2$$

Which gives for no radial function for fixed $r$;

$$\hat H =- \frac{\hbar^2}{2mr^2} \left [ {1 \over \sin \theta} {\partial \over \partial \theta} \left ( \sin \theta {\partial \over \partial \theta} \right ) + {1 \over {\sin^2 \theta}} {\partial^2 \over \partial \varphi^2} \right]$$This leads to the Eigenvalue equation written with the presumptive solutions, the spherical harmonics $Y_\ell^m (\theta, \varphi )$;

$$\hat H Y_\ell^m (\theta, \varphi ) = \frac{\hbar^2}{2mr^2} \ell(\ell+1) Y_\ell^m (\theta, \varphi )$$

Which gives the energy Eigenvalues where $m$ is the mass.;

$$ E_\ell = {\hbar^2 \over 2mr^2} \ell \left (\ell+1\right) ~~~ \ell=0,1,\dots$$

The lowest normalized angular Eigenfunction is $~~~ Y_{0}^{0}(\theta,\varphi)={1\over \sqrt{4\pi} R}$ where $R$ is fixed which means a constant probability density over the sphere. Interestingly, the ground state energy is zero for $\ell=0$ but the wavefunction is not. Rather than contradict the Heisenberg Uncertainty Principle, we can consider that the momentum and position of the particle are completely uncertain in the ground state. It is also instructive to choose the sphere radius to be equal to the Bohr radius $a_0$ which gives us after a little algebra;

$$ E_\ell = {e^2 \over 2 a_0} \ell \left (\ell+1\right) ~~~ \ell=0,1,\dots$$

Starting with the full Helium Hamiltonian which we will reduce to that of 2-Spherium; I'll leave off $r$ and put it in at the end for convenience.

$$\hat H(\vec{r}_1,\, \vec{r}_2) = - \frac{\hbar^2}{2m} (\nabla^2_1 + \nabla^2_2) - \frac{2 e^2}{r_1} - \frac{2e^2}{r_2} + \frac{e^2}{r_{12}} $$

Since the values of $r$ is fixed, we actually have a 4 dimensional space over the surface and 6 is we consider the actual volume of the spheres. We could add a positive $Z = 2$ charge fixed in the center of the sphere to create constant potential terms $- \frac{2e^2}{a_0}$ but that just adds a constant to the energy and is unnecessary. So for $unperturbed$ 2-Spherium (particles on a 2 dimensional sphere in 3 dimensional space) we are left with the Hamiltonian;

$$\hat H(\vec{r}_1,\, \vec{r}_2) = - \frac{\hbar^2}{2m} (\nabla^2_1 + \nabla^2_2) $$ where the vectors represent a fixed radius and variable angles $\theta, \varphi$.

This is separable and the solutions for each separate Hamiltonian are the the spherical harmonics $Y_\ell^m (\theta, \varphi )$ from which we can construct the total wavefunction which has to be anti-symmetric for the two electrons;

$$ \psi^{(total)}_\pm(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}} [\psi_{l_1,m_1}(\vec{r}_1) \psi_{l_2,m_2}(\vec{r}_2) \pm \psi_{l_2,m_2}(\vec{r}_1) \psi_{l_1,m_1}(\vec{r}_2)][S_{1,2}>$$

Where $[S_{1,2}>$ equals for the minus sign;
$$
\left.\begin{cases}
|1,1\rangle & =\;\uparrow\uparrow\\
|1,0\rangle & =\;(\uparrow\downarrow + \downarrow\uparrow)/\sqrt2\\
|1,-1\rangle & =\;\downarrow\downarrow
\end{cases}\right\}\quad s=1\quad\mathrm{(triplet)}
$$
And $[S_{1,2}>$ equals for the plus sign;$$
\left.\begin{cases}
|0,0\rangle & =\;(\uparrow\downarrow - \downarrow\uparrow)/\sqrt2\\
\end{cases}\right\}\quad s=0\quad\mathrm{(singlet)}
$$

For the spatial part we let $\Phi_a(\vec r_i)$ represent the $Y_\ell^m (\theta_i, \varphi_i)$ for a set of quantum numbers ${\ell, m}$ labeled $a$ or $b$ and coordinate set $i$ labeled 1 or 2;
$$\Psi_S(\vec r_1,\vec r_2)= \frac{1}{\sqrt{2}}[\Phi_a(\vec r_1) \Phi_b(\vec r_2) + \Phi_b(\vec r_1) \Phi_a(\vec r_2)]$$
$$\Psi_A(\vec r_1,\vec r_2)= \frac{1}{\sqrt{2}}[\Phi_a(\vec r_1) \Phi_b(\vec r_2) - \Phi_b(\vec r_1) \Phi_a(\vec r_2)]$$

Now the ground state $Y_{0}^{0} (\theta_1, \varphi_1) = {1\over 2\sqrt{\pi} }$ is a special case where the wavefunctions are not orthogonal and we can write the possible ground state wavefunctions.;
$$\Psi_S(\vec r_1,\vec r_2)[Spin_A>= \frac{1}{{2}}[ {1\over 2\sqrt{\pi} } {1\over 2\sqrt{\pi} }+ {1\over 2\sqrt{\pi}} {1\over 2\sqrt{\pi} }][singlet>$$$$\Psi_A(\vec r_1,\vec r_2)[Spin_S>= \frac{1}{{2}}[ {1\over 2\sqrt{\pi} } {1\over 2\sqrt{\pi} }- {1\over 2\sqrt{\pi}} {1\over 2\sqrt{\pi} }][triplet>$$

Which gives us;

$$\Psi_S(\vec r_1,\vec r_2)[Spin_A>= \frac{1}{{2}}[ {1\over 2{\pi}}][singlet>$$
$$\Psi_A(\vec r_1,\vec r_2)[Spin_S>= \frac{1}{{2}}[0][triplet>$$

Leaving the ground state of the $unperturbed$ Hamiltonian as;
$$\Psi_{GROUND}=\Psi_S(\vec r_1,\vec r_2)[Spin_A>= {1\over {4 {\pi}{R}^2}}[{1\over \sqrt{2} }(\uparrow\downarrow - \downarrow\uparrow)>$$The total energy is zero for the ground state up to an arbitrary fixed potential;

$$ E_{total} = {\hbar^2 \over {2 m R^2}} [ \ell_1 \left (\ell_1+1\right) + \ell_2 \left (\ell_2+1\right)]=0~~~ \ell_{1,2}=0$$

Interestingly, according to Loos and Gill* the ground state wavefunction with the Coulomb interaction in the Restricted Hartree-Fock formulation is the same however the energy is changed by the interaction with the addition of a term $e^2\over R$.Next, I'll look at some excited states before attempting to add the interaction term to resolve the degenerate states.

* PHYSICAL REVIEW A 79, 062517 (2009)

 

[bob012345]

The excited states for the $unperturbed$ Hamiltonian are constructed from basis one-electron states;
$$\psi=N|\ell,m\rangle_i={{Y_\ell^m (\theta_i, \varphi_i )}\over R}~~~ i={1,2}$$

Where there are twelve possible states for the lowest possible transition from $\Psi(\ell_1,\ell_2)$ going from $\Psi(0,0) \rightarrow \Psi(0,1)$.

Written in $|ket\rangle$ notation;
$$
\left.\begin{cases}
{\frac{1}{\sqrt{2}}}(|0,0\rangle_1|1,1\rangle_2+|0,0\rangle_2|1,1\rangle_1)|singlet\rangle ~~para\\
\frac{1}{\sqrt{2}}(|0,0\rangle_1|1,0\rangle_2+|0,0\rangle_2|1,0\rangle_1)|singlet\rangle~~para\\
\frac{1}{\sqrt{2}}(|0,0\rangle_1|1,\bar1\rangle_2+|0,0\rangle_2|1,\bar1\rangle_1)|singlet\rangle~~para\\
\frac{1}{\sqrt{2}}(|0,0\rangle_1|1,1\rangle_2-|0,0\rangle_2|1,1\rangle_1)|triplet\rangle~~ortho\\
\frac{1}{\sqrt{2}}(|0,0\rangle_1|1,0\rangle_2-|0,0\rangle_2|1,0\rangle_1)|triplet\rangle~~ortho\\
\frac{1}{\sqrt{2}}(|0,0\rangle_1|1,\bar1\rangle_2-|0,0\rangle_2|1,\bar1\rangle_1)|triplet\rangle~~ortho\\
\end{cases}\right\}
$$

Where $[S,m_s\rangle$ equals;
$$
\left.\begin{cases}
|1,1\rangle & =\;|\uparrow\uparrow\rangle\\
|1,0\rangle & =\;\frac{1}{\sqrt{2}}|\uparrow\downarrow + \downarrow\uparrow\rangle\\
|1,-1\rangle & =\;|\downarrow\downarrow\rangle
\end{cases}\right\}\quad S=1\quad\mathrm{(triplet)}
$$
And;
$$
\left.\begin{cases}
|0,0\rangle & =\;\frac{1}{\sqrt{2}}|\uparrow\downarrow - \downarrow\uparrow\rangle
\end{cases}\right\}\quad S=0\quad\mathrm{(singlet)}
$$
The energy of these twelve degenerate lowest excited states is given by;

$$ E_{total} = {\hbar^2 \over 2 m R^2} [ \ell_1 \left (\ell_1+1\right) + \ell_2 \left (\ell_2+1\right)]= {\hbar^2 \over m R^2}~~~ \ell_{1,2}=0,1$$

The next step is adding the Coulomb interaction for the ground and then the lowest excited states.