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bob012345

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I'm having no luck at all generating LaTex. I'm using the Safari web browser on an iPad.

Example:

\frac a b

I can see the LaTex formatting on other posts though.

Note to observers; I'm practicing LaTeX in this older message.

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 Eigenfunction is ##~~~ Y_{0}^{0}(\theta,\varphi)={1\over 2\sqrt{\pi} }## 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;

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

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;

$$\Psi_S(\vec r_1,\vec r_2)[Spin_A>= {1\over 4{\pi}}[{1\over \sqrt{2} }(\uparrow\downarrow - \downarrow\uparrow)>$$

The total energy is zero for this state up to an arbitrary fixed potential;

$$ E_{total} = {e^2 \over 2 a_0} [ \ell_1 \left (\ell_1+1\right) + \ell_2 \left (\ell_2+1\right)]=0~~~ \ell_{1,2}=0$$

Example:

\frac a b

I can see the LaTex formatting on other posts though.

Note to observers; I'm practicing LaTeX in this older message.

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 Eigenfunction is ##~~~ Y_{0}^{0}(\theta,\varphi)={1\over 2\sqrt{\pi} }## 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;

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

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;

$$\Psi_S(\vec r_1,\vec r_2)[Spin_A>= {1\over 4{\pi}}[{1\over \sqrt{2} }(\uparrow\downarrow - \downarrow\uparrow)>$$

The total energy is zero for this state up to an arbitrary fixed potential;

$$ E_{total} = {e^2 \over 2 a_0} [ \ell_1 \left (\ell_1+1\right) + \ell_2 \left (\ell_2+1\right)]=0~~~ \ell_{1,2}=0$$

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