Electronic Properties of Materials/Quantum Mechanics for Engineers/Hydrogen

This brings us to looking at atoms in materials. This sections provides the complete derivation of the Hydrogen Atom with relevant insights. Currently, we can solve the Hydrogen atom and the ionized Helium atom, as single electron systems. Higher order atoms, or many body problems, can be further observed through perturbation methods as discussed in the next chapter.

In this chapter, you should begin to see where the mathematics of atoms comes from, as much of the conversation around atoms is in terms of atomic orbitals which are initially derived here.

** INCOMPLETE **

Simple Hydrogen Atom
Imagine a "simple" Hydrogen atom. This simple Hydrogen atom has a nucleus with some momentum, the proton has some mass, and both have some position relative to the origin. The electron also has some position relative to the origin, some momentum and some mass.

If we want to talk about this we have to write the Hamiltonian. Given that $$T_p$$ is the kinetic energy of the proton, $$T_e$$ is the kinetic energy of the electron and $$V$$ is the coulomb potential then the Hamiltonian classically looks like:

$$H = T_p + T_e + V   = {{P_p}^2 \over 2m_p} + {{P_e}^2 \over 2m_e} - {|c|^2 \over |r_p - r_e|}$$

The problem with this equation is that it's complicated. Here, we are dealing with Cartesian space and two particles with unique $\vec{r}$, $\vec{p}$ , and $\vec{m}$ which is a many-body problem and quite complicated. That said, the many-body accounts for the movement of the hydrogen atom, which isn't that relevant here. We want to change to a more natural coordinate system which will allow us to distinguish hydrogen atom translation from  interaction, and which focuses on the movement of the electron around the proton. To accomplish this we will change our static coordinate system with the particle system moving in space to a center of mass system. Ideally, we also want a simplified version of $$\hat{V}$$.

 "Center of Mass System" (The first step to shifting the coordinate system is to identify the center of mass.)

Looking at , we can define a new coordinate system, where we have some center of mass, which we want to use as our new origin, and a distance from the original origin to this new origin, called $$R$$. Additionally, $$r$$ is the distance between the nucleus and the electron, and this combined system has some momentum for the center of mass, $$p_R$$, and some momentum for the electron, $$p_r$$. In doing this transform, we have identified the center of mass as:$$\begin{align} R &= {m_e r_e \over m_e + m_p} + {m_p r_p \over m_e + m_p} \\ &= {m_e r_e + m_p r_p \over m_e + m_p} \end{align}$$There are two approaches to this transformation from $(\vec{r}_e, \ \vec{r}_p) $ to $(\vec{R}, \ \vec{r})$.

The Physical Approach
 "Relative Velocity in a Two Body System" (The relative velocity of these two particles is $$\nu = \nu_1 - \nu_2$$.)

The first approach is to appeal to the physics of the situation utilizing relative and total momentum. As such, $$P = p_e + p_p $$ would be the total momentum of the system, and the relative velocity of the system is $$\nu = \nu_1 - \nu_2$$. Similarly, the relative momentum of the system, $$p=m \nu$$, where $$m$$ is the reduced mass ($$\mu$$):

$$\mu={m_p \ m_e \over m_p + m_e}$$

Applying this to our momentum equation and simplifying gets us:$$\begin{align} p &= m \ \nu = \mu \ \nu \\ p &= {m_p \ m_e \over m_p + m_e}\nu; \qquad \mu={m_p \ m_e \over m_p + m_e} \\ &= {m_p \ m_e \over m_p + m_e} (\nu_e - \nu_p) \\ &= {m_p \ m_e \ \nu_e \over m_p + m_e} - {m_p \ m_e \ \nu_p \over m_p + m_e} \\ &= {m_p \over m_p + m_e}p_e - {m_e \over m_p + m_e}p_p \end{align}$$

This equation for relative momentum is then combined with our earlier equations for the classical Hamiltonian, and for total momentum ($P = p_e + p_p $ ) to find the momentum of the electron ($$p_e$$) and the momentum of the proton ($$p_p$$), in terms of the relative and absolute momentums. $$\begin{align} H &= {P^2 \over 2 \ M} + {p^2 \over 2 \ \mu}; \qquad M=m_p + m_c \\ H &= {P^2 \over 2 \ M} + {p^2 \over 2 \ \mu} - {|e|^2 \over |r|} \end{align}$$

Now, this is a classical Hamiltonian. Let's turn it into a quantum Hamiltonian! Remembering that the quantum version of momentum is $p = (i\hbar)^2 {\partial \over \partial x}$, we get: $$H = {-\hbar^2 \over 2 \ M} \nabla_R^2 - {\hbar^2 \over 2 \ \mu} \nabla_r^2 - {|e|^2 \over |r|}$$

While this first approach is physically intuitive, it's not purely quantum. Physics people tend to do a lot of really bad math which happens to work out, and this solution has issues as it is not generalizable.

The General Approach
The second, better, approach is to use our already derived quantum Hamiltonian. This approach has the benefit of being generalizable and can be used to solve other many body systems.$$H = {- \hbar^2 \over 2 \ m_p} \nabla_{r_p}^2 - {\hbar^2 \over 2 \ m_e} \nabla_{r_e}^2 - {|e|^2 \over |r_p - r_e|}$$

Remember that the Laplacian operator is: $\nabla_{r_p}^2 = {\partial^2 \over \partial r_{p_x}^2} + {\partial^2 \over \partial r_{p_y}^2} + {\partial^2 \over \partial r_{p_z}^2}$

Remember that the chain rule for the transformation of differential operators.$$\begin{align} (x, \ y) \longrightarrow (p, \ q) \quad \ \\ {\partial \over \partial x} = {\partial \over \partial p} {\partial p \over \partial x} + {\partial \over \partial q} {\partial q \over \partial x} \\ {\partial \over \partial y} = {\partial \over \partial p} {\partial p \over \partial y} + {\partial \over \partial q} {\partial q \over \partial y} \ \end{align}$$

Using the relations $$r = r_e - r_p$$, and $R = {r_p \ m_p + r_e \ m_e \over m_p + m_e}$ to perform the coordinate transformation, $(\vec{r}_p, \ \vec{r}_e)$  to $(\vec{r}, \ \vec{R})$. With some algebra, it results in the same Hamiltonian as the earlier equation. $$H = \underbrace{{-\hbar^2 \over 2 \ M} \ {\nabla_R}^2}_{Pure \ R} - \underbrace{{\hbar^2 \over 2 \ \mu} \ {\nabla_r}^2 - {|e|^2 \over |r|}}_{Pure \ r}$$

We now have $H=H(R)+H(r)$, and since each section is dependent only on one parameter, this means our solution is separable as $\Psi = \psi(R) \ \psi(r)$. Looking specifically at $$H(R)$$, this is just a free particle, $$\nu =0$$. We've already solved this problem and found $$\psi(R)$$ is planewaves. Alternatively, looking at $$H(r)$$ shows us that this is an $$e$$, with mass correction in the kinetic energy term, moving around a central potential: ${-|e|^2 \over |r|}$ $$E_{TOT}=E_r+E_R$$The total energy is just the sum of the energy due to the translation of the atom on the whole, plus the energy associated with interactions between the proton and electron.

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How big is the reduced mass correction? Tiny. $$m_{proton} \approx 2000 \times m_{electron}$$

Experimentally, it is possible to distinguish hydrogen and deuterium by spectral shifts. As more particles are added this method can be extended. For three particles:

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 (related to figure)

The problem we want to solve is: $$\left ( {-\hbar^2 \over 2 \ \mu} \ \nabla^2 - {|e|^2 \over r} \right ) \ \psi(r) = E \ \psi(r)$$

But this is a spherically symmetric potential. Cartesian coordinates aren't the best choice. Switch to spherical coordinates.

 "Spherical Axis" (Description)

Spherical coordinates are highly relevant to the following mathematical calculations, now is a good time to pause and familiarize yourself if needed.

In Spherical Coordinates:$$\nabla^2 = {1 \over r^2} {\partial \over \partial r} \left (r^2 {\partial \over \partial r} \right ) + {1 \over r^2 \ \sin \theta} {\partial \over \partial \theta} \left ( \sin \theta \ {\partial \over \partial \theta} \right ) + {1 \over r^2 \sin^2 \! \theta } \ {\partial^2 \over \partial \phi^2}$$

Wow! What a mess! How do we solve this? #SeparationOfVariables

Let $$\psi(r, \ \theta, \ \phi)=R(r) \ Y(\theta, \ \phi)$$

$$\begin{align} ERY &= \left [ {-\hbar^2 \over 2 \mu} \left [ \underbrace {{1 \over r^2} \ {\partial \over \partial r} \left ( r^2 {\partial \over \partial r} \right )}_A + \underbrace{{1 \over r^2 \sin \! \theta} \ {\partial \over \partial \theta} \left (\sin \! \theta \ {\partial \over \partial \theta} \right )}_Q + \underbrace{{1 \over r^2 \sin^2 \! \theta} \ {\partial^2 \over \partial \phi^2}}_F \right ] - {e^2 \over r} \right ] RY \\ &= {-\hbar^2 \over 2 \mu} \left [ ARY + {1 \over r^2} QRY + {1 \over r^2} FRY \right ] \\ {-\hbar^2 \over 2 \mu} YAR - RY {e^2 \over r} - RYE &= {1 \over r^2} {\hbar^2 \over 2\mu} RQY + {1 \over r^2} {\hbar^2 \over 2 \mu} RFY \end{align}$$

Multiply Left By: ${r^2 2 \mu \over YR}$

$$\underbrace{{r^2 \hbar^2 \over R} AR + 2r \mu e^2 + r^2 2 \mu}_{Only \ r} = \underbrace{-{\hbar ^2 \over 2 \mu}QY - {\hbar^2 \over Y}FY}_{Only \ \theta \phi}$$

So both sides equal a constant, $$K$$. Thus:

$$\begin{align} K &= {r^2 \hbar \over R(r)} {1 \over r^2} {\partial \over \partial r} \left (r^2 {\partial \over \partial r} \right ) R(r) +2r \mu e^2 + r^2 2 \mu E \\ &= {-\hbar^2 \over Y(\theta, \ \phi)} \ {1 \over \sin \! \theta} \ {\partial \over \partial \theta} \ \left ( \sin \! \theta \ {\partial \over \partial \theta} \right ) \ Y(\theta, \ \phi) - {\hbar^2 \over Y(\theta, \ \phi)} \ {1 \over \sin^2 \! \theta} \ {\partial^2 \over \partial \phi^2} Y(\theta, \phi) \end{align}$$

Let's look closer at the second one:

As it happens, this is an operator. First I'll tell you the answer, and then I'll show you where it originates.

The operator is $$\hat{L}^2$$, where $$L$$ is angular momentum. The eigenequation is:

$$\hat{L}^2 \ Y(\theta, \ \phi) = \hbar^2 \ l (l+1) \ Y(\theta, \ \phi)$$, where $$l$$ is an integer and $$K=\hbar^2 l (l+1)$$.

Angular Momentum
Before we can proceed with studying Hydrogen, we need to learn a little about angular momentum. In Quantum Mechanics there are two types of angular momentum. "Orbital" momentum is analogus to the classically understood angular momentum where $$\vec{L} = \vec{r} \times \vec{P}$$, and "Spin" momentum which has no classical equivalent. We will talk about spin later, and for now we will focus on orbital angular momentum.

Orbital Angular Momentum
Classically: $$\vec{L} = \vec{r} \times \vec{P}$$ is a vector. Thus:$$\begin{align} \vec{L} &= \vec{r} \times \vec{P} \\ &= \langle x, \ y, \ z \rangle \times \langle p_x, \ p_y \ p_z \rangle \\ &= \langle yp_z - zp_y, \ zp_x - xp_z, \ xp_y - yp_x \rangle \\ \\ &= \langle L_x, \ L_y, \ L_z \rangle \begin{cases} \hat{L}_x = -i\hbar \left ( y{\partial \over \partial z} - z{\partial \over \partial y} \right ) \\ \hat{L}_y = -i\hbar \left ( z{\partial \over \partial x} - x{\partial \over \partial z} \right ) \\ \hat{L}_x = -i\hbar \left ( x{\partial \over \partial y} - y{\partial \over \partial x} \right ) \end{cases} \\ \\ \vec{L} &= -i\hbar (\vec{r} \times \nabla ) \end{align}$$

Consider the commutation of the $$\hat{L}$$ operator.$$\begin{align} \left [ \hat{L}_x, \hat{L}_y \right ] &= \left [ yp_z - zp_y, \ zp_x - xp_z \right ] \\ &= (yp_z - zp_x)(zp_x - xp_z) - (zp_x - xp_z)(yp_z - zp_y) \\ &= yp_z \ zp_x + zp_y \ xp_z - zp_y \ zp_x - yp_z \ xp_z - (zp_x \ yp_z + xp_z \ zp_y - zp_x \ zp_y - xp_z \ yp_z) \\ &= yp_z \ zp_x + zp_y \ xp_z + zp_x \ yp_z - xp_z \ zp_y \\ &= y p_x p_z z - y p_x z p_z + x p_y z p_z - x p_y p_z z \\ &= y p_x [p_z, \ z] + x p_y [z, \ p_z] \\ &= y p_x (-i\hbar) + x p_y (+ i\hbar) \\ &= i\hbar \hat{L}p_z \\ \\ \left [ L_x, \ L_y \right ] &= i\hbar L_z \begin{cases} \left [ L_y, \ L_z \right ] = i\hbar L_x \\ \left [ L_z, \ L_x \right ] = i\hbar L_y \end{cases} \end{align}$$

Which means that we cannot simultaneously measure all components of $$\vec{L}$$. Very odd properties for a vector!

What about the magnitude of $$\vec{L}$$?$$\begin{align} L^2 &= L \ \cdot \ L \\ &= {L_x}^2 + {L_y}^2 + {L_z}^2 \\ \\ \left [ L^2, \ L_x \right ] &= [{L_x}^2 + {L_y}^2 + {L_z}^2, \ L_x] \\ &= [{L_x}^2, \ L_x] + [{L_y}^2, \ L_y] + [{L_z}^2, \ L_z] \\ &= L_y (L_y L_x) - (L_x L_y)L_y + L_z (L_z L_x) - (L_x L_z) L_z \\ & \qquad \qquad 0 = -L_y (L_x L_y) + (L_y L_x) L_x \\ &= L_y (L_y L_x) - (x L_y)L_y + L_z (L_z L_x) - (L_x L_z) L_z \\ &= i\hbar \ [-L_y L_z - L_z L_y + L_z L_y + L_y L_z] \end{align}$$

Therefore simultaneous eigenfunctions of $$L^2$$ and any one component of $$\vec{L}$$ can be known. For this discussion to be useful we need to switch from cartesian coordinates to spherical. This requires a bunch of algebra, simple yet tedious. Here I will skip it and just provide the equations:$$\begin{align} L_x &= -i\hbar \ \left ( -\sin \theta \ {\partial \over \partial \theta} \ - \cot \theta \cos \phi \ {\partial \over \partial \phi} \right ) \\ L_y &= -i\hbar \ \left ( \cos \phi \ {\partial \over \partial \theta} - \cot \theta \sin \phi \ {\partial \over \partial \phi} \right ) \\ L_z &= -i\hbar \ {\partial \over \partial \theta} \end{align}$$

... and substituting into $$L^2 = {L_x}^2 + {L_y}^2 + {L_z}^2$$ provides:$$L^2 = -\hbar^2 \left [ {1 \over \sin \theta} \ {\partial \over \partial \theta} \left ( \sin \theta \ {\partial \over \partial \theta} \right ) + {1 \over \sin^2 \theta} \ {\partial ^2 \over \partial \phi^2} \right ]$$

It is noteworthy that in many problems the solution is invariant to rotation, so any direction we point we can define as $$z$$ and use the simple operator $$L_z$$ and $$L^2$$. Let's start by looking at the eigensolutions for $$L_z$$.$$L_z \phi = \ell_z \phi$$

What are good solutions for $$L_z = -i \hbar \ {\partial \over \partial \phi}$$ ?

$$\underbrace{-i\hbar \ {\partial \over \partial \phi}}_{L_z} \underbrace{\alpha \ e^{im \phi}}_{\Phi(\phi)} = \underbrace{\hbar m}_{\ell_z} \ \underbrace{\alpha e^{im\phi}}_{\Phi(\phi)}$$

$$\begin{align} \int_0 ^{2\pi} \Phi^* \Phi \ \operatorname{d}\!\phi=1 \longrightarrow \alpha = {1\over \sqrt{2\pi}} \\ \Phi={1\over \sqrt{2\pi}} \ e^{im\phi}; \ell_z = \hbar m \end{align}$$

Boundary Conditions: $$\Phi(0) = \Phi(2m)$$; implying that $$m=0, \ \pm1, \ \pm2, \ ...$$

$$L^2 \ Y(\theta, \ \phi)=\ell_{SQR} \ Y(\theta, \ \phi)$$

Since $$[L^2 L_z]=0$$, they share an eigenfunction.

$$L_z \ Y(\theta, \ \phi)= \ell_z' \ Y(\theta, \ \phi)$$ -> What solution?

The same:

$$L_z \ Y(\theta, \ \phi)=m \hbar \ Y(\theta, \ \phi)$$

$$L_z$$ depends only on $$\phi$$ so the separable solution is:

$$Y(\theta, \ \phi)=\Theta(\theta) \ \Phi(\phi)$$

which results in the same eigenvalues with $$\Phi(\phi)=e^{im\phi}$$

Returning to $$L^2$$...
$$\left [ {1 \over \sin(\theta)} \ {\partial \over \partial \theta} \left ( \sin \theta \ {\partial \over \partial \theta} \right ) + {1 \over \sin^2 \theta} \ {\partial^2 \over \partial \theta^2} \right ] Y(\theta, \ \phi) = \ell_{sq} \ Y(\theta, \ \phi)$$

Separating $$\theta$$ and $$\phi$$ and substituting $$\Phi$$$$\begin{align} \ell_{sq} \ \Theta \Phi &= {1 \over \sin \theta} \ {\partial \over \partial \theta} \left ( \sin \theta \ {\partial \over \partial \theta} \right ) \Theta \Phi \ + {1 \over \sin^2 \theta} \ {\partial^2 \over \partial \theta^2} \ \Theta \Phi \\ \ell_{sq} \ \Theta &= {1 \over \sin \theta} \ {\partial \over \partial \theta} \left ( \sin \theta \ {\partial \over \partial \theta} \right ) \Theta \ + {1 \over \sin^2 \theta} \ {\partial^2 \over \partial \theta^2} \ \Theta \\ 0 &= \left [ {1 \over \sin \theta} \ {\partial \over \partial \theta} \left ( \sin \theta \ {\partial \over \partial \theta} \right ) - \ell_{sq} - {m^2 \over \sin^2 \theta} \right ] \Theta (\theta) \end{align}$$

This is what we have to solve. With the appropriate substitutions and even more algebra, this can be transformed into the Legendre equation. With even more algebra, we can get a solution in terms of the associated Legendre functions.$$\Theta_{\ell m}(\theta)= \begin{cases} \begin{array}{lcl} (-1)^m \left [ {(2 \ell +1)(\ell - m)! \over 2 (\ell + m)! } \right ]^{1 \over 2} P_{\ell}^{ \ m} (\cos \theta) & \quad if \ \ m \geq 0 \\ (-1)^m \theta_{\ell |m| } (\theta) & \quad if \ \ m < 0 \end{array} \end{cases}$$Where $$\ell = 0, \ 1, \ 2, \ ...$$, and $$m=0, \ \plusmn 1, \ ..., \ \plusmn \ell$$.

$$P_{\ell}^{\ m}(x)$$ are the associated Legendre functions:$$P_{\ell}^{\ m} (x) = (-1)^m (1-x^2)^{m \over 2} \ {\operatorname{d}^m \over \operatorname{d}\!x^m} \ P_\ell (x), \ where \ P_\ell (x) = {1\over 2^\ell \ell !} \left [ {\operatorname{d}^\ell \over \operatorname{d}\!x^\ell } (x^2-1)^\ell \right ]$$

Most mathematical software packages (mathematica, maple, MATLAB, etc...) have these built in.\

$$Y_{\ell m}(\theta, \ \phi) = \begin{cases} \begin{array}{lcl} (-1)^m \left [ {(2\ell +1)(\ell - m)! \over 4\pi (\ell + m)!} \right ]^{1 \over 2} P_{\ell}^{\ m} (\cos \theta) \ e^{im\phi} & \quad if \ \ m \geq 0 \\ (-1)^m Y_{\ell, \ -m}^{\ *} (\theta, \ \phi) & \quad if \ \ m < 0 \end{array} \end{cases}$$Where $$\ell = 0, \ 1, \ 2, \ ...$$, and $$m=-\ell, \ -\ell + 1, \ ..., \ 0, \ \ell$$.

These are called "Spherical Harmonics" which are either normalized on a sphere with a unity radius, or orthonormal as:$$\begin{align} \int Y_{\ell ' m'}^{\ *} Y_{\ell m} \operatorname{d}\!\Omega &= \int_{0}^{2\pi} \operatorname{d}\!\phi \ \int_o^\pi \operatorname{d}\!\theta \ \sin \theta \ Y_{\ell ' m'}^{\ *} Y_{\ell m} \\ &= \delta_{\ell \ell'} \delta_{m m'} \end{align}$$

... where $\int \operatorname{d}\!\Omega $ means "to integrate over a sphere", and $\operatorname{d}\! \Omega =\sin\! \theta \operatorname{d}\! \theta \operatorname{d}\! \phi$. Note that $Y_{\ell m}$ are a "complete set" meaning that any function $f(\theta, \ \phi)$  can be written. $$f(\theta, \ \phi) = \sum_{\ell=0}^\infty \sum_{m=-\ell}^{+\ell} a_{\ell m} Y_{\ell m} (\theta, \ \phi)$$

This is analogous to planewaves in a cartesian space and all around, a good use full function.

As a matter of notation we designate states with: $$\begin{align} \ell =& \ 0, \ 1, \ 2, \ 3, \ 4, \ 5, \ 6, \ ... \\ =& \ s, \ p, \ d, \ f, \ g, \ h, \ i, \ ... \end{align}$$

For many-body systems we denote the total orbital angular momentum with a capital $$L=\sum_{i-1}^n \ell_i$$, where $$\begin{align} L =& \ 0, \ 1, \ 2, \ 3, \ ... \\ =& \ S, \ P, \ D, \ F, \ ... \end{align}$$.

Here, $$\ell$$ is called the "orbital angular momentum quantum number", and $$m$$ is called the "magnetic quantum number". Furthermore, $$S = \operatorname{Sharp}$$, $$P = \operatorname{Principal}$$, $$D=\operatorname{Diffuse}$$, $$F=\operatorname{Fundamental}$$, and operators $$G, \ H, \ \operatorname{and} \ I$$ all lack creative names.



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$$\begin{align} \hat{H} &= {-\hbar^2 \over 2 \mu} \nabla^2 - {|e|^2 \over r} \\ &= {-\hbar^2 \over 2\mu} \left [ {1 \over r} {\partial \over \partial r} \left ( r^2 {\partial \over \partial r} \right ) + {1 \over r^2 \sin\! \theta } {\partial \over \partial \theta} \left ( \sin \! \theta {\partial \over \partial \theta} \right ) + {1 \over r^2 \sin ^2 \! \theta} {\partial^2 \over \partial \phi^2} \right ] - {|e|^2 \over r} \\ &= {-\hbar^2 \over 2 \mu} {1 \over r^2} {\partial \over \partial r} \left ( r^2 {\partial \over \partial r} \right ) + {1 \over 2 \mu r^2} \left [ {-\hbar^2 \over \sin \! \theta} {\partial \over \partial \theta} \left ( \sin \! \theta {\partial \over \partial \theta} \right ) + {-\hbar^2 \over \sin^2 \! \theta} {\partial^2 \over \partial \phi^2} \right ] - {|e|^2 \over r} \\ &= {-\hbar^2 \over 2 \mu} {1 \over r^2} {\partial \over \partial r} \left ( r^2 {\partial \over \partial \theta} \right ) + {1\over 2 \mu r^2} \hat{L}^2 - {|e|^2 \over r} \end{align}$$ ====== Since $$\hat{H} \psi = E \psi$$, if we let $$\psi = Y(\theta, \ \phi) \ R(r)$$, and plug our previous equation in we get:$$\left [ {-\hbar^2 \over 2 \mu} {1 \over r^2} {\partial \over \partial r} \left ( r^2 {\partial \over \partial r} \right ) + {\hbar^2 \ell (\ell +1) \over 2 \mu r^2} - {|e|^2 \over r} \right ] R(r) = E \ R(r)$$

Here, $${|e|^2 \over r}$$ is the true Coulumb potential, and $${\hbar^2 \ell (\ell + 1) \over 2 \mu r^2}$$ is called "the angular momentum barrier.

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Simplifying $$\hat{H}$$ further...$${1 \over r^2} {\partial \over \partial r} \left ( r^2 {\partial \over \partial r} \right ) R \ - {2 \mu \over \hbar^2} \left ( V_{eff} - E \right ) R = 0$$ ====== Looking at the first term...$$\begin{align} {1 \over r^2} {\partial \over \partial r} \left ( r^2 {\partial \over \partial r} \right ) R =& {1 \over r^2} \left ( 2r {\partial R \over \partial r}\right ) + r^2 {\partial ^2 \! R \over \partial r^2} \\ =& {1 \over r} \left ( 2 \ {\operatorname{d} \! R \over \operatorname{d} \! r} + r {\operatorname{d}^2 \! \! R \over \operatorname{d} r^2} \right ) \\ =& {1\over r} {\operatorname{d}^2 \! u \over \operatorname{d} r^2} \end{align}$$

Substituting $$u(r) =rR(r)$$ in we get... $$\begin{align} {\operatorname{d} \! u \over \operatorname{d} \! r} r R &= {\operatorname{d} \! \over \operatorname{d} \! r} r \\ &= R + r {\operatorname{d} \! R \over \operatorname{d} \! r} \\ {\operatorname{d}^2 \! u \over \operatorname{d} \! r^2} &= 2 \ {\operatorname{d} \! R \over \operatorname{d} \! r} + r {\operatorname{d}^2 \! R \over \operatorname{d} \! r^2} \end{align}$$

Which combines with the initial equation to get: $${1 \over r} {\operatorname{d}^2 \! u \over \operatorname{d} \! r^2} - {2 \mu \over \hbar^2} \left ( V_{eff} - E \right ) {u \over r} = 0$$

This equation is nice an compact, but not solvable. To get something solvable we substitute $$V_{eff}$$ back into the equation...$$\left ( {\operatorname{d}^2 \over \operatorname{d} \! r^2} - {\ell (\ell + 1) \over r^2} + {2 \mu \over \hbar^2} { |e|^2 \over r} + {2 \mu \over \hbar} E \right ) u = 0$$

Introduce two dimensionless variables to substitution in to equation.$$\begin{align} \rho &= \left ( - {8 \mu E \over \hbar^2} \right )^{1 \over 2} r \\ \lambda &= {|e|^2 \over \hbar} \left ( {-\mu \over 2E} \right )^{1 \over 2} \\ {\operatorname{d}^2 \over \operatorname{d} \! r^2} &= {-8 \mu E \over \hbar^2} {\operatorname{d} \over \operatorname{d} \! \rho} \end{align}$$

This gets us:$$\left ( {\operatorname{d}^2 \over \operatorname{d} \! \rho^2} - {\ell(\ell + 1) \over \rho^2} + {\lambda \over \rho} - {1 \over 4} \right ) u = 0$$

Consider solutions for $$u$$. In the limit of the large $$\rho$$ (Large $$r$$), the equation simplifies to:$$\left ( {\operatorname{d}^2 \over \operatorname{d} \! \rho^2} -{1 \over 4} \right ) u = 0$$

Here, our solution is $$u = A e^{-P \over 2} + B e^{P \over 2}$$, but as $$P$$ approaches infinity, $$u$$ will approach zero. Therefore, $$B$$ equals zero, which means that:$$u \sim e^{-P \over 2} \ as \ \rho \rightarrow \infty$$

Consider the other limit where $$\rho$$ approaches zero. The problem becomes:$${\operatorname{d}^2 \over \operatorname{d} \! \rho} u - {\ell (\ell + 1) \over \rho^2} \rho^q = 0$$

Guessing the solution $$u=p^q$$, which gives us:$$\begin{align} 0 &= {\operatorname{d}^2 \over \operatorname{d} \! \rho^2} u - {\ell (\ell + 1) \over \rho^2} u \\ q (q - 1) \rho^{8 - 2} &= {\ell (\ell + 1) \over \rho^2} p^{8 - 2} \\ q^2 - q &= \ell^2 +\ell \\ q &= \ell \plusmn 1 \end{align}$$

Which means that $$u \sim \rho^{\ell + 1}$$ as $$\rho \rightarrow 0$$, and we're going to want a solution that looks like this:$$u(p) \sim e^{-\rho \over 2} \rho^{\ell + 1}$$

 "Title" description

Search for solutions as a polynomial expansion: $$u(\rho) = e^{\rho \over 2} \rho^{\ell + 1} \ \sum^{\infty}_{j = 0} c_j \ \rho^j$$

The substitution for which results in:$$\begin{align} 0 &= \left ( \rho {\operatorname{d} \over \operatorname{d} \! \rho^2} + (2 \ell + 2 - \rho) {\operatorname{d} \over \operatorname{d} \!\rho} + \lambda - \ell - 1) \right ) g(\rho) \\ g(\rho) &= \sum_{j = 0}^{\infty} c_j \ \rho^j \end{align}$$This is the Laguerre Differential Equation and the solution is known. The polynomial expansion is found to be finite. As it turns out, we tend to find exact solutions in quantum mechanics is manipulate the problem until it is a known PDE with an existing solution.

The Solution: $$\begin{align} E_n &= {-1 \over m} \reals_\mu \\ \reals_mu &= {\hbar^2 \over 2 \mu a^2} \\ a_\mu &= {\hbar^2 \over \mu e^2} \end{align}$$

When $$\mu = m$$, $$\reals$$ equals the Rydberg constant (~13.6 eV), and $$a_0$$is the Bohr radius (~ 0.529 Å). When $$\reals_n$$ and $$a_n$$ are substituted:$$E_n = {-1 \over n^2} {\mu e^4 \over 2 \hbar^2}$$

Exactly the energy from Bohr's Atom (Lecture 1). Note that Bohr's idea of quantized angular momentum is important since it is the angular momentum barrier that prevents the electron from spiraling into the nucleus.

The Radial Wave Function
$$\begin{align} R_{\eta \ell} (r) = - \left [ \left ( {2 \over \eta a_u} \right )^3 {(\eta - \ell - 1)! \over 2 \eta \left [(\eta + \ell )! \right ]^3} \right ]^{1 \over 2} e^{-\rho \over 2} \rho^\ell L_{\eta + \ell}^{ \ 2 \ell +1} (\rho) \end{align}$$

Where $$\rho = {2 \over \eta a_u} r$$, and $$a_u = {\hbar^2 \over \mu e^2}$$ to within an arbitrary phase factor (selected form of the solution).

=
$$L_{\eta + 1}^{ \ 2 \ell + 1} (\rho) = \sum_{k=0}^{n - \ell - 1} (-1)^{k + 1} {\left [ (\eta + \ell)! \right ]^2 \over (\eta - \ell - 1 - k )! (2 \ell + 1 + k)!} {\rho^k \over k!}$$ ======

If we assume that $$\ell \leq \eta - 1$$, we get the quantum numbers of hydrogen:$$\begin{align} \eta &= 1, \ 2, \ 3, \ ... \\ \ell &= 0, \ 1, \ 2, \ ..., \ n - 1 \\ m &= -\ell, \ ..., 0, \ ..., \ +\ell \end{align} \qquad \qquad \begin{align} & R_{\eta \ell} \\ & Y_{\ell m} \\ & \eta \ell m = Y_{\ell m} R_{\eta \ell} \end{align}$$  Bransden & Joachain 1983

$$\int \underbrace{r^2 R^* (r)}_{\Rho_{R} (r) = r^2 R^2(r)} \operatorname{d} \! r = 1$$ (since our $$R(r)$$ is real)

 "The eigenfunctions of the bound states." (Description)

The number of nodes in $$\Rho_R(r) = \eta - 1$$... As $$\eta$$ increases, $$R(r)$$ gets pushed out from the origin.

 "Radial Nodes" (Description)

The Solution so Far....
$$\psi_{\eta \ell m} (\vec{r}) = R_{\eta \ell} \ Y_{\ell m} (\theta, \ \phi)$$

Where: $$\begin{align} \eta &: \ Principle \ Quantum \ Number \ (the \ energy \ only \ depends \ on \ this.) \\ \ell &: \ Angular \ Momentum \ Quantum \ Number \\ m &: \ Z-Component \ of \ Quantum \ Number \end{align}$$

Also:   $$\begin{align} \hat{H} \ \psi_{\eta \ell m} &= {-\reals \over \eta^2} \ \psi_{\eta \ell m} \\ \hat{L}^2 \psi_{\eta \ell m} &= \hbar^2 \ell (\ell + 1) \ \psi_{\eta \ell m} \\ \hat{L}_z \psi_{\eta \ell m} &= \hbar m \ \psi_{\eta \ell m} \end{align}$$

which provides: $$[\hat{H}, \ L^2] = [L^2, \ L_z] = [L_z, \ \hat{H}] = 0$$

This result shares eigenfunctions and can be simultaneously measured!

 "Title" (Description)

Highly degenerate energy levels. $$\eta^2$$ eigenfunctions per level.

<FIGURE> "Title" (Description)

We've solved for the states of the given $$\hat{H}$$. Now what? We want to think about an atom. The nucleus is sitting at the origin. We then want to put electrons in. Hydrogen has one electron, but we could also make an $$H^-$$ or a $$H^{--}$$ ion. This is a very simplistic view, but we'll use it for out thought experiment.

Electrons are Fermions so no two can have the same Quantum Numbers.

<FIGURE> "Title" (Comparing a standard Hydrogen atom with an $$H^-$$ ion. Notice how both $$H^-$$ electrons have $$\eta = 1$$, $$\ell = 0$$, and $$m=0$$.)

Notice how both $$H^-$$ electrons in FIGURE have $$\eta = 1$$, $$\ell = 0$$, and $$m=0$$. This is seemingly contradictory to what we've already learned, but if we apply an electrical field to the ion we get something akin to FIGURE. The electron energies slit in what is called the Zeeman Effect.

<FIGURE> "Zeeman Effect" (Description)

Magnetic fields interact with angular momentum called "spin". There is no physical meaning to the word "spin", rather it is just an intrinsic angular momentum of purely quantum nature. Electrons have spin quantum numbers where $m_s = \plusmn {1 \over 2}$. We'll call $m_s = + {1 \over 2}$ spin up $(\uparrow)$, and $m_s = - {1 \over 2}$  spin down $(\downarrow)$. Including spin, there are now four quantum numbers ($$\eta$$, $$\ell$$, $$m$$, $$m_s$$). The wave function is given ($\psi_{\eta \ell m m_s} = R_{\eta \ell} Y_{\ell m} X_{m_s}$ ). And the levels of the system fill according to FIGURE.

<FIGURE> "Title" (Description)

Including spin, the degeneracy is now $$2 \eta^2$$. Relativistic corrections called "fine structures" in part lift this degeneracy. These "fine structures" include relativistic correction to $$\vec{T}$$, $$\vec{L} \cdot \vec{S}$$ coupling, and Darwin Term.

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