# Electronic Properties of Materials/Quantum Mechanics for Engineers/Hydrogen

This is the seventh chapter of the first section of the book Electronic Properties of Materials.

** INCOMPLETE **

Imagine a "simple" Hydrogen atom.

<FIGURE "Title" (Description)

Two particles with unique ${\vec {r}},{\vec {p}},{\vec {m}}$ . This is a many-body problem and quite complicated, but the many-body accounts for translation of the hydrogen atom which we're not really interested in. If $V$ is the couplumb potential then the Hamiltonian classically looks like:

$H=T_{p}+T_{c}+V={{P_{p}}^{2} \over 2m_{p}}+{{P_{c}}^{2} \over 2m_{c}}-{|c|^{2} \over |r_{p}-r_{c}|}$ We want to change to a more natural coordinate system which will allow us to distinguish hydrogen atom translation from <FIGURE> interaction. We also want a simplified version of ${\hat {V}}$ .

<FIGURE> "Title" (Description)

Identify the center of mass:

{\begin{aligned}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{aligned}} There are two approaches to this transformation: $({\vec {r}}_{e},\ {\vec {r}}_{p})\longrightarrow ({\vec {R}},\ {\vec {r}})$ The first approach is to use relative and total momentum. (classical picture) $P=p_{c}+p_{p}$ <FIGURE> "Title" (Description)

The relative momentum is: $p=m\nu$ Which m? Use the "Reduced Mass":

{\begin{aligned}p&=m\ \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_{c}\ \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_{c} \over m_{p}+m_{e}}p_{p}\end{aligned}} Solve for $p_{c}+p_{p}$ in terms of $P$ and $p$ . Substitute into ${\textstyle {{p_{p}}^{2} \over 2\ m_{p}}+{{p_{e}}^{2} \over 2\ m_{e}}}$ , and rearrange...

{\begin{aligned}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{aligned}} This is classical, let's turn it into a quantum Hamiltonian.

$H={-\hbar ^{2} \over 2\ M}\nabla _{R}^{2}-{\hbar ^{2} \over 2\ \mu }\nabla _{r}^{2}-{|e|^{2} \over |r|}$ This approach is physically intuitive, but not purely quantum. Also, it is not generalizable.

The second, better, approach is to use quantum ${\hat {H}}$ .

$H={\hbar ^{2} \over 2\ m_{p}}\nabla _{r_{p}}^{2}-{\hbar ^{2} \over 2\ m_{c}}\nabla _{r_{e}}^{2}-{e^{2} \over |r_{p}-r_{c}|}$ Remember the Laplacian operator:

$\nabla _{r_{p}}^{2}={\partial ^{2} \over \partial r_{p_{x}}^{2}}+{\partial ^{2} \over \partial r_{p_{y}}^{2}}+{\partial ^{2} \over \partial r_{p_{z}}^{2}}$ Using the relations $r=r_{e}-r_{p}$ , and ${\textstyle R={r_{p}\ m_{p}+r_{e}\ m_{e} \over m_{p}+m_{e}}}$ to perform the coordinate transformation, ${\textstyle (r_{p},\ r_{e})\longrightarrow (r,\ R)}$ . Remember that the chain rule for the transformation of differential operators.

{\begin{aligned}(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{aligned}} With some algebra, it results in the same $H$ .

$H=\underbrace {{-\hbar ^{2} \over 2\ M}\ {\nabla _{R}}^{2}} _{Pure\ R}-\underbrace {{\hbar ^{2} \over 2\ \mu }\ {\nabla _{r}}^{2}-{|e|^{2} \over |r|}} _{Pure\ r}$ We now have: $H=H(R)+H(r)$ This means the solution is separable: $\Psi =\psi (R)\ \psi (r)$ Look at $H(R)$ . This is just a free particle, $\nu =0$ . We've already solved this problem and found $\psi (R)$ is planewaves.

Look at $H(r)$ . This is an $e$ , with mass correction in the kinetic energy term, moving around a central potential: ${-|e|^{2} \over |r|}$ $E_{TOT}=E_{r}+E_{R}$ 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:

<FIGURE> "Title" (Description)

<MATHY STUFF> (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.

<FIGURE> "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{aligned}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{aligned}} Multiply Left By: ${\textstyle {r^{2}2\mu \over YR}}$ $\underbrace {{r^{2}\hbar ^{2} \over R}AR+2r\mu e^{2}+r^{2}2\mu } _{Only\ r}=\underbrace {-{\hbar ^{2} \over 2\mu }QY-{\hbar ^{2} \over Y}FY} _{Only\ \theta \phi }$ So both sides equal a constant, $K$ . Thus:

{\begin{aligned}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{aligned}} Let's look closer at the second one:

<MATH>

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{aligned}{\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{aligned}} Consider the commutation of the ${\hat {L}}$ operator.

{\begin{aligned}\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}\\&=yp_{x}p_{z}z-yp_{x}zp_{z}+xp_{y}zp_{z}-xp_{y}p_{z}z\\&=yp_{x}[p_{z},\ z]+xp_{y}[z,\ p_{z}]\\&=yp_{x}(-i\hbar )+xp_{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{aligned}} 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{aligned}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_{y}(L_{y}L_{x})-(L_{x}L_{y})L_{y}+L_{z}(L_{z}L_{x})-(L_{x}L_{z})L_{z}\end{aligned}} 