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Electronic Properties of Materials/Quantum Mechanics for Engineers/Degeneracy

Degeneracy is often talked about in electronics and quantum mechanic in reference to electrons which have the same energy level. In this case, since energy is an eigenvalue, you end up with two electrons with different eigenfunctions which still share the same eigenvalue. We call a quantum state "degenerate" if two or more eigenfunctions have the same eigenvalue as in the case of the electrons, but how does this happen? Well, there are three separate ways; symmetry, exchange, and accidental.

Degeneracy by Symmetry
This is the form of degeneracy associated with the hybridization of orbitals. Atoms behavior in the x, y and z-directions is the same assuming a spherical potential which generally applies to atoms in isolation.

 "Particle in a 2D Box" (Description)

Imagine a particle in another box, once again with infinite potential on all sides, but this time, in a 2D, giving us the Hamiltonian equation:$$\hat{H}={{\hat{p}_x}^2 \over 2m}+{{\hat{p}_y}^2 \over 2m}+V(x)+V(y)$$

This problem easily breaks into component parts: $$\hat{H}=\hat{H}_x+\hat{H}_y$$

Substituting in the Schrödinger equation and working through the math finds that:$$\begin{align} \phi_{n_1 n_2} &= \beta \sin \left ( {n_x \pi \over L_x} x \right ) \sin \left ( {n_y \pi \over L_y} y \right ) \\ E_{n_1 n_2} &= {\hbar^2 \over 2m} \left ( {n_x \pi \over L_x} \right )^2 + {\hbar^2 \over 2m} \left ( {n_y \pi \over L_y} \right )^2 \end{align}$$

Looking at these time-independent eigenfunctions, when $$L_x = L_y$$, then we find that $$E_{\alpha \beta} = E_{\beta \alpha}$$ but $$\phi_{\alpha \beta} \neq \phi_{\beta \alpha} $$ when $$\alpha \neq \beta$$. These are different planewave eigenfunctions with the same energy or eigenvalue, which completes the definition of degeneracy by symmetry.

The below math has no home :(

When $$\hat{H}=\hat{H}(x)+\hat{H}(y)$$, the solution to the eigenvalue problem is separable. Thus:$$\begin{align} \phi(x,y) &= \phi(x) \ \phi(y) \\ \hat{H} \ \phi(x,y) &= \varepsilon \ \phi(x,y) \\ H(x) \ \phi(x,y) +H(y) \ \phi(x,y) &= H(x) \ \phi(x) \ \phi(y) + H(y) \ \phi(x) \ \phi(y) \\ &= \phi(y) \ \underbrace{H(x) \ \phi(y)}_{\varepsilon_x \ \phi(x)} \ + \ \phi(x) \ \underbrace{H(y) \ \phi_y}_{\varepsilon_y \phi_y} \\ &= \phi(y) \ \varepsilon_x \ \phi(x) + \phi_(x) \ \varepsilon_y \ \phi(y) \\ &= (\varepsilon_x + \varepsilon_y) \phi(x) \phi(y) \\ &= \varepsilon \ \phi(x,y) \end{align}$$

Accidental Degeneracy
Occasionally, two eigenfunctions will just happen to have the same eigenvalue for energy making them "accidentally" degenerate. Accidental degeneracy simply refers to when two planewaves, $$\phi$$, share the same eigenvalue by accident and not due to symmetry or exchange.

Degeneracy by Exchange
 "Two Particles in a 1D Box" (Description)

Now let's look at a one dimensional box with two non-interacting particles. This box will once again have infinite potentials outside of the box, as seen in . As these particles are non-interacting, their potentials never see each other.

Our separable Hamiltonian for this situation will look like:$$H = H_1 + H_2 = \underbrace{T_1 + V_1(x)}_{Particle \ 1} + \underbrace{T_2 + V_2(x)}_{Particle \ 2}$$

The Hamiltonian solution for each particle solves to:$$\begin{align} \phi_n(x) &= \sqrt{2 \over L} \sin \left ( {n \pi x \over L} \right ) \\ E_n(x) &= {\hbar^2 \over 2m} \left ( {n\pi \over L} \right )^2 \end{align}$$

Combining these equations with our two particle Hamiltonian gives us:$$\begin{align} \phi_{n_1 n_2}(x_1,x_2) &= {2 \over L} \sin \left ( {n_1 \pi x_1 \over L} \right ) \sin \left ( {n_2 \pi x_2 \over L} \right ) \\ E_{n_1 n_2} &= {\hbar^2 \over 2m_1} \left ( {n_1 \pi \over L} \right )^2 + {\hbar^2 \over 2m_2} \left ( {n_2 \pi \over L} \right )^2 \end{align}$$

With this notation, $n_1$ refers to particle one with quantum number $n_1$, while $n_2$ refers to particle two with quantum number $n_2$. Similarly, $x_1$ and $m_1$  refer to particle one at position $x_1$  or mass $m_1$, while $x_2$  and $m_2$  refer to particle two at position $x_2$  or mass $m_1$. Explicitly, these particles can be at different positions and have different masses, but when the two particles end up having the same eigenvalue for position or mass, $m_1 = m_2$ or $x_1 = x_2$, there is degeneracy due to exchange.

 "Title" (Snapshots of classical mechanics)

This has to do with the loss of determinism. In a classical picture, if we take a snap-shot of the system, then wait a second and take another, we can tell which particle is which as the system is deterministic.  Quantum Mechanics is not this way. In the quantum world there is uncertainty which means I can't really tell you that there are "two particles in this box located here, at $x_1$, and there, at $x_2$ ." Even if it is known that the wavefunctions are centered on those points, in reality that is just the center of a probabilistic distribution of where the particle might be.

 "Title" (Note that these are really probability distributions as opposed to positions.)

Applying our classical method to this case, even if by some capacity we could look at the box and define each particle, looking away and looking back means that we can't tell which particle is which anymore, because the wave functions are overlapping. At any point both particles one and two are present in a superposition of particles.

Going back to our equations, this means is that $\phi_{n_1 m_2}$, and $\phi_{n_2 m_1}$ have the exact same energy regardless of if you switch $$n_1$$ and $$n_2$$. This is how exchange leads to degeneracy. We have lost that deterministic world view making it impossible to definitively tell the particles apart if they have the same eigenvalues, as if they had different masses to begin with, we would be able to tell them apart regardless.

Mathematically speaking, $E = \left ( {\hbar \pi \over L} \right )^2 {1 \over 2m} ({n_1}^2 + {n_2}^2)$, and switching the n values is the same as switching the particles.

Implications of Exchange
Comparative to degeneracy by symmetry and accidental degeneracy, degeneracy by exchange has several long reaching implications which should be considered. Imagine we have a system of $$N$$ non-interactive identical particles. We know wave functions define an interchange operator that operates on our wave function ($$\psi$$) to switch the variables of two of the particles.$$\psi(q_1, \ q_2, \ \dots, \ q_i, \dots, \ q_j, \ \dots, \ q_n)$$

Each of these q's represent a collection of variables and quantum numbers that represent a given particle. The written order of these variables further corresponds to each particle, precisely representing them and the state that these particles are in. Given this expression we can identify an interchange operator ($\hat{P}_{ij}$ ). This interchange operator takes two particles and switches them, essentially switching the $$q$$ for one particle with the $$q$$ of another particle. Applying an interchange operator to the wavefunction looks like:

$$P_{ij} \ \psi(q_1, \dots, \ q_i, \dots, \ q_j, \dots, \ q_n) = \psi(q_1, \dots, \ q_i, \dots, \ q_j, \dots, \ q_n)$$

The interchange operator doesn't change much as it simply moves around the parameter which were already in the system. Furthermore, the interchange operator doesn't change the energy of the system which means that our interchange operator and the Hamiltonian operator commute, where $$[P_{ij}, \ H]=0$$. While this was already implied by the non-interacting nature of our particles, this further proves that we can solve for each of these elements independently, thus the wavefunctions ($$\psi$$) are eigenfunctions of $$P_{ij}$$ and $$H$$.

As such, when this operator, the interchange operator, operates on some wavefunction, we know that it has to return an eigenvalue ($$P_{ij} \ \psi=\xi \ \psi$$). All good operators in Hilbert space will obey this relationship.

As it stands to reason, applying the same interchange operator to the same wavefunction twice will result in the original wave function. The operator switches all the particle parameters, and then switches them back.

What are the values of $$\xi$$? swapping $$i$$ and $$j$$ twice returns to initial state. $$\begin{align} P_{ij}(P_{ij} \ \psi) &= P_{ij} \ \xi \ \psi \\ &= \xi^2 \ \psi =1 \ \psi \\ \xi^2 &= 1 \ \psi \\ \xi &= \plusmn \ 1 \end{align}$$In this case, we give these eigenvalues names and when $$\xi = +1$$ we say that the wavefunction is "symmetric under exchange," and when $$\xi=-1$$, we say that the wavefunction is "antisymmetric under exchange." It's also worth noting that even though each version of the interchange operator commutes with the Hamiltonian, they don't necessarily commute with each other. It's not universally true that for any exchange that the two are equivalent. This is important because it limits the way that we can express the wavefunctions.

Let's define another operator, $$\hat{P}$$ as a permutation operator which is a series of interchange operators, $$\hat{P}_{ij}$$, that rearrange the variables in the wavefunction ($$\psi$$). This gives us:  $$P \psi(q_1, \ q_2, \ \dots, \ q_n) = \psi (q_{1}', \ q_{2}', \ \dots, \ q_{n}')$$

When $$P$$ operates, it returns a new wave function with the variables, $$q$$, with different ordering. If $$\hat{P}$$ contains an even number of swaps, we call it an even permutation, and if has an odd number of swaps, we call it an odd permutation. It's worth nothing that there exists an $$N!$$ number of total permutations, and these permutation operators will not commute with each other just as the interchange operators don't commute with each other.

Unfortunately, this doesn't make sense. In general, $$\hat{P}$$ operators of different arrangements do not commute because $$[P_{ij}, P_{lm}]\neq 0$$. This means that the wavefunction may be an eigenstate of the Hamiltonian and some permutation, $$\hat{P}_1$$, but not be an eigenstate of another permutation, $$\hat{P}_2$$. This means that the nature of our wavefunction is now limited. This means that we must express our wavefunction is such a way that is is either totally symmetric or totally antisymmetric. These are two special wavefunctions that commute with the Hamiltonian AND with all $$N!$$ possible $$\hat{P}$$.

$$\begin{align} P_{ij} \ \psi_S &= \psi_S \\ P_{ij} \ \psi_A &= \begin{cases} \psi_A \ & for \ even \ P \\ -\psi_S \ & for \ odd \ P \end{cases} \end{align}$$ In the case of the symmetric wavefunction the permutation operator acting on the symmetric is equal to the symmetric wavefunction for all permutations. Similarly, in the case of the antisymmetric operator acting on the antisymmetric wavefunction is equal to the antisymmetric wavefunction for even permutation operators and equal to the negative wavefunction for odd permutation operators. This also serves as our definition for symmetric and antisymmetric operators.

Postulate Zero
From what we know of the world, $$\psi_S$$ and $$\psi_A$$ are sufficient to describe all systems of identical particles. Particles that are totally symmetric are called Bosons and obey Bose-Einstein statistics (photon, phonon, cooper pair), while particles that are totally are called Fermions and obey Fermi-Dirac statistics (electrons, neutrino, quarks). This is called "Postulate Zero" since we can't actually prove this as fact, it is simply a pattern that every known particle follows, and happens to work out through the relationships of the commutators and our operators.

This means that when we write our solution, we have to make certain that our solution is either totally symmetric or totally antisymmetric. If for example, we wrote a solution to two non-interactive particles in a 1D box that wasn't either totally symmetric or totally antisymmetric, we would need to add different variations of our solution together to achieve either total symmetry or total antisymmetry. This might look like:

$$\begin{align} & \ \ \left [ \begin{align} \psi_S &= \phi(x_1 x_2) + \phi(x_2 x_1) \\ \psi_A &= \phi(x_1 x_2)-\phi(x_2 x_1) \end{align} \right ] \\ \\ & P_{12}\psi_A =P_{12} ( \phi(x_1 x_2) - \phi(x_2 x_1)) \\ & \qquad \ \ \ = \phi(x_2 x_1) - \phi(x_1 x_2) \\ & \qquad \ \ \ = -\phi_A \end{align}$$As you can see, operating on the wavefunction ($$\phi$$) returns the negative wavefunction. You can do the same with a symmetric wavefunction. If I tell you that the particles are Fermions or Bosons then you know right away which $$\psi_S$$ or $$\psi_A$$ they reside in. That said, these particles are supposed to be non-interacting and thus the Hamiltonians can't "see" each other, yet they become entangled.$$\begin{align} \psi_A &= \phi(x_1 x_2) - \phi(x_2 x_1) \\ &= \phi(x_1) \phi(x_2) - \phi(x_2) \phi(x_1) \end{align}$$

These are NOT a simple product of single particle states. In quantum mechanics we say that the states are said to be "entangled", and even though the particles do not interact their wavefunctions are entangled. This means that a measurement of one particle has consequences on the other. By looking at the $$\psi_{A,S}$$ you can see that there is a superposition.

Expressing Wave Functions
Imagine $$n_1 =2$$ and $$n_2=3$$. Which particle has energy $$E_2$$? We don't know. Each particle is a superposition of being in state $$E_2$$ and $$E_3$$. (Many body physics is neat stuff, but also extremely complicated.)

How do we express a general symmetric or antisymmetric wavefunction for any system? Symmetric wavefunctions are simple, it's just a sum of interchanges ($$\phi(\gamma_i)$$). On the other hand, antisymmetric wavefunctions are a little more complicated. To express an antisymmetric wavefunction we use a Slater determinant:

Say $$H = h_1 + h_2 + \dots + h_n$$, so $$\Phi = \phi_1(x_1) \ \phi_2(x_2) \ \phi_3(x_3) \dots \phi_n(x_n) $$.

$$h_j \ \phi_j(x_j) = E_j \phi_j (x_j)$$$$\psi_A (x_1 \ x_2 \ x_3 \dots \ x_n) = {1 \over \sqrt{N!}} \left | \begin{array}{lcl} \phi_1(x_1) & \phi_2(x_1) & \phi_3(x_1) & \cdots & \phi_n(x_1) \\ \phi_1(x_2) & \phi_2(x_2) & \phi_3(x_2)& \cdots & \phi_n(x_2) \\ \quad \ \vdots & \vdots & \quad \vdots & \ & \quad \vdots \\ \phi_1(x_n) & \phi_2(x_n) & \phi_3(x_n) & \cdots & \phi_n(x_n) \end{array} \right |$$

This is an N x N determinant. For an example, look at some three particle system of Fermions:$$\psi_A(q_1 \ q_2 \ q_3) = {1 \over \sqrt{6}} [ \phi(q_1 \ q_2 \ q_3) - \phi(q_2 \ q_1 \ q_3) + \phi(q_2 \ q_3 \ q_1) - \phi(q_3 \ q_2 \ q_1) + \phi(q_3 \ q_1 \ q_2) - \phi(q_1 \ q_3 \ q_2) ]$$

A symmetric wavefunction would have the same determinant, but all the terms are added together for the summation.

Pauli Exclusion Principle
In the above example, each particle is in a unique state $$q_1, q_2, q_3$$. What would happen if $$q_1 = q_2$$? Then all of the terms would cancel out and the wavefunction would equal zero. This is the crux of the Pauli Exclusion Principle in that Fermions must have unique quantum numbers, because if they share quantum numbers then their antisymmetric wavefunction disappears. This comes from the fact that electrons are indistinguishable particles.

 "Quantum Teleportation Between Distant Matter Qubits"


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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:

 "Title" (Description)

 (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.

=
$$\begin{align} \hat{H} &= {-\hbar^2 \over 2 \mu} \nabla^2 - {|e|^2 \over r} \\ &= {-\hbar^2 \over 2\mu} \left [ {1 \over r} {\partial \over \partial r} \left ( r^2 {\partial \over \partial r} \right ) + {1 \over r^2 \sin\! \theta } {\partial \over \partial \theta} \left ( \sin \! \theta {\partial \over \partial \theta} \right ) + {1 \over r^2 \sin ^2 \! \theta} {\partial^2 \over \partial \phi^2} \right ] - {|e|^2 \over r} \\ &= {-\hbar^2 \over 2 \mu} {1 \over r^2} {\partial \over \partial r} \left ( r^2 {\partial \over \partial r} \right ) + {1 \over 2 \mu r^2} \left [ {-\hbar^2 \over \sin \! \theta} {\partial \over \partial \theta} \left ( \sin \! \theta {\partial \over \partial \theta} \right ) + {-\hbar^2 \over \sin^2 \! \theta} {\partial^2 \over \partial \phi^2} \right ] - {|e|^2 \over r} \\ &= {-\hbar^2 \over 2 \mu} {1 \over r^2} {\partial \over \partial r} \left ( r^2 {\partial \over \partial \theta} \right ) + {1\over 2 \mu r^2} \hat{L}^2 - {|e|^2 \over r} \end{align}$$ ====== 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.

 "Title" (Description)

 "Title" (Descripiton)

=
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.

Content

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Electronic Properties of Materials/Quantum Mechanics for Engineers/Variational Methods

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

** INCOMPLETE**

An extremely useful fact is that the time-independent Schrödinger Equation is equivalent to a variation principle. The energy is a functional, or a function of functions, of the wavefunction.

$$\langle E \rangle \ = \ \langle \psi \ H \ \psi \rangle \ = \ E[\psi]$$

The $$E[\psi]$$ is minimized when $$\psi$$ is the ground state wavefunction. This can be proven by calculus of variation methods or by methods of Lagrange multipliers. Here, we're going to show this by a practical example.

A Practical Example
Say $$\psi_n$$ are a complete set of orthonormal eigenfunctions of $$H$$.

$$H \psi_n = E_n \psi_n$$

$$\phi$$ is an arbitrary square-integrable function, in that you can take the integral of $$\phi^* \phi$$ without singularity.

We can write $$\phi = \sum_n a_n \psi_n$$ as:

$$\begin{align} E[\phi] &= {\int \phi^* H \phi \over \int \phi^* \phi} \\ &= {\sum_n \sum_{n'} \int a_n^* \ \phi_n^* \ H \ a_{n'} \ \psi_{n'} \over \sum_n \sum_{n'} \int a_n^* \ \psi_n^* \ \psi_{n'} \ a_{n'}} \\ &= {\sum |a_n|^2 E_n \over \sum |a_n|^2} \end{align}$$

Further subtracting the lowest possible energy, called the ground state ($$E_o$$), from both sides gets us:

$$E[\phi] - E_o = {\sum_n |a_n|^2 E_n \over \sum_n |a_n|^2} - E_o$$

Since $$E_n$$ is always greater than or equal to $$E_o$$ for all $$n$$, the right side of this equation must always be greater than zero.

This equality has a very practical importance. It means that if $$\phi$$ is not the ground state wave function the energy will be larger than $$E_o$$. As it also happens, if $$\phi H = E \phi$$ and $$E=E_o$$, then $$\phi$$ is $$\psi_o$$. (for many non-degenerate $$\psi$$)

So... say you have a difficult $$H$$ and can't solve it, but you have a "good" guess for $$\Psi$$, say $$\phi$$. If you can find some way to tweak $$\phi$$ to minimize $$E$$ then $$\phi \rightarrow \psi$$. This allows for Rayleigh-Ritz Variational Method

Rayleigh-Ritz Variational Method

 * 1) Guess: $$(\vec{\phi}, \ \vec{\alpha})$$ which has a good form, where $$\vec{\alpha}$$ are a set or variational parameters.
 * 2) Calculate $$\langle E \rangle = {\int \phi^* H \phi \over \int \phi^* \phi}$$
 * 3) Solve $${\partial \langle E \rangle \over \partial \alpha_i} = 0$$, for each $$\alpha_i$$

To find the set of $$\vec{\alpha}$$ that minimizes $$\langle E \rangle$$ and returns the best $$\phi$$ given the chosen form of $$\phi(x)$$.

Rayleigh-Ritz Example
Take as an Example, the Hydrogen atom. What if we can't solve for it? We can try making a good guess. Let's see how close a reasonable guess is.

Excellent Guesses will get you close to the true ground state. Good guesses will still do "ok" but not great. The shortcoming of this method is that you can't know if your guess for $$\phi$$ is close unless you already know the general solution. The best approach is to make several educated guesses based on asybiotic behavior of $$\psi$$ in the extreme limits.


text:
Electronic Properties of Materials/Quantum Mechanics for Engineers/Perturbation Methods

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

** INCOMPLETE**

Most operators (Hamiltonians) are not simple. Fortunately, with a bit of effort, we can sometimes rewrite the operator ($$\hat{H}$$) as $$\hat{H} = \hat{H}_o + \lambda H'$$, where $$\hat{H}$$ is a Hamiltonian for which we know the solution.

$$\hat{H}_o \ \psi^{(o)}_\eta = E^{(o)}_\eta \ \psi^{(o)}_\eta$$

Here, $$\psi^{(o)}_\eta$$ are non degeneratory orthogonal eigenfunctions, and $$H'$$ is a small perturbation to the $$H_o$$. Additionally, $$\lambda$$ is a real arbitrary parameter, and when $$\lambda=0$$, we have:

$$\begin{array}{lcl} H = H_o  & \ \ when \ \lambda=1 \\ H = H_o + H' & \quad \end{array}$$

The problem we want to solve is: $$H \psi_\eta = E_\eta \psi_\eta$$

The perturbation is small and in the limit $$\lambda $$ goes to zero.

$$E_\eta = E^{(o)}_\eta + \psi_\eta = \psi^{(o)}_\eta $$

We will assume that $$E_\eta$$ and $$\psi_\eta$$ can be written as powers of $$\lambda$$.

$$\begin{align} E_\eta &= \sum^{\infty}_{j=0} \lambda^j E^{(j)}_\eta = E^{(o)}_\eta + \lambda E^{(1)}_\eta + \lambda^2 E^{(2)}_\eta + \cdots + \lambda^n E^{(n)}_\eta \\ \psi_\eta &= \sum^{\infty}_{j=0} \lambda^k \psi^{(j)}_\eta \end{align} $$

Substituting:

$$\left ( H_o + \lambda H' \right ) \left ( \psi^{(o)}_\eta + \lambda \psi^{(1)}_\eta + \lambda^2 \psi^{(o)}_\eta + \cdots + \lambda^{n} \psi^{(n)}_\eta \right ) = \left ( E^{(o)}_\eta + \lambda E^{(1)}_\eta + \lambda^2 E^{(o)}_\eta + \cdots + \lambda^{n} E^{(n)}_\eta \right ) \left ( \psi^{(o)}_\eta + \lambda \psi^{(1)}_\eta + \lambda^2 \psi^{(o)}_\eta + \cdots + \lambda^{n} \psi^{(n)}_\eta \right ) $$

Multiply through & collect common properties to form equations for each power of $$\lambda $$:$$\begin{align} \lambda^o :& \quad H_o \psi_n^{(o)} = E_n^{(o)} \psi_n^{(o)} \\ \lambda^1 :& \quad H_o \psi_n^{(1)} + H' \psi^{(o)} = E_n^{(o)} \psi_n^{(1)} + E_n^{(1)} \psi_n^{(o)} \\ \lambda^2 :& \quad H_o \psi_n^{(2)} + H' \psi^{(1)} = E_n^{(0)} \psi_n^{(2)} + E_n^{(1)} \psi_n^{(1)} + E_n^{(2)}\psi_n^{(o)} \end{align} $$

The powers of $$\lambda^o $$ are just our unperturbed $$H_o $$. We will begin by looking at powers of $$\lambda^1 $$.

Rearrange:

$$(H_o - E_n^{(o)}) \psi_n^{(1)} + (H' - E_n^{(1)}) \psi_n^{(o)} =0 $$

multiply left by $$\psi_n^{*(o)} $$ and integrate

$$\int \psi_n^{*(o)} (H_o - E_n^{(o)}) \psi_n^{(1)} + \int \psi_n^{*(o)} (H' - E_n^{(1)}) \psi_n^{(o)} =0 $$

Begin with left term. These operators are Hermitian. They have special properties, namely that they obey the postulates of quantum mechanics, including a few revations that are useful for proofs. One such property is:

$$\int \Psi^* H \Phi = \left ( \int \Phi^* H \Psi \right )^* $$

Which we will use here:

$$\begin{align} \int \psi_n^{(o)^*}H_o \psi_n^{(1)} =& \left ( \int \psi_n^{(1)^*} H \psi_n^{(o)} \right )^* \\ =& \left ( \int \psi_n^{(1)^*} E_n^{(o)} \psi_n^{(o)} \right )^* \\ =& E_n{(o)} \left ( \int \psi_n^{(1)^*} \psi_n^{(o)} \right )^* \\ =& E_n^{(o)} \int \psi_n^{(o)^*} \psi_n^{(1)} \end{align} $$

Thus, our entire term equals zero. As a result:

$$\int \psi_n^{(o)} H' \psi_n^{(o)} =E_n^{(1)} \int \psi_n^{(o)^*} \psi_n^{(o)} = E_n^{(1)} $$

Therefore, the first order correction to the eigenvalue is:

$$E_n^{(1)}=\int \psi_n^{(o)^*} H' \psi_n^{(o)} $$

Following the same steps we can find the higher order perturbations:

$$\begin{align} E_n^{(2)} &= \int \psi_n^{(o)^*} \left (        H' - E_n^{(1)} \right ) \psi_n^{(1)} \\ E_n^{(3)} &= \int \psi_n^{(1)^*} \left (       H' - E_n^{(1)} \right ) \psi_n^{(1)} - 2E_n^{(2)} \int \psi_n^{(o)^*} \psi_n^{(1)} \end{align} $$

Most simple theories do not require these higher order corrections, but how do we get the wavefunctions in the first place? Lets assume that $\psi_n^{(1)} = \sum_k {a_{nk}}^{(1)} {\psi_k}^{(o)} $ where $$a_{nk} $$ coefficient is the projection of $\psi_n^{(1)} $  onto $\psi_k^{(o)} $. Returning to our original $$\lambda' $$ term, gather:

$$H_o \psi_n^{(1)} + H' \psi_n^{(o)} = E_n^{(o)} \psi_n^{(1)} + E_n^{(1)} \psi_n^{(o)} $$

Rearranging and substituting gives us:

$$0= \left ( H_o -E_n^{(o)} \right ) \sum a_n^{(1)} \psi_k^{(o)} + \left ( H' - E_n^{(1)} \right ) \psi_n^{(o)} $$

Multiplying the right side by $$\psi_\ell^{(o)^*} $$and integrating gets us:

$$\begin{align} 0 &= \int \psi_\ell^{(o)^*} E_n^{(o)} \sum a_{nk}^{(1)} \psi_k^{(o)} - E_n^{(o)} \int \psi_\ell^{(o)^*} \sum_k a_{nk}^{(1)} \psi_k^{(o)} + \int \psi_\ell^{(o)^*} H' \psi_n^{(o)} - E_n^{(1)} \int \psi_\ell^{(o)^*} \psi_n^{(o)} \\ &= E_\ell^{(o)} a_{n \ell}^{(1)} - E_n^{(o)} a_{n \ell}^{(1)} + \int \psi_\ell^{(o)^*} H' \psi_n^{(o)} - E_n^{(1)} \delta_{n \ell} \end{align} $$

When $n=\ell $, we loose all $a_{n \ell}^{(1)} $ terms, giving us:     $$E_n^{(1)} = \int \psi_{\ell}^{(o)^*} H' \psi_n^{(o)} $$

However, when $n \neq \ell $ we get:

$$\begin{align} 0 &= \left ( E_\ell^{(o)} - E_n^{(1)} \right ) a_{n \ell}^{(1)} + \int \psi_\ell^{(o)^*} H' \psi_n^{(o)} \\ a_{n \ell} &= {-\int \psi_\ell^{(o)^*} H' \psi_n^{(o)} \over E_\ell^{(o)} - E_n^{(o)}} \end{align} $$

Since $$a_{nn}^{(1)} $$ does not seem to be determined from these equations, there is an uncomfortable degree of arbitrariness in selecting $$a_{nn}^{(1)} $$. Require normalization:

$$\begin{align} 1 &= \int \Psi^* \psi \\ 1 + 0 \left ( \lambda ^{j + 1} \right ) &= \int \left ( {\psi_n^{*}}^{(o)} + \lambda {\psi_n^*}^{(1)} + \lambda^2 {\psi_n^*}^{2}       + \cdots \right ) \left ( \psi_n^{(o)} + \lambda \psi_n^{(1)} + \lambda^2 \psi_n^{(2)} + \cdots \right ) \end{align} $$

Where: $$\begin{align} \lambda^o: \ 1 &= \int {\psi_n^{(o)}}^* \psi_n^{(o)} \\ \lambda^1: \ 0 &= \int {\psi_n^{(0)}}^* \psi_n^{(1)} + \int {\psi_n^{(1)}}^* \psi_n^{(o)} \\ \lambda^3: \ 0 &= \int {\psi_n^{(0)}}^* \psi_n^{(2)} + \int {\psi_n^{(1)}}^* \psi_n^{(1)} + \int {\psi_n^{(2)}}^* \psi_n^{(o)} \end{align} $$

Thus:

$$0 =\underbrace{\int {\psi_n^{(o)}}^* \psi_n^{(1)}}_{= a_{nn}} + \underbrace{\int {\psi_n^{(1)}}^* \psi_n^{(o)}}_{{=a_{nn{^{(1)}}}}^*} $$

which is a projection of $$\psi_n^{(1)} $$ onto $$\psi_n^{(o)} $$.

$$a_{nn}^{(1)} + {a_{nn}^{(1)}}^* =0 $$ is a complex number.

Complex number formula: $$\begin{align} 0 &= a+bi + a-bi \\ 0 &= a \end{align} $$

What is $$b $$? Here, we choose $$b=0 $$.

 "Title" (Description)

In quantum mechanics usually, but not always, $$\Psi $$ can have arbitrary phase $$\phi $$, so long as magnitude of $$\Psi $$ is correct. Here we choose $\phi =0 $. Therefore:

$$a_{nn}^{(1)} = 0 $$

This dictates that all of $$\psi_n^{(1)} $$ is orthogonal to $$\psi_n^{(o)} $$.

As an example, consider adding a correction to the hydrogen atom, what is actually a fairly common occurrence.

$$\begin{align} H &= H_o + H' \\ H_o &= {- \hbar^2 \over 2m} \Delta^2 - {e^2 \over r} \\ H' &= -G {m M_p \over r} \end{align} $$

This last equation is the influence of gravitational attraction between the positive ion and the negative ion. This is a first order energy correction.

$$\begin{align} E_{1s}^{(1)} &= \int {\psi_{1s}^{(o)}}^* \left ( {-G m M_p \over r} \right ) \psi_{1s}^{(o)} \operatorname{d} \! right \\ \psi_{1s}^{(o)} &= {1 \over \sqrt{\pi}} {1 \over a_o^{3 \over 2}} \exp \left [ {-2r \over a_o} \right ] {-G m M \over r} \\ \\ E_{1s}^{(1)} &= \int_0^\pi \sin \! \theta \operatorname{d} \! \theta \int_0^{2\pi} \operatorname{d} \! \phi \int_0^\infty r^2 \operatorname{d} \! r {1 \over \pi} {1 \over a_o} \exp \left [ {-2r \over a_o} \right ] {-G m M \over a_o^2} \left ( {a_o^2 \over 4} \right ) \\ &= 4\pi {-G mM \over \pi a_o^3} \int_0^\infty \operatorname{d} \! r r   \exp \left [ {-2r \over a_o} \right ] \\ &= -4 {G m M \over a_o^3} \left ( {a_o^2 \over 4 } \right ) \end{align} $$

When working with degenerate wavefunctions, the problem becomes slightly more complicated because the interactions amongst the degenerate wavefunctions must be carefully accounted for. That said, this is just bookkeeping. The general procedure for Rayleigh-Schrodinger perturbation theory is as outlined here.


text:
Electronic Properties of Materials/Quantum Mechanics for Engineers/Many Electron Atoms

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

We now have a solutions for Hydrogen.$$\begin{align} H &= T_{electron} + V_{ion-electric} \\ &= {- \hbar^2 \over 2m} \nabla^2 - {|e|^2 \over r} \\ H \psi_{nlmm_s} (r) &= E_n \psi_{nlmm_s} \\ E_n &= {-1 \over n^2} \reals \\ \psi_{nlmm_s} &= R_{nl} Y_{lm} \chi_{m_s} \end{align}$$

In undergraduate chemistry class, we were shown this solution and told "... and so it goes for the rest of the periodic table..." However, the truth is, it isn't so simple.

Consider Lithium:

 "Diagram of Lithium Ion" (Description)

We can apply center of mass corrections but for simplicity, assume that the nucleus has $$M_{noc} = \infty$$.

What does hydrogen look like?

$$H = T_{electron} + V_{ion-electron} + V_{electron-electron}$$$$\begin{align} T_{electron} &= T_{e_1} + T_{e_2} + T_{e_3} \\ &= {-\hbar^2 \over 2m} \left (    {\nabla_1}^2 + {\nabla_2}^2 + {\nabla_3}^2 \right ) \\ \\ V_{ion-electron} &= {-3e^2 \over r_1} + {-3e^2 \over r_2} + {-3e^2 \over r_3} \\ V_{electron-electron} &= {e^2 \over |r_1 - r_2|} + {e^2 \over |r_2 - r_3|} + {e^2 \over |r_3 - r_1|} \end{align}$$

Without the electron-electron interaction the problem would be much simpler.$$\begin{align} H &= T_1 + V_{ie1} + T_2 + V_{ie2} + T_3 + V_{ie3} \\ &= h_1 + h_2 + h_3 \end{align}$$

This makes for a separable solution:

$$\begin{align} H\Psi &= E\Psi \\ \Psi &= \psi_1 \psi_2 \psi_3 \end{align}$$

Here, each $$\psi_i $$ terms is just the hydrogen wave function with a slight modification for $$V= {3 e^2 \over r} = {z e^2 \over r}$$.

$$\begin{align} E = \varepsilon_1 + \varepsilon_2 + \varepsilon_3 \\ \varepsilon_i = -{Z^2 \mathbb{R} \over \eta^2} \end{align}$$

Here $$Z$$ is the atomic number.

Since electrons are fermions, $$\Psi$$ must be totally anti-symmetric as seen in 2 Particles in a Box in Chapter 4.

$$\Psi(r_1, r_2, r_3) = {1 \over \sqrt{6}} \left | \begin{align} \psi_1(r_1) \quad \psi_2(r_1) \quad \psi_3(r_1) \\ \psi_1(r_2) \quad \psi_2(r_2) \quad \psi_3(r_2) \\ \psi_1(r_3) \quad \psi_2(r_3) \quad \psi_3(r_3) \end{align} \right |$$

This is known as the Slater Determinant.

Approximating Lithium
Unfortunately, we can't just discard electron to electron interactions. The magnitude of $V_{ion-electron}$ and $V_{electron-electron}$  are equivalent. So, what do we do? Solve the problem as is? Well, after you solve $$H_{Li}$$, you could quite possibly earn a Nobel Prize and/or a Fields Medal which makes this a wholly impractical method for this course. Instead we will use approximate methods for quantum mechanical problems more complicated than our earlier Hydrogen solution, and we currently have two primary methods for doing this.

Thomas-Fermi Model
The Thomas-Fermi Model is a semi-classical statistical method where we replace the exact potential $V_{ion-electron}$ and $V_{electron-electron}$  with an effective potential, $V_{eff}$, the screened coulomb potential.

Consider an electron gas with so many electrons that you don't have to count them one at a time, but instead just consider a continuous charge distribution. If we put a positive charge in the electron cloud then the positive charge would attract the electrons so the electron distribution would no longer be uniform. That said, the negative charges nearest the positive charge "screen" the electrons further out such that the fringe electrons don't feel as strong of an attraction.

Working out the details of this method requires a discussion of electron gasses, provided later in this text. Meanwhile, you should know that this hypothetical situation turns into a spherically distributed electron gas where the charge density, $$\rho$$, is slowly varying. Although $$\rho(r)$$ does change with $$r$$, at each $$r$$, we can treat $$\rho$$ as a uniform gas. Additionally, a point charge $$+Q$$, placed in a uniform electron gas has a screened potential of:

$$\Phi(r)= {Q \over r} e^{-k_or}$$

This is sometimes called a Yukawa Potential. Also, ${1 \over k_o}$ is the Thomas-Fermi screening length and is related to the density of quantum states of the highest filled state, called the density of states at the Fermi Level.

Finally, for charge neutral atoms, where the number of electrons equals the atomic number, with a slowly varying electron gas in a spherically symetric central potential:

$$V(r) \simeq \underbrace{-{Ze^2 \over r}}_\text{Attractive Potential} + \underbrace{1.794 \ {Z^{4 \over 3} \over a_o}}_\text{Repulsive Potential} + \ \cdots$$

 "Electron Gas Potential" (Description)

The truth of the matter is that the Thomas-Fermi Model isn't a great solution for the multi-electron atom. Back in the 1920's it was quite successful and is still useful for simple approximations and the initial input to more advanced methods.

Hartree-Fock Method
A better method is the iterative Hartree-Fock Method, or the Self-Consistent Field Method. For this method, we start by assuming that we can write:

$$\Psi = \prod^n_{i=1} \psi_i$$This assumption further implies that the can say that $H = \sum^N_{i=1} h_i$. So, what is a reasonable $h_i$ ?

$$h_i = {-\hbar^2 \over 2m} \nabla^2_i + V_o (r_i)+{1 \over 2} \sum_{j \neq i} {e^2 \over |r_i - r_j|}$$

Now, this gives us:

$$H = \sum^N_{i=1} \left [ {-\hbar^2 \over 2m} \nabla^2_i + V_o (r_i)+{1 \over 2} \sum_{j \neq i} {e^2 \over |r_i - r_j|} \right ]$$

$V_o(r_i)$ is all the potential terms not involving electron-electron interaction. For the single atom, $V_o(r_i) = {Ze^2 \over r_i}$, but this method is generally applicable to more complex systems such as molecules. Hartree-Fock is the intellectual forbearer of several modern quantum chemical techniques.

 "Primer on Calculus of Variation and Lagrange Multipliers"

Energy Function
The functional we are interested in is the energy, where $E \left [ \Psi(r) \right ] = \int \Psi^* H \Psi \ dr$. We want to find the extrema subject to the constraints $\int \psi^*_i \psi_i \ dr =1$ for all $$i$$. We will do this by using Lagrangian multipliers $$\varepsilon_i$$ such that the integral we're interested in is:

$$\int \Psi^* H \Psi \ \operatorname{d}\!r - \sum_i \varepsilon_i \int \psi^*_i \psi_i \ \operatorname{d}\!r$$

Applying Calculus of Variation
We want to use calculus of variation to find the functions that make this function stationary (an extrema). Consider this function:$$I[\phi_1(x) \ \phi_2(x)] = \int f(\phi_1(x), \ \phi_2(x), \ x) \ \operatorname{d}\!x $$We want to vary the function to look for the stability condition. In other words, we want to change $$\phi_i(x) $$ at an arbitrary constant $$x $$.$$\Delta I = I[\phi_1(x) + \varepsilon_1 \eta_2 (x), \ \phi_2(x) + \varepsilon_2 \eta_2 (x)] - I[\phi_1, \ \phi_2] $$

Here $$\varepsilon_i $$ is a small constant (we take the limit of $$\varepsilon_i $$ as it goes to zero), and $$\eta_i $$ is an arbitrary function that deviates $$\phi_i $$ (this must be continuous over the limits of integration. This gives us:$$\Delta I =    \int f(\phi_1(x) + \varepsilon_1 \eta_2 (x), \ \phi_2(x) + \varepsilon_2 \eta_2 (x), \ x] - \int f(\phi_1(x), \ \phi_2(x), \ x) $$

The Maclaurin expansion in powers of $$\varepsilon_i $$ is:$$f(x) = f(0) + f'(0) + {1 \over 2!}f(0) + {1 \over 3!} f'(0) + \cdots $$$$\begin{align} \Delta I &= \int f(\phi_1(x), \ \phi_2(x), \ x) + {\partial f \over \partial \phi_1} \ \varepsilon_1 \eta_1 (x) + {\partial f \over \partial \phi_2} \ \varepsilon_2 \eta_2 (x) + \text{Higher Order Terms} - f(\phi_1(x), \ \phi_2(x), x) \\ &= {\partial f \over \partial \phi_1} \ \varepsilon_1 \eta_1 (x) + {\partial f \over \partial \phi_2} \ \varepsilon_2 \eta_2 (x) \\ \delta I &= \int {\partial f \over \partial \phi_1} \ \delta \phi_1 + {\partial f \over \partial \phi_2} \ \delta \phi_2 \\ &= \int \delta f \qquad \text{(The variation of the functional} \ f ) \end{align} $$

$$\delta \phi_i = \varepsilon_i \eta_i $$ in the limit as $$\varepsilon_i $$ goes to zero.

Let's return to our problem:

$$I = \int \Psi^* H \Psi - \sum \varepsilon_j \psi^*_j \psi_j \operatorname{d}\!r $$

Writing this function out in detail gives us:$$I = \int \underbrace{(\psi^*_1, \ \cdots, \ \psi^*_n)(h_1, \ \cdots, \ h_n)(\psi_1, \ \cdots, \ \psi_n) - \varepsilon_1 \psi^*_1 \psi_1 - \cdots - \varepsilon_n \psi^*_n \psi_n}_{f(\psi^*_1, \ \cdots, \psi^*_n)} \operatorname{d}\!r $$

Note that you could alternatively carry through $$\psi $$ in this step, but in the end you don't need them. I chose to utilize $$\psi^* $$ instead of $$\psi $$ for purely aesthetic reasons. This gives us:

$$\delta I = 0 = \int \sum^n_{q=1} {\partial f \over \partial \psi^*_q} \delta \psi^*_q \ \operatorname{d}\!r $$

Looking at the $$q^{th} $$ term in the summation, ${\partial f \over \partial \psi_q^*} \ \delta \psi_q^* $, we can expand to:

$$\begin{align} & \sum_{j=1}^n \left [ (\psi_1^*, \ \cdots, \ \delta \psi^*_q, \ \cdots, \psi_n^*)(h_j) (\psi_1, \ \cdots, \ \psi_q, \ \cdots, \psi_n) \right ] - \delta \psi_q^* \varepsilon_q \psi_q \\ & \sum_{j=1}^n \left [ (\psi_1^*, \ \cdots, \ \delta \psi^*_q, \ \cdots, \psi_n^*) \left[ \underbrace{{-\hbar^2 \over 2m} \nabla_j^2}_{T_j} + \underbrace{V_o (r_j)}_{V_j^o} + \underbrace{{1 \over 2} \sum_{k \neq j}^n {e^2 \over |r_j - r_k|}}_{V_{jk}^{ee}} \right ] (\psi_1, \ \cdots, \ \psi_q, \ \cdots, \psi_n)\right ] - \delta \psi_q^* \varepsilon_q \psi_q \end{align} $$

For the $$T_j $$ and $$V_j^o $$ terms, there exists two conditions:

 * 1) When $$j \neq q $$, the terms inside the summation become $$math $$, but integrating this will go to zero because $$\delta \psi_q^* $$ and $$\psi_q $$ are orthogonal.
 * 2) The only non-zero terms come about for $$j = q $$

Thus, dividing through by $$\Psi^* \Psi $$ gives us $$\delta \psi_q^* (T_q + V_q^o) \psi_q $$. Here, the $$V_{jk}^{ee} $$ term is again zero for $$j \neq q $$, because the orthogonal functions come out from $$\delta \psi_q^* \psi_q $$. The resulting term is:

$$\delta \psi_q^* \left (    {1 \over 2} \sum _{k \neq q} \psi_k^* {e^2 \over |r_q - r_k|} \psi_k \right ) \psi_q $$

Because $$V_{jk}^{ee} $$ depends on both $$j_q $$ and $$k $$, we retain the wavefunctions $$\psi_k $$ inside the summation.

The functional variation now looks like:

$$\begin{align} \delta I &= \int \sum_{q=1}^n \delta \psi_q^* (T_q + V_q^o) \psi_q + \delta \psi_q^* \left ( {1 \over 2} \sum_{k \neq q} \psi_k^* {e^2 \over |r_q - r_k|} \psi_k \right ) \psi_q - \delta \psi_q^* \varepsilon_q \psi_q \\ &=0 \end{align} $$

Which can be solved as $$n $$ equations.

$$\left [ {- \hbar^2 \over 2m} \nabla^2 + V_o (r) + V_d(r) \right ] \psi_k(r) = \varepsilon_k \psi_k(r) $$

Where $$V_d $$ is the Hartree Term which is:   $$V_d = \sum_{j \neq k}^n \int \psi_j^* (r_j) \psi_j(r_j) {e^2 \over |r_j - r_k|} \ \operatorname{d}\!r $$

This looks like a Schrodinger equation, but is it? Yes, sort of... $$\psi_i $$ are Hartree wave functions. Remember that these are single electron $$\psi_i $$ since we started with $$H = \sum_i h_i $$. $$\varepsilon_i $$ are Lagrangian multipliers for enforce normalization. However, we can use these $$\{ \psi_i \} $$ and $$\varepsilon_i $$ to study the system.

Hartree Equation
The Hartree equation has a physically intuitive form (if anything in quantum mechanics is physically intuitive). THe first term is kinetic energy, the second term is the electron-ion or any external potential energy (such as E or B), and the third term is the electron-electron potential energy defined by the sum:

$$\sum_{j \neq k} \int \underbrace{\psi^*_j(r_j) \psi_j(r_j)}_{\rho_j} {e^2 \over |r_j - r_k|} \ \operatorname{d}\!r$$

Here $$\rho_j$$ is the charge density of the $$j^{th}$$ electron. So this sum can be thought of as an electron at $$r_k$$ interacting with some charge at $$r_j$$.

What's wrong with this picture? Well, excluding our single particle assumption, we're putting several electrons into this system, which are fermions, but our wavefunction is not totally anti-symmetric. What we need to do is write $$\Psi$$ using a Slater determinant.

$$\begin{align} \Psi &= {1 \over \sqrt{n!}} \left | \begin{matrix} \psi_1(r_1) & \cdots & \psi_n(r_1) \\ \cdots \\ \psi_1(r_n) & \cdots & \psi_n(r_n) \end{matrix} \right | \\ &= {1 \over \sqrt{n!}} \sum_{\rho} (-1)^\rho \psi_1(r_1) \ \cdots \ \psi_n(r_n) \end{align}$$

Put this back into our original energy function and apply calculus of variation methods to the whole expression to find our answer. This was done by Fock (and Slater) in 1930. The resulting set of expressions is exactly the same except for one additional term. $$\left [ {-\hbar^2 \over 2m} \nabla^2 + V_o(r) + V_d(r) + V_{\text{exchange}}(r) \right ] \psi_k(r) = \varepsilon_k \psi_k(r)$$

The evaluation of $$V_{\text{exchange}}$$ is quite tricky. The expression for the expectation value is:$$\langle E_{\text{exchange}} \rangle = {1 \over 2} \sum_{1\neq j} \int \int \psi_i^*(r_i) \psi_j^*(r_j) {e^2 \over |r_i - r_j|} \underbrace{\psi_i(r_j) \psi_j(r_i)}_{\operatorname{d}\!{r_i} \operatorname{d}\!{r_j}} $$

In many cases it is acceptable to approximate using the so-called "free electron exchange" $V_{\text{exchange}}(r) = -e^2 \left ( {3 \over \pi} \right )^{1 \over 3} \rho(r)^{1 \over 3}$

Self-Consistent Field Loop
These equations are not trivial to solve. Notice that the operator depends on $$\rho(r)$$, but to get $$\rho(r)$$ you need $$\psi_i(r)$$ which requires solving the set of equations. This is where we solve by iterative, self-consistent methods. This technique is the basis for all of modern quantum mechanical methods.

Write $(T + V(r)) \psi_k(r_k) = \varepsilon_k \psi_k(r_k)$, where $V(r)$ is all the terms, not $$T$$. It will depend on $$\rho(r)$$ or $$\psi(r)$$ or both.

 * 1) Guess $V^{(1)}(r)$ . You can use known solutions, random guesses, the Thomas-Fermi, or any other method of guessing you wish.
 * 2) Solve the system of equations to find $$\{\psi_i^{(1)}(r)\}$$.
 * 3) Calculate  from $$\{\psi_i^{(1)}(r)\}$$.
 * 4) If $V^{(2)} \simeq V^{(1)}$  end. Otherwise return to step 2 and keep looping until $V^{(n)} \simeq V^{(n+1)}$  to within a given tolerance.

This is called a self-consistent field loop (S.C.F.).

These type of SCF methods can be applied to all sorts of systems, but since this chapter is about many-electron problems, we will look at the Herman-Skillman atomic data, calculated from the Hartree-Fock method in 1963.

 "Herman-Skillman" atomic data"


text:
Electronic Properties of Materials/Quantum Mechanics for Engineers/Density Functional Theory

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


text:
User:Whocare888/sandbox

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title: User:Whocare888/sandbox