# Studying quantum mechanics: From atomic energy levels to bands in solids

Vinay Tumuluru

Sardar Vallabhbhai National Institute of Technology (SVNIT), Surat 395007

Jan-e Alam

Variable Energy Cyclotron Centre (VECC), Kolkata 700064

## Abstract

Quantum Physics deals with the study of the microscopic particles which build the universe. The elementary particles and the interaction among them can help to unravel the mysteries of our universe. The study of quantum physics begins with a mathematical construct, where the classical dynamical variables are made into operators. In quantum mechanics, the Schrödinger differential equation describes the dynamics of particles. The solution of the Schrödinger equation gives the wave function for given interactions and boundary conditions. The Schrödinger equation has been solved for the case of hydrogen atom, Positronium, and one dimensional harmonic oscillator. The wave function and energy eigenvalues for these systems have been obtained. The degeneracy of the hydrogen atom due to underlying symmetry has been studied. The removal of degeneracy owing to relativistic and spin-orbit interactions has also been investigated. The one-dimensional harmonic oscillator problem has been solved and the physics of zero-point energy has been understood. The origin of band formation in solids has been investigated as energy levels of several atoms begin to overlap.

Keywords: schrödinger equation, wave function, fine structure, hydrogen atoms, harmonic oscillator, enrgy bands

## Abbreviations

Enlisted are a few abbreviations that have been used in the report :

Abbreviations
 QM Quantum Mechanics CM Classical Mechanics $e^-$ Electron eqn. Equation K. E. Kinetic Energy LRL Laplace Runge Lenz Vector fs Fine Structure Ps Positronium

## Background/Rationale

Phenomena like the Black Body Radiation, Photoelectric Effect and several others cannot be simply explained by the postulates of classical mechanics, which invariably hint at the inadequacy of CM. We therefore need a new theory that would account for all the facts noted from these phenomena. Quantum Mechanics addresses this need.

The application of CM is restricted to physical objects encountered in day-to-day life, while an application at the microscopic level yields incorrect results. With the advent of QM, the microscopic phenomena could be easily explained. The differing feature of QM from CM lies in the fact that physical quantities such as energy, momentum, angular momentum etc. are available in discrete quantitites (a phenomenon called quantisation). QM also poses a restriction on measuring certain quantities (such as position and velocity (or momentum)) simultaneously; as seen from the Uncertainty Principle.

The symmetry of physical systems is also of wide interest to physicists for it helps simplify many problems. The symmetry in hydrogen atom is of particular interest as there exists an accidental symmetry which accounts for the degeneracy in energy levels.

These phenomena are seen for individual systems such as atoms. However, when several atoms are brought close together for formation of a solid, the energy levels of all atoms are affected. The nature of the force of interaction between the constituent atoms and the type of atoms determine the physical properties of the solid such as density, tensile strength, elasticity, electrical behaviour etc.

The concept of temperature, pressure and volume, which has no meaning for isolated atoms, now has a meaning for solids. This report shall focus on the need for QM, how it explains mathematically the energy levels of atoms, and how eventually these energy levels are affected during the formation of solids.

## Statement of the Problems

• BACKGROUND: We are mostly surrounded by solids and not simple isolated atoms. Hence, a study of the formation of these solids has to be systematically investigated starting from the energy levels of isolated atoms (microscopic) and going all the way up to bulky solids (macroscopic).
• RATIONALE: As we interact with solids on a large scale in our daily lives, it is essential to know the properties of these solids. This can gradually be extended to higher studies focussing on the harnessing of these properties for use to the common man.

## Objectives of the Research

This report aims to study the properties and energy levels of isolated atoms. The same can be studied with the help of QM. How energy bands in solids are formed from the constituent atoms shall also be thoroughly investigated.

## Scope

As solids are the most familiar phase of matter, it is imperative to say that the study of the same is of huge importance. Quantum Physics and Solid State Physics are two of the most widely researched fields with immense scope for future and their enormous potential for producing new devices, techniques, technologies etc. for the betterment of mankind.

## Information

This report has consulted the works of several authors; a credit to their work is duly provided at the references section. Several researches have been carried out in the field of QM and Solid State Physics. Many physicists have proposed their theories to explain the features observed experimentally. The well accepted theories have been studied and suitably integrated to understand how solids are formed, right from the atomic level.

## Failure of classical mechanics; the need forquantum mechanics

A simple experiment of electron diffraction through two slits (considered homogeneous and rarefied beam) through crystal produces alternating maxima and minima. A classical explanation of the phenomenon would expect a simple superposition of the beams from each slit. This is incorrect as observed experimentally. Several such experiments confirm the failure of classical theory.

CM deals with the behaviour of macroscopic particles, and introduces the concept of ‘path of the particle’. QM is not independently constructed; objects that obey QM are assumed to obey CM to a sufficient degree of accuracy.

In QM, classical objects (not necessarily macroscopic), are called apparatus, while their interaction with any quantum object is called measurement. This interaction occurs independently of the observer. The measuring process, however affects the quantum object (such as an $e^-$) subject to it. It is impossible to make this effect arbitrarily small; as the dynamical characteristics of the $e^-$ appear only as a result of the measurement itself. If the effect of measuring process can be minimised, then the measured quantity would be independent of the measurement.

Of various measurements that can be made, the most common one is the coordinates of an $e^-$. When measured with a low degree of accuracy, a fairly smooth path is obtained; else a discontinuous and disordered variation of results are observed. If the time interval between successive measurements ∆t is made very small, (i. e. ∆t → 0); the successive coordinates are more irregularly distributed.

We can thus conclude the simultaneous inexistence of coordinates and velocity as an does not possess a definite path. The problem in QM involves determining the probability of a given result of measurement. Completely described states occur as a simultaneous measurement of a complete set of physical quantities. From the results of such measurement, we can then obtain the probability of various results of any subsequent measurement, regardless of the prior to the first measurement.

• PHOTOELECTRIC EFFECT

Before proceeding further, two experiments that irrevocably prove the failure of classical mechanics are outlined below :

1. Wave theory requires that the oscillating electric field vector E of light waves increase in amplitude as the intensity of light beam is increased. As the force applied on the electron is equal to $-e\mathbf{E}$, this implies that an increase in intensity of the light beam should consequently increase the kinetic energy of the ejected electrons. On the contrary, the kinetic energy of these electrons is found to be independent of the intensity of incident radiation.

2. As the K. E. of the ejected electrons is only intensity – dependent, the phenomenon should be invariant to the frequency of the incident light, assuming the intensity is enough to eject the electrons from the metal. It has however been experimentally established that there exists a cut-off frequency, called the threshold frequency ${\nu }_{0}$, below which no photoelectric effect takes place.

3. According to the wave theory, there is a measurable time lag between the time when the electron strikes the surface and the ejection of the photoelectrons. But no detectable time lag exists.

These observations can be explained on the basis of the fact that the energy of an electromagnetic wave is given by $E=hv$. Since the wave is continuous, the eletron emission should also be continuous.

In reality, the electrons in the metal surface are continuously absorbing energy from external electromagnetic radiations which are present even if there is no external light source present for irradiation, as we can never attain completely dark surroundings. Therefore, the loosely bound electrons get continuously energised, until they receive enough energy to get ejected from the metal surface. As it is impossible to know when the electron starts absorbing energy, there is no way to detect the said time lag.

An object with a temperature above absolute zero emits light of all wavelengths. If the object is perfectly black, (so that it does not reflect any light), then the light that comes from it is called black body radiation. The energy of blackbody radiation is not shared evenly by all  wavelengths of light. The spectrum of the black body radiation shows that some wavelengths get more energies than the rest.

The black body spectrum depends only on the temperature of the object, and not on the nature of its material at a particular temperature. Also, as the temperature of the metal increases, it emits more radiations as seen from the curve. With an increase in temperature, the peak of the curve becomes shorter (or equivalently, bluer). For example, blue stars are hotter than red stars.

As per classical mechanics, light is an electromagnetic wave that is produced when an electric charge oscillates. As heat is nothing but kinetic energy of molecules at random motion, in a hot object, electrons vibrate in random directions and produce light. More light is emitted by hotter objects as a consequence of stronger vibrations. But the spectrum cannot be explained by classical mechanics.

The electrons in a hot object can vibrate with a range of frequencies, ranging from a very few vibrations per second to a huge number of vibrations in a second. There is no limit to the range of frequencies and consequently, there is no limit to the energy of the oscillators. However, there is a maximum characteristic energy for a particular temperature that any oscillator can have. Such an observation cannot be explained by classical mechanics. This was resolved by Max Planck in the beginning of the twentieth century, by introducing the idea that energy came in discrete packets and not in a continuum.

## The Schrödinger Equation

A transition from CM to QM involves the conversion of dynamical physical variables to operators, which are represented by a hat (^) over the term. For instance, the momentum operator is given by (in three dimensions with $\hat x, \hat y, \hat z$ being unit vectors along the x, y and z axes respectively.)

$\widehat p\rightarrow-i\hslash\left(\hat x\frac{\displaystyle\partial}{\displaystyle\partial x}+\hat y\frac{\displaystyle\partial}{\displaystyle\partial y}+\hat z\frac{\displaystyle\partial}{\displaystyle\partial z}\right)$

The Hamiltonian operator whose eigenvalue gives the total energy of the quantum object, for a free particle is given by the relation

$\widehat H=-\displaystyle\frac{\hslash^2}{2m}\left(\frac{\displaystyle\partial^2}{\displaystyle\partial x^2}+\frac{\displaystyle\partial^2}{\displaystyle\partial y^2}+\frac{\displaystyle\partial^2}{\displaystyle\partial z^2}\right)=-\frac{\hslash^2}{2m}\nabla^2$

where m is the mass of the particle, $i=\sqrt{-1}$, and $\hslash=h/2\pi$where $h$ is the Planck's constant given by $6\cdot626\times10^{-34}\mathrm{ Js}$. Also, $\nabla^2\equiv\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$ is called the Laplacian operator.

If $V(x, y,z)$ denotes the potential, then the Hamiltonian will be given by $\widehat H=-\frac{\hslash^2}{2m}\nabla^2+V(x,y,z)$. If E denotes the total energy of the particle, which is the eigenvalue of the Hamiltonian operator, then for stationary states we have : $\hat H\psi=E\psi$, where $\psi$ denotes the wavefunction of the particle.

From the above equations, we have

$\boxed{\nabla^2\psi+\displaystyle\frac{2m}{\hslash^2}(E-V)\psi=0}$

The above equation is called the Time Independent Schrödinger Equation. Further, the Hamiltonian operator is given by : $\hat{H}=i\hslash\frac{\partial}{\partial t}.$Incorporarting the same in the above equation with some rearrangements, we obtain what is called the Time Dependent Schrödinger Equation :

$\boxed{i\hslash\displaystyle{\frac{\partial \psi}{\partial t}}=\displaystyle-\frac{\hslash^2}{2m}\nabla^2\psi+V(x,y,z)\psi}$

## Mathematical introduction to quantum mechanics

Linear Vector Spaces

A linear vector space $\mathrm{\textbf{V}}$ is a collection of objects $|1\rangle$, $|2\rangle$ , . . . , $|V\rangle$ , . . . , $|W\rangle$ , . . . , called vectors, for which there exists :

1.   A definite rule for forming the vector sum, denoted $|V\rangle + |W\rangle$ .

2.   A definite rule for multiplication by scalars a, b, . . . denoted by $a|V\rangle$ with the features : (a) scalar multiplication is associative, distributive, while addition is associative, commutative and distributive. Also, there exists null vector $|0\rangle$ such that $|V\rangle + |0\rangle=|V\rangle$. Finally, for every vector $|V\rangle$ there exists an inverse vector under addition $|-V\rangle$ such that $|V\rangle + |-V\rangle=|0\rangle$. ​

The numbers a, b, . . . are called the field over which the vector space is defined. If these numbers are real, then we have a Real Vector Space. Else if they are complex, then we have a Complex Vector Space.

Linear Dependence of Vector sets

Consider the relation

$\displaystyle\sum_{i = 1}^{n} a_{i} |i\rangle = |0\rangle$

We assume that the L.H.S. does not contain any $|0\rangle$ term; if it did, it can be taken to the R.H.S. The set of vectors that satisfy this relation are called Linearly Independent Vectors. Commonly, all null vectors are simply represented by 0.

A vector space has a dimension n if it can accommodate a maximum of n linearly independent vectors. It will be denoted by $\mathrm{\textbf{V}}^n(R)$ if the field is real; else it is represented by $\mathrm{\textbf{V}}^n(C)$ if it is complex.

Any vector $|V\rangle$ in an n-dimensional space can always be written as a linear combination of n linearly independent vectors $|1\rangle$, $|2\rangle$ , . . . , $|n\rangle$. A set of n linearly independent vectors in an n dimensional space is called a basis. Therefore, any vector $|V\rangle$ can always be written as

$|V\rangle=\displaystyle\sum_{i = 1}^{n} v_{i} |i\rangle$

where the vectors $|i\rangle$ form the basis. The coefficients $v_i$ of the vector $|V\rangle$ in terms of a linearly independent basis $|i\rangle$ are called components of the vector in that basis.

We assume that the L.H.S. does not contain any $|0\rangle$ term; if it did, it can be taken to the R.H.S. The set of vectors that satisfy this relation are called Linearly Independent Vectors. Commonly, all null vectors are simply represented by 0.

Matrices used as vectors do not have a length and a certain direction associated with them. Consider two vectors $\overrightarrow A$and $\overrightarrow B$. The inner product of these vectors is given by $\overrightarrow A . \overrightarrow B=AB \mathrm{cos} \theta$, where θ is the angle between them.

In the ket notation, the inner product of the two vectors $|V\rangle$and $|W\rangle$, is given by $\langle V|W\rangle.$It has the properties : (a) skew-symmetry $\langle V|W\rangle = \langle W|V\rangle^*$ (b) Positive semi= definiteness : $\langle V|V\rangle \geq 0$ and (c) Linearity in ket : $\langle V|a|W\rangle + b|Z\rangle \rangle = a\langle V|W\rangle + b\langle V|Z\rangle$. Using the definition of vectors, we have

$\langle V|W\rangle = \displaystyle \sum_{i} \sum_{j} v_i^* w_j \langle i|j\rangle$

## Fundamentals of classical mechanics

The Euler-Lagrange Equation

Consider a system of particles with n degrees of freedom, whose generalized coordinates and velocities are, respectively q and $\dot q$, characterised by the Lagrangian $L (q,\dot q,t)$, where q is short-hand for $q_1(t), q_2(t),... , q_n(t)$ with the dot representing the total time derivative; $\dot q = \frac{\mathrm{d}q}{\mathrm{d}t}$. From the principle of least action, the motion of the system from $t_1$ to $t_2$ is such that the action integral is :

$\displaystyle\int_{t_1}^{t_2}L\;(q,\dot q,t) \mathrm{d}t=0$

is a minimum for the path q (t) of motion. In other words, the variation $\delta S$ is zero for this path :

$\delta S=\displaystyle\int_{t_1}^{t_2}\delta L\;(q,\dot q,t) \mathrm{d}t$

Using the variation of Lagrangian and some mathematical steps, we obtain the Euler Lagrangian Equation, given as

$\boxed {\displaystyle \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial L}{\partial \dot q}\right)-\frac{\partial L}{\partial q}=0}$

Noether's Theorem

If the action of a given system is invariant under the infinitesimal transformation that changes q to δq, then corresponding to the transformation, there exists a conserved quantity, and this conserved quantity (called current) is J, and can be obtained from the Lagrangian and the infinitesimal transformation, and using the above Euler-Lagrange Equation :

$\boxed{\displaystyle J=\frac{\partial L}{\partial\dot q}\delta q - F}$​​

where $\delta L=\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot q}\delta \dot q = \frac{\mathrm{d}F}{\mathrm{d}t}$. The quantity $J$ is called the Noether's current.

## The hydrogen atom

Faliure of Classical Mechanics in the Explanation of Stability of the Hydrogen Atom

Hydrogen atom consists of an electron of mass m and charge rotating around a heavy proton of mass M, and charge in a circular orbit of radius R with a speed v. For this circular orbit, the centrifugal force should equal the Coloumbic force of attraction :

$\displaystyle \frac{mv^2}{R}=\frac{e^2}{4\pi\epsilon_0R^2}$

The kinetic and potential energies are respectively given by $E_{\mathrm{kin}}=\frac{1}{2}mv^2$ and $E_{\mathrm{pot}}=-\frac{e^2}{4\pi\epsilon_0 R}$. The period of rotation is given by

$\displaystyle T=\frac{2\pi R}{v}=\frac{4\pi^{3/2}R}{|e|}\sqrt{\epsilon_0mR}$

But this rotating electron radiates power, given by the Larmor's formula $P=\mu_0\frac{e^2R^2}{6\pi c}$, where c is the speed of light in vacuum, and $\mu_0$ is the permeability of free space. Let us suppose that the loss of energy due to radiation over one period is much smaller than the total energy; then we can consider $R \equiv R(t)$ as a slowly varying function of time t. After one revolution, the energy radiated is :

$\displaystyle PT=\mu_0\frac{e^2R^2}{6\pi c}T=\mu_0\frac{e^2}{6\pi c}\frac{v^2}{R^4}T$

The change of $E=E_{\mathrm{kin}}+E_{\mathrm{pot}}$ is $E(R(t+T))-E(R(t))=-\frac{e^2}{8\pi\epsilon_0R(t+T)}+\frac{e^2}{8\pi\epsilon_0R(t)}$. Since $R(t+T)=R(t)+T\frac{\mathrm{d}R(t)}{\mathrm{d}t}$, we have

$\displaystyle E(R(t+T))-E(R(t)) \approx -\frac{e^2T}{8\pi\epsilon_0R^2(t)}\frac{\mathrm{d}R(t)}{\mathrm{d}t}$

We therefore obtain a differential equation $\frac{\mathrm{d}R(t)}{\mathrm{d}t}=\frac{\mu_0ce^4}{12\pi^2m^2R^2(t)}$. We integrate both sides and use the condition $R=R_0$ at $t=0$, where $R_0$is the Bohr radius. Say at time $t=\tau$, the electron spirals and falls into the nucleus, i.e. $R(\tau)=0$, we have

$\displaystyle \tau=\frac{4\pi^2m^2}{\mu_0ce^4}R_0^3$

Making necessary substitutions, we have

$\boxed {\tau=1\cdot3\times 10^{-11}\mathrm{ sec}}$

According to Maxwell theory of radiation, a continuously accelerating electron (which is in circular motion) must continuously radiate energy. After expending all its energy, it would finally fall into the nucleus following a spiral path. The lifetime of the hydrogen atom has been computed as seen above.

On the contrary, hydrogen atom has been found stable. The classical theory thus fails and consequently calls for an alternative explanation. This problem has been resolved by quantum mechanics, as discussed in the subsequent sections

Mathematical Formulation for the Hydrogen Atom

As seen earlier, the hydrogen atom is a two body problem concerning an electron of charge $-e$ and mass m, along with a proton of charge $+e$ and mass M. The reduced mass $\mu$ shall therefore be given by :

$\displaystyle \mu=\frac{mM}{m+M}$

As $\frac{m}{M}\approx \frac{1}{2000}$, we have $\mu \approx m$, which is nothing but the mass of the electron. The Schrödinger equation is now given by $i\hslash\frac{\partial \psi}{\partial t}=\hat H \psi$, where $\hat H$is the Hamiltonian operator. Referring to section 2.2.2, we can write the time independent Schrödinger equations as $E\psi=-\frac{\hslash^2}{2m}\nabla^2\psi+V\psi$. Typically, the potential V is a function of the distance from origin only. It is therefore convenient to adopt the spherical coordinates in the expansion of the Laplacian operator :

$\displaystyle \nabla^2\equiv\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)+\frac{1}{r^2\mathrm{sin} \theta}\frac{\partial}{\partial \theta}\left(\mathrm{sin} \theta\frac{\partial}{\partial \theta}\right)+\frac{1}{r^2\mathrm{sin^2} \theta}\left(\frac{\partial^2}{\partial \phi^2}\right)$

We can incorporate the same in the Schrödinger equation. Then, we use the method of Separation of Variables to solve the resultant equation.

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

Making the above substitution in the Schrödinger equation, dividing by RY throughout and further dividing by $-2mr^2/\hslash^2$, we obtain an equation having two parts, one of which depends only on r while the other contains θ and ϕ. Since they add up to zero, they must each be equal to the same constant (and with opposite signs, of course). Therefore,

1. $\displaystyle \frac{1}{R}\frac{\mathrm{d}}{\mathrm{d}r}\left(r^2\frac{\mathrm{d}}{\mathrm{d}r}\right)-\frac{2mr^2}{\hslash^2}\left[V(r)-E\right]=l(l+1)$

2. $\displaystyle \frac{1}{\mathrm{sin} \theta}\frac{\partial}{\partial \theta}\left(\mathrm{sin} \theta\frac{\partial Y}{\partial \theta}\right)+\frac{1}{\mathrm{sin^2} \theta}\left(\frac{\partial^2 Y}{\partial \phi^2}\right)=-l(l+1)$

where $l$ is a constant, and so is $l(l+1)$; called the separation constant. Eqn. 1 is the radial equation, while eqn. 2 is called the angular equation.

SOLVING THE ANGULAR EQUATION :

Eqn. 2; the angular equation may be re-written as :

$\displaystyle \mathrm{sin} \;\theta \; \frac{\partial}{\partial \theta} \left(\mathrm{sin}\; \theta \; \frac{\partial Y}{\partial \theta} \right)+\frac{\partial^2Y}{\partial \phi^2}=-l(l+1)\;\mathrm{sin}^2\theta \; Y$

Again, using the method of separation of variables, we write $Y(\theta,\phi)=\Theta(\theta) \Phi(\phi)$. Making this substitution in the above eqn., followed by division with $\Theta \Phi$, we can separate the parts containing Θ and Φ, and as done earlier, we equate each to a separation constant, $m^2$.

1. $\displaystyle \frac{1}{\Theta} \left \{\frac{\mathrm{d}}{\mathrm{d} \theta} \left( \mathrm{sin} \theta \frac{\mathrm{d}\Theta}{\mathrm{d}\theta} \right) \right \} + l(l+1) \mathrm{sin^2 }\theta = m^2$

2. $\displaystyle \frac{1}{\Phi} \; \frac{\mathrm{d^2}\Phi}{\mathrm{d}\phi^2}=-m^2$

A thing to note here is that m does not denote the mass of the particle! What m denotes has to be understood from the context. We find that eqn. 2 is an ordinary differential equation of second order that can easily be solved. The solution is $\Phi (\phi)=e^{im\phi}=\mathrm{cos} m\phi + \mathrm{sin} m\phi$, where m is either positive or negative. When Φ advances by 2π, we find that the wavefunction remains invariant due to spherical symmetry; $\Phi (\phi + 2\pi) = \Phi (\phi)$. We therefore conclude that $m=\pm1, \pm2,\pm 3,...$.

The equation in θ is : $\mathrm{sin} \theta \frac{\mathrm{d}}{\mathrm{d} \theta} \left( \mathrm{sin} \theta \frac{\mathrm{d}\Theta}{\mathrm{d}\theta} \right) + \left[ l(l+1) \mathrm{sin^2 }\theta - m^2 \right]\Theta=0$. The solution to this differential eqn. is given by : $A=P_l^m (\mathrm{cos}\; \theta)$, where ${P}_{l}^{m}$called the associated Laguerre Polynomial is defined as

$\displaystyle P_l^m \equiv \left(1-x^2 \right)^{|m|/2} \left(\frac{\mathrm{d}}{\mathrm{d}x} \right)^{|m|} P_l (x)$

with $P_l (x) = \frac{1}{2^l l!} \left(\frac{\mathrm{d}}{\mathrm{d}x} \right)^l \left(x^2-1 \right)^l$ called the Laguerre Polynomial. We may separately normalise the radial and angular parts of the wavefunction to obtain the total solution. Normalising the angular part, we obtain what are called the spherical harmonics :

$\boxed{Y_l^m (\theta, \phi)=\epsilon \sqrt{\frac{(2l+1)}{4\pi} \frac{(l-|m|)!}{(l+|m|)!}}\;e^{im\phi} P_l^m (\mathrm{cos} \theta)}$

where $\epsilon = (-1)^m$ for all $m \geqslant 0$, and $\epsilon = 1$ for all $m \leqslant 0$. The above eqn. is referred to as spherical harmonics because they are defined over the surface of a sphere, with polar angle θ and azimuth angle ϕ.

l is called the azimuthal quantum number, while m is called the magnetic quantum number. Further, since factorials are defined only for natural numbers, it follows that $- l\leqslant m\leqslant l$.

The radial equation may be written as $\frac{\mathrm{d}}{\mathrm{d}r} \left(r^2 \frac{\mathrm{d}R}{\mathrm{d}r} \right)-\frac{2mR^2}{\hslash^2} \left[V(r)-E \right]R=l(l+1)R$. Changing variables, $u(r)=rR(r)$ we find that $R=\frac{u}{r}; \; \frac{\mathrm{d}R}{\mathrm{d}r}=\left[r\frac{\mathrm{d}u}{\mathrm{d}r}-u \right]\frac{1}{R^2}$. Making the necessary substitutions, we have

$\displaystyle -\frac{\hslash^2}{2m}\frac{\mathrm{d^2}u}{\mathrm{d}r^2}+\underbrace{\left[V+\frac{\hslash^2}{2m}\frac{l(l+1)}{r^2} \right]}_{V_{\mathrm{eff}}}u=Eu$

Comparing the same to the one-dimensional Schrödinger equation, we find that the effective potential ${V}_{\text{eff}}$is given by

$\boxed {V_{\mathrm{eff}}=V+\frac{\hslash^2}{2m}\frac{l(l+1)}{r^2}}$

The term $\frac{\hslash^2}{2m}\frac{l(l+1)}{r^2}$ is called the centrifugal term; This tends to throw the particle outward, i. e. away from the origin.

SOLVING THE HYDROGEN ATOM PROBLEM

The (Coloumb's) potential V for a heavy, motionless $e^-$ is given by

$\displaystyle V(r)=-\frac{e^2}{4\pi \epsilon_0}\frac{1}{r}$

Making this substitution in the radial equation, we have

$\displaystyle -\frac{\hslash^2}{2m} \frac{\mathrm{d^2}u}{\mathrm{d}r^2} + \left[-\frac{e^2}{4\pi \epsilon_0} \frac{1}{r}+ \frac{\hslash^2}{2m}\frac{l(l+1)}{r^2} \right]u=Eu$

We need to solve this eqn. in order to obtain $u(r)$, as well as the allowed energies. Let us denote $K \equiv \frac{\sqrt{-2mE}}{\hslash}$(for bound states, E is negative; consequently K is real)

$\displaystyle \frac{1}{K^2} \frac{\mathrm{d^2}u}{\mathrm{d}r^2} =\left[1-\frac{me^2 }{2\pi \epsilon_0 \hslash^2 K} \frac{1}{Kr}+\frac{l(l+1)}{(Kr)^2} \right]u$

Defining $\rho = Kr$ and $\rho_0 = \frac{me^2}{2\pi \epsilon_0 \hslash^2 K}$, we have

$\displaystyle \frac{\mathrm{d^2}u}{\mathrm{d}\rho^2} = \left(1-\frac{\rho}{\rho_0}+\frac{l(l+1)}{\rho^2} \right)u$

Now, as $\rho \rightarrow \infin$, we find that $\frac{\mathrm{d^2}u}{\mathrm{d}\rho}=u$(the constant term dominates). The general solution to the above equation is given by $Ae^{-\rho} + Be^{\rho}$. But as $\rho \rightarrow \infin$, the term $e^\rho$ blows up. As a necessary consequence, $B=0$. Therefore, for larger values of $\rho$, we have $u(\rho) \sim Ae^{-\rho}$. On the other hand, as $\rho \rightarrow 0$, the centrifugal term dominates. Therefore, we find $\frac{\mathrm{d^2}u}{\mathrm{d}\rho^2}=\frac{l(l+1)}{\rho^2}u$.

The general solution therefore is $u(\rho) = C\rho^{l+1}+D\rho^{-l}$. But as $\rho \rightarrow 0$, $\rho^{-l}$ blows up. We therefore conclude $D=0$. For small $\rho$, $u(\rho)\sim C\rho^{l+1}$. We next try to eliminate the asymptotic behaviour by introducing the new function $v(\rho)$: $u(\rho)=\rho^{l+1}e^{-\rho}v(\rho)$. Making this substitution in the above eqn., we have

$\displaystyle \rho \frac{\mathrm{d^2}v}{\mathrm{d}\rho^2}+2(l+1-\rho) \frac{\mathrm{d}v}{\mathrm{d}\rho}+\left[\rho_0-2(l+1) \right]v=0$

The above equation has a series solution of the form :

$\displaystyle v(\rho)=\sum_{j = 0}^{\infin}c_j \rho^{\;j}$

Differentiating the above eqn. with respect to ρ, we find

$\left.\begin{array}{r}\displaystyle \frac{\mathrm{d}v}{\mathrm{d}\rho}=\sum_{j = 0}^{\infin} jc_{j}\rho^{ j-1}=\sum_{j = 0}^{\infin} (j+1)c_{j+1} \rho^{ j}\\\displaystyle \frac{\mathrm{d^2}v}{\mathrm{d}\rho^2}=\sum_{j = 0}^{\infin} j(j+1)c_{j+1}\rho^{ j-1}\end{array}\right\}$

Substituting these expressions for v, $\frac{\mathrm{d}v}{\mathrm{d}\rho}$ and $\frac{\mathrm{d^2}v}{\mathrm{d}\rho^2}$ in the differential equation of ρ, and further equating the like powers of ρ, we obtain the recurrence relation :

$j(j+1)c_{j+1}+2(l+1)(j+1)c_{j+1}-2jc_j (l+1)(j+1)c_{j+1}+\left[\rho_0 - 2(l+1) \right]c_j = 0$

Rewriting the above, we finally obtain

$\displaystyle c_{j+1}= \left\{\frac{2(j+l+1)-\rho_0}{(j+1)(j+2l+2)} \right\}c_j$

We may begin with $c_0$, and then determine the remaining coefficients by normalising each. For large values of j, $\frac{c_{j+1}}{c_j}\approx \frac{2}{j + 1}$. This implies $c_{j+1}=\frac{2^j}{j!}c_0$. Therefore, we find $v(\rho)=c_0 \sum_{j=0}^{\infin} \frac{2^j}{j!} \rho^{ j} = c_0 e^{2\rho}$. We can now obtain $u(\rho)$ as $u(\rho) = c_0 \rho^{l+1} e^\rho$. But this indicates that $u(\rho)$ must blow up at large values of ρ, which is undesirable. Therefore, the series (solution) obtained for u (ρ) must terminate. This puts an upper limit for the expansion coefficients : $c_{j_{\mathrm{max}}}$. All higher coefficients should be zero. This is ensured by setting $c_{j_{\mathrm{max}}+1}=0$. This impliexs :

$2(j_{\mathrm{max}}+l+1)-\rho_0=0$

Denote $j_{\mathrm{max}}+l+1=n$. The term n is referred to as the principal quantum number. It follows that $\rho_0 = 2n$. Rewriting, we have $j_{\mathrm{max}} = n-l-1$. As $j_{\mathrm{max}} \geqslant 0$, it follows that $l \leqslant n-1$.

As $K \equiv \frac{\sqrt{-2mE}}{\hslash}$, we can finally write E in terms of K and also $\rho_0$.

$\displaystyle E = - \frac{\hslash^2 K^2}{2m} = - \frac{me^4}{8\pi^2 \epsilon_{0}^{2} \hslash^2 \rho_{0}^{2}}$

Therefore, the allowed energies are given by

$\boxed {E_n = - \left(\frac{m}{2\hslash^2} \left(\frac{e^2}{4\pi \epsilon_0} \right)^2 \right) \frac{1}{n^2} = \frac{E_1}{n^2}; n=1,2,3 ...}$

where $E_1$ denotes the energy for $n = 1$. The above relation is called the Bohr's Formula. Further, $K=\left(\frac{me^2}{4\pi \epsilon_{0}\hslash^2} \right)\frac{1}{n}\equiv\frac{1}{an}$, where a denotes the Bohr radius, given by

$\displaystyle a=\frac{4\pi\epsilon_0\hslash^2}{me^2}=0\cdot529\;\overset{\mathrm{o}}\mathrm{A}$

We find $R_{nl}(r)=\frac{1}{r}\rho^{l+1}e^{-\rho}v(\rho)$, where $v(\rho)$ is a polynomial of degree $j_{\mathrm{max}}=n-l-1$ in ρ, whose coefficients are given by

$\displaystyle c_{j+1}=\frac{2(j+l+1-n)}{(j+1)(j+2l+2)}c_j$

Also, the energy as seen above is given by

$\boxed {E_1=-\frac{m}{2\hslash^2}\left(\frac{e^2}{4\pi \epsilon_0} \right)^2 =- 13\cdot6 \mathrm{eV}}$

This $13\cdot6 \mathrm{eV}$ is called the binding energy of the hydrogen atom. We now need to determine the coefficient ${c}_{0}$. For this, we may evaluate $\psi_{nlm}$, and use the condition of normalisation :

$\psi_{100}(r,\theta,\phi)=R_{10}(r)Y_{0}^{0}(\theta, \phi)$

Using the definition of ρ and K, we can easily evaluate ${R}_{10}$ and ${Y}_{0}^{0}$ :

$\displaystyle R_{10}=\frac{c_0}{a} e^{-r/a}$ and $\displaystyle Y_{0}^{0}=\frac{1}{\sqrt{4\pi}}$

${c}_{0}$ is obtained by normalising as follows :

$\displaystyle \int_{0}^{\infin} \left|R_{10}(r) \right|^2r^2\mathrm{d}r=1$

​Making necessary substitutions, we find that $c_0 = \frac{2}{\sqrt{a}}$. Now, we can finally put all these together to obtain the expression for the wavefunction $\psi_{100}(r,\theta,\phi)=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}$.

We can obtain a general expression for $v(\rho)$ in terms of what are called the associated Laguerre Polynomials : $v(\rho) = L_{n-l-1}^{2l+1}(2\rho)$, where

$\displaystyle L_{q-p}^{p}(x) = (-1)^p \left(\frac{\mathrm{d}}{\mathrm{d}x} \right)^q L_q(x)$

where ${L}_{q}$ denotes the q th Laguerre Polynomials defined as : $L_q(x) = e^x \left(\frac{\mathrm{d}}{\mathrm{d}x} \right)^q \left(e^{-x}x^q \right)$. We can now finally write the wavefunction for the hydrogen atom as :

$\boxed{\psi_{nlm}(r,\theta,\phi)=\sqrt{\frac{8}{(na)^3}\frac{(n-l-1)!}{2n\left((n+l)! \right)^3}} e^{-r/na} \left(\frac{2r}{na} \right)^l \left[L_{n-l-1}^{2l+1}\left(\frac{2r}{na} \right) \right] Y_{l}^{m}(\theta, \phi) }$

The above relation is of immense importance in Quantum Mechanics and explains the stability of Hydrogen atom while the classical counterpart fails, as seen earlier in section of Failure of CM in explaining the stability of the hydrogen atom.

Further, we find that the energy depends only on n, and not on l and m, unlike the wave function which depends on all three of them. This is called degeneracy of the orbitals. In other words, for a given value of n, the enrgy values are the same regardless of the values of l and m.

Due to rotational symmetry, the energy levels are found to be independent of m. But the independence of energy levels on l is linked with a different and deeper symmetry, related to the conservation of a special vector called the Laplace Runge Lenz vector, discussed below.

ACCIDENTAL SYMMETRY - THE LAPLACE RUNGE LENZ (LRL) VECTOR

As seen earlier, the energy levels of the hydrogen atom are independent of m (due to rotational symmerty) as well as on l, due to conservation of a special vector called the Laplace Runge Lenz vector, which is defined by :

$\overrightarrow A = \overrightarrow p \times \overrightarrow L - mk\widehat r$

Central inverse square law force problem exhibits a great deal of symmetry. Conservation of the LRL vector implies a hidden symmetry not associated with any cyclic coordinate.

Consider a central force. This would imply $\frac{\mathrm{d}\overrightarrow p}{\mathrm{d}t}=f(r)\widehat r = f(r)\frac{\overrightarrow r}{r}$.

Post-multiplying by $\overrightarrow L$, we have :

$\displaystyle \frac{\mathrm{d}\overrightarrow p}{\mathrm{d}t}\times \overrightarrow L = f(r)\frac{\overrightarrow r}{r} \times \left(\overrightarrow r \times\overrightarrow p \right) = \frac{mf(r)}{r}\overrightarrow r \times \left(\overrightarrow r \times \frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} \right)$

$\displaystyle = \frac{mf(r)}{r} \left(\overrightarrow r \left(\overrightarrow r \cdot \frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} \right) - r^2 \frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} \right)$

Now, $\overrightarrow r \cdot \frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} = \frac{1}{2} \left(\frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} \cdot \overrightarrow r + \overrightarrow r \cdot \frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} \right) = \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \left(\overrightarrow r \cdot \overrightarrow r \right) = \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}r^2=r\frac{\mathrm{d}r}{\mathrm{d}t}$. Therefore,

$\displaystyle \frac{\mathrm{d}\overrightarrow p}{\mathrm{d}t}\times \overrightarrow L = \frac{mf(r)}{r} \left(\overrightarrow r r\frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} - r^2\frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} \right) = mf(r)r^3 \left(\frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} \frac{\overrightarrow r}{r^2} - \frac{1}{r}\frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} \right)$

Since $\overrightarrow L$ is a consereved quantity, we can write

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\overrightarrow p \times \overrightarrow L \right) = -mf(r)r^2 \left(\frac{1}{r} \frac{\mathrm{d}\overrightarrow r}{\mathrm{d}t} - \frac{\mathrm{d}r}{\mathrm{d}t} \frac{\overrightarrow r}{r^2} \right) = - mf(r)r^2 \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\overrightarrow r}{r} \right)$

If the force follows the inverse-square law, i.e. $f(r) = -k/r^2$, then

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\overrightarrow p \times \overrightarrow L \right) = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{mk\overrightarrow r}{r} \right) = \frac{\mathrm{d}}{\mathrm{d}t}\left(mk\widehat r \right)$

Rearranging, we find

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\overrightarrow p \times \overrightarrow L - mk\widehat r \right) = \frac{\mathrm{d}\overrightarrow A}{\mathrm{d}t} = 0$

Thus, we find that the LRL vector is conserved, as its time derivative is found to be equal to zero.

In a one body Kepler problem, there are 6 degrees of freedom. But now, we already have $E$ and $\overrightarrow L$, and now $\overrightarrow A$, suggesting that there are $1+3+3=7$ conserved quantities. Also in this problem, there are no mentions of the initial conditions of motion. So atmost, there are 5 degrees of freedom, that can be conserved. Therefore, there must be relations between $E$, $\overrightarrow A$ and $\overrightarrow L$.

It can be shown that

$\left.\begin{array}{r}\overrightarrow A \cdot \overrightarrow L = 0\\A^2 = m^2k^2 + 2mEL^2\end{array}\right\}$

So if we take $E$ and $\overrightarrow L$ as primary conserved quantities, we find that $\overrightarrow A$ gives an additional single degree of freedom and this appears as a consequence of the force following the inverse square law.

SPIN

There are certain quantum phenomena which cannot be explained with a direct application of the postulates of quantum mechanics. These involve an additional quantum degree of freedom called the spin. There is no classical analogue of spin. Most particles such as electrons, protons, neutrons, photons etc. have this spin.

A simple experiment in which an electron taken in a state of zero linear momentum (i.e. constant wave function) would predict that the angular momentum, say in z direction would expect that a zero result would be obtained, as ${L}_{x}$, ${L}_{y}$ and ${L}_{z}$ operators would give zero acting on the wave function. An experimental approach however, shows that the result is ±ћ/2.

Therefore, we find that an electron has an intrinsic angular momentum not associated with the orbital motion. This angular momentum is called spin.

In order to determine the expression of the spin angular momentum, we will first determine the quantisation relation of orbital angular momentum :

The K. E. of the $e^{-}$ has two parts, $\mathrm{KE}_{\mathrm{radial}}$, due to its motion toward/away from the nucleus and $\mathrm{KE}_{\mathrm{orbital}}$ the due to its motion around the nucleus. The potential energy U is given by

$\displaystyle U=-\frac{e^2}{4\pi \epsilon_0 r}$

If E denotes the total energy, then

$\displaystyle E=\mathrm{KE_{\mathrm{radial}}} + \mathrm{KE_{\mathrm{orbital}}} + U$

$\displaystyle =\mathrm{KE_{\mathrm{radial}}} + \mathrm{KE_{\mathrm{orbital}}} - \frac{e^2}{4\pi \epsilon_0 r}$

If we insert this value of E in the radial part R(r) of $\psi$, as in :

$\displaystyle \frac{1}{r^2} \frac{\mathrm{d}}{\mathrm{d}r} \left(r^2 \frac{\mathrm{d}R}{\mathrm{d}r} \right) + \left\{\frac{2m}{\hslash^2}\left(\frac{e^2}{4\pi \epsilon_0 r} + E \right) - \frac{l(l+1)}{r^2} \right\}R=0$

Then, we can set the value $\mathrm{KE_{\mathrm{orbital}}}$ of in the above equation in such a manner so that R(r) would exclusively be a function of r :

$\displaystyle \mathrm{KE_{\mathrm{orbital}}}=\frac{\hslash^2 l(l+1)}{2mr^2}$

Now, as $\mathrm{KE_{orbital}} = \frac{1}{2}mv_{\mathrm{orbital}}^{2}$ and $L=mv_{\mathrm{orbital}}r$, we can write the above as :

$\displaystyle \mathrm{KE_{orbital}}=\frac{L^2}{2mr^2}=\frac{\hslash^2 l(l+1)}{2mr^2}$

​We can therefore conclude

$\boxed{L=\sqrt{l(l+1)}\; \hslash }$

We finally conclude that angular momentum is quantised.

As an analogy to orbital angular momentum, the spin angular momentum is given by :

$\boxed{S=\sqrt{s(s+1)}\; \hslash }$

where s denotes the spin of $e^{-}$, and is taken to be $\frac{1}{2}$. The spin angular momentum of the electron is therefore found to be equal to $\frac{\sqrt{3}}{2}\hslash$. The z-component of the angular momentum is : ​

$\displaystyle s_z = s\hslash = \pm \frac{1}{2} \hslash$