Part 1. The Electronic Structure of Atoms

The wave function of an atom contains all the information about the properties of the atom. The wave functions for the hydrogen atom and other one-electron atoms (such as He+ and Li2+) can be calculated exactly by solving Schrödinger’s equation. Approximate methods must be used to calculate the wave functions for atoms containing two or more electrons. These numerical calculations can be very accurate, but require complicated computer calculations.

So we start with (brief review from Chem 331):

Hydrogen-Like Atoms

What are they? He+ (Z = 2) , Li2+ (Z = 3), Be3+ (Z = 4), etc. Ions with one nucleus and one electron (two bound particles), atomic number Z > 1.

Also a few rare, but interesting, “exotics”, such as muonium (more on that in the tutorial!).

Why are they important? Schrödinger’s equation can be solved exactly for hydrogen-like atoms, using the reduced mass concept to simplify the two-body problem to an effective one-body problem. This greatly expands the number of chemical systems that can be treated analytically by quantum mechanical principles. (For molecules and multi-electron atoms, cumbersome and less accurate numerical methods must be used to solve Schrödinger’s equation.)

For hydrogen-like atoms, the electrostatic potential energy and the reduced mass in Schrödinger’s equation for the hydrogen atom

e2 V (r) = − 40r

memp  = me + mp are changed to (notice the factor of Z)

Ze 2 V (r) = − 40r

memN  = me + mN me is the electron mass and mN is the nucleus mass.

Schrodinger Equation for Hydrogen-Like Atoms  H (r,,) = E (r,,) with Hamiltonian operator

 h2 Ze 2 H = − 2 − 2 8  40r

The Laplacian operator 2 can be separated into radial and angle-dependent terms

ˆ2 2 1   2   1 L  = r  − r 2 r  r  r 2  2

The angle-dependent terms can be represented by the operator Lˆ2 for the square of the angular momentum.

ˆ2 ˆ2 ˆ2 ˆ2 L = Lx + Ly + Lz

 1     1 2  = 2 sin +    2 2  sin     sin   

Partial separation of variables leads to solutions of Schrödinger’s equation of the form

 nm (r,,) = Rn (r)Ym (,)

Rnℓ(r) is called the hydrogen-like radial wave function and Yℓm(θ,) are spherical harmonics. As for the hydrogen atom,

principal quantum number n = 1, 2, 3, 4, … orbital angular momentum quantum number ℓ = 0, 1, 2, …, n − 1 (ℓ < n) magnetic quantum number m = 0, ±1, ±2, … , ± ℓ

A Few Solutions to Schrödinger’s Equation for Hydrogen-Like Atoms: 2 2 (the Bohr radius is a0 = ε0h /πμe ) 3/ 2 1  Z    n = 1, ℓ = 0, m = 0  100 =   exp(−Zr / a0 )   a0 

3/ 2 1  Z   Zr   =   2 − exp(−Zr / 2a ) n = 2, ℓ = 0, m = 0 200     0 32  a0   a0 

3/ 2 1  Z  Zr  =   cos exp(−Zr / 2a ) n = 2, ℓ = 1, m = 0 210   0 32  a0  a0

3/ 2 1  Z  Zr  =   sin exp(−Zr / 2a )exp(i) n = 2, ℓ = 1, m = ±1 211   0 64  a0  a0

Okay, but how can a chemist or physicist use these wave functions?

1. Energies of Hydrogen-Like Atoms

For an isolated hydrogen atom (no externally applied magnetic fields or other fields), the energy depends only on the principal quantum number n. The same result obtained for hydrogen-like atoms:

e4Z 2 En = − 2 2 2 (no externally applied fields) 8 0 h n

Why is the energy negative and proportional to Z2?

The energy equation can be simplified by lumping the constants together to form a single parameter. If the energy is expressed in cm−1, the unit favored by spectroscopists, this factor  called the Rydberg constant.

Photon energies hv are inversely proportional to wavelengths

hv = hc/ so energies are converted from units of Joules to wavenumbers by dividing by hc (and don’t forget to convert from m−1 to cm−1!)

4 2 ~ En e Z En = = − 2 3 2 hc 8 0 h cn

~ Using energies En in wavenumbers defined by

~ Z 2 E = − n n2 the Rydberg constant (in wavenumbers) is e4  = 2 3 8 0 ch The Rydberg constant depends on fundamental physical constants and the mass of the nucleus.

5 −1 For the hydrogen atom, H = 1.096 775 856  10 cm

5 −1 For the deuterium atom, D = 1.097 074 275  10 cm

5 −1 In the limit me/mN → ,  = 1.097 373 153  10 cm

Notice the extraordinary precision of Rydberg constants. This is possible because frequencies can be measured very precisely and accurately.

2. Ionization Energies: AZ-1 → AZ + e–

The ionization energy of a hydrogen-like atom is defined as the energy required to take the atom from its ground state (ni = 1) to the dissociated state nf =  where the electron and nucleus are infinitely far apart. In wavenumbers:

2 2 ~ ~ ~ Z  Z  2 E = E − E = − − −  = Z IONIZATION f i 2  12 

~ -1 For the hydrogen atom (Z = 1) EIONIZATION = H = 10967.75 856 cm = 2.178722  10−19 J = 13.605698 eV

For hydrogen-like atoms, ignoring small changes in the reduces mass, the ionization energies are

~ 2 2 EIONIZATION  Z H = 13.606Z eV

13.606  22 = 54.424 eV for He+, 13.606  32 = 122.45 eV for Li++, etc.

3. Electronic Spectra of Hydrogen-Like Atoms

The selection rules for transitions between atomic energy levels are

Δn unrestricted Δℓ = ±1 Δm = ±1, 0

Transitions from any initial ni value to any final nf are therefore “allowed”.

For emission spectra: nf < ni For absorption spectra, nf > ni

The photon energy ΔE = hv = Ef − Ei in wavenumbers is

~ ~  1 1  E − E = Z 2  −  f i  n 2 n 2   i f 

Hydrogen Atom Spectra

(in Wavenumbers)

Electronic Transitions for Hydrogen Atoms (in nm)

4. Electron Density – Important for Atomic Structure

For a hydrogen-like atom, the probability of finding the electron in the volume element dτ = dxdydz is (nℓm)* nℓm dτ. (Recall that the asterisk denotes the complex conjugate of nℓm, where all values of i in nℓm are converted to −i.)

Transforming to more-convenient spherical r, θ,  coordinates with dτ = r2sinθdrdθd gives

2 electron density = r sinθnℓm(r,θ,)* nℓm(r,θ,)

Finding the maximum value of the electron density gives the most probable locations of the electron. The locations of zero electron density are called nodes. Nodes are required to make the wave functions orthogonal:

  ( nm ) * nm d = 0 if n  n or    or m  m

Electron Densities for n = 1, 2, 3 and ℓ = 0 (s), 1 (p), 2(d)

(graphical display)

Stereo plots of probability electron probability densities

See:

D. T. Cromer, Stereo plots of hydrogen-like electron densities, Journal of Chemical Education, 45, 626−631 (1968)

Online Views of Hydrogen Orbitals

www.falstad.com/qmatom

• shows real wave functions (used by chemists)

• also complex wave functions

• phase information can be color coded

• views can be rotated in three dimensions

• radial probability distribution can be plotted

• time evolution also simulated

Plotting probability distributions How?

Monte Carlo Method (“Algorithm”)

Compute values of the probability distribution function (pdf) at a large number of randomly selected points in space. Plot points in proportion to the value of the pdf.

Step 1 Calculate the maximum value of the probability distribution function:

(*nlmnlm)max

Step 2 Use a random number generator (0 < RAND < 1) to calculate random values of r, ,  on the intervals 0 to rmax, 0 to , 0 to 2

r = RAND  rmax

 = RAND  

 = RAND  2

The radial extent rmax is chosen so that *nlmnlm  0 at r = rmax.

Step 3 At the randomly selected point r, ,  calculate

 *(r,,) (r,,) p = [ *(r,,) (r,,)] max

Step 4 Calculate a fourth random number q = RAND between 0 and 1. Plot the point at r, ,  if p > q.

repeat (go to Step 2) … Example: Probability Density Plot for the Hydrogen 2s Orbital

hydrogen 2s orbital: n = 2 l = 0 m = 0

1 1  r   (r) = 2 −  e−r / 2a0 200 3/ 2   4 2 a0  a0 

maximum value at r = 0:

1 1

[ 200 (r)]max =  200 (0) = 3/ 2 2 4 2 a0

At randomly selected point r, ,  in three-dimensions, calculate

2  r  2 −  e−r / a0  * (r) (r)  a  p = 200 200 =  0  * [ 200 (r) 200(r)]max 4

Plot a point at r, ,  if p > random number q between 0 and 1. Repeat many times:

at r = 0.223567a0, calculate p = 0.630876 and q = 0.389954. Plot this point.

at r = 2.07378, calculate p = 0.000684 and q = 0.440932. Do not plot this point.

Etc.

20

“Monte10 Carlo” Probability Density Plots

0 a 0

/ 1s 100*(r)100(r) z

-10 20

-20 10 200*(r)200(r) -20 -10 0 10 20

x / a0 0

a 0 2s view along y / / axis z

-10 10,000 points 200(r) > 0 red -20 200(r) < 0 blue

-20 -10 0 10 20

x / a0

20

10

0

a 3s 300*(r)300(r)

0 /

z

-10

-20

-20 -10 0 10 20

x / a0

Other “Monte Carlo” Applications

1. Quantum Computing

Conventional “silicon-based” computers use strings of binary bits

01100101011101000 … (two possible values, 0 or 1)

A string of n binary bits exists as one of 2n possible values.

But quantum computer bits (qbits) are superpositions of 0 and 1.

A qbit can take an infinite number of possible values, making quantum computing, in theory, very fast and very powerful. Still under development.

2. Monte Carlo Error Analysis

Use random numbers to simulate random experimental errors.

Important for the design of experiments.

3. Secure Ciphers and Encryption

Use random numbers to convert messages or data (organized sequences of letters and numbers) to random sequences. Provides secret codes that cannot be broken.

FLASH ORDER SINK CRUISER BELGRANO

→ BSGPEUHJHPCWMFPUUWHJZAEOMLQPU

4. Monte Carlo Numerical Integration

Useful for evaluating multi-dimensional integrals that are too complicated to be treated using trapezoidal and other conventional integration techniques.

Example: Calculate the area of circle of radius r = 1.

Place N randomly located point on a 2r  2r square of known area 4r2 = 4.

The ratio of the number of points inside the circle to the total number of points on the square gives the ratio of the circle are to the area of the square =  r2/(2r)2 =  /4.

1.0 10,000 random points on square area 2  2 = 4.

0.5 7843 points are located inside the circle.

1 to 1)

− 0.0 circle area / square area

= 7843/10000

(random, (random, y -0.5

circle area = (7843/10000)  4

-1.0 = 3.137

-1.0 -0.5 0.0 0.5 1.0 (close to the exact value ) x (random, −1 to 1)

Back to applications of atomic hydrogen wave functions: the average distance between the nucleus and the electron

 r = ( )*r r 2sindrdd nm  nm nm

the average potential energy

Ze 2 Ze 2 1  V  = − = − ( )*  r 2sindrdd  nm nm 40r 40 r

the average kinetic energy (because the total energy is E = + )

e4Z 2  K  = E − V  = − − V  n 2 2 2 8 0 h n

It is not too difficult to show that the average potential energy is twice the total energy.

= 2En Virial Theorem

This result gives is known as the Virial Theorem. Remarkably, the virial theorem is valid not only for the hydrogen atom and hydrogen-like atoms, but all systems with coulombic potential energy (all atoms and molecules!).

From the Virial Theorem, it follows that

= −2

= −

5. Total Orbital Angular Momentum of Hydrogen-Like Atoms

Like the hydrogen atom, hydrogen-like atoms can have orbital angular momentum. Classically, this kind of momentum corresponds to the rotational motion of the electron around the nucleus.

Operating on the wave function with the Lˆ2 operator

ˆ2 L nm (r,,) = ( +1) nm (r,,) gives the square of the total orbital angular momentum

L2 = ( +1)2 and therefore L = ( +1) 

ℓ is the orbital angular momentum quantum number. Historically, atoms with ℓ values of 0, 1, 2, 3, … are assigned the symbols s, p, d, f, …

6. Orbital Angular Momentum Along the z Axis, Lz

ˆ ˆ ˆ ˆ ˆ ˆ Operators Lx and Ly , Lx and Lz , and Ly and Lz do not commute.

ˆ ˆ 2 But all three , Ly , and Lz operators commute with Lˆ .

From the postulates of quantum mechanics, this means that the total angular momentum and only one of its components ( , by convention) can be measured simultaneously.  nm (r,,) is also an eigenfunction of with

ˆ Lz nm (r,,) = m nm (r,,) m and eigenvalue m . The orbital angular momentum along the z axis is

Lz = m m = 0, ±1, ±2, … , ± ℓ m is called the magnetic quantum number.

7. Magnetic Fields Can Split the Energy Levels of H Atoms

(yet another reason to know about angular momentum!)

What happens when an external magnetic field is applied to an electron? (Relevant to electric motors, nmr spectroscopy, and many other applications.)

An applied magnetic field “splits” (removes the degeneracy) of certain lines in the spectrum of the hydrogen atom:

What’s going on?

First, a brief review of magnetic fields, forces, dipoles, and potential energy.

An electric current I flowing in a closed loop of area A produces a magnetic dipole of magnitude .

 = IA

The current is in units of amperes (Coulombs per second).

The magnetic dipole has units C s−1 m2 = ampere m2.

Example Charge q moving at velocity v in a circular loop of radius r (area A =  r2), generates the electric current I = qv/2 r and a magnetic dipole of magnitude  = IA = qrv/2

Magnetic Dipoles and Angular Momentum

If an electric charge moves in an orbit that is not circular, the magnetic dipole (a vector) can be calculated using the more general cross-product relation    q(r v)  =   2 r and v are the position and velocity of the moving charge.

The linear momentum and the angular momentum are   p = mv

     L = r  p = r (mv)

So the magnetic dipole moment is directly proportional to the angular momentum.  q   = L 2m

The magnetic dipole and the angular momentum are both perpendicular to the plane of motion of the electron. z The force acting on the moving charge is L    F = q(v  B )  y

p r x

The magnetic field, velocity and force acting on the moving charge are mutually perpendicular. Can you think of any applications? (Mass spectroscopy, TV, electron microscopes, …)

Magnetic Potential Energy

Just like an electric dipole interacts with an electric field, the magnetic potential energy V of a magnetic dipole in a magnetic field B is the scalar product

  Vmagnetic = −  B

The SI units of the magnetic field are J ampere−1 m−2 = N ampere−1 m−1. (one ampere = one coulomb per second = one Tesla)

The strength of the magnetic field near the surface of the earth is about 50 x 10−6 Tesla. Superconducting magnets can generate magnetic field strengths of about 10 Tesla.

If the magnetic field is presumed to act along the z axis (this is the usual convention), then the magnetic potential energy is calculated from the z-components of the magnetic dipole and the magnetic field.

Vmagnetic = −z Bz

Hydrogen Atoms in Magnetic Fields

For hydrogen or hydrogen-like atoms with a single electron, the magnetic dipole generated by the electron is

 e   = − L

2me

Using the expression for the magnetic dipole of an electron gives

e B V = z L magnetic z 2me

For a hydrogen atom in an externally-applied magnetic field, the term in the Hamiltonian operator that accounts for the magnetic potential energy is therefore

e B z Lˆ z 2me

For the hydrogen atom, we already know that the eigenvalues of the Lz operator are mh/2. So the magnetic field gives the electron the magnetic potential energy:

eB L eB m V = − B = z z = z magnetic z z 2me 2me

This expression can be simplified by defining the Bohr magneton

e  = = 9.27410−24 J T−1 0 2me

The equation for the energy levels of a hydrogen-like atom in a magnetic field become

e4Z 2 E = − +  mB n 2 2 2 0 z 8 0 h n with

n = 1, 2, 3, 4, … ℓ = 0, 1, 2, …, n − 1 (ℓ < n) m = 0, ±1, ±2, … , ± ℓ

Notice that a state with a given ℓ value is split into 2ℓ + 1 levels by an externally applied magnetic field.

For example, a 1s state (n = 1, ℓ = 0, m = 0) has no z angular momentum and is not split at all.

A 2p state (n = 2, ℓ = 1, m = –1, 0, 1) is split into 3 levels:

e4 Z 2 e4 Z 2 e4 Z 2 − 2 2 −  0 Bz − 2 2 − 2 2 + 0 Bz 8 0 h 4 8 0 h 4 8 0 h 4

In spectroscopy parlance, the E2p to E1s transition that occurs in the absence of an externally applied magnetic field becomes a triplet in the presence of a field.

This is an illustration of the Zeeman effect discovered in 1896 by the Dutch physicist Pieter Zeeman.

What is the magnitude of the Zeeman splitting? The energy difference between the 1s and 2p levels for a hydrogen atom (Z = 1) in the absence of an externally applied magnetic field is

e4Z 2 e4Z 2 E − E = − + =1.6310−18 J 2p 1s 2 2 2 2 2 2 8 0 h 2 8 0 h 1

In a magnetic field of strength 10 T (typical for a superconducting magnet), for comparison, the energy difference between the split 2p levels is

–24 –1 –23 Β0Bz = (9.274  10 J T ) (10 T) = 9.274  10 J

This splitting is very small, only about 0.006 % of the E2p – E1s energy difference, but it is easily measured by high resolution spectroscopy.

8. Electron Spin: a Fourth Quantum Number

In the early days of quantum mechanics, the belief of the times was that whole atoms and molecules possessed magnetic moments. This was supported by a huge body of spectroscopic evidence, including the Zeeman effect.

But nobody could explain the doublet in the spectra of single-electron atoms and alkali metal atoms with a single unpaired electron.

Much more significantly, it was not at all understood why all the electrons in atoms and molecules did not drop into the lowest energy level!

In a legendary 1922 experiment, German physicists Otto Stern and Walter Gerlach passed a beam of silver ions through an inhomogeneous magnetic field to separate the beam into 2ℓ + 1 quantized components. Unexpectedly, the beam separated sharply into two components, suggesting a non-integer ℓ value of ½.

In 1925, Wolfgang Pauli showed that all of these experimental observations could be explained by postulating that the electron can exist in two distinct states. Pauli introduced the fourth quantum number, now called the electron spin quantum number ms restricted to values –1/2 and +1/2 for electron spin along the z axis.

Pauli provided no explanation or interpretation of this postulate. But it worked! For a while, the spin quantum number was an ad hoc construction and remained a mystery.

The two states of the electron were soon identified with two intrinsic (“internal”) angular momentum values of the electron, called spin states. It is as if a spinning electron generates an internal electric current and intrinsic magnetic moment. This internal current is not real or observable. For most quantum chemistry applications, electron spin can simply be grafted onto quantum postulates already employed.

Fortunately, the quantum mechanics extended by Paul Dirac and others to include relativity came to the rescue. In relativistic quantum mechanics, spin appears naturally as a well defined and natural variable. Unfortunately, relativistic quantum mechanics is a course in itself and beyond the scope of this course.  Just as the orbital angular momentum vector L has quantum number ℓ and magnitude

L = ( +1) ℓ = 0, 1, 2, 3, …

 the spin angular momentum vector S has quantum number s and magnitude

S = s(s +1) s = 1/2

But unlike ℓ , which can take values from 0 to , the spin quantum number s can only have the value ½ and therefore

S = ( 3 / 2) s can never take large values, so the spin angular momentum never assumes classical behavior. Spin is strictly a non-classical property.

Just as the z-component of the orbital angular momentum is

Lz = m m = 0, ±1, ±2, …, ± ℓ the spin angular momentum has the z-component

Sz = ms ms = ±½

The case where ms = +½ is sometimes called “spin up”, and the case where ms = –½ is called “spin down”. ˆ 2 ˆ It is customary to define the spin operators S and S z and their eigenfunctions  and β by the formal equations

ˆ 2 1  1  2 S  =  +1  2  2 

ˆ 2 1  1  2 S  =  +1  2  2 

1 Sˆ  =  z 2

1 Sˆ  = −  z 2

The spin eigenfunctions  and β are orthonormal

 * d =   * d = 1

 * d =  *d = 0 and do not involve spatial coordinates, such as x, y or z. The spin variable  has no classical analog. 9. Complete Wave Functions Must Include Spin

Why is spin important? For hydrogen-like atoms, a complete wave function must include the spin state of the electron. Because there are two possible spin states, there are twice as many wave functions as we assumed earlier, as indicated by

nℓm(r,θ,) nℓm(r,θ,)β

Notice that including spin increases the degeneracy of level n for hydrogen and hydrogen-like atoms from n2, assumed previously, to 2n2.

Furthermore, the electron has a spin magnetic dipole of magnitude

s = −g0 s(s +1) s = 1/2 with z-component

sz = −g0ms ms = ±1/2 with

g = 2.002 322

Interestingly, because the value of g is very close to 2, the magnetic splitting of the state with m = 1 and ms = –1/2 and the state with m = –1 and ms = +1/2 are almost degenerate.

In the early days of quantum mechanics, the g factor was evaluated empirically, from Stern-Gerlach data, for example, and called the “anomalous” spin factor because its origin was unknown. Happily, the experimental g value was soon justified and correctly predicted by Dirac’s relativistic quantum mechanics. You can be excused if you think electron spin is mysterious. We will treat it as just another postulate, but it “works” and can be justified by more advanced relativistic theory. More importantly, you might remember from high school or first-year general chemistry that no two electrons in an atom can have the same values of all four n, ℓ, m and ms quantum numbers. This will give us the periodic table of the elements.

Summary of Orbital Angular Momentum and Spin Angular Momentum for a Single Electron

______Orbital Spin Angular Angular Momentum Momentum ______  Angular momentum vector L S

Magnitude of the angular L = ( +1) S = s(s +1) momentum vector

z-Component of the Lz = m Sz = ms angular momentum

Operator for the square of Lˆ2 Sˆ 2 the angular momentum

ˆ ˆ Operator for the z-component Lz S z of the angular momentum

Quantum number for the  = 0,1, 2,... s =1/ 2 angular momentum

Quantum number for the m = 0, 1,  2,..., ms = 1/ 2 z-component

  Magnetic dipole moment  s vector

More on Electron Spin: the Stern-Gerlach Experiment

In the early days of quantum mechanics, whole atoms and molecules were believed to have magnetic moments, supported by a large body of spectroscopic evidence, including the Zeeman effect. Electrons orbiting nuclei …

But wait! In a legendary 1922 experiment, Stern and Gerlach passed a beam of silver ions through a magnetic field gradient to separate the beam into 2ℓ + 1 quantized components. Unexpectedly, the beam separated into two components, suggesting a non-integer ℓ value of ½.

Recall that an externally applied magnetic field removes the degeneracy of the energy of levels of hydrogen atoms with ℓ = 1, 2, 3, … and produces m = 2ℓ + 1 = 3, 5, 7, … different states. But how do you get two different energy states from orbital angular momentum? Something is missing! Temporary solution:

Additional “intrinsic” electron spin angular momentum (up or down) was assumed.

Also … “Fine Structure” in Electronic Spectra

Schrodinger’s equation accurately describes the H atom. But not quite exactly!

High resolution spectroscopy reveals mysterious “fine structure” splitting, even in the absence of externally applied magnetic fields.

According to Schrodinger’s treatment, the energy of a hydrogen atom depends only on principal quantum number n.

 e4 1 e2 1 E (H atom, Z =1) = − = − n 8 2h2 n2 8 a n2 0 0 0

For example, n = 2 to n = 1 transitions should be a singlet. But a narrowly spaced doublet is observed:

n = 2 to n =1 Transition for Atomic Hydrogen

n = 2

Schrodinger Experiment Prediction singlet doublet

−1 v = 82257 cm−1 v = 82259 cm 1 −1 v2 = 82260 cm

n = 1

Similar small splittings are observed for many other transitions and for many other atoms, commonly attributed to “Spin-Orbit Coupling” (the interaction between spin angular momentum and orbital angular momentum). Problem: spin is not included in the Schrodinger equation. Other Points to ponder:

1. Hydrogen 1s and 2s orbitals (ℓ = 0) have no orbital angular momentum for spin angular momentum to couple with!

2. Electron spin can’t be angular momentum caused by an electron rotating about an axis through it (like the daily rotation of planet earth around the axis between the north and south geographical poles). Such an electron rotation would be faster than the speed of light. Not allowed!

WRONG !

3. Spin can be included empirically in by multiplying Schrodinger hydrogen orbitals by a spin-up () or spin-down factor (). Not very informative!

nlm  nlm 

4. Other elementary particles have spin:

Fermions Particles with half-integer spins (such as electrons, protons, neutrons) called fermions obey the Pauli exclusion principle: no two fermions can have identical quantum numbers. Important for the periodic table and all of chemistry!

Bosons Particles with integer spins (such as photons carrying electromagnetic forces and gluons carrying nuclear forces) can have identical quantum numbers and “bunch together”. Important for superconductivity, superfluidity, nuclear structure and nuclear reactions.

bosons fermions

5. Spin is an intrinsic property of elementary particles, as fundamental as mass and electrical charge.

But what is “Spin”?

Spin can be understood by combining quantum mechanics and special relativity to give relativistic quantum mechanics

Special Relativity Describes particles moving at very high speeds (v) approaching the speed of light (c) using theory developed by Einstein

m0 relativistic mass m → m0 = rest mass v2 increase 1 − c2

v2 length contraction x → 1 − 2 x c

t time dilation t → v2 1 − c2

Why “special” relativity? Applies to non-accelerating systems, gravity negligible.

Why needed for quantum chemistry? Electrons in atoms and molecules are moving at high speeds (e.g., 0.7 % of c for ground state H atoms, much faster in high Z atoms).

Tests of Special Relativity

Many experiments have verified the theory of special relativity.

Atomic Clocks

Set two high-precision atomic clocks to read exactly the same time.

Take one of the clocks for a ride in a high-speed jet.

Find that the clock taken for a ride is running slightly slower.

Muon Decay

Unstable elementary particles called muons have the same charge as the electron, but are 207 times heavier

Muons “at rest” (not moving, v = 0) have a half-life of 2.2 s

The measured half-life of muons produced by cosmic rays and moving at high speeds is significantly longer than 2.2 s.

Example The measured half-life of muons moving at 99.5 % of the speed of light is 22 s

t 2.2 μs 1/ 2(0) = = 22μs v2 1 − 0.9952 1 − c2

The problem for quantum chemistry:

The Schrodinger equation we’ve used is non-relativistic

2  2 2 2  −  2 + 2 + 2  + V = E 2m x y z 

(does not include the theory of relativity), so it’s “only” an approximation (but a very good one!) for atoms and molecules in which electrons are moving at high speeds.

Non-relativistic energy:

total energy = kinetic energy + potential energy

1 2 1 2 1 2 E = mvx + mv y + mvz + V 2 2 2

2 2 2 px py pz = + + + V 2m 2m 2m

Non-relativistic Hamiltonian (total energy) operator:

2 pˆ 2 pˆ pˆ 2 Hˆ = x + y + z + Vˆ 2m 2m 2m

ˆ Using the non-relativistic momentum operators pˆ x = − i   /  x ,p y = − i   /  y

and p ˆ z = − i   /  z gives

2 2 2 2 ˆ       H = −  2 + 2 + 2  + V 2m x y z  

How can a relativistic Schrodinger equation be developed?

Einstein’s theory of special relativity gives

E 2 = p2c2 + p2c2 + p2c2 + m2c4 x y y

2 for the energy (reducing to the famous equation E = m0c for a stationary particle).

Answer : The Dirac equation

In one of the most important discoveries in the history of science, Dirac developed relativistic quantum mechanics by using a Hamiltonian operator of the form

ˆ 2 H = −x pˆ xc − y pˆ yc −z pˆ zc − mc

Why? When squared, this Hamiltonian gives the relativistically correct result

for the energy, and the relativistic Schrodinger equation

    2  ic  x + y + z −  mc  +V = E  x y z 

But what are x , y , z ,  ???

Clues are obtained by squaring the expression for the energy E:

2 2 2 2 2 3 E =  x x px c − x y px pyc − x z px pzc − x pxmc

2 2 2 2 3 −  y x px p yc + y y p y c − y z p y pz c − y p y mc

2 2 2 2 3 −  z x px pz c − z y p y pz c + z z pz c − z pz mc

3 3 3 2 4 − x px mc −  y p y mc + z pz mc − m c

Comparing this result with the relativistic energy squared

2 2 2 2 2 2 2 2 4 E = px c + p y c + p y c + m c

shows that x, y, z, and  must give unity when squared.

xx = 1 yy = 1 zz = 1  = 1

Also, to cancel the cross-terms in the equation for E2, the product of any two different operators must change sign when their order of multiplication is changed:

xy = −yx xz = −zx x =−x

How can this be? Everyone knows a times b equals b times a!

But wait! ab does not have to equal ba for matrices! (Another reason to know something about linear algebra!)

Dirac, in one of the most important scientific papers of all time, showed that the ’s and  for the electron are the 4  4 matrices:

0 0 0 1   0 0 1 0  =   x 0 1 0 0   1 0 0 0

0 0 0 − i

0 0 i 0     y = 0 − i 0 0    i 0 0 0 

0 0 1 0    0 0 0 −1  =   z 1 0 0 0    0 −1 0 0 

1 0 0 0  0 1 0 0   =   0 0 −1 0    0 0 0 −1

The Dirac relativistic wave function  is a four-component column vector.

1     =  2   3     4 

The position probability distribution is obtained by pre-multiplying  by its complex conjugate row vector

1    * * * * *  2  * * * *   = 1  2  3  4  = 11 + 2 2 + 3 3 + 4 4  3     4 

Solving the Dirac equation

Solving the Dirac equation for the hydrogen atom

    2  ic  x + y + z −  mc  +V = E  x y z  for the hydrogen atom gives relativistic wave functions with four quantum numbers

n, ℓ, j, m ( not the Schrodinger n, ℓ, m, ms !)

The Dirac Hydrogen Atom: The Quantum Numbers principal quantum number n = 1, 2, 3, …

similar meaning as for the Schrodinger atom, but n no longer completely decides the energy

azimuthal angular momentum number ℓ = 1, 2, 3, …, n − 1

but ℓ no longer gives orbital angular momentum because orbital angular momentum is not a constant of motion angular momentum number j

always positive, can have at most two values   1 / 2 , gives the total angular momentum (a constant of motion)

magnetic quantum number m

all half-odd integer values from –j to + j

Notice there is no spin quantum number ms. Also, orbital angular momentum and spin angular momentum are not separately distinguished.

Examples of Dirac Quantum Numbers for the Hydrogen Atom

n = 1 ℓ = 0 j = 1/2 m = −1/2, +1/2

n = 2 ℓ = 0 j = 1/2 m = −1/2 or 1/2 n = 2 ℓ = 1 j = 1/2 m = −1/2 or 1/2 n = 2 ℓ = 1 j = 3/2 m = −3/2, −1/2, 1/2, or 3/2

Comparison of Schrodinger and Dirac Models of the Hydrogen Atom

To compare non-relativistic Schrodinger wave functions and relativistic Dirac wave functions, it is informative to look at the ground state hydrogen atom and a few excited states.

Schrodinger Ground-State Hydrogen Atom (Two States)

( Spin Included Emprically, Up or Down )

1 1 − r / a 0 n = 1, 100 = 3/ 2 e  ℓ = 0, m = 0, ms = +1/2  a0

1 1 − r / a 0 n = 1,  100 = 3/ 2 e  ℓ = 0, m = 0, ms = −1/2  a0

n is the principal quantum number

ℓ gives the orbital angular momentum

m is the magnetic quantum number

ms is the spin quantum number

Important! The spin quantum number ms is included empirically and is not derived from the Schrodinger equation.

Dirac Ground-State Hydrogen Atom (Two States)

 1     0  n = 1, ℓ = 0, j = ½, m = ½ 1 1 −r / a 1 i 0 1,0,1/ 2,1/ 2 = 3/ 2 e  cos   a0  2 137  (the chemist’s 1s spin up state) 1 i  −i   sin e  2 137 

 0 

 −1    n = 1, ℓ = 0, j = ½, m = −½ 1 1 −r / a0 1 i 1,0,1/ 2,−1/ 2 = 3/ 2 e − sin   a0  2 137   1 i  (the chemist’s spin down state) cos  2 137 

Important! No distinction is made between orbital and spin angular momentum in the Dirac equations. There is just angular momentum.

The non-relativistic Schrodinger ground-state hydrogen atom has zero angular momentum, but the Dirac ground state hydrogen atom has angular momentum j  =  / 2 . .

1/137 is the fine structure constant, the ratio of the average speed of the electron to the speed of light.

e 2  v  1 =  for ground-state H atoms (n = 1) 2 hc c 137 0

Schrodinger 2s Hydrogen Atom (Two States)

1 1  r   = 2 −  e−r / 2a0  200 3/ 2   n = 2, ℓ = 0, m = 0, ms = 1/2 4 2 a0  a0 

1 1  r   = 2 −  e−r / 2a0  200 3/ 2   n = 2, ℓ = 0, m = 0 ms = −1/2 4 2 a0  a0 

Dirac 2s Hydrogen Atoms (Two States)

 r  2 −  a   0  0 n = 2, ℓ = 0, j = ½, m = ½ 1 1   −r / 2a0  i  r    2,0,1/ 2,1/ 2 = 3/ 2 e 1− cos 4 2 a0     137 4a0 (chemist’s 2s spin up state)     i  r   1− sine−i      137 4a0  

 0   r  2 −  a   0  1 1 −r / 2a 0  − i  r  −i  n = 2, ℓ = 0, j = ½, m = −½  2,0,1/ 2,−1/ 2 = 3/ 2 e 1− sine 4 2 a0     137 4a0    (chemist’s 2s spin down state)  i  r   1− cos  137 4a     0  

Important Features of the Dirac Hydrogen Atom

1. No nodes! The four components of the relativistic Dirac wave functions for the hydrogen atom ( 1, 2, 3, 4 ) are never simultaneously zero. This means the electron is free to move throughout the whole atom, not trapped between nodes.

2. Explains the empirical  and  spin functions appended to non-relativistic wave functions. Vector components 3 and 4 are relatively small compared to main components 1 and 2 , so the approximate Dirac wave functions for the H atom are:

1  1   nm   0       0     2   2     nm   =   or  3   0   0   0           4   0   0   0 

Ignoring the small third and fourth vector components 3 and 4 gives

1 0    nm   or  nm   0 1

The 2  1 column vectors

1 0  =   and  =   0 1 are the so-called “spin up” or “spin down” functions included empirically in the Schrodinger (non-relativistic) nℓm wave functions.

nℓm(r,θ,) nℓm(r,θ,)β

3. Non-degenerate energy levels of the hydrogen atom.

The non-relativistic Schrodinger equation gives the energy

e2 Z 2 E = − non-relativistic n 8 a n2 0 0 that depends only on quantum number n. (Z = 1 for H, Z = 2 for He+, Z = 3 for Li2+…)

The Dirac equation shows that relativistic energy depends on n and j

e2 Z 2  Z 2  n 3  E = − 1 +  −  relativistic n, j 2  2 4   80a0 n  137 n  j + 1/ 2 4 

and is slightly more negative than the Schrodinger energy by a fraction of its value equal to (Z/137)2 multiplied by the following fractions:

1/4 for the 1s1/2 state

1/64 for the 2p3/2 state

5/64 for the 2s1/2 and 2p1/2 states

1/324 for the 3d5/2 state

3/324 for the 3p3/2 and 2d3/2 states

11/324 for the 3s1/2 and 2p1/2 states

Etc.

Energy Levels of the H atom

(not to scale)

Relativistic effects remove the degeneracy of the n = 2, n = 3, n = 4, … states

non-relativistic

relativistic

For the 1s and 2s Dirac atoms, j = 1/2, each with m equal to −1/2 or 1/2. These states are given the spectroscopic designations 1s1/2 and 2s1/2.

For the 2p Dirac atoms, there are two j values, 1/2 and 3/2, with the corresponding sets of m values

−1/2, 1/2 and −3/2, −1/2, 1/2, 3/2

These states are given the spectroscopic designations 2p1/2 and 2p3/2, respectively.

Predicted Energies of the Schrodinger and Dirac H Atoms

States with the same values of n but different values of j have different energies.

Schrodinger H atom Dirac H atom 0.02 cm 1 n = 2 2p 1 3/2 0.36 cm 2s and 2p 1/2 1/2

n = 1

1 1.46 cm Why is the 1s1/2 1s shift largest? 1/2

The Schrodinger equation predicts a single frequency for the n = 2 to n = 1 transition for the hydrogen atom.

The Dirac equation accounts for the observed doublet caused by the relativistic splitting of the n = 2 energy level.

Fine structure explained!

But it’s not spin-orbit coupling! The 1s state (zero orbital angular momentum according to the nonrelativistic wave function) is shifted the most!

OK, But What is Electron Spin ?

Solving the Dirac equation for a free electron reveals strange behavior:

electrons do not move in straight lines!

Instead, an electron behaves like a “jitterbug”.

It zig-zags through space with a “trembling” motion at the enormously high frequency

2 21 −1 4m0c /h  10 s

(too large to observe) with a very tiny amplitude << Bohr radius (too small to observe).

Picture an electron not moving in a straight line (zero angular momentum) but instead in a helical “corkscrew” path.

Electron spin can be visualized as the lower limit of circulatory motion and angular momentum below which the electron cannot drop, analogous to the zero-point energy of a particle in a box .

The Dirac equation provided the explanation for fine structure in atomic spectra. The story continues …

Hyperfine Structure

Additional small splittings are produced by the interaction of the electron angular momentum and the nuclear angular momentum!

Application: High Precision Atomic Clocks

International Coordinated Time

Since 1967, the SI definition of the second is the duration of exactly 9,192,631,770 cycles of radiation corresponding to the transition between two hyperfine energy levels of 6s1 electrons in cesium-133 atoms.

A network of atomic clocks keeps International Time with an accuracy of about 1 ns per day.

To accurately check the time, go to the National Research Council of Canada website:

http://www.nrc-cnrc.gc.ca/eng/services/time/web_clock.html

At midnight Universal Time on 31 December 2016 a leap second was added to International Time. Why?

What is a leap second?

What is the connection to angular momentum?

Wider Significance of the Dirac Equation - Antimatter!

Dirac took the positive roots of the energy equation

2 2 2 2 2 2 2 2 4 E = px c + pyc + pyc + m c to calculate the energies of the hydrogen atom

The negative energies were ignored at first for being physically unrealistic.

Further investigation led Dirac to suggest (in 1928) the negative energy states represented positrons: antielectrons with positive charge.

Positrons were discovered in 1932!

Significance of the Dirac Equation (Summary)

1. Gives the physical origin of the four quantum numbers need to describe the hydrogen atom.

2. There are no nodes in the hydrogen atom. The electron is free to move throughout the atom without being “trapped” in a particular region.

3. Hydrogen atom states with identical electronic configuration (i.e., identical values of n and ℓ, such as 2p1) can have different energies.

4. Accounts for the fine structure in atomic spectra and quantitatively predicts the magnitude of the splittings.

5. All of the angular momentum of the H atom is described by the circulatory motion of the electron. Orbital angular momentum and spin angular momentum are not separately defined and not separately conserved.

6. The Dirac equation provides a physical explanation for the  (spin up) and  (spin down) formalism used by chemists.

More generally:

The Dirac equation links quantum mechanics and relativity, two of the most important theories of science.

Correctly predicted the existence of antimatter.

Fine and Hyperfine Structure in Spectroscopy

• such tiny frequency splittings (ppm, ppb, …) are unimportant

• interesting only for academics dotting i’s and crossing t’s?

• high-precision atomic clocks, but who cares?

• no relation to the real world …

• another big yawn, ZZZzzzzzz …

NO!

Fine and hyperfine atomic spectra provide:

• fine and hyperfine energy levels

• and therefore precise (fine and hyperfine!) tests of atomic theory

• detailed and fundamental information about atomic structure

• many practical applications

Example Satellite Location Systems

Use atomic-clock timing of radio signals received from satellites orbiting at different distances to provide precise location information.

American GPS system

Russian Glonass system

Applications:

• mapping

• surveying

• locating aircraft, ships, submarines, cars …

• all-weather navigation

• search and rescue

• tracking