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PHGN300/310: Quantum MechanicsFred Sarazin ([email protected])Physics Department, Colorado School of Mines

Quantum Mechanics


Part I: Wave Properties of Matter

Louis De Broglie1892-1987Nobel Prize in Physics 1929


• Electromagnetic radiations shown to behave like:– Waves: Interferences, diffraction…– Particles: Photoelectric & Compton effects…

• Particles shown to behave like:– Particles (duh!): Rutherford scattering, …– Waves ???: stationary orbits in the Bohr model ?

• By the 1920’s, nobody made the fateful leap to assume that particles may also have a wave-like behavior

So…


• Suggests that particles could have wave-like properties.

• Uses Einstein’s special theory of relativity together with Planck’s quantum theory to establish the wave properties of particles.

1924: De Broglie defends his thesis


De Broglie wavelength

• Energy in the special theory of relativity:

• Photoelectric Effect:

• Combining both expression (with 𝜐 = #$):

• De Broglie suggest this relation for photons extends to ALL particles à “Matter Waves”

𝐸& = 𝑝𝑐 & + 𝑚𝑐& &

𝑚 = 0 → 𝐸 = 𝑝𝑐

𝐸 = ℎ𝜐

Photon:

𝜆 =ℎ𝑝


De Broglie WavelengthsMicroscopic / macroscopic objects

• What is the De Broglie wavelength of a 50 eV electron (non-relativistic approach OK)?– l = h/p = 0.17 nm ~ 10-10 m (interatomic distance in crystals)

• What is the De Broglie wavelength of a Tennis Ball (57g) at 25 m/s (about 56 mph) [assuming the tennis ball to be a point particle!]?– l = h/p = 4.7 x 10-34 m (about 1019 times small than the size of a nucleon !)

Exercise


• Bohr Quantization:– Bohr’s hypothesis of stationary orbits:

• Angular momentum: 𝐿 = 𝑟×�⃗� = 𝑚𝑣𝑟 = 𝑛ℎ/2𝜋

• With momentum 𝑝 = 𝑚𝑣 and 𝑝 = 9$, we get 𝐿 = 𝑝𝑟 = 9:

$

With De Broglie’s wavelength, Bohr’s quantization appears “naturally”

A new look at Bohr’s model of the atom

• If electron has a wave-like behavior(stationary orbit ↔ standing wave)

• Orbit: 𝐷 = 2𝜋𝑟 = 𝑛𝜆 → 𝑟 = =$&>

• 𝐿 = 9:$

with 𝑟 = =$&>

:

𝐿 =𝑛ℎ2𝜋 = 𝑛ℏ


• 1925: C. Davisson and L.H.Germer observe electron scattering and interference effects

--> Experimental evidence of the wave-likeproperties of the electron

Electron scattering


120-keV electron on quasicrystal Diffraction on a powder

Electron diffraction (transmission)


• De Broglie Wavelength: l = h/p

• In the non-relativistic approximation, p = √(2mK) with K, the kinetic energy of the electronà The higher K, the higher p

• Higher electron energies (smaller wavelength) à Sharper probe of the material.

A word on particle energies


The Spallation Neutron Source (SNS) facility – Oak Ridge National Laboratory

Neutron diffraction


Part II: Wave or Particle ?

Werner Heisenberg1901-1976

Nobel Prize in Physics 1932

Niels Bohr1885-1962



• A particle is a localized object, a wave is not.

• How does “it” go from waves to particles ?

• Wave-particle duality?

Wave or particle?


Consider a single wave: y(x,t) = A cos(kx-wt)

• POSITION:Where is the wave ? Everywhere (within the limit of the “definition” of the wave)

• MOMENTUM:The wave number k=2p/l and the De Broglie wavelength l = h/p à k = 2pp/h = p/ħIn this case: k is well defined à the momentum p =ħk is also very well defined

x (arbitrary units)

Location of a wave?


Consider 2 waves: y(x,t) = A cos(k1x-w1t) + A cos(k2x-w2t) with k1 close to k2 [BEATS!]= 2 A cos(½[(k1-k2)x-(w1-w2)t]) cos(½[(k1+k2)x-(w1+w2)t])

Posing Dk=k1-k2 and Dw=w1-w2: y(x,t) = 2 A cos(½[Dkx-Dwt]) cos(½[(k1+k2)x-(w1+w2)t])

• POSITION:Where is the ”particle” ? Still not really localized, but now some “regions of space” seem more likely to contain the particle than others.

lenvelope/2

x1 x2

The particle is between x1 and x2:½Dk x2- ½Dk x1 = pPosing Dx = x2-x1: DxDk=2p

• MOMENTUM:The momentum is now only known within Dp=ħDk

We gained a better knowledge of x, butwe are losing our “perfect” knowledge of p

Location of two superposed waves?


1 WavesDkmax=0 • Adding more waves:

Dx decreasesDk increases

Dx.Dk~1

• One can show similarly that:

Dw.Dt~1(Energy: E = hn = ħ(2p)n = ħw)

• Uncertainty relations between:• Position and Momentum• Energy and Time

• Particles can be represented by wave packets

2 WavesDkmax=2.5

5 WavesDkmax=5

10 WavesDkmax=10

20 WavesDkmax=20

k=10 More waves?


• What about sending one photon at a time ?

INTERFERENCE EFFECT !

From: Quantum (J. Al-Khalili)

The double-slit experiment with photons


• 1961: C.Jonsson manages to produce very narrow slits to observe the interference effects due to the wave-like behavior of the electrons.

The double-slit experiment with electrons


One slit blocked No interference

Pictures from: Quantum (J. Al-Khalili)

Which slit? (I)


A apparatus is placed todetect when one photon goes through the top slit.

Detector On

Detector Off

NO INTERFERENCE !

INTERFERENCE !

Pictures from: Quantum (J. Al-Khalili)

Which slit? (II)


• Use a subscript “ph” to denote variables for light (photon). Therefore the momentum of the photon is:

• The momentum of the electrons will be on the order of

• The difficulty is that the momentum of the photons used to determine which slit the electron went through is sufficiently great to strongly modify the momentum of the electron itself, thus changing the direction of the electron! The attempt to identify which slit the electron is passing through will in itself change the interference pattern.

• To determine which slit the electron went through: We set up a light shining on the double slit and use a powerful microscope to look at the region. After the electron passes through one of the slits, light bounces off the electron; we observe the reflected light, so we know which slit the electron came through.

Measurements and quantum systems

𝑝@9 =ℎ𝜆@9

>ℎ𝑑

𝑝CD =ℎ𝜆CD

~ℎ𝑑


Mach-Zender interferometer


• Bohr’s principle of complementarity: it is not possible to describe physical observables simultaneously in terms of both particles and waves

• Consequence: once you measure the wave-(particle-) like behavior of a phenomenon, you cannot measure a property linked to its particle- (wave-) like behavior.– e.g. once you determine that the

photon/electron has made it through a given slit, you reveal its particle-like behavior, therefore you cannot observe the interference phenomenon anymore (linked to the wave behavior)! Niels Bohr’s Coat of Arms (1947)

“Opposites are Complementary”

Principle of complementarity


• It is impossible to measure simultaneously, with no uncertainty, the precise values of k and x for the same particle. The wave number k may be rewritten as

• For the case of a Gaussian wave packet we have:

• Thus for a single particle we have Heisenberg’s uncertainty principle:

Heisenberg uncertainty principle (I)

k =2𝜋𝜆 =

2𝜋ℎ 𝑝⁄ = 𝑝

2𝜋ℎ =

𝑝ℏ

∆𝑘∆𝑥 =∆𝑝Kℏ ∆𝑥 =

12

∆𝑝K∆𝑥 ≥ℏ2


• If we are uncertain as to the exact position of a particle, for example an electron somewhere inside an atom, the particle can’t have zero kinetic energy.

• The energy uncertainty of a Gaussian wave packet is:

combined with the angular frequency relation:

• Energy-Time Uncertainty Principle:

Heisenberg uncertainty principle (II)

with ∆𝑥~ ℓ&

[we know the particle is located between 0 and ℓ] ∆𝑝K∆𝑥 ≥ℏ2

∆𝐸∆𝑡 ≥ℏ2

∆𝜔∆𝑡 =∆𝐸ℏ ∆𝑡 =

12

∆E = ℎ∆𝜈 = ℎΔ𝜔2𝜋 = ℏΔ𝜔

𝐾UV= =𝑝UV=&

2𝑚 ≥∆𝑝 &

2𝑚 ≥ℏ&

2𝑚ℓ&


• A proton is confined in a Uranium nucleus (diameter: d=16fm). Determine the minimum kinetic energy [non-relativistic] of the proton confined within the diameter of the uranium nucleus.

Exercise


• It is possible for some fundamental particles to “violate” conservation of energy by creating and quickly reabsorbing another particle. For example, a proton can emit a p+ according to p à n + p+ when the n represents the neutron. The p+ has a mass of 140 MeV/c2. The re-absorption must occur within a time Dt consistent with the uncertainty principle.– Considering that example, by how much DE is energy conservation

violated? [Ignore kinetic energy]– For how long Dt can the p+ exist?– Assuming that the p+ is moving at nearly the speed of light, how far from

the nucleus could it get in the time Dt?

Exercise


• “No one knows what happens behind the quantum curtain… but we should not care!”

• Physics only depends on the outcomes of measurements (in other words: only the results count… )

From: Quantum (J. Al-Khalili)

The Copenhagen interpretation


• De Broglie - Bohm• Many-Worlds• …

Other interpretations…


Part III: The Schrödinger Equation

Erwin Schrödinger1887-1961



• The Schrödinger wave equation in its time-dependent form for a particle of energy E moving in a potential V in one dimension is:

• The extension into three dimensions is

where 𝑖 = −1� is an imaginary number.

The Schrödinger Wave Equation

𝑖ℏ𝜕Ψ(𝑥, 𝑡)𝜕𝑡 = −

ℏ&

2𝑚𝜕&Ψ 𝑥, 𝑡𝜕𝑥& + 𝑉Ψ(𝑥, 𝑡)

𝑖ℏ𝜕Ψ𝜕𝑡 = −

ℏ&

2𝑚𝜕&Ψ𝜕𝑥& +

𝜕&Ψ𝜕𝑦& +

𝜕&Ψ𝜕𝑧& + 𝑉Ψ(𝑥, 𝑦, 𝑧, 𝑡)


• The general form of the wave function is:

which also describes a wave moving in the x direction. In general the amplitude may also be complex.

• The wave function is also not restricted to being real. Notice that the sine term has an imaginary number. Only the physically measurable quantities must be real. These include the probability, momentum and energy.

General solution

Ψ 𝑥, 𝑡 = 𝐴𝑒V dKefg = 𝐴 cos 𝑘𝑥 − 𝜔𝑡 + 𝑖 sin 𝑘𝑥 − 𝜔𝑡


• People gave some long thoughts to what was Y, the wave-function (even Schrödinger himself):

“Erwin [Schrödinger] with his psi can doCalculations quite a fewBut one thing has not been seenJust what does psi really mean”

From: Walter Hückel, translated by Felix Bloch

• The Schrödinger equation allows to calculate analytically [exactly] quantum problems, but it does not reveal the nature of Y. Schrödinger called it “field scalar”.

“Shut up and Calculate !” from R.Feynman (and many others!)

Y: The Wave Function


Solvay Congress 1927


• When squared, the wave function is a probability density (Max Born – 1926). The probability P(x) dx of a particle being between x and x+dx is given in the equation:

• The probability of the particle being between x1 and x2 is given by

• The wave function must also be normalized so that the probability of the particle being somewhere on the x axis is 1.

Properties of the wave function

𝑃 𝑥 𝑑𝑥 = Ψ∗ 𝑥, 𝑡 Ψ 𝑥, 𝑡 𝑑𝑥

𝑃 = o Ψ∗Ψ𝑑𝑥Kp

Kq

o Ψ∗ 𝑥, 𝑡 Ψ 𝑥, 𝑡 𝑑𝑥 = 1rs

es


• In order to avoid infinite probabilities, the wave function must be finite everywhere.

• In order to avoid multiple values of the probability, the wave function must be single valued.

• For finite potentials, the wave function and its derivative must be continuous. This is required because the second-order derivative term in the wave equation must be single valued. (There are exceptions to this rule when V is infinite.)

• In order to normalize the wave functions, they must approach zero as xapproaches infinity.

• Solutions that do not satisfy these properties do not generally correspond to physically realizable circumstances

Boundary conditions


• The dependence on time and position can then be separated in the Schrödinger wave equation. Let:

which yields:

Now divide by the wave function:

• The left side depends only on time, and the right side depends only on spatial coordinates. Hence each side must be equal to a constant. The time dependent side is:

Simplification possible if the potential is not time-dependent [happens in many cases]

Time Independent Schrödinger Wave Equation (I)

Ψ 𝑥, 𝑡 = 𝜓 𝑥 𝑓(𝑡)

𝑖ℏ𝜓 𝑥𝜕𝑓(𝑡)𝜕𝑡 = −

ℏ&𝑓 𝑡2𝑚

𝜕&ψ 𝑥𝜕𝑥& + 𝑉 𝑥 ψ 𝑥 𝑓(𝑡)

𝑖ℏ1𝑓(𝑡)

𝑑𝑓(𝑡)𝑑𝑡 = −

ℏ&

2𝑚ψ 𝑥𝑑&ψ 𝑥𝑑𝑥& + 𝑉 𝑥

𝑖ℏ1𝑓(𝑡)

𝑑𝑓(𝑡)𝑑𝑡 = B


• We integrate both sides and find:

where C is an integration constant that we may choose to be 0. Therefore

This determines f(t) to be

• Renaming the constant 𝐵 → 𝐸:

This is known as the time-independent Schrödingerwave equation, and it is a fundamental equation in quantum mechanics.

(Energies E are the eigenvalues of the equation)

Time Independent Schrödinger Wave Equation (II)

𝑖ℏo𝑑𝑓𝑓

�

�

= o𝐵𝑑𝑡�

�

→ 𝑖ℏ ln 𝑓 = 𝐵𝑡 + 𝐶

ln 𝑓 =𝐵𝑡𝑖ℏ

𝑓 𝑡 = 𝑒{g Vℏ⁄ = 𝑒eV{g ℏ⁄

𝑖ℏ1𝑓(𝑡)

𝑑𝑓(𝑡)𝑑𝑡 = E

−ℏ&

2𝑚𝑑&ψ 𝑥𝑑𝑥& + 𝑉 𝑥 ψ 𝑥 = 𝐸ψ 𝑥


• The wave function can be written as:

• The probability density becomes:

• The probability distributions are constant in time. This is a standing wave phenomena that is called the stationary state.

Solution – stationary state

Ψ 𝑥, 𝑡 = 𝜓(𝑥)𝑒eVfg

Ψ∗ 𝑥, 𝑡 Ψ 𝑥, 𝑡 = 𝜓& 𝑥 𝑒rVfg𝑒eVfg = 𝜓& 𝑥


• Because we are dealing with probabilities, we are likely to get different results for many measurements of the physical observables [position, momentum, energy etc…]

• The average measurement [called expectation value] of a given quantity can be calculated using wave functions. The expectation value of quantity 𝑥 is denoted 𝑥 in quantum mechanics.

Probabilities and expectation values


• We can change from discrete to continuous variables by using the probability 𝑃(𝑥, 𝑡) of observing the particle at a particular 𝑥.

• Using the wave function, the expectation value is:

• The expectation value of any function 𝑔 𝑥for a normalized wave function:

Discrete:

Continuous:

Continuous expectation values

�̅� =∑ 𝑁V𝑥V�V∑ 𝑁V�V

�̅� =∫ 𝑥𝑃 𝑥 𝑑𝑥rses

∫ 𝑃 𝑥 𝑑𝑥rses

𝑥 =∫ 𝑥Ψ∗ 𝑥, 𝑡 Ψ 𝑥, 𝑡 𝑑𝑥rses

∫ Ψ∗ 𝑥, 𝑡 Ψ 𝑥, 𝑡 𝑑𝑥rses

𝑔(𝑥) = o Ψ∗ 𝑥, 𝑡 𝑔(𝑥)Ψ 𝑥, 𝑡 𝑑𝑥rs

es

Abbreviated notation: 𝜓 𝑔 𝜓


Operators

Operator Expectation value

Momentum �̂� = −𝑖ℏ𝜕𝜕𝑥 𝑝 = −𝑖ℏo Ψ∗ 𝑥, 𝑡

𝜕Ψ 𝑥, 𝑡𝜕𝑥 𝑑𝑥

rs

es

Energy 𝐸� = 𝑖ℏ𝜕𝜕𝑡 𝐸 = 𝑖ℏo Ψ∗ 𝑥, 𝑡

𝜕Ψ 𝑥, 𝑡𝜕𝑡 𝑑𝑥

rs

es

ExerciseShow that the operators �̂� and 𝐸� indeed lead to a measurement of the

momentum and energy respectively.


Part IV: Solving the Schrödinger equation

Erwin Schrödinger1887-1961



• The simplest such system is that of a particle trapped in a box with infinitely hard walls that the particle cannot penetrate. This potential is called an infinite square well and is given by:

1D-infinite square well potential

𝑉 𝑥 = �+∞𝑥 ≤ 0, 𝑥 ≥ 𝐿00 < 𝑥 < 𝐿


• The wave function must be zero where the potential is infinite.

• Where the potential is zero inside the box, the Schrödinger waveequation becomes:

• The general solution is:

Solving the Schrödinger equation

𝑉 𝑥 = �+∞𝑥 ≤ 0, 𝑥 ≥ 𝐿00 < 𝑥 < 𝐿

𝑑&𝜓𝑑&𝑥 = −

2𝑚𝐸ℏ& 𝜓 = −𝑘&𝜓 with 𝑘 = 2𝑚𝐸 ℏ&⁄�

𝜓 𝑥, 𝑡 = 𝐴 sin 𝑘𝑥 + 𝐵 cos 𝑘𝑥


• Boundary conditions of the potential dictate that the wave function must be zero at x = 0 and x = L. This yields valid solutions for integer values of n such that kL = nπ.

• The wave function is now

• We normalize the wave function

• The normalized wave function becomes

• These functions are identical to those obtained for a vibrating string with fixed ends.

(with n=1,2,3…)

Quantization

𝜓= 𝑥 = 𝐴 sin𝑛𝜋𝑥𝐿

o 𝜓=∗ 𝑥 𝜓= 𝑥 𝑑𝑥 = 1rs

es𝐴& o 𝑠𝑖𝑛&

𝑛𝜋𝑥𝐿 𝑑𝑥 = 1

�

�

𝜓= 𝑥 =2𝐿

�sin

𝑛𝜋𝑥𝐿


• The special case of n = 1 is called the ground state energy:

wave functions probabilitydensity

position energy

Quantized energy

• The quantized wave number now becomes

• Solving for the energy yields

• Note that the energy depends on the integer values of n. Hence the energy is quantized and nonzero.

𝑘= =𝑛𝜋𝐿 =

2𝑚𝐸=ℏ&

�

𝐸= = 𝑛&𝜋&ℏ&

2𝑚𝐿& with n=1,2,3…

𝐸� =𝜋&ℏ&

2𝑚𝐿&


• Study of a particle trapped between x=0 and x=L in a realistic potential (no “infinite” walls).

Finite square-well potential

−ℏ&


𝑉 𝑥 = �𝑉�𝑥 ≤ 0

00 < 𝑥 < 𝐿𝑉�𝑥 ≥ 𝐿

Region IRegion IIRegion III


• In regions I and III, the Schrödinger equation becomes:

• By posing 𝛼& = 2𝑚 𝑉� − 𝐸 ℏ&⁄ , the equation can be written:

• Solutions has to tend to 0 when x tends ±∞:

• In region II, the Schrödinger equation is:


−ℏ&


−ℏ&

2𝑚1𝜓𝑑&ψ𝑑𝑥& = 𝐸 − 𝑉�

𝑑&𝜓𝑑𝑥& = 𝛼&𝜓

𝜓� 𝑥 = 𝐴𝑒�K

𝜓�� 𝑥 = 𝐵𝑒e�KRegion I, x<0

Region III, x>L

𝑑&𝜓��𝑑&𝑥 = −𝑘&𝜓�� with 𝑘 = 2𝑚𝐸 ℏ&⁄�

𝜓�� 𝑥 = 𝐶𝑒VdK + 𝐷𝑒eVdK Region II, 0 ≤ x ≤ LSolution:


• Boundary Conditions

– Continuity:

– “Smoothness”:

The wave function is non-zero outside of the box


𝜓� 𝑥 = 0 = 𝜓�� 𝑥 = 0 𝜓�� 𝑥 = 𝐿 = 𝜓�� 𝑥 = 𝐿

𝑑𝑑𝑥 𝜓� 𝑥 = 0 =

𝑑𝑑𝑥 𝜓�� 𝑥 = 0

𝑑𝑑𝑥 𝜓�� 𝑥 = 𝐿 =

𝑑𝑑𝑥 𝜓�� 𝑥 = 𝐿&

&


• Simple harmonic oscillators describe many physical situations: springs, diatomic molecules and atomic lattices.

• Substituting V(x)=½k(x-x0)2 [assuming x0=0 for simplification] into the wave

equation:

• Introducing: 𝛼& = U�ℏp

and 𝛽 = &U�ℏp

, we get:

A more realistic potential: the harmonic oscillator

𝑑&𝜓𝑑𝑥& = −

2𝑚ℏ& 𝐸 −

𝜅𝑥&

2 𝜓 = −2𝑚𝐸ℏ& +

𝑚𝜅𝑥&

ℏ& 𝜓

𝑑&𝜓𝑑𝑥& = 𝛼&𝑥& − 𝛽 𝜓


Hn(x): Hermite polynomial functions

Energy Solutions:

Wave function solutions:

Solutions

𝜓= 𝑥 = 𝐻= 𝑥 𝑒e�Kp &⁄

𝐸= = 𝑛 +12 ℏ 𝜅 𝑚⁄� = 𝑛 +

12 ℏ𝜔

𝐸� =12ℏ𝜔


• Barrier: region II (V=V0>0)

Assuming the energy of the particle E larger than V0

Barriers and tunneling

𝑑&𝜓𝑑𝑥& = −𝑘&𝜓

𝑘� = 𝑘�� =2𝑚𝐸�

ℏ

𝑘�� =2𝑚 𝐸 − 𝑉�

�

ℏ

𝑑&𝜓�𝑑𝑥& +

2𝑚ℏ& 𝐸𝜓� = 0

𝑑&𝜓��𝑑𝑥& +

2𝑚ℏ& 𝐸𝜓�� = 0

𝑑&𝜓��𝑑𝑥& +

2𝑚ℏ& 𝐸 − 𝑉� 𝜓�� = 0

Region I (V=0):

Region II (V=V0):

Region III (V=0):


• From the 3 preceding equations:

• Then boundary conditions, etc…

Solutions

Region I (V=0):

Region II (V=V0):

Region III (V=0):

𝜓� 𝑥 = 𝐴𝑒Vd�K + 𝐵𝑒eVd�K

𝜓�� 𝑥 = 𝐶𝑒Vd��K + 𝐷𝑒eVd��K

𝜓�� 𝑥 = 𝐹𝑒Vd��K + 𝐺𝑒eVd��K


• The probability of the particles being reflected R or transmitted T is:

And after much calculations:

Transmitted / reflected waves

𝜓� 𝑥, 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 = 𝐴𝑒Vd�K

𝜓� 𝑥, 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 = 𝐵𝑒eVd�K

𝜓�� 𝑥, 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 = 𝐹𝑒Vd��K = 𝐹𝑒Vd�K

𝑅 =𝜓� 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 &

𝜓� 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 & =𝐵×𝐵𝐴×𝐴

𝑇 =𝜓�� 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 &

𝜓� 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 & =𝐹×𝐹𝐴×𝐴

𝑇 = 1 +𝑉�&𝑠𝑖𝑛& 𝑘��𝐿4𝐸 𝐸 − 𝑉�

e�


• What if the energy of the particle considered is smaller than the potential energy (E<V0) ?

• Quantum Mechanics ?

What if…


• Region II solution:

• Transmission:

If 𝜅𝐿 ≫ 1, T can be simplified:

Exponential

Tunneling

𝑇 = 1 +𝑉�&𝑠𝑖𝑛ℎ& 𝜅𝐿4𝐸 𝑉� − 𝐸

e�

𝜓�� 𝑥 = 𝐶𝑒�K + 𝐷𝑒e�K 𝜅 =2𝑚 𝑉� − 𝐸

�

ℏwith

𝑇 = 16𝐸𝑉�

1 −𝐸𝑉�

𝑒e&��


a-particle

Example: a-decay


• Electrons tunnel through the gap to be collected by the tip.

• Current collected is very sensitive to the distance between the tip and the surface.

Heinrich Rohrer & Gerd BinnigNobel Prize in Physics 1986

Example: scanning tunneling microscope


Part V: The Hydrogen atom


• Hydrogen = proton + electron system

• Potential:

• The 3D time-independent Schrödinger Equation:

Schrödinger equation and the Hydrogen atom

𝑉 𝑟 = −𝑒&

4𝜋𝜀�𝑟

−ℏ&

2𝑚1

𝜓 𝑥, 𝑦, 𝑧𝜕&𝜓 𝑥, 𝑦, 𝑧

𝜕𝑥& +𝜕&𝜓 𝑥, 𝑦, 𝑧

𝜕𝑦& +𝜕&𝜓 𝑥, 𝑦, 𝑧

𝜕𝑧& = 𝐸 − 𝑉 𝑟


• The Coulomb potential has a radial symmetry V(r): switch to the spherical polar coordinate system.

• Schrödinger Equation (in spherical polar coordinate system):

Radial symmetry of the potential

𝑉 𝑟 = −𝑒&

4𝜋𝜀�𝑟

1𝑟&

𝜕𝜕𝑟 𝑟&

𝜕𝜓𝜕𝑟 +

1𝑟& sin 𝜃

𝜕𝜕𝜃 sin 𝜃

𝜕𝜓𝜕𝜃 +

1𝑟&𝑠𝑖𝑛&𝜃

𝜕&𝜓𝜕𝜙& +

2𝜇ℏ& 𝐸 − 𝑉 𝜓 = 0

𝜓 𝑟, 𝜃, 𝜙Wave function:


• In the spherical polar coordinate, the wavefunction:

• Solution may be the product of three functions:

• Derivatives of y (the ones useful to solve the Schrödinger equation):

Separation of variables (I)

𝜓 𝑟, 𝜃, 𝜙

𝜓 𝑟, 𝜃, 𝜙 = 𝑅 𝑟 𝑓 𝜃 𝑔 𝜙

𝜕𝜓𝜕𝑟 = 𝑓𝑔

𝜕𝑅𝜕𝑟

𝜕𝜓𝜕𝜃 = 𝑅𝑔

𝜕𝑓𝜕𝜃

𝜕&𝜓𝜕𝜙& = 𝑅𝑓

𝜕&𝑔𝜕𝜙&


• Into the Schrödinger equation:

• × 𝑟&𝑠𝑖𝑛&𝜃 𝑅𝑓𝑔⁄ :

Depends on r and q Depends on f only

Separation of variables (II)

𝑓𝑔𝑟&

𝜕𝜕𝑟 𝑟&

𝜕𝑅𝜕𝑟 +

𝑅𝑔𝑟& sin 𝜃


𝜕𝑓𝜕𝜃 +

𝑅𝑓𝑟&𝑠𝑖𝑛&𝜃

𝜕&𝑔𝜕𝜙& +

2𝜇ℏ& 𝐸 − 𝑉 𝑅𝑓𝑔 = 0

−𝑠𝑖𝑛&𝜃𝑅

𝜕𝜕𝑟 𝑟&

𝜕𝑅𝜕𝑟 −

2𝜇ℏ& 𝑟

&𝑠𝑖𝑛&𝜃 𝐸 − 𝑉 −sin 𝜃𝑓


𝜕𝑓𝜕𝜃 =

1𝑔𝜕&𝑔𝜕𝜙& = 𝐶


• One can solve the right hand part of the equation

• Assuming 𝐶 = −𝑚ℓ& (with a little bit of clairvoyance!):

• Solution: 𝑒VUℓ£ (with 𝑚ℓ = integer (positive/negative) or zero)

Azimuthal equation

𝑑&𝑔𝑑𝜙& = −𝑚ℓ

&𝑔


• Now looking at the left side of the equation:

• And rearranging the equation by separating the variables (again):

Depends on r Depends on q

Left side of the equation

−𝑠𝑖𝑛&𝜃𝑅

𝜕𝜕𝑟 𝑟&

𝜕𝑅𝜕𝑟 −

2𝜇ℏ& 𝑟

&𝑠𝑖𝑛&𝜃 𝐸 − 𝑉 −sin 𝜃𝑓


𝜕𝑓𝜕𝜃 = −𝑚ℓ

&

1𝑅𝜕𝜕𝑟 𝑟&

𝜕𝑅𝜕𝑟 +

2𝜇𝑟&

ℏ& 𝐸 − 𝑉 =𝑚ℓ&

𝑠𝑖𝑛&𝜃 −1

𝑓 sin 𝜃𝜕𝜕𝜃 sin 𝜃

𝜕𝑓𝜕𝜃 = 𝐶


• Setting the constant to ℓ ℓ + 1 [with even more clairvoyance!]:

Radial equation

Angular equation

Radial and angular equations

1𝑟&

𝑑𝑑𝑟 𝑟&

𝑑𝑅𝑑𝑟 +

2𝜇ℏ& 𝐸 − 𝑉 −

ℏ&

2𝜇ℓ ℓ + 1𝑟& 𝑅 = 0

1sin 𝜃

𝑑𝑑𝜃 sin 𝜃

𝑑𝑓𝑑𝜃 + ℓ ℓ + 1 −

𝑚ℓ&

𝑠𝑖𝑛&𝜃 𝑓 = 0


• Solving for ℓ = 0 𝑚ℓ = 0 :

becomes

Solving ¥¥:

𝑟& ¥¦¥:

and introducing 𝑉 𝑟 :

Solution of the radial equation (I)

1𝑟&

𝑑𝑑𝑟 𝑟&

𝑑𝑅𝑑𝑟 +

2𝜇ℏ& 𝐸 − 𝑉 −

ℏ&

2𝜇ℓ ℓ + 1𝑟& 𝑅 = 0

1𝑟&

𝑑𝑑𝑟 𝑟&

𝑑𝑅𝑑𝑟 +

2𝜇ℏ& 𝐸 − 𝑉 𝑅 = 0

𝑑&𝑅𝑑𝑟& +

2𝑟𝑑𝑅𝑑𝑟 +

2𝜇ℏ& 𝐸 +

𝑒&

4𝜋𝜀�𝑟𝑅 = 0


• Solution:

• Solving the equation with:

= 0 = 0

The Bohr RadiusGround State Energy Level

Solution of the radial equation (II)

𝑑&𝑅𝑑𝑟& +

2𝑟𝑑𝑅𝑑𝑟 +

2𝜇ℏ& 𝐸 +

𝑒&

4𝜋𝜀�𝑟𝑅 = 0

𝑅 𝑟 = 𝐴𝑒e: §¨⁄

1𝑎�&+2𝜇ℏ& 𝐸 +

2𝜇𝑒&

4𝜋𝜀�ℏ&−2𝑎�

1𝑟 = 0

𝑎� =4𝜋𝜀�ℏ&

𝜇𝑒&𝐸 = −

ℏ&

2𝜇𝑎�&= −𝐸�


Radial Wave Functions Rnℓ(r)• Boundary conditions lead to:

• ℓ = 0, 1, 2, 3, …• 𝑚ℓ ≤ ℓ


Yℓm(q,f): Spherical Harmonics

• Combining the angular and azimuthal solutions:

Solution of the angular and azimuthal equations

𝑌ℓU 𝜃, 𝜙 = 𝑓 𝜃 𝑔 𝜙


• The three quantum numbers:– n Principal quantum number– ℓ Orbital angular momentum quantum number– 𝑚ℓ Magnetic quantum number

• The boundary conditions:– n = 1, 2, 3, 4, . . . Integer– ℓ = 0, 1, 2, 3, . . . , n − 1 Integer– 𝑚ℓ = − ℓ, − ℓ + 1, . . . , 0, 1, . . . , ℓ− 1, ℓ Integer

• The restrictions for quantum numbers:– n > 0– ℓ < n– | 𝑚ℓ| ≤ ℓ

Solution of the Schrödinger equation for the Hydrogen atom

𝜓=ℓUℓ 𝑟, 𝜃, 𝜙 = 𝑅=ℓ 𝑟 𝑌ℓUℓ 𝜃, 𝜙


3 different electron states

Probability distribution functions


Part VI: Quantum numbers


• The three quantum numbers:– n Principal quantum number– ℓ Orbital angular momentum quantum number– 𝑚ℓ Magnetic quantum number

• The boundary conditions:– n = 1, 2, 3, 4, . . . Integer– ℓ = 0, 1, 2, 3, . . . , n − 1 Integer– 𝑚ℓ = − ℓ, − ℓ + 1, . . . , 0, 1, . . . , ℓ− 1, ℓ Integer

• The restrictions for quantum numbers:– n > 0– ℓ < n– | 𝑚ℓ| ≤ ℓ

Solution of the Schrödinger equation for the Hydrogen atom

𝜓=ℓUℓ 𝑟, 𝜃, 𝜙 = 𝑅=ℓ 𝑟 𝑌ℓUℓ 𝜃, 𝜙


• In the hydrogen atom, this is the number of the Bohr orbit (n=1,2,3… no upper limit)

• Associated with the solution of 𝑅(𝑟)

• Quantized energy:

• (-) sign: proton-electron system bound

Principal Quantum Number n

𝐸= = −𝜇2

𝑒&

4𝜋𝜀�ℏ

& 1𝑛& = −

𝐸�𝑛&


• Associated with the solutions of 𝑅 𝑟 and 𝑓(𝜃)– Boundary Conditions: ℓ = 0, 1, … , 𝑛 − 1

• Classical Orbital Momentum:

• Quantum Orbital Momentum:

• Note: ℓ = 0 state à quantum orbital momentum L=0

This disagrees with Bohr’s semi-classical “planetary” model of electrons orbiting a nucleus 𝐿 = 𝑛ℏ.

Orbital angular momentum quantum number ℓ

𝐿 = 𝑟×�⃗�

𝐿 = ℓ ℓ + 1� ℏ


• Energy is independent of the quantum number ℓ, we say the energy level is degenerate with respect to ℓ. Note: this is only true for the Hydrogen atom.

• States:

– ℓ = 0 1 2 3 4 5 …– Letter s p d f g h …

• Atomic states are referred to by their n and ℓ.

• A state with n = 2 and ℓ = 1 is called a 2p state. Note: the boundary conditions require ℓ < n.

(sharp)

(principal)

(diffuse)

(fundamental)

More on quantum number ℓ


• The angle 𝜙 is a measure of the rotation about the z axis.

• The solution for 𝑔 𝜙 specifies that 𝑚ℓ is an integer and related to the z component of 𝐿.

Note: One cannot know 𝐿 exactly, as this would violate the uncertainty principle.

The magnetic quantum number 𝑚ℓ

𝐿 = 𝑚ℓℏ

• Figure: the relationship of 𝐿, 𝐿, ℓ and 𝑚ℓfor ℓ = 2

• 𝐿 = ℓ ℓ + 1� ℏ = 6� ℏ is fixed by the value of ℓ.

• Only certain orientations of 𝐿 are possible however. Those are given by the projection of 𝐿 on the quantization axis 𝐿. This is called space quantization.


• Samuel Goudsmit and George Uhlenbeck in Holland proposed that the electron must have an intrinsic angular momentum and therefore a magnetic moment (1925)

• Paul Ehrenfest showed that the surface of the spinning electron should be moving faster than the speed of light!

• In order to explain experimental data, Goudsmit and Uhlenbeckproposed that the electron must have an intrinsic spin quantum numbers = ½. [ Number of possible values: 2s+1 = 2 à ms=-½ or ms=½]

Intrinsic spin quantum number 𝑚®


• Does not appear from the solutions of the Schrödinger equation

• Appears when solving the problem in a relativistic way

• For the electron: ms = +½ or ms = -½

• The spinning electron reacts similarly to the orbiting electron in a magnetic field.

Intrinsic spin quantum number 𝑚®


• If the principal quantum number n for a certain electronic state is equal to 3, what are the possible values of the orbital (angular momentum) quantum number ℓ?

• If the orbital quantum number ℓ for a certain electronic state is equal to 2, what are the possible values for the magnetic quantum number 𝑚ℓ?

• How many distinct electronic states are there with n=2?

Exercise


• Experimentally: By the 1920s, a fine structure in the spectra lines of Hydrogen and other atoms has been observed. Spectra lines appeared to be split in the presence of an external magnetic field.

INTERPRETATION:

• Energy is independent of the quantum number ℓà the energy level is degenerate with respect to ℓ

• Example: Considering n=2 and ℓ =1à 𝑚ℓ = -1,0,1 e.g. 3 quantum states are degenerate at the same energy

These 3 magnetic states would behave differently under a magnetic field resulting in the degeneracy being lifted !

Atomic fine structure


• Model:

– electron circulating around the nucleus à Loop of current 𝐼 = ¥°¥g= ±

²• T, time it takes for the electron to make one rotation: 𝑇 = 2𝜋𝑟 𝑣⁄• Introducing 𝑝 = 𝑚𝑣 à 𝑇 = 2𝜋𝑚𝑟 𝑝⁄

– Magnetic Moment induced: 𝜇 = 𝐼𝐴 = ±@&>U:

𝜋𝑟&

• Simplification: 𝜇 = ±&U

𝑟𝑝, introducing 𝐿 = 𝑟𝑝 gives 𝜇 = ±&U

𝐿

Magnetic moment

Electron magnetic moment: �⃗� = −𝑒2𝑚𝐿


• Potential energy of the dipole created by the electron orbiting around the nucleus (under a magnetic field B):

• One can only know one component of 𝐿: 𝐿 = 𝑚ℓℏ

• Along z, the magnetic moment becomes:

• Quantization:Bohr MagnetonµB = 9.274 x 10-24 J/T

The normal Zeeman effect (I)

𝑉{ = −�⃗� ³ 𝐵

𝜇 =𝑒ℏ2𝑚𝑚ℓ = −𝜇{𝑚ℓ

𝑉{ = −𝜇𝐵 = +𝜇{𝑚ℓ𝐵


• When a magnetic field is applied, the 2p level of atomic hydrogen is split into three different energy states with energy difference of ∆𝐸 = 𝜇{𝐵∆𝑚ℓ.

mℓ Energy

1 E0 + μBB

0 E0

−1 E0 − μBB

Fine Structure

Potential energy of the dipole:

The normal Zeeman effect (II)

𝑉{ = −𝜇𝐵 = +𝜇{𝑚ℓ𝐵


Transition 2p à 1s

Fine structure


• An atomic beam of particles in the ℓ = 1 state pass through a magnetic field along the z direction.

• The 𝑚ℓ = +1 state will be deflected down, the 𝑚ℓ = −1 state up, and the 𝑚ℓ = 0 state will be undeflected.

• If the space quantization were due to the magnetic quantum number 𝑚ℓ only, and since the number of 𝑚ℓ states is always odd (2ℓ + 1), the experiment should produce an odd number of lines à But it doesn’t. Why?

z

Stern & Gerlach experiment (1922)

𝑉{ = −𝜇𝐵 𝐹 = − 𝑑𝑉{ 𝑑𝑧⁄ = 𝜇 𝑑𝐵 𝑑𝑧⁄

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