4d3zz-1

1Å

Applied Quantum Mechanics

lecture site: http://www.cond-mat.de/teaching/QM/ slides, links, exercises

register for course through CAMPUS Office! exercises and some material through L2P

course policies: • exercises:

• hand in before lecture! • need to solve at least 50% of problems • prepare to present your solutions at blackboard

• exam: • final exam planned two weeks after last lecture

textbooks

D.A.B. Miller: Quantum Mechanics for Scientists and Engineers

Cambridge Univ. Press

D.J. Griffiths: Introduction to

Quantum Mechanics Cambridge Univ. Press

F. Schwabl: Quantum Mechanics

Springer

If, in some cataclysm, all of scientific knowledge were to be destroyed, and only one sentence passed on to the next generations of creatures, what statement would contain the most information in the fewest words? I believe it is the atomic hypothesis (or the atomic fact, or whatever you wish to call it) that all things are made of atoms – little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. In that one sentence, you will see, there is and enormous amount of information about the world, if just a little imagination and thinking are applied.

Lecture 1 of The Feynman Lectures on Physics, Vol. I (1961)

Matter is made of atoms

periodic table

H

Li

Na

K

Rb

Cs

Fr

Be

Mg

Ca

Sr

Ba

Ra

Sc

Y

Lu

Lr

Ti

Zr

Hf

Rf

V

Nb

Ta

Db

Cr

Mo

W

Sg

Mn

Tc

Re

Bh

Fe

Ru

Os

Hs

Co

Rh

Ir

Mt

Ni

Pd

Pt

Cu

Ag

Au

B

Al

Ga

In

Tl

C

Si

Ge

Sn

Pb

N

P

As

Sb

Bi

O

S

Se

Te

Po

F

At

Ne

Xe

Rn

Ce

Th

Pr

Pa

Nd

U

Pm

Np

Sm

Pu

Eu

Am

Gd

Cm

Tb

Bk

Dy

Cf

Ho

Es

Er

Fm

Tm

Md

Yb

No

La

Ac

He

Zn

Cd

Hg

I

Kr

ArCl

Br

atomic radii

typical size 10-10 m = 1 Å

exercise

given Ne electrons, Ni atomic nuclei of mass Mα und charge Zα,

solve:

The underlying laws necessary for the mathematical theory of a large part of physics and the whole of chemistry are thus completely known, and the difficulty is only that exact applications of these laws lead to equations which are too complicated to be soluble. It therefore becomes desirable that approximate practical methods of applying quantum mechanics should be developed, which can lead to an explanation of the main features of complex atomic systems without too much computation. P.M.A Dirac, Proceedings of the Royal Society A123, 714 (1929)

H = ��22m

Ne�

j=1

⇤2j �Ni�

�=1

�22M�

⇤2� �1

4⇥�0

Ne�

j=1

Ni�

�=1

Z�e2

|rj � R�|+1

4⇥�0

Ne�

j<k

e2

|rj � rk |+1

4⇥�0

Ni�

�<⇥

Z�Z⇥e2

|R� � R⇥ |

i~ ��t�(⇥r1, . . . ,⇥rNe ; ⇥R1, . . . , ⇥RNi ; t) = H�(⇥r1, . . . ,⇥rNe ; ⇥R1, . . . , ⇥RNi ; t)

giant magnetoresistance

Peter Grünberg (Jülich) and Albert Fert (Paris), 1988 Nobel prize in Physics 2007

scanning tunneling microscope

Gerd Binnig and Heinrich Rohrer, IBM Rüschlikon, 1981 Nobel Prize in Physics 1986

scanning tunneling microscope

http://www2.fz-juelich.de/ibn/microscope_e

seeing and manipulating atoms

Microelectronics: 14nm process14 nm Tri-gate Transistor Fins

11

8 nm Fin Width

42 nm Fin Pitch

Si Substrate

Gate

14 nm Tri-gate Transistor Fins

How Small is 14 nm?

10

Very small

1 1 1 1 10 100 10 100

meter millimeter micrometer nanometer

10 100

Fly 7 mm

Mite 300 um

Virus 100 nm

Blood Cell 7 um

Silicon Atom 0.24 nm

Mark 1.66 m

10 0.1

14 nm Process

double-slit experiment with electrons

http://www.hitachi.com/rd/research/em/doubleslit.html

see also: R.P. Feynman: Feynman Lectures on Physics, Vol. 3 Ch. 1: Quantum Behavior

Physik Journal 9 Okt. 2010, p. 37

not just electrons behave as waves ...

quantum vs. classical behavior

time-dependent Schödinger equation

particle wavep = ~kE = ~!

dispersion relation

photons ! = ck

electrons E =p2

2m

time-dependent Schrödinger equation

solution e i(kz�!t)

1

c2@2�

@t2=

@2�

@z2

i~@ @t= �

~22m

@2

@z2

1st derivative: complex waves,

initial-value problem:

(~r , t + �t) ⇡ (~r , t) +@ (~r , t)

@t�t

i~��(⇥r , t)�t

=

✓�~22m⇥r2 + V (⇥r , t)

◆�(⇥r , t)

separation of variables

time-independent potential

ansatz: (~r , t) = A(t) (~r)

A(t) = A0 e�iEt/~

general solution: linear combination of eigenstates

�(⇥r , t) =X

n

an e�iEnt/~ �n(⇥r)

i~��(⇥r , t)�t

=

✓�~22m⇥r2 + V (⇥r)

◆�(⇥r , t)

i~⇥A(t)⇥t

�(⇤r) = A(t)E �(⇤r) = A(t)

✓�~22m⇤r2 + V (⇤r)

◆�(⇤r)

time-independent Schrödinger equation (eigenvalue problem)

✓�~22m⇥r2 + V (⇥r)

◆�(⇥r) = E �(⇥r)

exercise

given Ne electrons, Ni atomic nuclei of mass Mα und charge Zα,

solve:

The underlying laws necessary for the mathematical theory of a large part of physics and the whole of chemistry are thus completely known, and the difficulty is only that exact applications of these laws lead to equations which are too complicated to be soluble. It therefore becomes desirable that approximate practical methods of applying quantum mechanics should be developed, which can lead to an explanation of the main features of complex atomic systems without too much computation. P.M.A Dirac, Proceedings of the Royal Society A123, 714 (1929)

H = ��22m

Ne�

j=1

⇤2j �Ni�

�=1

�22M�

⇤2� �1

4⇥�0

Ne�

j=1

Ni�

�=1

Z�e2

|rj � R�|+1

4⇥�0

Ne�

j<k

e2

|rj � rk |+1

4⇥�0

Ni�

�<⇥

Z�Z⇥e2

|R� � R⇥ |

i~ ��t�(⇥r1, . . . ,⇥rNe ; ⇥R1, . . . , ⇥RNi ; t) = H�(⇥r1, . . . ,⇥rNe ; ⇥R1, . . . , ⇥RNi ; t)

particle in a box

discrete energies zero-point energy

increasing number of nodes

symmetry of potential symmetry of solutions (density)

even/odd eigenfunctions

En =~22m

✓n⇡

Lz

◆2boundary conditions ⇒ quantization

⇥n(z) =

r2

Lzsin

✓n� z

Lz

◆

particle in a box

En =~22m

✓n⇡

Lz

◆2boundary conditions ⇒ quantization

⇥n(z) =

r2

Lzsin

✓n� z

Lz

◆

typical units

http://physics.nist.gov/cuu/Constants/index.html h = 6.626068 10-34 Js mel = 9.109382 10-31 kg e = 1.602176 10-19 C E =

~2 k22mel

why use Å and eV?1 Å = 10-10 m 1 eV = 1.602176 10-19 J

E [in J] = 6.10 10-39 (k [in m-1])2

E [in eV] = 3.81 (k [in Å-1])2

from math import pihbar = 1.0546e-34 # h/2pi in Jsme = 9.1094e-31 # electron mass in kge = 1.6022e-19 # electron charge in C

const=hbar**2/(2*me) # print(const) --> 6.10457966496e-39L = 1e-9 # in m (1 nm = 10 \AA)k1=pi/LE1=const*k1**2 # ground-state energy --> 6.024979e-20 J

const=hbar**2/(2*me)/(1e-10**2*e) # print(const) --> 3.81012337097L = 10 # in \AAk1=pi/LE1=const*k1**2 # ground-state energy --> 3.760441e-01 eV

calculating with Google

atomic unitsatomic units

! = 1.0546 ·10−34 Js [ML2T−1]me = 9.1094 ·10−31 kg [M ]e = 1.6022 ·10−19 C [Q]

4πϵ0 = 1.1127 ·10−10 F/m [M−1L−3T2Q2]

http://physics.nist.gov/cuu/

solve

! = 1 a20 me / t0

me = 1 mee = 1 e

4πϵ0 = 1 t20 e2 / a30 me

to obtain

1 a.u. length = a0 = 4πϵ0!2

mee2≈ 5.2918 ·10−11 m

1 a.u. mass = me = ≈ 9.1095 ·10−31 kg

1 a.u. time = t0 = (4πϵ0)2!3

mee4≈ 2.4189 ·10−17 s

1 a.u. charge = e = ≈ 1.6022 ·10−19 C