• Nem Talált Eredményt

PETER PAZMANY CATHOLIC UNIVERSITYConsortiummembersSEMMELWEIS UNIVERSITY, DIALOG CAMPUS PUBLISHER

N/A
N/A
Protected

Academic year: 2022

Ossza meg "PETER PAZMANY CATHOLIC UNIVERSITYConsortiummembersSEMMELWEIS UNIVERSITY, DIALOG CAMPUS PUBLISHER"

Copied!
37
0
0

Teljes szövegt

(1)

**Molekuláris bionika és Infobionika Szakok tananyagának komplex fejlesztése konzorciumi keretben

***A projekt az Európai Unió támogatásával, az Európai Szociális Alap társfinanszírozásával valósul meg.

Development of Complex Curricula for Molecular Bionics and Infobionics Programs within a consortial* framework**

Consortium leader

PETER PAZMANY CATHOLIC UNIVERSITY

Consortium members

SEMMELWEIS UNIVERSITY, DIALOG CAMPUS PUBLISHER

The Project has been realised with the support of the European Union and has been co-financed by the European Social Fund ***

(2)

PHYSICS FOR NANOBIO-TECHNOLOGY

Principles of Physics for Bionic Engineering

Chapter 8.

Many Body Problem and Statistical Model

(A nanobio-technológia fizikai alapjai )

(Soktest probléma és statisztikai modellek)

Árpád I. CSURGAY,

Ádám FEKETE, Kristóf TAHY, Ildikó CSURGAYNÉ

(3)

Table of Contents

8. Multiple Body Problem and Statistical Models

1. Multiple Body System with Negligible Interaction between Identical Particles

2. Equilibrium in Multiple Body Systems 3. FermiDirac Statistics

(4)

1. Multiple Body System with Negligible Interaction between Identical Particles

If the interactions between particles are negligible, then the

potential energy does not depend on the relative position of the particles, thus

The solution of the time-independent Schrödinger equation can be given as

This is the product of N single-electron functions.

  



 

  

N

i

i i

i i

i E

1 m

pot 2

2 r

H

H

r1, r2, , rN

   

r1r2

 

rN .

(5)

This expression holds if and only if every member is constant, thus

       

 

1 1 1 ,

 

.2 2 2 2 2 2 , ,

1 1 1

N N N N

N

N E

E E

r r

H

r r

H r

r H

 

   

1

   

1

   

.

1

2 2 2 2

2 1

1 1 1

1

N E

N N N

N

H r

r r r H

r

r H

 

 

2 .

1 1

2 1

2 2 3

1 1

1 2

N N

N N

N N

E  

H H H

(6)

The eigenvalue problem for N particles has been reduced to N single-electron eigenvalue problems

The energy eigenvalue of the system of N particles is

and the system‟s eigenfunction can be composed as a linear combination of the products of single particle eigenfunctions

EN

E E

E12 

r1, r2, , rN

   

r1 r2

 

rN .

(7)

Multiple body systems of identical particles

There are plenty of identical particles (e.g. electrons) in macroscopic bodies.

In case of two identical particles the eigenvalue problem reads as

If we replace the two particles by each other (instead of r1 we write r2, and instead of r2 we write r1 ), the eigenvalue-equation remains the same, thus the replacement does not change the E eigenvalue.

H1H2

 

 r r1, 2

Epot

r r1, 2

 

 r r1, 2

 E

r r1, 2

.

(8)

From this fact it follows that r1, r2 cannot be else, but constant  r2, r1

it follows that a2 = 1.

Two possibilities can be distinguished

Symmetric Anti-symmetric

r2, r1

1

r1, r2

 

a

r2, r1

 1

r1, r2

a

2, 1

a

1, 2

a2

2, 1

,

r r   r r   r r

(9)

Results of measurements depend only on  .

In quantum mechanics the exchange of identical particles does not change the results of measurements.

Identical particles are indistinguishable.

In classical physics we can follow the „history‟ of every individual particle, independently of each other.

In quantum physics (i.e. for matter-waves) it is impossible to distinguish individual particles.

As soon as the wave functions of two particles overlap, the two particles become indistinguishable.

(10)

The wave functions of many body systems are either symmetric or anti-symmetric for the exchange of identical particles.

There are two types of particles in nature : a particle is either a

“fermion”, or a “boson”.

Symmetric Anti-symmetric

BOSON

symmetric wave function Integer „spin‟ (0, 1, 2)

FERMION

anti-symmetric wave function spin 1/2

photon electron, proton, etc.

r2, r1

r1, r2

 

r2, r1

r1, r2

(11)

Two Identical Particles

Indistinguishable

Symmetric Bosons Anti-symmetric

Fermions Classically

distinguishable

Single electron states ) 1 ( )

2 ( )

1 , 2 (

), 2 ( )

1 ( )

2 , 1 (

b a

b a

) 1 ( )

2 2 (

1

) 2 ( )

1 2 (

1

) 1 , 2 ( )

2 , 1 (

b a

b a

 (1, 2) 

1 (1) (2) 2

1 (2) (1) 2

a b

a b

 

 

 

 ...

), ( ),

(r b r

a

 

(12)

In-distinguishability  Pauli principle  Slater determinant N-Particle System

N Single Particle wave function

N-Particle wave function (Slater determinant)

Normalization

 

     

     

     

1 1 1

2 2 2

1 2

1 2

1, 2, , 1

!

1 2

N N N

N N N

N

N

  

  

  

   

rrN

 

r

1 , 2 ,,

(13)

2. Equilibrium in Multiple Body Systems

The microstate of a system is specified by its wave function

In general the energy eigenstates are degenerate. More than one microstate belongs to an energy E.

The distribution of electrons along energy is called macrostate of a system.

A macrostate is realized by many different microstates.

Macrostate Microstates

.

 

1 , 2

 

2 , 3

 

3 ,

1 E N E N E

N

 

E ,

N1, 2, 3, , Z.

(14)

Two fundamental laws of statistical physics

1. In a closed system the probability of every microstate is the same;

2.The closed system quickly approaches the most probable macrostate;

The most probable macrostate is called equilibrium.

Microstate Ψ Macrostate N (E) Energy eigenvalues E1, E2, ... ,En, ....

Number of microstates Z1, Z2, ... , Zn, ....

Number of electrons at E N1, N2 …, Nn, ...

(15)

Closed system: The total number of electrons and the total energy of the system is constant (does not change)

How many microstates realize this macrostate in case of molecules (MB), bosons (BE) and fermions (FD)?

1 2

0

, .

i i

i i i

i

N N N N N

N E E

     

(16)

In-distinguishable

Bosons (BE) Fermions (FD)

any number of bosons at a microstate

only one fermion at a microstate (Pauli)

Classical particles are distinguishable (M-B) Number of microstates that realize a macrostate

1 2

F-D

1 2

i i

Z Z Z

w N N N

 

  

     

    

M-B

! .

!

Ni

i i i

w N Z

N

B-E

i i 1

i i

N Z

w N

 

 

  

 

(17)

Number of microstates that realize a macrostate

The most probable macrostate is called equilibrium.

The probability of a macrostate is proportional to the number of microstate that realize the macrostate.

Bosons (BE) Fermions (FD)

 



 

 

 

 

 

 



 

 

i i i i

i

i i

i

i i

N Z

N

Z N

Z

N Z N

Z N

w Z

!

!

!

2 2 1

1 D

-

F

 

 

 

 

 

  

i i i

i i

i i

i i

Z N

Z N

N Z w N

! 1

!

! 1

1

E - B

(18)

Equilibrium ~ maximum number of microstates under closed system conditions.

Number of particles and total energy do not change In case of fermions (FermiDirac statistics)

Stirling approximation: ln n! nln nnnln n

, 0 .

i i i

i i

N

N E

N E

 

!.

!

!

D -

F



 

 

i i i i

i

i i

i

N Z

N

Z N

w Z

 

 

   

ln ln ln

.

ln

! ln

! ln ln F-D

i i

i i

i i

i i

i

i i

i i

N N

N Z

N Z

Z Z

N N

Z Z

w

(19)

In a closed system there are N fermions with total energy E0. From the solution of the quantum mechanical single electron problem we know, that the fermions can have discrete energies E1, E2, …, En,… At the energies there are Z1, Z2, …, Zn, …

microstates.

Determine the equilibrium macrostate N1(E1), N2(E2), …, Nn(En), …

The maximum of wF-D has to be found, under the conditions and

,

 0

i

Ni

N 0

 0.

i

i iE N E

(20)

We are looking for the conditional maximum of

Let us introduce  and β as unknown parameters. According to the method of Lagrange multiplicators, the extremum problem can be formulated as

   

ln ln ln

.

ln F-D

i

i i

i i

i i

i

i Z Z N Z N N N

Z w

 , 2 , 1 for

, 0

ln 0   

 

 

 

 



 

 

 

w N

N E

N E i

Nii ii i i

ln i 0, for 1, 2,

i

w E i

N  

    

(21)

Let us introduce:

With these new notations

   

ln ln ln

.

ln F-D

i

i i

i i

i i

i

i Z Z N Z N N N

Z w

 

F-D

F-D

ln 0 ln 1 1 ln 0, for 1, 2,

ln 0, .

e i 1

i i i

i

i i i

i i E

i

w Z N N i

N

Z N Z

E N

N    

        

     

F 23

B

B B

1 Ws

, , k 1.38 10 .

k k K

E

T T

      

 

F B

F-D F-D

k

1 .

e 1

e 1

i i

i

i E i i E E

T

N ZNZ

 

(22)

Instead of the Lagrange constants α, β we introduced two new constants T, EF.

We shall see later that T will be identical to the absolute temperature, and the energy EF will play significant role in material science as the well known Fermi energy level.

The total energy E0 and the number of particles N determine the new parameters T, EF

0 F

, , , .

N E     T E

(23)

Population in Equilibrium

Fermi function

at T = 0

F

kB

( ) 1

e 1

E E

T

f E

0

0

F 0

F

1, if

( ) .

0, if

T

E E

f E E E

 

  

F B

F-D

k

1 ( ),

e 1

i i Ei E i

T

NZ Zf E

T 0 T 0

EF

EF

0 1 0 1

Fermi level

Population

Energy

(24)

At T = 0 every microstate is occupied for and no electrons in microstates

If we increase the total energy, then the population of the electrons will change according to the Fermi function.

A heuristic model can be established:

1. from the solution of the single-electron problem we identify the energy eigenvalues and microstates;

2. at absolute zero temperature we fill all the microstates below the Fermi level (Pauli principle);

3. At T = 0 the population of electrons is determined by the Fermi function.

F0 , EE

F0. EE

(25)

Example 1: Electrons in a one-dimensional potential box

At T = 0 the first and second electrons we drop into a one-dimensional

potential box, occupy the ground state with opposite spins. The third and fourth electrons occupy the first excited state with opposite spins, and so on. The N- body wave function can be constructed as a Slater determinant, and the Fermi level of the system will be the highest energy level populated.

Example 2: Periodic system of elements

If we start with a hydrogen-like nucleus of positive charge Z  e, and consider the possible orbitals of a single electron, we can construct the atomic orbitals at T = 0. (We assume that the interaction between the electrons can be neglected compared to the interaction of the electrons with the nucleus).

(26)

We take the nucleus of charge +Z  e, and add to the system electrons one-by- one 1,2 …, Z, construct the orbitals, e.g. with the help of the Slater determinant.

We get the well known periodic system of elements.

1 H (1s)1 2 He (1s)2 3 Li (He)(2s)1 4 Be (He)(2s)2

5 B (He)(2s)2(2p)1 6 C (He)(2s)2(2p)2 7 N (He)(2s)2(2p)3 8 O (He)(2s)2(2p)4 9 F (He)(2s)2(2p)5 10 Ne (He)(2s)2(2p)6

11 Na (Ne)(3s)1 12 Mg (Ne)(3s)2

13 Al (Ne)(3s)2(3p)1 14 Si (Ne)(3s)2(3p)2 15 P (Ne)(3s)2(3p)3 16 S (Ne)(3s)2(3p)4 17 Cl (Ne)(3s)2(3p)5 18 Ar (Ne)(3s)2(3p)6 19 K (Ar)(4s)1

20 Ca (Ar)(4s)2

(27)

Example 3: Population of electrons in a potential box

Considering a cubic potential box, of size a and at T = 0 let us drop a large number (in the order of Avogadro number) of electrons. What will be the equilibrium macrostate?

We know the single-electron microstates.

Different quantum numbers belong to each degenerate microstate

 

1 2 3

2

2 2 2

, , 2 1 2 3

1 2 3

h ,

8

1, 2, ; 1, 2, ; 1, 2, ;

n n n

E n n n

ma

n n n

  

  

2 . , 1 ,

, 2 3

1

 

n n n

(28)

Let us visualize the microstates in a coordinate system.

For positive n1, n2 és n3

We drop the electrons into the box one-by-one.

Each takes a microstate of the smallest possible energy.

In our coordinate system, they occupy a little box of size as close to the origin as possible.

At T = 0 all little boxes in the 1/8 of the sphere of radius will be occupied (2 electrons in each box, because of the spin).

0

1 E

n

0

2 E

n

0

3 E

n

E r

E0

1 . 2 2

h

0 m a

E

12 22 32

2,

0 ,

, 2 3

1 E n n n r

En n n

E

(29)

If the number of electrons in a unit volume is denoted by n, the total number of electrons with energy smaller than E is

From the known density of electrons n, in a cubic potential box, we can calculate the Fermi energy level

The ideal „large‟ box is a good approximate model of the electron gas in a metal, and the density is known. E.g. in cupper electron/m3, thus

 

 

3 1 , where 8h .

3 π 4 8 1

2 2

2 0

2 3 2 3 0 3

0

3 3

E ma E E

E E na

N

0

2 23

F

h .

E 8 n

m

1028

4 , 8 n

. eV 7 Ws 10

1 . π 11

3 8

h 3 19

2 2

F0

n

E m

(30)

At absolute zero temperature the energy of the fastest electron is 11,110 J, and classically this energy could be reached by a temperature

(The classical result is absurd!) In the box the energy levels are “quasi-continuous”.

In a three dimensional box the distance between the neighboring energy levels can be guessed as

If we consider a box of dimension a = 1 cm then

In a cupper box of dimension a = 1 cm the Fermi level is ~7 eV. The energies of 8,241022 electrons in the energy interval 0 and 7 eV form a „quasi-continuous‟

set.

19 B

3 k 11.1 10 56 600 K.

2 T

. 8

/ h

~ E0 2 ma2

E

eV 10

4 J 10 6 ,

0 33 15

E

(31)

3. FermiDirac Statistics

The equilibrium macrostate of an electron gas in a „big box‟ just introduced, is an excellent approximate model of the electron gas in many applications. The distributions of electrons as the function of energy, velocity and velocity-coordinates is called FermiDirac statistics. We saw that at T = 0 the energy of every electron is

and

The number of electrons between E and E + dE is proportional to dE: where is a distribution.

3 3/2

3/2 0

π 1 .

N na 3 E

  E

F0

E E

 

E E

N d

d   

 

E

(32)

For T > 0

   

0

0

1 2 3 2

0

1 2 2

3 2 F

0 0 2

F

d π

d d d E d .

d 2

π E d , if , h

d 2 .

0, if , 8

N N E E N E N E E

E E

E E E

N E E

E E ma

    

 

  

 

E

 

E

F0

E

 

F F

B B

1 2 3 2 1 2

3 2 3

0 k k

4π 2

d ( )d π d d

2 h

e 1 e 1

E E E E

T T

V m

E E

N E E E E

E

  

 

 

F

3 2 1 2

3

4π 2

( ) .

h E E

V m E

E

(33)

The distribution as a function of energy can be expressed as a function of the velocity absolute value, because

   

F

B

2

3 2

k

1 , d d ,

2

d d , 8π .

h e 1

E E T

E mv E mv v

m v

Nv vv V

 

     

v

 

v

F0

v

(34)

From the number of electrons with velocity between v and

v + dv , we can calculate the number of electrons in an interval (vx, vx+dvx), (vy, vy+dvy) and (vz, vz+dvz)

Distribution of the electrons depends on the velocity components as

 

F

B

3

k

, , 2 1 .

h e 1

x y z E E

T

v v v V m

     

 

v 4πv2

vx,vy,vz

dvxdvydvz.

 

vx,vy,vz

ρ

vx F

vx F

vx

(35)

Distribution of electrons in the phase-space

The coefficient of the Fermi function is the number Zi:

Let us define a six dimensional (abstract) space with coordinates x, y, z as three spatial coordinates, and px=mvx, py=mvy, pz=mvz as three momentum coordinates.

We call this six dimensional space „phase-space‟.

The elementary cell of the phase-space is d d d d d dx y z p p px y zVm v v v3d d d .x y z

z y x

i m v v v

V

Z d d d

2 h

3



 

 

(36)

Note that Zi is equal to two times those number of cells with size h3 which can be placed into the phase-space.

In the phase space we can put two electrons (one with spin +1/2 and the other with spin –1/2 ) in each cell of size h3 .

The cell of the phase space is not smaller than h3 .

(37)

Average energy of electrons in an electron gas

Average energy of electrons at absolute zero temperature

 

 

F F

3/ 2

total 0 0

F F

1/ 2

0 0

d d

3 . 5

d d

E E

EF E

E E E E E

E E E

N

E E E E

 

 

 

 

 

F B

3/ 2 3/ 2 total 0

3

0 k

0

d 2

4π d .

d h e

E E T

E E E

E V m E

E E

N N

E E

 

 

Hivatkozások

KAPCSOLÓDÓ DOKUMENTUMOK

The force depends only on r. It does not depend on velocity.. Newton equation is second order, thus two initial conditions are to be given: initial position and initial

If at t = t 0 , we know the fields inside a volume covered by a closed surface, and at each point of the covering surface we know either the tangential component of the electric or

If we know the radiating currents, we can calculate the vector potential, utilizing the Lorentz gauge we know the scalar potential as well, thus we know

 The experimental results of partial reflection can be predicted by assuming that the photon explores all paths between emitter and detector, paths that include single and

A free particle (electron in a force-free space) moves in a constant (can be zero) potential energy space, thus the Schrödinger!. equation

If the number of atoms in a single atom solid state crystal is N, and the number of protons of the atom is Z, then the total number of electrons in a neutral crystal are Z·N.. Let

large number of holes in the valence band behave similar to the free electrons in the conduction band of the metals.. Holes are virtual particles with positive charge and positive

Stimulated emissions are proportional to the number of atoms at the higher energy level and to the energy density of photons. 12 1