by Reinaldo Baretti Machín







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Chapter 1. Some experimental facts






 1. Principles of quantum mechanics; nonrelativistic wave mechanics with illustrative applications.  

by W. V. Houston


2. Introduction to Quantum Mechanics with Applications to Chemistry by Linus Pauling and E. Bright Wilson Jr.


3. Quantum Mechanics: Non-Relativistic Theory, Volume 3, Third Edition (Quantum Mechanics) by L. D. Landau and L. M. Lifshitz





A summary of experimental facts shows the importance of Planck constant , the quantization of energy and the corpuscular nature of electromagnetic radiation and the wavelike behavior of particles. The order is not chronological and no attempt is given to establish  historical priority among scientists.



Chapter 1 . Some experimental facts

The ratio of charge over mass (e/m).


The ratio of charge over mass (e/m) can be determined combining a beam of  electrons  accelerated to a speed v by a potential Ф with the magnetic deviation in a field B.

The electrons acquire a kinetic energy


                     (1/2) m v2 = e Ф                                    (1)

In the field B they experience a force

                       m v2/r   = e v B                                   . (2)

It follows that


                      (e/m) = 2 Ф/(r2 B2)                            . (3)


 Fig 1.1 Set up for measuring e/m



The errors in B and r are doubled since these quantities enter to the second power in the formula. The beam has to be “thin” compared to the error in r.

The best value is about -1.759E+11 coulombs/kg.



The scattering of alpha particles

The scattering of alpha particle (charge = +2e) by a thin metallic film are consistent with the assumption that  the atom has a positive nuclei with a radius of the order of 1.E-15 meters.

 Some particles with say  1 Mev are scattered “right back” making and angle of nearly 180.


Their approach to the nuclei must have been of the order of (in SI units)


r =  k e2 /( 1Mev) ~ 9E9 (1.6E-19)2 /(1.6E-13) ~ 1E-15 m



Compton scattering


Fig 1.2 Compton scattering

A photon of wavelength λ0 bounces from an electron at rest making an angle θ and with larger wavelength λ1 . We seek a relationship between the change in wavelength (λ1 – λ0 ) and the scattering angle θ.

From momentum conservation the square of the electron momentum is 

pe 2 =   p02 + p12 – 2 p0 p1 cos (θ)    . Using  p0 = hf0/c , p1= h f1/c one can




(cpe)2 = (hf0)2 +(hf1)2 -2(hf0)(hf1) cos(θ)                          ( 4 )


From energy conservation and  assuming that the electron is initially at “rest” ,


hf0 + mc2 = hf1 + { (cpe)2 + (mc2)2 }1/2                                  ( 5 )

and again solving for (cpe)2 one obtains


(cpe)2 = ( hf0 + mc2 -hf1)2 - (mc2)2        .                               (6 )


Substract  (4  ) from ( 6) and define a function


G(f0 ,f1 ,θ) = ( hf0 + mc2 -hf1)2 - (mc2)2  

                        -{(hf0)2 +(hf1)2 -2(hf0)(hf1) cos(θ) }            . (  7  )

Let h=1 ,.m=1 , c=1 , which mean that the unit of length is h/(mc).


G(f0 ,f1 ,θ) = ( f0 + 1 -f1)2 - (1)  

                        -{ f0 2 + f12 -2f0f1 cos(θ) }                   (8)


Let the initial frequency be fixed at f0 =1.

Then vary the angle θ   from 0 to 2π . For each value of   θ find the frequency of the recoil photon f1 employing the Newton’s Raphson method, that is the nth iteration of f1 is


 f1 (n) = f1 (n-1)  -  G(f0 ,f1(n-1) ,θ) / dG(f0 ,f1(n-1) ,θ )/df1     .      (9)


Plotting  ( 1/f1 – 1/f0) = (λ’-λ0)  vs   θ (degrees) produces Fig 1.2.


Fig. 1.3 Compton scattering by an electron.


The plot shows that in our adimensional units


                       λ1 -  λ0 =  (1- cos(θ) )   or re-introducing the unit of length


( h/(mc) ) = 2.43E-12 m,   


gives Compton’s formula,


                        λ1 -  λ0 =  (h/(mc))(1- cos(θ) ).                     (10)


theta(deg),Lprime-L0,1.-cos(theta)=    0.000E+00    0.000E+00    0.000E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.120E+02    0.219E-01    0.219E-01

theta(deg),Lprime-L0,1.-cos(theta)=    0.240E+02    0.865E-01    0.865E-01

theta(deg),Lprime-L0,1.-cos(theta)=    0.360E+02    0.191E+00    0.191E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.480E+02    0.331E+00    0.331E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.600E+02    0.500E+00    0.500E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.720E+02    0.691E+00    0.691E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.840E+02    0.895E+00    0.895E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.960E+02    0.110E+01    0.110E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.108E+03    0.131E+01    0.131E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.120E+03    0.150E+01    0.150E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.132E+03    0.167E+01    0.167E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.144E+03    0.181E+01    0.181E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.156E+03    0.191E+01    0.191E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.168E+03    0.198E+01    0.198E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.180E+03    0.200E+01    0.200E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.192E+03    0.198E+01    0.198E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.204E+03    0.191E+01    0.191E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.216E+03    0.181E+01    0.181E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.228E+03    0.167E+01    0.167E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.240E+03    0.150E+01    0.150E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.252E+03    0.131E+01    0.131E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.264E+03    0.110E+01    0.110E+01

theta(deg),Lprime-L0,1.-cos(theta)=    0.276E+03    0.895E+00    0.895E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.288E+03    0.691E+00    0.691E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.300E+03    0.500E+00    0.500E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.312E+03    0.331E+00    0.331E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.324E+03    0.191E+00    0.191E+00

theta(deg),Lprime-L0,1.-cos(theta)=    0.336E+03    0.865E-01    0.865E-01

theta(deg),Lprime-L0,1.-cos(theta)=    0.348E+03    0.219E-01    0.219E-01

theta(deg),Lprime-L0,1.-cos(theta)=    0.360E+03    0.000E+00    0.000E+00                      





Fortran code for Compton scattering

c Compton scattering by electrons     References Eisberg page 84

c wikipedia http://en.wikipedia.org/wiki/Compton_scattering

      real m

      data f,nf/1., 30/

      data m,c,h ,epsi/1.,1.,1.,1.e-3/


     $ -(h*f+m*c**2-h*fprime)**2 + m**2*c**4







      do 20 jf=1,nf+1


      do 10 i=1,10

c      theta1=theta0-g(theta0)/derivg(theta0)


c      print*,'theta=',theta1


10    continue

      print 100,theta*180./pi,1./fprime0-1./f ,(1.-cos(theta))

c      fprime=fprime+df

      theta=theta + dtheta

20    continue

100   format('theta(deg),Lprime-L0,1.-cos(theta)=',3(3x,e10.3))







Plank’s Law of Blackbody Radiation


Max Planck postulated that the probability of an oscillator having a quantized  energy nhν is proportional to


P(n) ~  nhν exp(-nhν/kT)  or  nhν exp(-β nhν)    (β=1/(kT) ) ,     where k is Boltzmann constant and T is the absolute equilibrium temperature of the cavity where the radiation is enclosed. A most important point is that n assumes integral values n=0,1,2,3..and that a new physical constant h was introduced.


The classical expression for the probability of a system with energy ε is


                                 P(ε) ~ ε exp( -β ε)             .                            (11)


The average energy is given by


εave  = ∫ ε exp( -β ε) d ε /  ∫ exp( -β ε) d ε         ,   0 ≤  ε  ≤ ∞      .     (12)


Notice that the energy  variable ε is continuous and varies from zero to infinity. This is physically questionable.


We get from (12) 

εave  =  β-2 / β-1 =  β-1 = kT      .                                                   (13)



Some mode number accounting is necessary to calculate the total energy of the blackbody.


Take a cubic  box with  dimensions  ax = ay = az =L  ,enclosing the blackbody radiation. The wave numbers of the standing waves have

             kx = πn/L  ,  ky = πn/L   ,  ky = πn/L  .                      (14)


A cell in k space has an elementary  volume

           τ =   ∆kx  ∆ky  ∆kz  =  (π/L )3 =    π3 /V    .                (15)

 For each cell there are two modes of oscillation due to two degrees of polarization.


A spherical shell of width dk has a volume  4 π k2 dk . But only one eighth correspond to physical solutions since other parts correspond to negative values of one or more of the components kx ,  ky or ,kz .


Hence the number of states having a wave number between k and k+dk is

obtained dividing  4 π k2 dk by τ and multiplying by 2 (1/8) ,


          dN   =  2(1/8) (4 π k2 dk ) /( π3 /V)  = (V/ π2)  k2 dk    . (16)


Writing  k2 =(4 π2/c2) ν2   , dk = (2π/c) dν and inserting in (15) gives


                                    dN = ( 8 π V/c3 ) ν2 d ν .                     (17)


If the classical average value (13) is multiplied by (17) we have the energy contribution for a number of oscillators with frequencies in the range

  ν ,   ν   +   d ν.


               dE =    k T ( 8 π V/c3 ) ν2 d ν   ~ energy          .       (18)


Dividing by the volume V we obtain the density differential ,


                      dρ =  k T ( 8 π /c3 ) ν2 d ν   ~ energy/volume    . (19)


Some objections to formula (19) can be raised. Some of them were raised at the end of the 19 th century , at  a time when atomic physics was in a rudimentary stage and special relativity had not been formulated. These objections are still today repeated with no afterthought , see for example Ref 1 , page 10.


a) If we integrate    ∫ k T ( 8 π V/c3 ) ν2 d ν    ,    0  ≤ ν ≤ ∞ we get an infinite energy density.  However in experimental setups with blackbody radiation ν does not goes much beyond ultra violet light.


If ν was to become much higher than UV it would become first X-rays and then gamma rays. The walls of the enclosure would be permeable to such radiation. Also the temperature of the wall would have to be much higher and the enclosure would melt and vaporize.


From relativity  we know that a particle of mass m  if annihilated would give a quantum of energy  hν ~ m c2 . This would suggest in any case a cutoff of magnitude νcot off = mc2/h to the frequency range.


Nevertheless Planck’s resolution of the divergence problem opened the way for the quantum theory. The key steps are to substitute (11) and (12) with  expressions where the energy is quantized.


εaverage  = ( ∑n=0 nhν exp(-βnhν) / ( ∑n=0 exp(-βnhν) )


           = -   ∂ ln ( Z) /∂β                


Where Z= ∑n=0 exp(-βnhν) )  = ∑n0 xn     ;   ( x= exp(-βhν)   )


It reduces to Z= (1-x)-1   .


Then εaverage = 1/(1-x)  {∂ ( -x) /∂β} = (1/(1-x)) (hν) exp(-βhν ) 


           =   hν/( exp(βhν) -1)                                 .            (20)



A function like that in (20)  , f(E) = 1/(A exp(E/kT) -1) describes the average quantum number n .


   Fig. 1.4  Plot  f(u) =1 /(exp(u) -1).

Sage code


It is called the Bose-Einstein distribution function. In our example A=1 . It serves to describe the statistical behavior of integer spin (0,1,2..) particles, thereby called bosons. The photon is an example of a bosonic particle.


 Using    hν/( exp(βhν) -1)    instead of kT in (19) leads to Planck’s formula


         dρ =   ( 8 π /c3 )  {hν3 /(exp(hν/kT) -1) } d ν  ~ joules/m3   . (21)


Introducing the wavelength ν =c/λ , d ν = -(c/λ2) dλ  and recalling that an integral over the frequency range is related to the integral over the  lambda range by


∫ {    } d ν   , 0 <ν < ∞     equals ∫ {    } (-d λ)  ,  ∞ > λ < 0 , we write



 dρ =   ( 8 π  )  { (hc/λ5) /(exp(hc/kTλ ) -1) } d λ  .                   (22)


Define   u= hc/kTλ   ,  1/λ = (kT/(hc)) u   , hence


dρ =    ( 8 π  ) {(kT)5/(hc)4 ]  {u5  /(exp(u) -1) } d λ ~J/m3          (23) 

 The function f(u) = {u5  /(exp(u) -1) }

Fig . 1. 5 Plot of f(u).


A detailed view shows that the maximum of f(u) occurs at about

u  = 4.93 . It follows that the maximum of radiation occurs for a lambda satisfying

 (hc)/(kTλm) = 4.93   and therefore


                            T  λm ≈ 2.92E-3 kelvin-meters.             (24)



Returning to (23) , writing dλ =-(hc/kT) du/u2 gives


dρ =    ( 8 π  ) ( (k4/(hc)3 ) T4   {u3  /(exp(u) -1) } d u  ~J/m3

The density is


ρ =  (8 π ) ( (k4/(hc)3 )  T4 ∫ u3  /(exp(u) -1) } d u .


The integral   ∫ u3  /(exp(u) -1) } d u    is    6.49110222 .


      data nstep/10000/






      do 10 i=1,nstep



10    continue






Fig 1.6 Plot of  u3/(exp(u)-1) .





                                          ρ = 7.53E-16 T4  .




A deeper explanation for the sum rule lies in a new statistical physics that supplants the Boltzmann statistics.  Photons area part of a class of indistinguishable particles called bosons which can occupy a quantum state in a system without limitations in their numbers. An important principle asserts that the wave functions of bosons are symmetric.  Electrons are part of another family called fermions with an occupation restricted to one or zero particles per quantum state. Electrons have anti- symmetric wave functions.



by Reinaldo Baretti Machín


Chapter 1.  part b 









Para preguntas y sugerencias escriba a :







 1. Principles of quantum mechanics; nonrelativistic wave mechanics with illustrative applications.  

by W. V. Houston


2. Introduction to Quantum Mechanics with Applications to Chemistry by Linus Pauling and E. Bright Wilson Jr.


3. Quantum Mechanics: Non-Relativistic Theory, Volume 3, Third Edition (Quantum Mechanics) by L. D. Landau and L. M. Lifshitz





The photoelectric effect


The photoelectric effect, shows the  particle nature of light and bears on the importance of Planck’s constant h , see e.g. http://www.cas.muohio.edu/~marcumsd/p293/lab1/lab1.htm


Fig 1.   A schematic arrangement is shown.


A lamp of mercury can provide a variety of lines in the visible. By means of filters,  light of a single frequency  ν ( Hz)  (“color”) can be selected to be incident on the photocell . A stopping voltage is measured to require zero current. The plot of V vs frequency ν results in a linear form


                                          V = m ν - b                            .  ( 25   )

It truns out that the slope (m)  equals the ratio of Planck constant to the electron charge (h/e). The intercept b , depends on the particular metal from which the electrons are ejected. Equation (   ) leads to


hν = e( V+ b) ~energy of the light corpuscle, is spent in work done against the metal  e b  ~ joules and the work against the stopping potential eV ~ joules.


Taken from  http://rds.yahoo.com/_ylt=A0WTb_0apytKXC0BSn.jzbkF/SIG=127s0u909/EXP=1244461210/**http%3A//courses.umass.edu/chem111v/Photoelectric.jpg 




Emission lines of hydrogen




Fig 1. The Balmer series is composed of the visible lines of hydrogen .


The first four lines of this spectra going from Hα  to Hδ have wavelengths

6.56 E -7 m , 4.86E-7 m ,4.34E-7 m ,4.10E-7 . Other lines of shorter wavelength fall to the left of Hδ .


It was discovered that these lines satisfied the relation


                          1/λ = R( 1/4 - 1/n2 ) , n= 3,4,5….. (Balmer series),


where R =1.10E7 m-1 is called Rydberg constant.


Another  series that falls in the ultraviolet region satisfies


                              1/λ = R( 1/1 - 1/n2 )  , n=2,3,4… ( Lyman series).


There is no wavelength detected smaller than λ=1/R.


And there are still other series with a set  λ  greater than the Balmer series


To explain these spectra one can suppose that the energy is quantized according to the formula


                                               En = - /E1/ /n2                      , (   26 )


where /E1 / is the absolute value of hydrogen’s ground state.


The Lyman series , for example, is produced when the atom makes a transition from the excited n-th state to the first level. the Balmer series is produced from transitions from the  n-th level to the second level.

The differences occurring in the transitions  have to be related to the photon energies hν.


Take the Lyman series and equate the energy of a photon given off in the transition from lelel n to level 1.


The relation               hν = /E1 /( 1/1 – 1/n2 )  can be writte in terms of wavelength , ν =c/λ as ,


                          1/ λ  =   (E1/(hc) ) ( 1/1 – 1/n2 )         .  ( 27   )


It follows that      R= 1.10E7 m-1 = E1/(hc)    and



/E1 / = R (hc) = 2.19E-18 joules =  13.6 eV                   . ( 28     )       


This is consistent with eq(3) of the Introduction where the energy was obtained through dimensional analysis.






Davisson - Germer experiment



Fig 1. Davisson on the left and Germer on the right.


The  Davisson-Germer experiment confirmed the de Broglie hypothesis that electron can show  a wave-like nature. Electrons were diffracted in a crystal very much like X-rays. The observation of diffraction allowed the  measurement of a  λ  which agreed well with de Broglie's equation λ = h / p.


Taken from http://hyperphysics.phy-astr.gsu.edu/hbase/davger.html.


Bragg’s law for X- rays state that the angles of diffraction - θ- for a crystal with a known lattice spacing= d   satisfy


                      d sin(θ) = n λ      , n=1,2….


A beam of electrons was accelerated by 54 volts .

Thus for an individual electron ,   p2/(2m) = eV ; p = ( 2meV)1/2  and   

λ = h / ( 2meV)1/2  .

Using the data ,

data h,k,m, e/6.63e-34,1.38e-23,9.11e-31,1.602e-19/

      data V/54.0 /

Gives λ= 1.70 Angstrom  , and the first diffraction maxima should be detected at   θ = sin -1 ( 1.70/2.15) = 52.2 degrees. The above diagram shows the peak at 50 degrees.