BARETTI  LECTURES ON QUANTUM MECHANICS

by Reinaldo Baretti Machn

 

Chapter 2.  The Old Quantum Theory

 

Home page

 

http://www1.uprh.edu/rbaretti

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysics.htm

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysicsPart2.htm

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysicsPart3.htm

 

Lectures on Quantum Mechanics -Introduction  

Chapter 1 - Some experimental facts

 Chapter 2 - The Old Quantum Theory

Chapter 3 - The Postulates of Quantum Mechanics  

Chapter 4- Potential Barriers

Chapter 5 - Potential Wells

Chapter 6 -  Periodic Potentials

Chapter 7 - Angular Momentum

Chapter 8 - The Hydrogen Atom

Chapter 9 - The Electron Spin

Chapter 10 - The Two Electron Atom

Chapter 11- Ground State Energies of Lithium and Beryllium

Chapter 12- The Self Consistent Field Method

Chapter 13-   The H2 molecule 

Chapter 14- Density Functional Theory  

 

 

Para preguntas y sugerencias escriba a :

reibaretti2004@yahoo.com

  Free counter and web stats

 

 

References:

 

 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

 

4.Wikipedia   http://en.wikipedia.org/wiki/Old_quantum_theory

 

5. The Rise of the New Physics ** 2 Volumes Complete** by A. D'Abro

 

A line can be drawn and quantum  theories coming  before Schrodinger equation are called the old quantum theory. The old quantum theory is a variety of results spanning  the years 19001925.

 

These attempts went as far as they could with the classical notion of trajectories and added quantum postulates to them. The uncertainty principle had not been enunciated .

 

Stated by the inequality  ∆x∆p  ≥ h /(4π) we see  that zero uncertainty in position implies infinite uncertainty in the momentum.

 

In classical mechanics one has, in one dimension, Newtonlaw ,a second order differential equation. For example suppose that on a particle of mass m  acts the total  force F= F(x,v). The equation of motion is

 

                              d2x/dt2 = (1/m)  F(x,v)           .                  (1)

 

A second order DE require two initial  conditions , position and velocity , x0 and dx0/dt . They are assumed to be known exactly x0 and v0 or  x0 and p0/m.  An numerical solution is provided by a Taylor series about t0=0 and has of the form,

 

x(t)  =x0 + v0 t  + (1/2) t2 F(x0 ,v0) /m  + ( 1/6) t3 d{ F(x0 ,v0) /m}/dt  ( 2 )

 

But the uncertainty principle set limitations on the simultaneous knowledge of x0 and v0 . Therefore the first two terms in the series , eq (2) , can not be known with absolute precision.

          

That the  classical notion of trajectory had to be done away with as a starting point in quantum mechanics was not apparent to the pioneers.

 

One of the early approaches to quantum mechanics consisted in applying a t postulate that quantized the integral 

 

                 J  =   ∫ p dq  = nh  where the variables p and q form a conjugate pair. J is called the action . For example if q=x then p=m dx/dt. But more important if p is the angular momentum of a particle going around in a circle of radius r- , around the Z axis ,   pφ = m r2 (dφ/dt)=mr2ω  ~ units of angular momentum ~ J-s.

WARNING THIS APPROACH IS SUPERSEDED BY SCHRODINGERS THEORY. The quantization of angular momentum is shown there to be given by another rule.

 

 

hydrogen atom

Going bak to the action J ,  dq = dφ= ωdt   and

 

                                     J = mr2ω (2π) = n h              .         (3)

 

Another equation is that of mechanical equilibrium (in CGS units), in the case of the hydrogen atom ,

 

                           e2/r2 = m r ω2                                         . (4)

 

One can view (3)-(4) as two equations for the variables r and ω or alternativelyfro the variables r and v ( v= ωr) .

 

From (3)          

                        m r v = n (h/2π) = n h    ,  (h=h/2π)                  . (5)

And from(4)

                        m r v2 = e2                                          .    (6)    

 

Dividing eq (6) by eq. ( 5)   one obtains a quantized orbital velocity

 

vn = e2 /(n h)   =  2.18E8/n2 ~ cm/s = 2.18E6 /n2 ~ m/s     (7)

   ≈ 7.27E-3 c /n2   , where c is the velocity of light.

 

This in turn leads to  quantized orbit radius

 

rn = (h2/(e2m) ) n2  = 5.30E-9 n2 (cm) =5.30E-11 n2 (m)  . (8)

 

Writing down the energy in the n-th quantum state ,

 

En = T + U = (1/2)m vn2  +  (-e2/rn

 

     =   (1/2) m e4 /(n2 h2 )  - m e4 /( n2 h2)

 

    = -(1/2)  m e4 /( n2 h2)                                       ,           (9)

with n=1,2,3   .

  

 

We give the energy in various systems of units;

En  = - 2.17163076E-011 /n2    ~ ergs   

       = -2.17 E-18  /n2   ~    joules

       = -13.6   /n2    ~ eV                                                       (10)                                         

 

This result was anticipated in the Introduction.

 

 

Particle in the box

Another elementary example is that of the particle in a one dimensional box.

A particle of mass m moves between the impenetrable walls at x=0 and x=L .

The kinetic energy T equals the total energy E ( i.e. U(x)=0)

                                 E= p2/(2m) ;  p=(2mE)1/2                 .  (11)

 

The action integral is J= 2∫ p dx = 2∫ (2mE)1/2 dx = 2L(2mE)1/2 = n h

and    

                 En =  { h2/(8mL2) } n2                                   ,    (11)

n=1,2,3..

A more complete treatment with quantum mechanics arrives at the same result of the energy of the particle in a box.

 

 

Comparing (11) with (10) we see two different dependence on the quantum number n. In (10) the coulomb attractive  potential gives rise to a negative energies with a dependence  1/n2 . In (11) the eigenvalues are positive and grow as ~ n2 .

 

 

Quantum harmonic oscillator

The total energy for an oscillator is

                                   E= p2/(2m) + (1/2) k x2             (12)

and  p(x) = [ 2m(E-(1/2)kx2)] 1/2 . Letting  m=1 ,k=1 , h =1 ,l gives the unit (scale) of energy      Escale =  h ( k/m)1/2 = hω .

 

So p =  [ 2(E-(1/2)x2)] 1/2     and the action integral takes the form

 

J= 2 ∫[ 2(E-(1/2)x2)] 1/2 dx  = nh= n(2π)h

                               = n(2π)  ,        -(2E)1/2  ≤ x  ≤ (2E)1/2     , (13)

 

with x1=-(2E)1/2  , x2 = +(2E)1/2  .

 

Define  a2 = 2E   , X = (a2 x2)  we find the integral

 

2 ∫ X1/2 dx = (x X1/2  +a2 arcsin (x/a)  )  │x2 x1 

                

                = 2E( arcsin(1) arcsin(-1) )=2E( π/2 (-π/2) )

                = 2πE    = n(2π)

 

Thus in our units E = n    (n=1,2,3 ..)

or                        E n=n hω                                               . (14) 

 

Eq (14) is Planck formula.

 

It is shown later that the qunatum  harmonic oscillator has a spectrum given by                     E n=(n +1/2)hω              n= 0,1,2,3       (15)

 

The minima   E 0 = (1/2) h ω is called the zero point energy and can have physical significance  , it is not an arbitrary constant to be swept away. 

 

Now we introduce an algorithm to find the eigenvalues by integrating

∫ p dx for different values of E. The eigenvalues are those for which the numerical integral equals n(2π) =  2π , 4π, 6π ,.   

 

FORTRAN code                                           

c Wilson Somnerfeld Quantization rule

      real k ,m

      data k,m,nstep,ne/1.,1.,1000,50/

      data epsi/1.e-5/

      v(x)=(1./2.)*k*x**2

      p(x)=sqrt(2.*m*(E-v(x)))

      pi=2.*asin(1.)

      ei=.001

      ef=5.

      de=(ef-ei)/float(ne)

      e=ei

      do 10 ie=1,ne

      x1=-sqrt(2.*e/k)

      x2=-x1

      dx=(x2-x1)/float(nstep)

      sum=0.

      do 20 ix=1,nstep

      x=x1+dx*float(ix)

      sum=sum+2.*(dx/2.)*(p(x-epsi)+p(x-dx+epsi))

20    continue

      print 100,e, sum, 2.*Pi,4.*pi

      e=e+de

10    continue

100   format('e,sum,2.*Pi,4.*pi=',4(2x,e10.3))

      stop

      end

 

 

 

Fig 2.1 Eigenvalues of the harmonic oscillator. The first two values are

 E= 1 and E=2.

 

 

Another example is the linear potential on the positive halfline, U = kx   , x > 0  , U= ∞  , x ≤ 0.

The result of differs from the full quantum mechanical treatment but for large quantum numbers (n) they converge.

Dimensionally   p2/m ~ k L  where L is the length scale.

Using p~ h/L   one gets

                    h2/(mL2) = k L   , from which follows that

                   L = {h2/(mk)}1/3                                  (15)

The scale of energy is ,

                     Escale = kL = { h2 k2/m}1/3              .   (16)

Letting m=1, k=1, h=1 ,h = 2π , the quantum condition is

2∫ { 2(E-x) }1/2 dx = n(2π)  ,      0  ≤   x ≤  E               .  (17)

The integral is 

           25/2 E 3/2 /3   = 2π n                                ,  (18)

thus     E = {(3π)2/3/2 } n2/3                          .   (19)

This simple linear potential , with proper values of k and m , can serve to estimate among other things such exotic themes as the nature of quark confinement.

In the three example given  we have found different dependencies of  the energy  E on the quantum number n.  

We  have E ~ 1/n2 for the coulombic potential , E ~ n for the harmonic oscillator , U= (1/2)kx 2 and E ~ n2/3 for the linear potential U = kx.    

The dependency of E on the quantum number n  is related to  dimensional analysis. Suppose the potential energy is of the form U = kxα   (α≠0).Introducing a length scale L, dimensional analysis says that,

p2/m ~ h2 /mL2 ~ k Lα  ~ E                                         .    (20)

Then L ~ (h2 /mk)1/(2+α)   and E ~ k (h2 /mk)2α/(2+α)

or E ~ factor1 * h 2α / (2+α)    or vice versa

h ~ factor2 * E(2+α)/2α                              .                         (21)

We need not ascertain at this point what factor1 and factor 2 are. 

What matters is  that  ∫ p dx has dimensions of h , thus

∫ p dx ~factor* E(2+α)/2α  ~ n (from the quantization condition).

Hence   the energy dependence on thequantum number n has the form  

                         En ~  n2α/(2+α)                    .                   (22)                       

 

 

 

α                                     U                                  E ~

-1 U=  -1/r ~ - n-2
 + 1 U = x ~  n2/3
+ 2 U = x2 ~ n

 

For the particle in the box one has U=0 inside the box and

h2 /mL2  ~  E        or   h ~ E1/2 .

Thus ∫ p dx ~ E1/2 ~ nh   and E ~ n2 .

 

A numerical integration of ∫p dx for the first two values of n (n=1 and n=2) can provide the energy expression in the form

                                         En = constant ( n power) .

 

Fig 2. 2  The action J/(2π) vs energy  .

The plot shows that En=1 ≈ 2.26  and En=2 = 2.26 (2)γ ≈ 3.59   . Therefore γ = ln(1.59)/ln(2) = 0.669

This leads to

                                                      En = 2.26 n0.669                          , (23)

 

which matches the analytical result given by eq(19).

Fortran code for Fig 2.

c Wilson Somnerfeld Quantization rule
real k ,m
data k,m,nstep,ne/1.,1.,1000,60/
data epsi/1.e-5/
c v(x)=(1./2.)*k*x**2
v(x)=k*x
p(x)=sqrt(2.*m*(E-v(x)))
pi=2.*asin(1.)
ei=.01
ef=5.
de=(ef-ei)/float(ne)
e=ei
do 10 ie=1,ne
x1=0.
x2=e/k
dx=(x2-x1)/float(nstep)
sum=0.
do 20 ix=1,nstep
x=x1+dx*float(ix)
sum=sum+2.*(dx/2.)*(p(x-epsi)+p(x-dx+epsi))
20 continue
print 100,e, sum/(2.*pi)
e=e+de
10 continue
100 format('e,sum =',2(2x,e10.3))
stop
end