UNIVERSIDAD DE PUERTO RICO EN HUMACAO
www.uprh.edu

 

 

 

LECTURES ON QUANTUM MECHANICS

by Reinaldo Baretti Machín

 

Chapter 3.  The Postulates of Quantum Mechanics

 

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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-The H2 molecule

 

 

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

 

6. Foundations of Physics by Robert B. Lindsay , Henry Margenau

 

7. LECTURES ON QUANTUM MECHANICS by Gordon Baym

 

 

We will present a series of postulates pertinent to what was initially called wave mechanics. This is the formulation that followed from Schrodinger equation in 1926.

 

However there are other formulations of QM with varying degrees of abstraction. One of them was formulated by Werner Heisenberg , Pascual Jordan and Max Born, called  matrix mechanics, one or two years before Schrodinger published his equation.

 

Eventually we will also touch upon matrix mechanics.

 

Postulate A. The state of a system is represented by a wave function Ψ (q1 , q2… qn ,t)    , here qi can be for example the independent cartesian coordinates of a system. Ψ may be a complex function. And it may also may have a “spinor” part when account is taken of the spin property of electrons. The spinor part is not a function of the coordinates qi . It represents an internal degree of freedom like for example the charge or mass of a particle.

 

A probability density is defined by    Ψ *, where Ψ * is the complex conjugate.

 

dP = Ψ * Ψ dq1 dq2 … dqn = Ψ * Ψ dτ  ( dτ is the volume element)  is the probability that a measurement of the coordinates will give values lying in the range (q1 , q1 +dq1 )….  (qn , qn +dqn) . The probability is usually normalized such that

                                                            ∫ dP = ∫ Ψ * Ψ dτ  = 1.                              (1)

 

 

Examples:

a) plane wave

A plane wave of the form  Ψ(x,t) = C * exp[ i( kx-ωt) ] can represent partially a beam of free particles with a well defined linear momentum p traveling to the right , i.e + X axis. Here k = 2π/λ = 2π p/h and  ω = v k . The expression for Ψ gives no details as to the spread or cross section of the beam in the Y-Z plane. So this is in an idealization. The constant C is determined by the normalization

 

       ∫ Ψ*  Ψ dx = C* C (xf-xi) =N  particles  . If (xf-xi ) = L   , the length of the space over which the particles manifest their wave motion , then C2 = N/L    and C= (N/L)1/2  . This option is called box normalization. There is another one , involving Dirac's delta function , but we won't go into that at this moment.

                The normalized plane wave would be   

                                                        Ψ (x,t) = (N/L)1/2  exp[ i(kx-ωt) ]  .           (2)

Since Ψ2 is constant , the particles have a uniform probability of being anywhere along the X axis.

 

Setting   N=1 and    h'k= p  , E= h'ω   in (2) the wave function takes the form

 

                      Ψ (x,t) = (1/L)1/2  exp[ (i/h') (px-Et) ]                                         .   (3)

 

The result of taking the following derivatives  of Ψ   , -ih'∂ /∂x    ,  i h'∂ /∂t , justifies calling them the momentum operator and the energy operator respectively.

 

                               

                                           -ih'∂ Ψ/∂x  =  p Ψ                                          ,      (4)

                                           i h'∂ Ψ/∂t   = E Ψ                                           .      (5)

 

Later we will use the notion of 

                                         p (variable)   → poperator =  -ih'∂ /∂x   ,     (6)

and

                                        E(variable) →   E operator  =  i h'∂ / ∂t                    .     (7)

The energy operator is called the hamiltonian and is denoted with the letter H.

 

The use of these  two operators will be the guiding principle in constructing Schrodinger equation.  In this very special case

 

Ψ is an eigenfunction of the linear momentum and of the energy. It  means that

                                  popΨ = p Ψ    and    H Ψ = E Ψ .                                   . (8)

 

                                         

b) particle in the box

A particle confined in a one dimensional box of length L has an un-normalized  wave function 

 

                            Φn(x)  = sin (nπx/L)       ,  0 ≤ x ≤ L     , n =1,2,3..                            . (9)

Notice the boundary conditions  Φn(0) =  Φn(L) =0.  The normalization requires that 

 

               ∫ N2  Φn(x) 2 dx   = 1     ,       0 ≤ x ≤ L .

The integral   ∫  {sin (nπx/L)}2 dx = L/2     , thus  N2 = (L/2)-1 and  N =(2/L)1/2 .The normalized wave function is then

 

                                                   Ψn (x) = (2/L)1/2 sin ( nπx/L )                      .    (10)

 

 

 When Ψ  is written as a function of coordinates it is said to be in the coordinate representation. This is most commonly

used.

 

 

The probabilty distribution in  p- space ( momentum space) is given by the rule

                        

                      Φ(p) = (1/h1/2 ) ∫ exp (- i px/h') Ψ(x) dx  ~ (momentum)-1/2    - ∞   ≤  x ≤  +  ∞     .  (11)

 

Notice the h' (h bar ) in the exponential but plain h in the the factor ( 1/h1/2  ).

 

To show that Φ(p) is normalized  , i.e.  

                                                                ∫ Φ* Φ dp  = 1                 - ∞   ≤   p  ≤  +  ∞             ,  (12)

use has to be made of the Dirac delta function.

 

                       δ( x-x') = (1/(2π) ) ∫ exp( i k(x-x')) dk                                         ,                      (13)

 

or since h'k= p  , of  the equivalent expression

 

                   δ( x-x') = (1/h ) ∫ exp {i p(x-x')/h'} dp                              .                             (14)

Eq. (12) takes the form

 

 ∫ Φ* Φ dp  = ( 1/h1/2 ) ∫ exp (+ i px/h') Ψ* (x') dx' (1/h1/2 ) ∫ exp (- i px/h') Ψ(x) dx  dp   .      (15)

Intgerate first over the p variable  and using (14) one gets

 

       ∫ ∫ Ψ* (x')  Ψ(x) dx' dx  (1/h) ∫ exp (+ i p(x-x')/h') dp = ∫ ∫    δ( x-x') Ψ* (x')  Ψ(x) dx' dx 

 

                                 =  ∫  Ψ* (x)  Ψ(x)  dx = 1.     (QED)                                  (16)

 

Example:  Find the momentum space distribution corresponding to the Fourier transform of the normalized gaussian

                                                              Ψ(x) = (α/π )1/4 exp ( - αx2 /2)  .                     (17)

 

Since Ψ is even the only contribution is from the real part of  (11)

 

         Φ(p) = (1/h1/2 ) ∫ cos(px/h') (α/π )1/4 exp ( - αx2 /2)  dx            - ∞   ≤  x ≤  +  ∞    .      (18)

       

    Fig. 3-1.  Plot of     g(x) = cos(px) (α/π )1/4 exp ( - αx2 /2)  with p=3 ,  α = π .

 

Code using SAGE

a=pi;p=3;
y=plot((a/pi)^(1/4)*cos(p*x)*exp(-a*x^2/2),-3,3)
show(y)    

 

 Fig. 3-2 . The momentum space distribution Φ(p) of the gaussian distribution Ψ(x).

 

It is not difficult to conclude from the plot that  Φ(p) ~ exp{ -p2 /(2αh'2 )}  . Requiring its normalization

 

we have   N2 ∫ exp{ -p2 /( αh'2 )} dp = 1 , or N = [ exp{ -p2 /( αh'2 )} dp ]-1/2 = ( απ h'2 )-1/4 .

Hence the normalized p space distribution is

 

                                                  Φ(p) = ( απ h'2 )-1/4 exp{ -p2 /(2αh'2 )}      .  (19)

 

 

 

 

 

FORTRAN code

c gaussian p distribution june 8 ,2009
real Lscale
data hbar, alfa,np,nstep/1.,1.,30,1000/
f(x) =(1./sqrt(h))*cos(p*x/hbar)*(alfa/pi)**(.25)*
$ exp(-alfa*x**2/2.)
pi=2.*asin(1.)
lscale=1./sqrt(alfa)
h=2.*pi
p1=0.
p2=35.*pi
dp=(p2-p1)/float(nstep)
do 10 ip=0,np
p=dp*float(ip)
xmax=40.*lscale
dx=xmax/float(nstep)
sum=0.
do 20 ix=1,nstep
x=dx*float(ix)
sum=sum+(dx/2.)*(f(x)+f(x-dx))
20 continue
c the integrand is an even function
sum=2.*sum
print 100,p,sum
print 100,-p,sum
10 continue
100 format(1x,'p,phi(p)=',2(4x,e10.3))
stop
end


 

 

 p,phi(p)= 0.000E+00 0.751E+00
p,phi(p)= 0.000E+00 0.751E+00
p,phi(p)= 0.110E+00 0.747E+00
p,phi(p)= -0.110E+00 0.747E+00
p,phi(p)= 0.220E+00 0.733E+00
p,phi(p)= -0.220E+00 0.733E+00
p,phi(p)= 0.330E+00 0.711E+00
p,phi(p)= -0.330E+00 0.711E+00
p,phi(p)= 0.440E+00 0.682E+00
p,phi(p)= -0.440E+00 0.682E+00
p,phi(p)= 0.550E+00 0.646E+00
p,phi(p)= -0.550E+00 0.646E+00
p,phi(p)= 0.660E+00 0.604E+00
p,phi(p)= -0.660E+00 0.604E+00
p,phi(p)= 0.770E+00 0.559E+00
p,phi(p)= -0.770E+00 0.559E+00
p,phi(p)= 0.880E+00 0.510E+00
p,phi(p)= -0.880E+00 0.510E+00
p,phi(p)= 0.990E+00 0.460E+00
p,phi(p)= -0.990E+00 0.460E+00
p,phi(p)= 0.110E+01 0.410E+00
p,phi(p)= -0.110E+01 0.410E+00
p,phi(p)= 0.121E+01 0.361E+00
p,phi(p)= -0.121E+01 0.361E+00
p,phi(p)= 0.132E+01 0.315E+00
p,phi(p)= -0.132E+01 0.315E+00
p,phi(p)= 0.143E+01 0.270E+00
p,phi(p)= -0.143E+01 0.270E+00
p,phi(p)= 0.154E+01 0.230E+00
p,phi(p)= -0.154E+01 0.230E+00
p,phi(p)= 0.165E+01 0.193E+00
p,phi(p)= -0.165E+01 0.193E+00
p,phi(p)= 0.176E+01 0.160E+00
p,phi(p)= -0.176E+01 0.160E+00
p,phi(p)= 0.187E+01 0.131E+00
p,phi(p)= -0.187E+01 0.131E+00
p,phi(p)= 0.198E+01 0.106E+00
p,phi(p)= -0.198E+01 0.106E+00
p,phi(p)= 0.209E+01 0.847E-01
p,phi(p)= -0.209E+01 0.847E-01
p,phi(p)= 0.220E+01 0.669E-01
p,phi(p)= -0.220E+01 0.669E-01
p,phi(p)= 0.231E+01 0.522E-01
p,phi(p)= -0.231E+01 0.522E-01
p,phi(p)= 0.242E+01 0.403E-01
p,phi(p)= -0.242E+01 0.403E-01
p,phi(p)= 0.253E+01 0.307E-01
p,phi(p)= -0.253E+01 0.307E-01
p,phi(p)= 0.264E+01 0.231E-01
p,phi(p)= -0.264E+01 0.231E-01
p,phi(p)= 0.275E+01 0.172E-01
p,phi(p)= -0.275E+01 0.172E-01
p,phi(p)= 0.286E+01 0.126E-01
p,phi(p)= -0.286E+01 0.126E-01
p,phi(p)= 0.297E+01 0.916E-02
p,phi(p)= -0.297E+01 0.916E-02
p,phi(p)= 0.308E+01 0.657E-02
p,phi(p)= -0.308E+01 0.657E-02
p,phi(p)= 0.319E+01 0.465E-02
p,phi(p)= -0.319E+01 0.465E-02
p,phi(p)= 0.330E+01 0.326E-02
p,phi(p)= -0.330E+01 0.326E-02
 

We show next that  for the gaussian function

                                    (  x2 average )1/2  ( p2 average )1/2  = h'/2 .                   ( 20 )

This means that the gaussian satisfies the uncertainty principle, ∆x ∆p ≥ h/2 , lower bound

 

Use is made of the definite integral

 

                                                 ∫ u2 exp(-a u 2 ) du = π1/2 /(2 a3/2 )   , -∞ ≤  u  ≤ +∞      .      (21) 

The average of x2 is       

              

x2 average = ( α/π)1/2 ∫ x2 exp( -α x2 ) dx = ( α/π)1/2 π1/2 /(2  α3/2 ) = 1/(2α)             ,        (22) 

and   

 

( x2 average)1/2 = { (x- xave )2 }1/2 ≡ ∆x =    1/(2α)1/2  , since xave =0 .                                    (23)

 

The average of p2 is, using (19)

 

    p2ave     =     ( απ h'2 )-1/2 ∫ p2 exp{ -p2 /(αh'2 )} dp = ( α/2) h' 2                   .              (24)

So                                           ( p2ave )1/2 ,    = ( α/2)1/2 h'  ,                                    (25)

which equals ∆pave  since    pave = 0.   Multiplying (23) and (25) we get,

 

                                                      ∆x ∆p = h'/2                                              .   (26)

 

    Average values:  For brevity we write out only one cartesian coordinate. We have already used the postulate that states  the average of  xn  is

 

                             < xn > =  ∫ Ψ* xn Ψ dx                  .                                        (27)

 

The average of pn can either be obtained from

 

                      <  pn   >  =  ∫ Ψ* (-ih' ∂/∂x )n Ψ dx                     ,                           (28)

or from the p - distribution,

 

                        <  pn   > =   ∫  Φ* pn   Φ  dp                            .                        (29)

                        

 

Postulate B:  A function of the coordinates and momenta in cartesian coordinates Q ( q,p) becomes an operator by the prescription

q → q ,  p → - ih'∂ /∂q such that Q ( q,p) → Q (q , - ih'∂ /∂q ) .

 

Example: a) The hamiltonian operator.The most important operator in quantum mechanics is the hamiltonian operator obtained from the hamiltonian function.

For a particle of mass m ,moving in a potential field  U(x,y,z) the hamiltonian function is

 

H = { px2  +  py+ pz} /(2m)   + U (x,y,z)   .                                                        (30)

The hamiltonian is

 

H = ( -h'2 /(2m) ) {∂2 /∂x2 +  ∂2 /∂y2   + ∂2 /∂z2 }  + U(x,y,z)              .                        (31)

 

 

 

 

Example b) The angular momentum

The cartesian components of the angular momentum are,

 

L x = ypz - z py   → Lx = -ih'(y ∂ /∂z - z ∂ /∂y )                                      ,               (32)

L y = zpx - x pz   → Ly = -ih'(z ∂ /∂x - x ∂ /∂z )                                       ,              (33)

L z = xpy - y px   → Lz = -ih'(x ∂ /∂y - y ∂ /∂x )                               .                      (34)  

 

 

 

 

 

 

 

 

 

Principle of indeterminism and noncommuting quantities.

 

The position (x) and momentum operator (-ih'∂/∂x) do not commute. Actually  for a properly normalized wave function

 

                                                 ∫ Ψ *  ( pop x - x pop )  Ψ dx = -ih'               . (35)   

Proof :  The firts term in (35) is

             ∫ Ψ *  ( pop x )  Ψ dx =  ∫Ψ * { -ih'∂ (xΨ)  /∂x} dx = ∫{ Ψ *-ih'Ψ  +  Ψ*x pop Ψ}  dx      . (36)  

Substitution in (35) gives

                        

                             ∫{ Ψ *-ih'Ψ  +  Ψ*x pop Ψ  -Ψ * x pop  Ψ }  dx = ∫ Ψ *-ih'Ψ x = -ih'    .

Born and Jordan following Heisenberg formulated the commutation rule in matrix form as

 

                                       P Q - Q P = -ih' 1                      ,                                           (37)

 

where Q and P are infinite order Hermitean matrices and  1 is the diagonal unit matrix. A detailed account is given ref 5 Volume 2 and also in    The 1925 Born and Jordan paper “On quantum mechanics” 
William A. Fedak and Jeffrey J. Prentis .

 

We will show a derivation of the uncertainty principle with respect to the coordinate x and linear momentum px usually written as

∆x ∆p ≥ h'/2 . But it keep in mind that any two non commuting operators A , B , satisfying

 

                         A BBA  = -ih '     will have the uncertainty (or dispersion)

 

                          ∆A ∆B ≥ h'/2                                            .                                 (38)

 

 

Suppose for reason of simplicity that  <x> =  and <p>= 0 .  Assume that the  wave function Ψ is real and normalized  ∫ Ψ 2 dx=1 with

vanishing Ψ( +∞) , Ψ( -∞) .

Consider the obvious inequality 

                                        

                                               ∫ (α xΨ + β h'dΨ/dx) 2 dx ≥ 0      ,  - ∞  ≤  x ≤ + ∞,                (39)

where α , β are two arbitrary  real parameters. The dimensions can be chosen to make the integral a-dimensional. One has

 

α 2 x 2 Ψ2 dx ~  L0 ~ α 2  L2 (1/L) (L)   , hence α ~ 1/L .  The required dimensions of  β are obtained from,

 

(β h'dΨ/dx) 2 dx ~ β2 h' 2 (1/L)(1/L2 ) L ~ L0 ; hence  β2 ~ (L/h')2 , β ~ (L/h')  . There are three terms in (39).

 

The first term is

                                                    I1 = ∫α 2 x 2 Ψ2 dx  = α 2 < x2 > .                            (40)

 

The second term is  I2 = 2 ∫ {α xΨ β h'dΨ/dx} dx. We must integrate  ∫ x Ψ (dΨ/dx) dx   twice by parts to get

 

               ∫ x Ψ (dΨ/dx) dx =  -(1/2)  ; hence

 

                      

                                                       I2 = -α β h'                        .                          (41)

The third contribution is intgrated by parts

I3 =  β2 h'2 ∫ (dΨ/dx)2 dx =   β2 h'2  Ψ (dΨ /dx)|-∞   + β2   ∫  Ψ (- h'2d2 Ψ/dx2 ) dx  ,

 

                                      = 0   +  β2 < p2 >                           .                       (42)

 

So

                    I1 + I 2+ I 3 = α 2 < x2 > - h'αβ + < p2 > β2 ≥ 0                    .    (43)

Eq (43) is symmetric in α and  β. Consider for example β fixed and α variable. Equation (43) can not have real roots and therefore the discriminant of the quadratic must be less than or equal to zero . This means that,

                                     h'2 β2 - 4 < x2 > < p2 > β2 ≤ 0       ,

       

                                                

                                              < x2 > < p2 >   ≥  (h'/2)2  .                         (44)

Taking the square root  and labeling  ∆x ≡ < x2 >1/2  , ∆x≡ < p2 >1/2  . The notation  ∆x and  ∆p do not refer to a single measurement but to a statistical measurement. With this notation the indeterminacy principle is written as ,

                                                     

                                                                  ∆x ∆p ≥  (h'/2)           .                       (45)

 

Examples. Taken from reference 1.

Problem 1. Find the coordinate and  p  distribution corresponding to the wave function

                                 Ψ(x) =   (α/π)1/4   exp( i k0 x) exp[-(α/2) (x-x0)2 ] ,    (46-a) 

or                              Ψ(x) =   (α/π)1/4   exp( i p0 x/h') exp[-(α/2) (x-x0)2 ]  ,  (46-b)

where   h'k0 = p0  and  α ~ 1/L2 . The coordinate wipes out the phase part , i.e. the exponential exp(ik0 x)

 

leaving    Ψ(x)* Ψ(x) =  (α/π)1/2   exp[-α (x-x0)2 ]  producing the gaussian already studied in a previous example, see eq(17) ,except this one is centered at x0 .

Let   α =1 ; k0 =1;  x0=1 . The real and imaginary parts of  Ψ(x) are shown.

 

Fig 3-3  . Real part of Psi vs x.

 

 

Fig  3-4.   Imaginary part of Psi  vs x. 

 

 

 

Fig 3-5 .   (Real Ψ)2 + ( Im Ψ)2

 

SAGE code

 # Ψ(x) = (α/π)1/4 exp( i k0 x) exp[-(α/2) (x-x0)2
alfa=1 ; k0=1; x0=1;
repsi=(alfa/pi)^(1/4)*cos(k0*x)*exp(-(alfa/2)*(x-x0)^2 );
impsi=(alfa/pi)^(1/4)*sin(k0*x)*exp(-(alfa/2)*(x-x0)^2 );
y= plot(repsi,x,-4*x0,5*x0)
#y= plot(impsi,x,-4*x0,5*x0)
show(y)                                    

                                                                                                  

                                   

Φ(p) = (1/h1/2 ) ∫ exp (- i px/h') Ψ(x) dx  ~ (momentum)-1 

 

The p -representation is

 

 Φ(p) = (1/h1/2 ) ∫ exp (- i px/h')  (α/π)1/4 exp( i k0 x) exp[-(α/2) (x-x0)2 ]  = (1/h1/2 ) ∫ exp (- i px/h') Ψ(x) dx 

       = (1/h1/2 ) ∫ exp (- i px/h')  (α/π)1/4   exp( i p0 x/h') exp[-(α/2) (x-x0)2 ]  dx.          (47)

In our units  h'=1, k0 = 1, thus p0 =1 , and we have selected  α =1 , x0 =1. The p- space wave function is ,

 

Φ(p) = (1/π)1/4 ∫ cos( (p-p0)x ) exp( -(1/2) (x-x0)2 dx  + i (1/π)1/4 ∫ sin ( (-p+p0)x ) exp( -(1/2) (x-x0)2 dx . (48)

 

  Fig 3- 6 . Real {Φ(p)}  and   {RealΦ(p)} 2 .    

 

 

 

Fig 3-7.           

 

 

 

 

Fig 3-8 . Φ2 = RealPart(Φ)2 + ImgPart(Φ)2 is  a gaussian distribution centered at p0 =1.

 

One can verify that Φ(p)2 = (1/(α h'2π))1/2 exp { -(p-p0)2 /(α h'2 ) } , with α =1 , h'=1. In conclusion the wave function

 

Ψ(x) =   (α/π)1/4   exp( i p0 x/h') exp[-(α/2) (x-x0)2 ] ,has a gaussian distribution Ψ(x)2 around x0 and a gaussian distribution

Φ2  about p0 .

 

 

FORTRAN code

c gaussian in p space centered at p=1
data p0,np,range/1.,60,2./
data alfa,hbar/1.,1./
phi(p)=(1./(alfa*hbar**2*Pi))**.25*
$ exp(-(p-p0)**2/(2.*alfa*hbar**2))
pi=2.*asin(1.)
dp=(2.*range)/float(np)
do 10 ip=0,np,2
p=p0-range+dp*float(ip)
print 100,p,phi(p)**2
10 continue
100 format('p,phi(p)**2=',2(4x,e11.4))
stop
end

 

p,phi(p)**2= -0.1000E+01 0.1033E-01
p,phi(p)**2= -0.8667E+00 0.1730E-01
p,phi(p)**2= -0.7333E+00 0.2796E-01
p,phi(p)**2= -0.6000E+00 0.4361E-01
p,phi(p)**2= -0.4667E+00 0.6565E-01
p,phi(p)**2= -0.3333E+00 0.9536E-01
p,phi(p)**2= -0.2000E+00 0.1337E+00
p,phi(p)**2= -0.6667E-01 0.1808E+00
p,phi(p)**2= 0.6667E-01 0.2361E+00
p,phi(p)**2= 0.2000E+00 0.2975E+00
p,phi(p)**2= 0.3333E+00 0.3617E+00
p,phi(p)**2= 0.4667E+00 0.4245E+00
p,phi(p)**2= 0.6000E+00 0.4808E+00
p,phi(p)**2= 0.7333E+00 0.5255E+00
p,phi(p)**2= 0.8667E+00 0.5542E+00
p,phi(p)**2= 0.1000E+01 0.5642E+00
p,phi(p)**2= 0.1133E+01 0.5542E+00
p,phi(p)**2= 0.1267E+01 0.5255E+00
p,phi(p)**2= 0.1400E+01 0.4808E+00
p,phi(p)**2= 0.1533E+01 0.4245E+00
p,phi(p)**2= 0.1667E+01 0.3617E+00
p,phi(p)**2= 0.1800E+01 0.2975E+00
p,phi(p)**2= 0.1933E+01 0.2361E+00
p,phi(p)**2= 0.2067E+01 0.1808E+00
p,phi(p)**2= 0.2200E+01 0.1337E+00
p,phi(p)**2= 0.2333E+01 0.9536E-01
p,phi(p)**2= 0.2467E+01 0.6565E-01
p,phi(p)**2= 0.2600E+01 0.4361E-01
p,phi(p)**2= 0.2733E+01 0.2796E-01
p,phi(p)**2= 0.2867E+01 0.1730E-01
p,phi(p)**2= 0.3000E+01 0.1033E-01
 

 

 

 

 

 FORTRAN code

c gaussian p distribution june 13 ,2009
real Lscale
data hbar, alfa,np,nstep/1.,1.,40,10000/
data p0,x0/1.,1./
f(x) =(1./sqrt(h))*sin((p-p0)*x/hbar)*(alfa/pi)**(.25)*
$ exp(-alfa*(x-x0)**2/2.)
pi=2.*asin(1.)
lscale=1./sqrt(alfa)
h=2.*pi
p1=p0-3.
p2= p0+3.
dp=(p2-p1)/float(np)
do 10 ip=0,np
p=p1+dp*float(ip)
range=10.*lscale
dx=2.*range/float(nstep)
sum=0.
do 20 ix=1,nstep,2
x=(x0-range)+dx*float(ix)
sum=sum+(dx/3.)*(f(x+dx) + 4.*f(x) + f(x-dx))
c sum=sum+(dx/2.)*( f(x) + f(x-dx))
20 continue
sum=sum
print 100,p,sum, sum**2
c print 100,-p,sum
10 continue
100 format(1x,'p,phi(p),phi**2=',3(4x,e10.3))
stop
end


 

Problem 2.( From ref 1. page 41). find the x and p distribution for the function

 

   Ψ(x) = (i/L)1/2 exp( i k0 x)          ,  0 < x < L

 

             =0                            ,      x <   0   , x >  L  .                    (49)

 

Since ,  i = exp(iπ/2)  ,    Ψ(x) = (1/L)1/2 exp( iπ/4) exp( i k0 x)  and  Ψ(x)* = (1/L)1/2 exp(- iπ/4) exp(- i k0 x) .

Hence   Ψ(x)* Ψ(x) = 1/L    and   we verify that  ∫ Ψ(x)* Ψ(x) dx = 1  .   

 

To find Φ(p) we employ a complex number FORTRAN code to integrate

                 

 

 

                     Φ(p) = (1/h1/2 ) ∫ exp (- i px/h')  (1/L)1/2 exp( iπ/4) exp( i p0 x/h')  dx   ,    -  ∞ < x <+∞   (50)

setting   k0 = p0/h'  ,and  p0=2  , h'=1 , h= 2π.

                    

   

 FORTRAN complex code fro eq. (50)

c complex phi(p) 17 june 2009
real L
complex psi ,rooti ,sum
data hbar,p0,L/1.,2.,1./
data nx,np/2000,60/
data rangep/12./
psi(x)=sqrt(1./L)*exp(rooti*pi/4.)*exp(rooti*p0/hbar)
pi=2.*asin(1.)
xi=0.
xf=L
dx=(xf-xi)/float(nx)
dp=2.*rangep/float(np)
h=2.*pi
rooti=cmplx(0.,1.)
do 10 ip =0,np
p=-rangep + dp*float(ip)
sum=0.
do 20 ix=1,nx
x=xi+dx*float(ix)
sum=sum+(1./sqrt(h))*(dx/2.)*(exp(-rooti*p*x/hbar)*psi(x) +
$exp(-rooti*p*(x-dx)/hbar)*psi(x-dx) )
20 continue
print 100 ,p,real(sum), aimag(sum),real(sum)**2+aimag(sum)**2
10 continue
100 format('p,Real,imag,Phi**2=',4(3x,e10.3))
stop
end
 

 


 

 


  

 

 

Fig 3-8. Momentum distribution corresponding to function (49).

 

 

Problem 3. (from ref 1. page 41). Find <x> , < x2 > , <p> , <p2 > of a particle in a state represented by

 

           Ψ(x) = N { sin(x) +(i/2) sin(3x) }     ,   0 < x < 2π ,Ψ(x) = 0  elsewhere and N is the normalization constant.

 

We can avoid calculating N by writing for example

 

                                      <x> = ∫ Ψ* x Ψ dx / ∫ Ψ* Ψ dx                                 .   (51)

In this manner the N cancels , albeit at the cost of another integral.

 

 

The numerator is ∫ Ψ* x Ψ dx = ∫0  x { sin(x)2 + (1/4)sin(3x)2 } dx = (5/4) π2 .

SAGE code

f=x*( sin(x)^2+(1/4)*sin(3*x)^2 );
integral(f,x,0,2*pi)

 
5/4*pi^2

The denominator is ∫ Ψ* Ψ dx =∫ {sin(x)2 + (1/4)sin(3x)2 } dx

SAGE code

f=sin(x)^2+(1/4)*sin(3*x)^2 ;
integral(f,x,0,2*pi)

5π/4

Hence  <x> = π .

Now      <x2 > =  ∫ Ψ* x2 Ψ dx / ∫ Ψ* Ψ dx = x2 {sin(x)2 + (1/4)sin(3x)2 } dx  (4/(5π))

The numerator is x2 {sin(x)2 + (1/4)sin(3x)2 } dx =   -(37/72)π + (5/3) π3

Sage code

f=x^2*( sin(x)^2+(1/4)*sin(3*x)^2 );
integral(f,x,0,2*pi)
-37/72*pi + 5/3*pi^3

Hence

<x2 >=  { -37/72*pi + 5/3*pi^3} ( 4/(5*pi) ) = 40.050/π = 12.7483616

The average

<p> =  ∫ { sin(x) -(i/2)sin(3x) }( -i ∂/∂x ) { sin(x) + (i/2)sin(3x)    } dx / ∫ Ψ*  Ψ  dx 

      =  ∫ { sin(x) -(i/2)sin(3x) }( -i ) { cos(x) + (3i/2)cos(3x)  } dx / ∫ Ψ*  Ψ  dx ,

      = 0 

because the numerator contains terms proportional to sin(x) cos(x) , sin(x) cos(3x) , sin(3x)cos(x) and sin(3x)cos(3x). All integrals are zero on the interval from 0 to 2π.

The average

<p2 > = ∫ { sin(x) -(i/2)sin(3x) }( -∂2/∂x2 ) { sin(x) + (i/2)sin(3x)} dx / ∫ Ψ*  Ψ  dx

        = ∫ { sin(x) -(i/2)sin(3x) }( -1 ) { - sin(x) - (9i/2) sin(3x)} dx /(5π/4) .

Due to the orthogonality this reduces to two terms,

<p2 > = ∫ { sin(x) -(i/2)sin(3x) }( -1 ) { - sin(x) - (9i/2) sin(3x)} dx /(5π/4)

<p2 > =  ∫ { sin(x)2 +(9/4)sin(3x)2 } dx /(5π/4)

         = 13/5

Sage code

f=(4/(5*pi))*(sin(x)^2 +(9/4)*sin(3*x)^2 ) ;
integral(f,x,0,2*pi)

13/5

We check now the product ∆x ∆p.

∆x ≡ {<x2> - (xave)2 }1/2  ={12.7 -π2 }1/2 = 1.68 

∆p ≡ {<p2> - (pave)2 }1/2 = {2.6-0}1/2 =1.61

So ∆x ∆p= 2.70 > h'/2     where h'=1  .

Postulate C : The time rate of change of Ψ(x,t) is related to the hamiltonian by

                                         i h' ∂ Ψ (x,t)/∂t = H Ψ(x,t)                     .   (52)

Eq (52) is Schrodinger's time dependent equation.

Why is this equation plausible. Can we get rid of the time derivative by separation of variables ?

In the case of plane waves we have   H= -(h'2 /(2m) ) ∂2 / ∂x2 , the potential energy U(x)=0  and

Ψ(x,t)= N exp( i (px -Et)/h') =N exp ( i (kx -ωt) ). Therefore  , the left hand of (52) is

                            i h' ∂ Ψ (x,t)/∂t = i h' Ψ(x,t) (-iE/h')  =E  Ψ(x,t).         (53)

On the right hand side of (52) we have

-(h'2 /(2m) ) ∂2 Ψ(x,t)/ ∂x2 = -{h'2 /(2m)} Ψ(x,t) ( ip/h')2 =  (p2 /(2m)) Ψ(x,t)    . (54)

Equations (53) and (54) state the obvious fact that the total energy in this case is the kinetic energy.

The time dependent Schrodinger equation is the necessary tool in scattering problems and time dependent perturbations.This is very much the situation in the quantum theory of radiation where the electromagnetic potentials may be functions of time i.e. Φ(t) and A(t). 

Eq (52) has the form of an initial value problem , for if  Ψ(x,t=0)= Ψ (x) (italics have been used) and if  H Ψ (x) = EΨ (x) then

                                          Ψ(x,t) = Ψ (x) exp( -iEt/h') is the solution.

Applying the simple Euler rule of numerical integration to eq. (52)    ( with Ψ(x,t=0)= Ψ (x) = g(x) ) one gets

                            Ψ(x,0+∆t) = Ψ(x,0) + (∆t/(ih') {[g(x+∆x)-2 g(x)+g(x-∆x) ]/∆x2 + U(x) g(x) } ,

or in general

                      Ψ(x,t+∆t) = Ψ(x,t) + (∆t/(ih') {[Ψ (x+∆x,t)-2Ψ (x,t) +Ψ (x-∆x,t) ]/∆x2 + U(x) Ψ(x,t) }. (55)

 

 

The equation                 

                                                               H Ψ (x) = EΨ (x)                          (56)

is Schrodinger's time independent equation. The hamiltonian operator is H = -(h'2 /(2m) )2 / ∂x2 + U(x). In general this is a boundary value problem. Both Ψ (x) and E are unknowns in stationary problems where the time parameter does not enter. Conditions have to be imposed on the allowed solutions Ψ (x) resulting in general in a set  Ψn (x) and the corresponding quantized energies En   , n is the quantum number. The allowed values of n emanate from the specific boundary conditions and are not imposed a priori.

                                  

Example: We give a numerical integration example of   i h' ∂ Ψ (x,t)/∂t = H Ψ(x,t) using the algorithm (55).

The following fact is unknown to us, at this stage of the Lectures . The function Ψ(x,t) =N exp(-x2 /2) exp(-iEt/h') is the time dependent solution  for the  harmonic oscillator ground state. N is a normalization constant obtained from  requiring  

     ∫+ ∞ -∞ Ψ * Ψ dx =1. The hamiltonian is H= -(h'2 /(2m) )2 / ∂x2 + (1/2) k x2 -(1/2)2 / ∂x2 + (1/2) x2 in the system of units where where k ≡1, m≡1, h≡1.

We are given Ψ (x,0)= g(x) = exp(- x2/2). a) find and plot  Ψ (x,t) for the instants  t= τ /4  ,  τ /2 ,  τ , where τ = 2π(m/k)1/2 = 2π.

b) find the energy E.

 

One finds by applying the hamiltonian that indeed exp(-x2 /2) is indeed an eigenfunction. The Sage code given below was used to corroborate that  { -(1/2)2 / ∂x2 + (1/2) x2 } exp(-x2 /2) = (1/2)  exp(-x2 /2) , thus E =1/2.

One can also find E from the time evolution of the numerical solution Ψ (x,t) = g(x) cos(Et/h') + i g(x) sin(Et/h') .

Using the real part of the solution one gets , (with h'=1)  cos(Et) = real part Ψ (x,t) /g(x) and

E=(1/t)  arc cos{ real part Ψ (x,t) /g(x)  }. A numerical example is provided below.Of course not all estimates of E are the same and as the numerical  solution becomes poor especially at the extremes so does the value of E.

 

Sage code

psi=exp(-x^2/2) ; v=(1/2)*x^2;
-(1/2)*diff(psi,x,2)+v*psi

1/2*e^(-1/2*x^2)

 

Fig 3-9. Time dependent solution of the harmonic oscillator  ground state, at t=0.  

 

 

Fig 3-10. Time dependent solution of the harmonic oscillator  ground state, at t= τ/4 .  

 

Estimate of E at  t= τ/4 . The best values occur for    -2 < x < 2. where     .494 < E < 5.20

 tf= 1.57079637
tau,dt,dx,nt,nx = 6.28318548 0.000900000043 0.300000012 1745 20

x,t,acos( ),E= -0.300E+01 0.157E+01 0.157E+01 0.100E+01
x,t,acos( ),E= -0.270E+01 0.157E+01 0.102E+01 0.652E+00
x,t,acos( ),E= -0.240E+01 0.157E+01 0.859E+00 0.547E+00
x,t,acos( ),E= -0.210E+01 0.157E+01 0.811E+00 0.517E+00
x,t,acos( ),E= -0.180E+01 0.157E+01 0.828E+00 0.527E+00
x,t,acos( ),E= -0.150E+01 0.157E+01 0.817E+00 0.520E+00
x,t,acos( ),E= -0.120E+01 0.157E+01 0.796E+00 0.507E+00
x,t,acos( ),E= -0.900E+00 0.157E+01 0.785E+00 0.500E+00
x,t,acos( ),E= -0.600E+00 0.157E+01 0.780E+00 0.497E+00
x,t,acos( ),E= -0.300E+00 0.157E+01 0.773E+00 0.492E+00
x,t,acos( ),E= 0.119E-06 0.157E+01 0.775E+00 0.494E+00
x,t,acos( ),E= 0.300E+00 0.157E+01 0.773E+00 0.492E+00
x,t,acos( ),E= 0.600E+00 0.157E+01 0.780E+00 0.497E+00
x,t,acos( ),E= 0.900E+00 0.157E+01 0.785E+00 0.500E+00
x,t,acos( ),E= 0.120E+01 0.157E+01 0.796E+00 0.507E+00
x,t,acos( ),E= 0.150E+01 0.157E+01 0.817E+00 0.520E+00
x,t,acos( ),E= 0.180E+01 0.157E+01 0.828E+00 0.527E+00
x,t,acos( ),E= 0.210E+01 0.157E+01 0.811E+00 0.517E+00
x,t,acos( ),E= 0.240E+01 0.157E+01 0.859E+00 0.547E+00
x,t,acos( ),E= 0.270E+01 0.157E+01 0.102E+01 0.652E+00
x,t,acos( ),E= 0.300E+01 0.157E+01 0.157E+01 0.100E+01
 

c Schrodinger time dependent equation 21 jun 2009
cReferences heat eq solution by forward difference h'^2*dt/(m*dx**2)<<1
c inhomogenous PDE see D. G. Duffy PDE page 91
c ih'd psi/dt= del^2 Psi +u(x)*psi
c BC u(-a/2,t)=u(a/2.,t)=0.
real k,m
complex psi, psinew, rooti, Hpsi
dimension psi(0:500) ,psinew(0:500)
data k,m ,hbar,nx /1.0,1.0,1.0,20/
V(x)=(1.0/2.0)*x**2
g(x)=exp(-.5*x**2)
Hpsi(u,i)=-(hbar**2/(2.0*m))*(psi(i+1)-2.0*psi(i)+
$ psi(i-1))/dx**2 + V(u)*psi(i)
pi=2.0*asin(1.0)
rooti= cmplx(0.0,1.0)
xi=-3.
xf=-xi
omega=sqrt(k/m)
tau=2.0*pi/omega
tf=tau/4.
c initial value of temp
dx=(xf-xi)/float(nx)
dt= .01*(m*dx**2)/hbar**2
nt=int(tf/dt)
print*,'tf=',tf
print*,'tau,dt,dx,nt,nx =' ,tau,dt,dx,nt,nx
print*,' '
do 30 i=0,nx
x=xi+dx*float(i)
psi(i)=g(x)
30 continue
c do 10 is the time loop
do 10 j=1,nt
t=dt*float(j)
do 20 i=1,nx-1
x=xi+dx*float(i)
psinew(i)=psi(i)+ (dt/(rooti*hbar))*HPsi(x,i)
20 continue
psinew(0)=0.0
psinew(nx)=0.0
c changes new and old
do 50 i2=0,nx
psi(i2)=psinew(i2)
50 continue
10 continue
do 40 i=0,nx
x=xi+dx*float(i)
psisq= Real(psinew(i))**2+aimag(psinew(i))**2
print 100, t,x,Real(psinew(i)),aimag(psinew(i)),psisq
40 continue
print 100,t,x,Real(psinew(i)),aimag(psinew(i)),psisq
100 format('t,x,Re,Im,psi2=',5(2x,e10.3))
stop
end

A continuity equation is constructed starting from the time dependent Schrodinger equation. For example in one dimension

multiply by Ψ*  and write its conjugate equation to get 

ih'Ψ* ∂ Ψ/∂t = - (h'2/(2m)) Ψ*2Ψ/∂x2 + U  Ψ*  Ψ ,

-ih'Ψ ∂ Ψ* /∂t = - (h'2/(2m)) Ψ ∂ 2Ψ* /∂x2 + U  Ψ  Ψ* . Substract the second from the first and after rearranging

    ∂ ( Ψ Ψ* ) /∂t  + (ih'/(2m)){ ∂ [Ψ ∂ Ψ* /∂x ] /∂x -  ∂ [Ψ* ∂ Ψ /∂x ] /∂x }  = 0 .                           (57)

 

 This equation is of the form

                                                                       ∂ ρ/∂t + ∂ J/∂x = 0 ,                                      (58)

where ρ = Ψ Ψ* ~ 1/L  and   J = (ih'/(2m)){ Ψ ∂ Ψ* /∂x  -  Ψ* ∂ Ψ /∂x } ~ 1/time ( in one dimnsion).

This a continuity equation analogous to the one obtained in electromagnetism for the charge and current densities. Lets generalize to three dimensions. The probability density is  ρ = Ψ Ψ* ~ 1/L3 , and define the probability current density by

                                                    J =  (ih'/(2m)){ Ψ del Ψ*   -  Ψ* del Ψ }                        . (59)

In cartesian coordinates the del operator is given by   del = i ∂/∂x  + j ∂ /∂y  + k ∂ /∂z . Equation (58) takes the form

                                                                      ∂ρ/∂t + del ∙ J = 0   .                            (60)

 Ψ  is a  stationary state  when it is represented by a function Ψ ~ g(x) exp(-iEt/h')  (with g(x)  real) , then  J = 0 and ρ is of course time independent.

Example : Let       Ψ(x,0) = g(x)=exp(-x2 ) . Is this a stationary state ? Certainly not because the application of    

H g(x) = {-(3/2)x2 +1 } g(x)   , does not yield a  ,   constant times g(x) , i.e. an eigenvalue times the original function.

SAGE code

psi=exp(-x^2) ; v=(1/2)*x^2;
-(1/2)*diff(psi,x,2)+v*psi
-3/2*x^2*e^(-x^2) + e^(-x^2)

 

The next two plots show a situation where ρ = Ψ* Ψ is time dependent.

Fig 3-11. Plot of Real psi , Im Psi and psi2  (t ≈ 0) when Ψ(x,0) =g(x)= exp(-x2 ) , is not an eigenfunction or

stationary function of H.

 

Fig 3-12. Plot of Real psi , Im Psi and psi2  (t ≈ 3τ/4 ) when Ψ(x,0) =g(x)= exp(-x2 ) , is not an eigenfunction or

stationary function of H. Psi is spreading to the sides as time passes.

 

 


 

Principle of Superposition:

The principle of superposition is exemplified in the following considerations. Suppose the time independent Schrodinger equation has been solved obtaining  the set of orthonormalized eigenfunctions

 φn and eigenvalues En  , i.e.

                          H φn = En φn        ,   ∫ φn* φm dx  = δ n m   .            (61)                            

Then an arbitrary  normalized function  Ψ (x) can be obtained  from a superposition

                   Ψ (x) = ∑n  cn φn       ,    where  cn =  ∫ φn* Ψ (x)  dx .    (62)

The sum in (62 ) may contain an infinite number of terms. Because Ψ  is mormalized ,  ∫ Ψ* Ψ dx =1 , one has

                                                                 ∑n  c*cn = 1.          .  (63)

The energy of a particle repsented by   Ψ (x) is given by

                            E = ∫ Ψ *(x)  H  Ψ (x)  dx  =    ∑n En c*cn                             .    (64)

The time dependent superposition follows from (62)

                                              Ψ (x,t) =  ∑n  cn φn exp(-i Ent/h')             .    (65)

Example:  Let the hamiltonian be    H= -(1/2) ∂2 /∂x2 +U(x)    , where U(x)=0 ,0 ≤ x ≤ 1, and infinite otherwise.

It will be shown later that the set of orthonormalized eigenfunctions is  φn (x) = (2)1/2 sin(nπx) .Given  the normalized function

Ψ (x) = (30)1/2 * x*(1-x)  find the coefficients cn , the energies En and the average energy ,

                                                    E=<Ψ (x)│ H│Ψ (x) >/<Ψ (x)│Ψ (x) >.

The energy eigenvalues are obtained from  -(1/2) ∂2 φn (x) /∂x2 = -(1/2) (-1)(nπ)2 φn (x) = En φn (x),

that is , En =  (1/2)(nπ)2 in our units where m=1 ,h'=1 , L =1(length of the inifinite well),  Escale =h' 2/(mL2 ).

Results

cn vanishes for even n

n,cn= 1 0.999E+00
n,cn= 2 -0.598E-07
n,cn= 3 0.370E-01
n,cn= 4 0.445E-08
n,cn= 5 0.799E-02
n,cn= 6 0.140E-07
n,cn= 7 0.291E-02
n,cn= 8 -0.457E-07
n,cn= 9 0.137E-02
n,cn= 10 -0.268E-07

 

A comparison of Ψ(x)   with a five term summation shows agreement to four digits.
x,Psi,Psinum= 0.0000E+00 0.0000E+00 0.0000E+00
x,Psi,Psinum= 0.1000E+00 0.4930E+00 0.4943E+00
x,Psi,Psinum= 0.2000E+00 0.8764E+00 0.8754E+00
x,Psi,Psinum= 0.3000E+00 0.1150E+01 0.1151E+01
x,Psi,Psinum= 0.4000E+00 0.1315E+01 0.1314E+01
x,Psi,Psinum= 0.5000E+00 0.1369E+01 0.1370E+01
x,Psi,Psinum= 0.6000E+00 0.1315E+01 0.1314E+01
x,Psi,Psinum= 0.7000E+00 0.1150E+01 0.1151E+01
x,Psi,Psinum= 0.8000E+00 0.8764E+00 0.8754E+00
x,Psi,Psinum= 0.9000E+00 0.4930E+00 0.4943E+00
x,Psi,Psinum= 0.1000E+01 0.0000E+00 -0.1463E-06

The average energy (eeave) is slightly higher than E1  .

 eeave,E1= 4.99919844 ,     4.93480253
 

FORTRAN code

c calculation of expansion coefficients
real L
data L,nc,nx /1.,10, 100/
dimension coef(20)
psi(x)=sqrt(30.)*x*(1.-x)
phi(x,n)=sqrt(2./L)*sin(float(n)*pi*x/L)
G(x)=psi(x)*phi(x,n)
pi=2.*asin(1.)
deltax=0.1
do 10 n=1,nc
dx=L/float(nx)
sum=0.
do 20 ix=1,nx
x=dx*float(ix)
sum=sum +(dx/2.)*(g(x)+g(x-dx))
20 continue
coef(n)=sum
10 continue
do 30 i=1,nc
print 100, i,coef(i)
30 continue
100 format('n,cn=',i3,3x,e10.3)
print*,' '
do 40 j=0,10
x=deltax*float(j)
print 200, x,psi(x),psinum(x,L,pi,nc,coef)
40 continue
call eave(L,pi,nc,coef,eeave)
print*,'eeave,E1=',eeave,(1./2.)*(pi/L)**2
200 format('x,Psi,Psinum=',3(4x,e11.4))
stop
end

function psinum(x,L,pi,nc,coef)
real L
dimension coef(20)
phi(x,n)=sqrt(2./L)*sin(float(n)*pi*x/L)
sum=0.
do 10 i=1,nc
sum=sum+ coef(i)*phi(x,i)
10 continue
psinum=sum
return
end

subroutine eave(L,pi,nc,coef,eeave)
real L
dimension coef(20)
eeave=0.
do 10 n=1,nc
en=(1./2.)*(pi/L)**2*float(n)**2
eeave=eeave+coef(n)**2*en
10 continue
return
end





 

 Amplitude Mechanics:  (see Chapters 1 , 2 , from ref. 7).

Two states systems provide, what in a sense may be, the simplest examples of the superposition principle.

Here a state vector    ,   │Ψ > is represented by the superposition of just two basis states, which are ortho normal,

call them │1> and │2> ,

                             │Ψ > =  c1│1 > + c2│2 >     .                                       (66)

 

This notation is called bracket or Dirac's notation. │Ψ > (the ket) represents a column vector while < Ψ│, (the "bra") represents a row formed by the complex conjugate of │Ψ > . The coefficients are given by , c1 = < 1│Ψ > , c2 = < 2│Ψ >.

 

At this stage Schrodinger's equation has not  been invoked  and no constant  h' is used . One does know for a physical fact, that  certain systems can show two states or a mixture of them. Such is the case of light with two planes of polarization X or Y or  right /left polarization. The electron spin  also shows two projections along the magnetic field called spin up or down.

Fig 3-13. A light wave propagating in the Z direction with the electric field polarized θ degrees with respect to X.

The situation shown in Fig 3-13 can be represented with the abstract vectors,

│1 > = │x > , │2> = │y > , < x│Ψ > = k Ex=kEcos(θ)  ,  < y│Ψ > = k Ey= kEsin( θ)  , where k ={V/(8πh'ω)}1/2 .

Thus   

          │Ψ > =  {V/(8πh'ω)}1/2 [   Ex   │x > +  Ey   │y >  ]                             .            (67)        

The normalization of  < Ψ│Ψ > = k2 E2 cos 2(θ ) + k2 E2 sin 2(θ ) =1  , requires that   E2 /(8π) = h'ω/V . This states that the energy density of the electromagnetic field (CGS units)  equals the energy of the photon ( h'ω ) divided by the volume V. It also ties a classical expression ,energy density of the field, to the quantum concept of the photon.