LECTURES ON QUANTUM MECHANICS

by Reinaldo Baretti Machín

Chapter 3.  The Postulates of Quantum Mechanics

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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… q_n ,t)    , here q_i 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 q_i . 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 … dq_n = 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 )….  (q_n , q_n +dq_n) . 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)   → p_operator =  -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

                                  p_opΨ = 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 

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

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

               ∫ N²  Φ_n(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

                                                   Ψ_n (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/h^1/2 ) ∫ 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/h^1/2  ).

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/h^1/2 ) ∫ exp (+ i px/h') Ψ* (x') dx' (1/h^1/2 ) ∫ 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/h^1/2 ) ∫ 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 = π^1/2 /(2 a^3/2 )   , -∞ ≤  u  ≤ +∞      .      (21) 

The average of x² is       

x² _average = ( α/π)^1/2 ∫ x² exp( -α x² ) dx = ( α/π)^1/2 π^1/2 /(2  α^3/2 ) = 1/(2α)             ,        (22) 

and   

( x² _average)^1/2 = { (x- x_ave )² }^1/2 ≡ ∆x =    1/(2α)^1/2  , since x_ave =0 .                                    (23)

The average of p² is, using (19)

    p²_{ave     =}     ( απ h'² )^-1/2 ∫ p² exp{ -p² /(αh'² )} dp = ( α/2) h' ²                   .              (24)

So                                           ( p²_ave )^1/2 ,    = ( α/2)^1/2 h'  ,                                    (25)

which equals ∆p_ave  since    p_ave = 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 = { p_x²  +  p_y ²  + p_z ²  } /(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 = yp_z - z p_y   → L_x = -ih'(y ∂ /∂z - z ∂ /∂y )                                      ,               (32)

L _y = zp_x - x p_z   → L_y = -ih'(z ∂ /∂x - x ∂ /∂z )                                       ,              (33)

L _z = xp_y - y p_x   → L_z = -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 *  ( p_op x - x p_op )  Ψ dx = -ih'               . (35)   

Proof :  The firts term in (35) is

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

Substitution in (35) gives

                             ∫{ Ψ *-ih'Ψ  +  Ψ^*x p_op Ψ  -Ψ * x p_op  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 p_x usually written as

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

                         A B -  BA  = -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/h^1/2 ) ∫ exp (- i px/h') Ψ(x) dx  ~ (momentum)^-1 

The p -representation is

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

       = (1/h^1/2 ) ∫ 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/h^1/2 ) ∫ 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 = ∫₀^2π  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

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.

 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 E_x=kEcos(θ)  ,  < y│Ψ > = k E_y= kEsin( θ)  _, where k ={V/(8πh'ω)}^1/2 .

Thus   

          │Ψ > =  {V/(8πh'ω)}^1/2 [   E_x   │x > +  E_y   │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.