LECTURES ON QUANTUM MECHANICS

by Reinaldo Baretti Machín   (reibaretti2004@yahoo.com)

Chapter 6 - Periodic Potentials

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

FREE DOWNLOAD FORTRAN-FORCE 2.0.8

References:

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

by W. V. Houston

2. Introduction to Solid State Physics (Hardcover) by Charles Kittel

Electron in metals can be treated to a first approximation as entirely free , except for the box confinement which is represented by an infinite wall potential. In this model the interaction between the free electrons is neglected as well as the interaction between the electron and the sea of positive ions.

The allowed energies are  those already known from the infinite box  E_n = (1/2) (n π /L_box )² where L_box = N d ; N being the number of atoms and d the separation in a linear chain.

The next best approximation would be to suppose that in  a linear chain of N atoms separated a distance -d- there are N electrons , if each atom contributes one electron to the free electron sea. Or instead one may suppose that there are 2N electrons if each atom allows two electrons to roam freely. The electrons interact with the periodic potential of the ions.

We integrate numerically Schrodinger equation using the FORTRAN code given below. A linear chain of  "6 atoms " is simulated with the potential

v=v0*(1-sin(pi*x/d)^2) , v0=10 , d= 1 .

The initial conditions on Ψ are Ψ (0)=0. , (dΨ/dx)_x=0 =1 .

Sage math

# plot periodic potential
v0=10 ; d=1 ; v=v0*(1-sin(pi*x/d)^2) ;
y=plot(v,x,0,6*d)
show(y)

 Fig 1 . Periodic potential V=v0*(1-sin(pi*x/d)^2) , v0=10, d=1, with boundaries at x=0, x=L_box = 6d .

  At the boundaries V(x)= ∞ , for x < 0   , x > L .

V(x+d) =V(x)   d  is the atomic separation. 

Bloch waves

A wave function is of the form

                                  Ψ(x) = u(x) exp(ikx)                                                    (1)

It is a wave function for a  chain of N atoms separated a distance d. The size of the box is Nd=L_box .

It is assumed that  Ψ(x) = Ψ(x + n L_box) . Then , in a mathematical sense the box is infinite with period L_box. On the other hand 

the function u(x)  is periodic with respect to lattice translations , u(x + nd) = u(x).

What we will asume here instead is that the ends of the box represent an infinite potential imposingΨ(0) = Ψ(L_box) =0 . Equation (1) will be modified to

                                                            Ψ(x) = u(x) sin(kx)        .         (2)  

The trigonometric part sin(kx) takes into account the boundary conditions of the box.  k =( n π /L_box  )   n=1,2,3....

The function u(x) is estimated by dividing Ψ(x) /sin(kx) . Of course near the zeros this would give wrong values. Nevertheless the shape of u(x) can be discerned from the plot confirming the periodicity of u(x).

Fig 2. First state  E1= 4.50  

Fig 3 . Second electron state , E =4.865    ,  k₂ = (2π /L_box )

Fig 4.  Second state. Shows Ψ₂ and  u (x) .The errors or distortion in u(x) at x=0 , x=3 and x= L_box is due to the fact that we are dviding Ψ₂ by sin(k₂x) to extract u(x) . At those point the sine function is zero giving invalid results and L'Hopital rule should be used.

Fig 5. E3 = 5.435 , k₃ =(3π /L_box )   again there is distortion in u(x) at the zeros of sin(k₃ x) .

Fig 6 . E(k) shows a gap at k= π/d .Another gap would show at k= 2π/d , 3π/d...etc .They are less pronounced when  E >> V(x). As k becomes larger

E(k) approaches (1/2)k² .

Fig 7. Wave function for the 8-th state , it is the second state in the second "band". Ψ₈ is un normalized and is

 practically identical to sin (k₈x). Compare to the ground state wave function in Fig 2.

c Periodic potential /Bloch waves
real Lbox , Lscale ,k1 ,k2 ,k3,k4,k5,k6, k
dimension psinf(100),energy(100)
data niter/1/
data v0,d, natoms /10.,1.,6/
v(x)=v0*(1.-sin(pi*x/d)**2 )
pi= 2.*asin(1.)
Lbox=float(natoms)*d
k1=pi/lbox
k2=2.*pi/lbox
k3=3.*k1
k4=4.*k1
k=k3
e=5.435
efinal=5.70
de=(efinal-e)/float(niter)
xi=0.
xlim=Lbox
lscale=1./abs(v0)**.5
dx=lscale/300.
nstep=int(xlim/dx)
e1box=(1./2.)*k1**2
print*,'e1box,e2box=', e1box,4.*e1box
print*,'xlim,dx,nstep=', xlim,dx,nstep
print*,' '
do 110 iter=1,niter
c dx=xlim/float(nstep)
kp=int(float(nstep)/90.)
kount=kp
c IC for even functions
c psi0=1.
c psi1=psi0
c IC for odd functions
psi0=0.
psi1=psi0+dx
if(niter.eq.1) print 200 ,0.,psi0,psi0/sin(k*(0.+dx)),sin(k*xi)
do 100 i=2,nstep
x=xi+dx*float(i)
psi2=2.*psi1-psi0 -dx**2*2.*(E-v(x-dx))*psi1
if(i.eq.kount)then
uk=psi2/sin(k*x)
if(niter.eq.1)print 200 ,x, psi2 ,uk,sin(k*x)
kount=kount+kp
endif
psi0=psi1
psi1=psi2
100 continue
energy(iter)=e
psinf(iter)=psi2
20 continue
e=e + de
110 continue
do 30 i=1,niter
If(niter.gt.1)print 120 ,energy(i),psinf(i)/abs(psinf(i))
30 continue
120 format(3x,'energy,psinf=',2(3x,e10.3))
200 format(3x,'x,psi,un(x),sin(k*x) =',4(3x,e10.3))
stop
end