Computational  modeling  

Lecture 8 : Quantum Mechanics

Physics:

•  Differential equations with boundary conditions •  Schrodinger’s cat •  Particle in a Box

Programming:

•  Modification of the Pendulum program

Instructor  :    Cedric  Weber  

Course        :    4CCP1000  

1.  Using functions to solve differential equations " "ý""

2.  Reading an initial condition from the keyboard ý"

3.  Plotting the solution of a differential equation " ý"

4.  Modifying a differential equation to introduce friction "ý

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What  did  you  learn  last  time?  

Class/Week   Chapter   Topic   Milestones  

1   Monte  Carlo   UNIX  system  /  Fortran  

2   Monte  Carlo   Fibonacci  sequence  

3   Monte  Carlo   Random  variables  

4   Monte  Carlo   Central  Limit  Theorem  

5   Monte  Carlo   “Monte  Carlo”  integration   Milestone  1  

6   Differential  equations   The  Pendulum  

7   Differential  equations   The  Taylor’s  method  

8   Differential  equations   A  Quantum  Particle  in  a  box  

9   Differential  equations   The  Tacoma  bridge   Milestone  2  

10   Linear  Algebra   Matrix  operations   Milestone  3  

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Schedule  

*  Lecture  1-­‐5  :  *  Statistics  and  integrals  

*  Lecture  6-­‐7  :    *  Differential  equations  with  

initial  conditions  

*  Lecture  8  :  *  The  particle  in  the  box  :  we  

will  combine  do  loops  functions,  and  arrays,  differential  equation  with  boundary  conditions  

 

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Where  are  we  going?  

0%  10%  20%  30%  40%  50%  60%  70%  80%  90%  100%  

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Milestone  2  :  Solving  differential  equations  

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Examination  

*  Examination:  

*  Sit-­‐in  :    -­‐December  12th  &  December  13th  (depending  on  your  group)        -­‐you  can  use  all  your  programs,  bring  all  your  notes,  books,  …        -­‐you  will  have  to  solve  problems  during  the  3h  

*  Report  :    -­‐Deadline  end  of  January        -­‐10  pages  max  (including  figures)        -­‐discuss  and  complete  the  problems  given  during  the            examination  (detailed  guidelines  will  be  posted  on  KEATS)  

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Today‘s  experiment  …!  

*  [  do  not  repeat  this  experiment  …  ]  

*  Erwin  Schrödinger,  1887-­‐1961,  born  in  Vienna  *  Developed  the  wave  equation  of  quantum  mechanics  *  Idea:  every  particle  (electron,  proton..)  can  be  described  by  a  function  

*  This  particle  has  a  probability  P(x)  to  be  observed  at  a  point  x  of  space,  this  probability  is  given  by  the  function:    

*  Condition:  the  probability  to  observer  the  particle  somewhere  (anywhere)  has  to  be  P=1  :  

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Schrodinger’s  equation  

ψ(x)

P (x) = |ψ(x)|2

�P (x)dx =

�|ψ(x)|2dx = 1

*  How  can  we  know  this  function  ?  To  make  the  connection  with  the  last  lecture,  we  simply  change  the  notation,  let’s  call  this  function  y(x)  

*  Answer:  the  function  is  the  solution  of  the  Schrodinger  equation  

*  m    :    mass  of  the  particle  *  E    :      energy  of  the  particle  *  x    :    the  coordinate,  the  position  in  space  of  the  particle    *  V(x)    :    a  potential  which  affects  the  particle  and  defines  the  problem  to  

   solve.  If  the  particle  is  a  rolling  ball,  the  potential  V(x)  would  be  the      potential  energy    

*                           :                                  Plank  constant  

Differential  equation  

ψ(x) → y(x)

d2y(x)

dx2=

2m(V (x)− E)

�2 y(x)

�

Let’s  be  more  specific:  *  Let’s  describe  the  problem  of  a  particle  resting  in  a  box  *  The  box  has  only  one  dimension  (to  simply  our  problem)  *  The  box  has  infinitely  hard  walls  on  each  side  *  The  particle  is  not  able  to  “enter”  into  the  wall,  so  the  probability  to  find  

the  particle  in  x=a  and  x=b  is  zero:                  y(a)=y(b)=0  

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Particle  in  a  box  

V (x) = 0Wall   Wall  

x

a = 0

b = 5V (x) = ∞ V (x) = ∞

x = a x = b

*  Shake  the  box  :  give  an  energy  E  to  the  particle  *  Where  is  the  particle?  Sitting  at  the  center  of  the  box?  *  To  answer  this  question,  we  need  to  find  y(x)  in  the  interval  a<x<b  *  In  this  interval  V(x)=0  

 *  With  boundary  conditions:  (the  ball  cannot  enter  the  walls)  

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Shake  the  box?  

y(a) = 0 y(b) = 0

d2y(x)

dx2= −2mE

� y(x)

*  Shake  the  box  :  give  an  energy  E  to  the  particle  *  Where  is  the  particle?  Sitting  at  the  center  of  the  box?  *  To  answer  this  question,  we  need  to  find  y(x)  in  the  interval  a<x<b  *  In  this  interval  V(x)=0  

 *  With  boundary  conditions:  (the  ball  cannot  enter  the  walls)  

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Shake  the  box?  

y(a) = 0 y(b) = 0

d2y(x)

dx2= −2mE

� y(x)

What  is  the  difference  with  the  pendulum?  

(last  lecture)  

*  For  simplicity,  we  set  the  mass  of  the  particle  such  that  :      *  This  is  the  equation  that  we  have  to  solve:  

 *  Reminder:  Pendulum  was  a  similar  problem,  but  with  initial  condition.    *  Here,  we  need  to  impose  y(a)  =y(b)=0.  We  have  a  free  parameter  E.  *  Idea:    can  I  start  from  y(a)=0,  and  find  E  such  that  y(b)=0  ?  

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Simplified  equation  

d2y(x)

dx2= −Ey(x)

2m

� = 1

y(a) = y(b) = 0

y(a) = 0 & E y(b) = 0

² Strategy:  aim  and  shoot  

² Solve  the  differential  equation  with  initial  condition  y(a)=0  

² …  and  with  an  energy  E,  see  what  you  obtain  for  y(b)    

² Obtain  y(b)  as  a  function  of  E,  when  y(b)=0  we  find  the  right  Energy  E.  

² Find  which  energies  E  satisfy  the  condition  y(b)=0  (finding  zeros  of  a  function,  check  where  the  plot  is  crossing  the  horizontal  axis)   14  

Solving  differential  equation  with  boundary  conditions  

y(x = a) = 0

y(x = b) = 0

² We  discussed  in  the  last  lecture  that  this  equation  can  be  decomposed  in  two  coupled  first  order  equations:  

² Simple  idea:  let’s  define  a  new  variable:      ² We  get  the  equations  :    

² We  are  trying  to  obtain  :  y(b=5)=0  ² Plot  what  you  obtain  for  y(b=5)  for  energies  E  ranging  from  0  to  20     15  

Reminder  :  coupled  first  order  equations  

Initial  conditions  

dy

dx= z

d2y(x)

dx2= −Ey(x) y(a) = 0 y�(a) = 1

z(0) = 1

y(0) = 0dz

dx= ....

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Energy  quantization  

Each  time  y(5)=0,  we  solved  our  problem!    Yay!    

d2y(x)

dx2= −Ey(x)

y(0) = y(5) = 0

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Energy  quantization  

Each  time  y(5)=0,  we  solved  our  problem!    Yay!    But  Wait….    The  energy  of  the  particle  only  can  take  SOME  particular  values  right?    Indeed,  this  is  what  we  call  the  quantification  of  the  energy  (hence  “quantum”  physics)    The  particle  can  only  take  some  quantized  energies  

d2y(x)

dx2= −Ey(x)

y(0) = y(5) = 0

*  For  each  obtained  energy  E,  we  can  also  obtain  the  solution  to  the  differential  equation  y(x),  0<x<5  

*  What  do  we  get?  *  For  each  allowed  energy:  

*  First  energy  (n=1)    *  Second  energy  (n=2)  *  Third  energy  (n=3)    

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Quantum  wave-­‐function  y(x)  

n=1   n=2  

n=3  

*  Probability  P(x)=y(x)2  

*  The particle with first energy has two maxima"

*  The particle with second energy has three maxima"

*  The particle with third energy has four maxima… "

*  When the particle has a high energy it is everywhere ! !

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Where  is  my  particle?  

n=1  

n=2  

n=3  

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1.  We  use  the  code  written  for  the  pendulum  (last  lecture)  

2.  The  function  F1  and  F2  need  to  be  updated  

3.  The  differential  equation  needs  to  be  solved  for  many  different  energies,  we  want  to  scan  energies  from  0  to  20  

4.  For  each  energy  E  we  obtain  y(x=5),  so  we  will  write  into  a  file  :                        E            ,            y(x=5)  

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Finding  the  energies  

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module library!"contains!" function F1( y , z , E)" implicit none" real(8) :: F1" real(8) :: y , z , E " F1= [ FILL IN ]" end function!" function F2(y , z , E)" implicit none" real(8) :: F2" real(8) :: y , z , E" F2= [ FILL IN ] " end function!"end module!

program quantumbox![FILL IN] !how can you use the module “library”?"implicit none""integer,parameter :: N=100000"integer :: j,i, iE"real(8) :: h,a,b ,energy"real(8) :: x(0:N),y(0:N),z(0:N)"" a = 0.0" b = 5.0" h = (b - a) / dble(N)" " do iE=1,1000"" energy = ... iE [FILL IN] "" x(0)=a; y(0)=0; z(0)=1." do j = 1, N" y( j ) = y( j-1 ) + h * F1( [Fill in] )" z( j ) = z( j-1 ) + h * F2( [Fill in] )" x( j ) = j*h" enddo"" write(100,*) energy, [FILL IN]"" end do""end program"

MODULE  

PROGRAM  Main  structure  of  the  code  

Solve  the  equ.  diff.  for  every  energy  

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program quantumbox![ FILL IN ]!implicit none"integer,parameter :: "N=100000"integer :: "j , i , iE "real(8) :: "h , a, b, energy"real(8) :: "x(0:N) , y(0:N) , z(0:N)"" a = 0.0" b = 5.0" h = (b - a) / dble(N)"" do iE=1,1000"" energy = ... iE [FILL IN]!" x(0)=a; y(0)=0; z(0)=1."" do j = 1, N"" y( j ) = y( j-1 ) + h * F1 ( [Fill in] )" z( j ) = z( j-1 ) + h * F2 ( [Fill in] )" x( j ) = j * h"" enddo"" write(100,*) energy, [FILL IN]!" end do""end program"

Main  structure  of  the  code  

Solve  the  differential  equation  for  every  energy  E  

Solve  the  differential  equation  one  energy  E  

The  solution  y(x)  is  stored  in  the  array  y(  j  )  ,  j=1,N  

*  We  modify  bits  in  the  previous  code,  now  additionally  we  want  to  detect  each  time  y(x)  satisfies  y(x=5)=0  

*  We  add  a  counter  countE,  which  is  incremented  each  time  we  obtain  :    y(x=5)=0  

   *  how  can  I  detect  if  y(x=5)  is  equal  to  zero  ?  

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Plotting  the  quantum  functions  

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program quantumbox!use library"implicit none" real(8) ":: yb( 0:N ) ……." yb=0.0! countE = 0!" do iE = 1 , 1000"" energy = ... iE [ FILL IN ] !" x(0)=a; y(0)=0; z(0)=1."" do j = 1, N" y(j) = y(j-1) + h * F1( [Fill in] )" z(j) = z(j-1) + h * F2( [Fill in] )" x(j) = j * h" enddo"" yb(i)=y(N)!! if( yb(i) * yb(i-1) [...FILL IN...] )then!! [Fill in]!! end if!"" end do""end program"

y(N)  is  the  last  point  obtained  for  the  solution  of  the  equation,  so  y(N)  is  y(x=5)  

countE  counts  how  many  times  we  obtain  y(x=5)=0  as  the  energy  is  increased.  It  will  hence  count  the  energy  levels  for  the  wave-­‐function  

Here  we  want  to  check  if  y(x=5)  is  equal  to  zero.  When  it  is,  we  want  to  write  into  a  new  file  the  array  “y”  which  contains  the  solution  y(x)  for  this  given  energy  E    

Problem  1  :      Particle  in  a  box  

Problem  2  :      Infinite  box    Problem  3  :    Particle  tunneling  through  a  barrier  (optional)  

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Problems  

Wall  

27  

Particle  in  a  box  /  Energy  barrier  

Wall  

V (x) = ∞ V (x) = ∞V (x) = 0

x = a x = bx

a = 0

b = 5V (x) = 0Wall  

V (x) = 5

*  Problem  1  :      Particle  in  the  box  

*  Problem  2  :      Quantum  free  particle  (plane  wave)  

*  Problem  3  :      Quantum  tunneling  (optional)  

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Friday,  problem  session