Free Particle
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Transcript of Free Particle
Free Particle
(x) = A cos(kx) or (x) = A sin(kx) (x)= A eikx = A cos(kx) + i A sin(kx)(x)= B e-ikx = B cos(kx) - i B sin(kx)
d
dxk
2
22
d
dxik Ae kikx
2
22 2
( )
d
dxik Be kikx
2
22 2
( )
Free Particle
(x)= A eikx +B e-ikx is a solution
• A and B are constants
• hence (x,t)= (x)e-it
• = A ei(kx- t) +B e-i(kx+ t)
Travelling wave to right Travelling wave to left
Free Particle(x,t)= A ei(kx- t) is matter wave travelling to the
right(along the positive x-axis) *(x,t)= A* e-i(kx- t)
• | (x,t)|2 = (x,t) *(x,t)= AA* =|A|2
• intensity of wave is constant!
• Probability is the same everywhere
• a free particle is equally likely to be found anywhere
• P(x,t)= |(x,t)|2 is probability of finding a particle at position x at time t
• total probability of finding it somewhere is
Free Particle
2( , ) ( , ) 1P x t dx x t dx
• consider a classical point particle moving back and forth
with constant speed between two walls located at x=0 and x=8cm
• particle spends same amount of time everywhere
• P(x)=P0 if 0< x < 8 cm
• P(x)=0 if x< 0 or x> 8cm
Free Particle• Since
• hence P0 = (1/8) cm-1 ===> probability/unit length is 1/8
• probability of finding particle in length dx is (1/8)dx
• probability of finding it at x=2cm is zero! (dx=0)• Probability of finding it in some range 1.9 to 2.1 is
(1/8)x = (1/8)(2.1-1.9)= .025
2( , ) ( , ) 1P x t dx x t dx
8
0 0
0
( ) 8 1cm
P x dx P dx P cm
Free Particle• Probability of finding it between x=0 and x=8cm is
(1/8)(8-0) = 1
• intensity of wave is constant!
• Probability is the same everywhere
• a free particle is equally likely to be found anywhere
• free particle has definite energy E=(1/2)mv2 and momentum p=mv but uncertain position
R+T=1
Barrier Tunneling• consider a barrier E < U0
U0
Schrodinger Solution• Consider the three regions : left of barrier, right of barrier and in
the barrier
• left:
• right:
• inside:
U0
1 2( ) 0ikx ikxx e e x
5( ) ikxx e x L
22
1 2( ) ( ) ikx ikxP x x e e
2
5( )P x
2 2
02 2
8( ) 0
d mE U
dx h
Barrier Tunneling• Solution inside barrier has form
• since
• P(x) is smaller as U0 increases
3 4( ) 0k x k xx e e x L
02 2 ( )m U Ek
h
2 23( ) k xP x e
3 4
Tunneling
Tunneling
• Transmission coefficient T ~ e-2kL
• k={82m(U0-E)/h2}1/2 Note: E < U0
• if T=.02 then for every 1000 electrons hitting the barrier, about 20 will tunnel
• extremely sensitive to L and k
• width and height of barrier