# EM Wave Equation by pptfiles

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• pg 1
```									Topic 5: Schrödinger Equation
• Wave equation for Photon vs. Schrödinger equation for Electron+ • Solution to Schrödinger Equation gives wave function 
– 2 gives probability of finding particle in a certain region

• Square Well Potentials: Infinite and Finite walls
–  oscillates inside well and is zero or decaying outside well, E  n2

• Simple Harmonic Oscillator Potential (or parabolic)
–  is more complex, E  n
Example Infinite Well Solution

Schrödinger Equation

Page 1

Schrödinger Equation
• Step Potential of Height V0
–  is always affected by a step, even if E > V0 – For E > V0,  oscillates with different k values outside/inside step. – For E < V0,  oscillates outside step and decays inside step.

• Barrier Potential of Height V0
–  oscillates outside and decays inside barrier.

• Expectation Values and Operators
• Appendix: Complex Number Tutorial

Schrödinger Equation

Page 2

Wave Equation for Photons: Electric Field E

2 E 1 2 E  2 2 2 x c t
2nd space derivative 2nd time derivative

Propose Solution:

E  x, t   E0 cos  kx  t 
2 E   k 2 E  x, t  x 2

Calculate Derivatives:
2 E   2 E0 cos  kx   t    2 E  x, t  t 2

After Substitution:
k 2  

2
c2

  kc  E  pc

where E   and p  k
Page 3

Schrödinger Equation

Schrödinger Eqn. for Electrons+: Wave Function 
2   x, t   2    x, t    V  x, t    x, t   i  2 2m x t

2nd space derivative

1st time derivative

Propose Simple Solution for constant V:

  x, t   Ae 

i kx t 

 A cos  kx   t   i sin  kx  t   

Calculate Derivatives:

 i kx  t   i Ae   i t
After Substitution: 2   k 2   V0   i  i   2m 2 k 2  V0   2m

2  2 i kx  t    ik  Ae   k 2  x 2







Ek  Vo  ETOT
where E   and p  k
Page 4

Schrödinger Equation

Schrödinger Equation: Applications
• Now, find the eigenfunctions  and eigenvalues E of the Schrödinger Equation for a particle interacting with different potential energy shapes. (assume no time dependence)

   ( x)  2m x 2
2 2

 V ( x) ( x)  E  ( x)

• Possible potential energies V(x) include: • Infinite and Finite square wells (bound particle). • Simple Harmonic or parabolic well (bound particle). • Step edge (free particle). • Barrier (free particle).
Phys320 - Baski Schrödinger Equation Page 5

Schrödinger Equation: Definitions
• Wave function  has NO PHYSICAL MEANING! • BUT, the probability to find a particle in width dx is given by:

P( x, t )dx  



 x, t   x, t  dx    x, t 

2

dx

• Normalization of  – Probability to find particle in all space must equal 1. – Solve for  coefficients so that normalization occurs.




 



 x   x dx  1
Schrödinger Equation Page 6

Infinite Square Well Potential: Visual Solutions
Wave and Probability Solutions
n=3

Energy Solutions

n(x)

n2(x)

n=2

n=1

2  nx   n x   sin  L  L 

  2 2  2 k 2 En   n2   2m 2mL2  
Page 7

Infinite Square Well: Solve general  from S.Eqn.
 2 d 2 x    E  x  2 2m dx
where

Inside Well: (V = 0)

 " x   k 2 x 

2mE  p  k  2    
2

2

Oscillatory

 x   A1 sin kx  A2 coskx
 x   0
 cannot penetrate barriers!

Outside Well: (V = )

Schrödinger Equation

Page 8

Infinite Square Well: Satisfy B.C. and Normalization  x   A1 sin kx  A2 coskx • Satisfy boundary conditions  0  0  L   0 n A2 cos0  0  A2  0 A1 sin kL  0  kL  n  kn 
L
2 2  2k 2 2     En  n  2  2mL 2m  
 

Quantized Energy Solutions
L

• Satisfy normalization
An  2 L


 nx    dx   An sin  L dx  1    0
2 2

using identity

 sin
0

2

md 


2

and



x
L

 n x  

2  nx  sin  L  L 

Wave Solutions
Page 9

Schrödinger Equation

Infinite Square Well : Probability Example
An electron is in the n = 3 excited state of a 1-D infinite square well (width L). Draw the wave function and probability distribution of the electron. Solve the probability of finding the electron at x = 0.5L in a width Dx = 0.02L.
(x)

P(x)

L/2 L

x
L/2 L

x

Because Dx is so small, use P(x)Dx =  2 (x)Dx with  for infinite well potential. Pn 3 ( x)Dx   2 3 ( x)Dx  n Pn 3 (x  L )Dx  2

2 2  3 x  sin   Dx L  L     0.04 
Page 10

2 2  3 L   3 sin  (0.02 L)   0.04  sin 2   L  2L   2
Schrödinger Equation

Finite Square Well Potential: Visual Solutions
Wave and Probability Solutions
n=3 “leaks” n(x) outside barrier

Energy Solutions
E

n

2(x)

High energy particles “escape”
Vo

E3

n=2

E2
E1

n=1

En 

n2  2 2  2 2m  L   k2  
2

Energy vs. width: Energy vs. height: Phys320 - Baski

http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_07a.html http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_07b.html Schrödinger Equation

Page 11

Finite Square Well: Solve general  from S.Eqn.
Inside Well: (V = 0)

 2 d 2 x    E  x  2 2m dx

"

 x   k1   x 
2

where

k12 

2mE 2

Oscillatory

  x   A1 sin  k1 x   A2 cos  k1 x 

Outside Well:  (V = Vo)
Decaying

"

 x   k2   x 
2

where

k2 
2

2m 2

Vo  E   0

  x   B1ek2 x  B2 ek2 x
x  a  x  a 
Schrödinger Equation

 can penetrate barriers!

Page 12

Finite Square Well: Example Problem
(a) Sketch the wave function (x) for the n = 4 state for the finite square well potential. (b) Sketch the probability distribution 2(x).

(x)

n=4
L/2 L

 (x)
2

n=4

x
L/2 L

x

Schrödinger Equation

Page 13

Finite Square Well: Example Problem
Sketch the wave function (x) corresponding to a particle with energy E in the potential well shown below. Explain how and why the wavelengths and amplitudes of (x) are different in regions 1 and 2.
(x)
Region 1 Region 2

V1 E V2
x1 x2

x

• (x) oscillates inside the potential well because E > V(x), and decays exponentially outside the well because E < V(x). • The frequency of (x) is higher in Region 1 vs. Region 2 because the kinetic energy is higher [Ek = E - V(x)]. • The amplitude of (x) is lower in Region 1 because its higher Ek gives a higher velocity, and the particle Schrödinger Equation less time in that region. 14 therefore spends Phys320 - Baski Page

Simple Harmonic Well Potential: Visual Solutions
Wave and Probability Solutions n=2 n(x) n2(x)
(different well widths)

Energy Solutions

n=1

n=0

1  En   n    2 
Schrödinger Equation Page 15

Simple Harmonic Well: Solve  from S.Eqn.
NEW!
Inside Well:  " x   k 2 x  where k 2  2m  E  V  x   2 

 (x) is not a simple trigonometric function.

Outside Well:  "  x    2  x  where

2m   2 [V ( x)  E ]  0 
2

(x) is not a simple decaying exponential.

 ( x)  (Hermite Polynomial) (Gaussian Function)
  Kx  o  Ao exp   2
2

   Kx 2 K   ,  1  A1    x exp       2
Schrödinger Equation

1 2

  , etc. 
Page 16

Step Potential: (x) outside step
2 2  d   x S. Eqn:   V ( x)  x   E  x  2 2m dx

Outside Step: V(x) = 0

 x 
"

2  k1 

x 

where

2 k1

(x) is oscillatory

 2m   2 E  

Case 1
Energy

Case 2

(x)

Schrödinger Equation

Page 17

Step Potential: (x) inside step
Inside Step: V(x) = Vo



"

 x   k2   x 
2

where

(x) is oscillatory for E > Vo (x) is decaying for E < Vo

 2m  k22   2   E  Vo   

Case 1
Energy

Case 2

E > Vo
(x)

E < Vo

Scattering at Step Up: http://www.kfunigraz.ac.at/imawww/vqm/pages/samples/107_06b.html Scattering at Well - wide: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_05d.html Scattering at Well Phys320 - Baski - various: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_05b.html 18 Page Schrödinger Equation

Step Potential: Reflection and Transmission
• At a step, a particle wave undergoes reflection and transmission (like electromagnetic radiation!) with probability rates R and T, respectively.

R(reflection) + T(transmission) = 1

 k1  k2  R    k1  k2 

2

T

 k1  k2 

4k1k2

2

• Reflection occurs at a barrier (R  0), regardless if it is step-down or step-up. – R depends on the wave vector difference (k1 - k2) (or energy difference), but not on which is larger. – Classically, R = 0 for energy E larger than potential barrier (Vo).
Phys320 - Baski Schrödinger Equation Page 19

Step Potential: Example Problem
A free particle of mass m, wave number k1 , and energy E = 2Vo is traveling to the right. At x = 0, the potential jumps from zero to –Vo and remains at this value for positive x. Find the wavenumber k2 in the region x > 0 in terms of k1 and Vo. In addition, find the reflection and transmission coefficients R and T. 2m  2Vo  4mVo 2mE k1    and   
k2  2m V  E 
2



2m Vo  2Vo 
2



2m  3Vo  



6mVo 

or

3 k1 2

 k1  k2   k1  R     k1  k2   k1  

3 2 k1

2  0.225       0.0102 3   2.225  2 k1 

(1% reflected)

T  1  R  1  0.0102  0.99 (99% transmitted)
Schrödinger Equation Page 20

Barrier Potential
Outside Barrier:   x   k1   x  V(x) = 0 (x) is oscillatory
"
2

where

 2m  k12   E  2 

Inside Barrier:   x   k2   x  where V(x) = Vo (x) is decaying
"
2

k2 
2

2m 
2

Vo  E   0

Energy

Transmission is Non-Zero!
(x)

Te

2 k2 a

http://www.sgi.com/fun/java/john/wave-sim.html Single Barrier: Phys320 - Baski http://www.kfunigraz.ac.at/imawww/vqm/pages/samples/107_12c.html Schrödinger Equation Page 21

Barrier Potential: Example Problem
Sketch the wave function (x) corresponding to a particle with energy E in the potential shown below. Explain how and why the wavelengths and amplitudes of (x) are different in regions 1 and 3.
 (x)
Region 2

E Vo
Region 1 Region 3

• (x) oscillates in regions 1 and 3 because E > V(x), and decays exponentially in region 2 because E < V(x). • Frequency of (x) is higher in Region 1 vs. 3 because kinetic energy is higher there. • Amplitude of (x) in Regions 1 and 3 depends on the initial location of the wave packet. If we assume a bound particle in Region 1, then the amplitude is higher there and decays into Region 3 (case shown above).
Non-resonant Barrier: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_resonance-5.html Resonant Barrier: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_resonance-6.html Double Barrier Phys320 - Baski+ : http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_resonance-0.html Page 22 Schrödinger Equation

x

Scanning Tunneling Microscopy: Schematic
Tip Bias voltage

VDC
eDistance s

Constant current contour e e

Sample

Tunneling current  e -2ks
• STM is based upon quantum mechanical tunneling of electrons across the vacuum barrier between a conducting tip and sample. • To form image, tip is raster-scanned across surface and tunneling current is measured.
Phys320 - Baski Schrödinger Equation Page 23

STM: Ultra-High Vacuum Instrument
Coarse Motion Sample Scanner Tip

• Well-ordered, clean surfaces for STM studies are prepared in UHV.

• Sample is moved towards tip using coarse mechanism, and the tip is moved using a 3-axis piezoelectric scanner.
Phys320 - Baski Schrödinger Equation Page 24

STM: Data of Si(111)7×7 Surface
STM
empty

7×7 Unit

18 nm

7 nm

• STM topograph shows rearrangement of atoms on a Si(111) surface.
• Adatoms appear as bright “dots” when electrons travel from sample to tip.
Phys320 - Baski Schrödinger Equation Page 25

Expectation Values and Operators
• By definition, the “expectation value” of a function is:

 f(x)   x  f x  x dx
 



• “Operate” on (x) to find expectation value (or average expected value) of an “observable.”
Observable Symbol Operator
x

Position

x p

Momentum

 i

 x

Kinetic energy K Hamiltonian

 2  2 2m x 2
 2  2  V  x 2 2m x

H

 i Total Energy E t Schrödinger Equation

Page 26

Expectation Values: Example Problem
• Find <p>, <p2> for ground state 1(x) of infinite well (n = 1)
0 L

   <p> =   x   i  1  x  x   0
L  1

 2 x     2 x    sin   i   sin  dx L L  x   L L 0
L

-L/2

+L/2

x   2  x     i   sin  cos  dx L  L L L 0
L

<p> = 0 by symmetry (odd function over symmetric limits)
Note: The average momentum goes to zero because the “sum” of positive and negative momentum values cancel each other out.
Phys320 - Baski Schrödinger Equation Page 27

Expectation Values: Example Problem, cont.
 x 2 2 x sin L L
2

<p2> =

 1  x    2 
0
L

L

 

  1  x  dx where  1  x   
 
2

   2      1  x   1  x     dx where 2  1  x       1  x    L x   L 0

     1  x  1  x dx  L    0
2 L

= 1 by normalization

<p2> =

      L 

2

Schrödinger Equation

Page 28

Complex Number Tutorial: Definitions
• Imaginary number i given by: i2 = –1 ( i3 = –i, i4 = 1, i–1 = –i ) • Complex number z is composed of a real and imaginary parts.
2x + iy 5(cos30º + isin30º) 5 ei/6 (2, i) 30º or /6
–1 –i 1 2

Cartesian Form: Polar Form: where r = (x2 +

z = x + iy z = r(cos + i sin y2)1/2 and tan = y/x

2i

i

Exponential Form:

z = rei 

Conjugate: z* = x – iy = rcos – i rsin  re– i  where (z*)(z) = (x – iy)(x + iy) = x2 + y2 (real!)
Phys320 - Baski Schrödinger Equation Page 29

Complex Number Tutorial: Taylor Series
• Proof of equivalence for polar and exponential forms:
x 2 x3 x 4 exp( x)  1  x     etc. 2! 3! 4! x2 x4 x3 x5 cos( x)  1    etc. and sin( x)  x    etc. 2! 4! 3! 5!     x2 x4 x3 x5 cos x  i sin x  1    etc.   i  x    etc.  2! 4! 3! 5!     x 2 ix3 x 4 ix5  1  ix      etc. 2! 3! 4! 5! x 2 ix3 x 4 exp(ix)  1  ix     etc. 2 ! 3! 4!  exp(ix)  cos x  i sin x
Phys320 - Baski Schrödinger Equation Page 30

where i 2  1

Schrödinger Eqn.: Derivation of Space & Time Dependence
2   x, t      x, t    V ( x, t )   x, t   i  2 2m t x 2

Schrödinger Equation is 2nd Order Partial Differential Equation
Assume  is separable [i.e. V(x) only]

  x, t     x   t 
  x    t  2    x   t    V ( x)  x    t   i  2 2m t x
2

Substitution of 

d 2  x  d  t  2   t   V ( x)  x    t   i   x  2 2m dt dx
2 2 1 d   x  1 d  t    V ( x)  i  2 2m   x  dx   t  dt

Partial derivatives are now ordinary derivatives Divide by (x)0(t)

Space dependence ONLY

Time dependence ONLY
Schrödinger Equation Page 31

Schrödinger Eqn.: Derivation of Space & Time Dependence
2 2 1 d   x  1 d  t    V ( x)  i  2 2m   x  dx   t  dt
2 2 1 d   x    V ( x)  C 2 2m   x  dx

Left and right sides have only space (x) and time (t) dependence now

Space:

Time:

1 d  t  i C   t  dt
2 2

Set each side of equation equal to a constant C

 d   x Space Equation   V ( x)  x   E  x  2  need V(x) to solve! 2m dx
Time Solution: