Transcript 幻灯片 1 - cqupt.edu.cn

1D systems
Solve the TISE for various 1D potentials
• Free particle
• Infinite square well
• Finite square well
• Particle flux
• Potential step
Transmission and reflection coefficients
• The barrier potential
Quantum tunnelling
Examples of tunnelling
• The harmonic oscillator
2015/7/17
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1
A Free Particle
Free particle: no forces so potential energy
independent of position (take as zero)
d 2

 E
2
2 m dx
Time-independent Schrödinger
equation:
2
Linear ODE with
constant coefficients
so try
  exp( x)
General solution:
Combine with time dependence to
get full wave function:
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2
Notes
• Plane wave is a solution (just as well, since our
plausibility argument for the Schrödinger equation
was based on this assumption).
 ( x)  ei  kxt 
• Note signs in exponentials:
– Sign of time term (-iωt) is fixed by sign adopted in
time-dependent Schrödinger Equation
– Sign of position term (±ikx) depends on
propagation direction of wave. +ikx propagates
towards +∞ while -ikx propagates towards –∞
• There is no restriction on k and hence on the allowed
energies. The states form a continuum.
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3
Particle in a constant potential
General solutions we will use over and over again
Time-independent Schrödinger
equation:
Case 1: E > V
(includes free particle with
V = 0 and K = k)
d 2

 V  E
2
2m dx
2
2m  E  V 
d 2
2
2
 K   0, K 
2
2
dx
Solution:
Case 2: E < V
(classically particle can not be here)
2m V  E 
d 2
2
2
 q   0, q 
2
2
dx
Solution:
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4
Infinite Square Well
Consider a particle confined to a finite
length –a<x<a by an infinitely high
potential barrier
V 
V 
V(x)
V 0
No solution in barrier region (particle
would have infinite potential energy).
In the well V = 0 so equation
is the same as before

x
d
 E
2 m dx 2
2
2
-a
a
General solution:
Boundary conditions:
Continuity of ψ at x = a:
Continuity of ψ at x = -a:
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Note discontinuity in
dψ/dx allowable, since
potential is infinite
5
Infinite Square Well (2)
Add and subtract
these conditions:
Even solution: ψ(x) = ψ(-x)
Odd solution: ψ(x) = -ψ(-x)
Energy
We have discrete states labelled by an integer quantum number
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6
Infinite Square Well (3)
Normalization
Normalize the solutions

Calculate the normalization integral N 

( x, t ) dx
2

Normalized solutions are
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Infinite Square Well (4)
Sketch solutions
Wavefunctions  ( x)
Probability density
 ( x)
2
3
1
Note: discontinuity of gradient of ψ at edge of well.
OK because potential is infinite there.
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Infinite Square Well (5)
Relation to classical probability distribution
Classically particle is equally likely to be anywhere in the box
Quantum probability distribution is
2
1
 n 
 n  x   cos 2 
x  , n  1,3,5 
a
 2a 
2
1 2  n 
 n  x   sin 
x  , n  2, 4, 6 
a
But
1
Pcl ( x) 
2a
 2a 
cos2   sin 2   1/ 2
so the high energy quantum states are consistent with the classical result when
we can’t resolve the rapid oscillations.
This is an example of the CORRESPONDENCE PRINCIPLE.
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Infinite Square Well (5) – notes
•
Energy can only have discrete values: there is no continuum of states
anymore. The energy is said to be quantized. This is characteristic of boundstate problems in quantum mechanics, where a particle is localized in a finite
region of space.
•
The discrete energy states are associated with an integer quantum number.
•
Energy of the lowest state (ground state) comes close to bounds set by the
Uncertainty Principle:
•
The stationary state wavefunctions are even or odd under reflection. This is
generally true for potentials that are even under reflection. Even solutions are
said to have even parity, and odd solutions have odd parity.
•
Recover classical probability distribution at high energy by spatial averaging.
•
Warning! Different books differ on definition of well. E.g.
– B&M: well extends from x = -a/2 to x = +a/2.
Our results can be adapted to this case easily (replace a with a/2).
– May also have asymmetric well from x = 0 to x = a.
Again can adapt our results here using appropriate transformations.
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10
Finite Square Well
Now make the potential well
more realistic by making the
barriers a finite height V0
V(x)
I
II
III
V0
Assume 0  E  V0
x
i.e. particle is bound
Region I:
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-a
Region II:
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a
Region III:
11
Finite Square Well (2)
Boundary conditions: match value and derivative
of wavefunction at region boundaries:
x  a
x  a
Match ψ:
Match dψ/dx:
Now have five unknowns (including energy) and five equations
(including normalization condition)
Solve:
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Finite Square Well (3)
Even solutions when
Odd solutions when
q  k tan  ka 
q  k cot  ka 
We have changed the notation
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Cannot be solved algebraically.
Solve graphically or on computer
 into q
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13
Finite Square Well (4)
Graphical solution
k0 = 4
a=1
Even solutions at intersections of blue and red curves (always at least one)
Odd solutions at intersections of blue and green curves
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Finite Square Well (5)
Sketch solutions
Wavefunctions  ( x)
Probability density  ( x )
2
Note: exponential decay of solutions outside well
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Finite Square Well (6): Notes
•Tunnelling of particle into “forbidden” region where V0 > E
(particle cannot exist here classically).
•Amount of tunnelling depends exponentially on V0 – E.
•Number of bound states depends on depth of well,
but there is always at least one (even) state
•Potential is even, so wavefunctions must be even or odd
•Limit as V0→∞:
n
k0    Solutions at ka 
2
We recover the infinite well solutions as we should.
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Example: the quantum well
Quantum well is a “sandwich” made of two different semiconductors in which the
energy of the electrons is different, and whose atomic spacings are so similar that
they can be grown together without an appreciable density of defects:
Material A
(e.g. AlGaAs)
Material B (e.g. GaAs)
Electron energy
Position
Now used in many electronic devices (some transistors, diodes, solid-state lasers)
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Summary of Infinite and Finite Wells
Infinite well
Even parity
Odd parity
Finite well
Infinitely many solutions
1
 n 
 n  x 
cos 
x  , n  1,3,5 
2
a
a


1
 n 
 n  x 
sin 
x  , n  2, 4, 6 
a
 2a 
Finite number of solutions
At least one solution (even parity)
Evanescent wave outside well.
Even parity solutions
Odd parity solutions
q  k tan  ka 
q   k cot  ka 
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n2 2 2
En 
8ma 2
Chang-Kui Duan, Institute of Modern Physics, CUPT
q2 
2m
2
k2 
V0  E 
2mE
2
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Particle Flux
In order to analyse problems involving
scattering of free particles, need to
understand normalization of free-particle
plane-wave solutions.


A exp[i(kx  t )] dx 
2



A dx  
2

Conclude that if we try to
normalize so that


 dx  1
2
we get A = 0.

This problem is related to Uncertainty Principle:
Momentum is completely defined
Solutions:
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Position completely undefined;
single particle can be anywhere
from -∞ to ∞, so probability of
finding it in any finite region is
zero
Normalize in a finite box
Use wavepackets (later)
Use a flux interpretation
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19
Particle Flux (2)
More generally: what is the rate of change of
probability that a particle is in some region
(say, between x=a and x=b)?
b
b
*


d Prab d


  *dx   *

dx

dt
dt a
t
t 
a
Use time-dependent Schrödinger equation:
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a
b
x
2

2
i

 V ( x, t )
2
t
2m x
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Particle Flux (3)
d Prab

dt
Interpretation:
Particle flux at position x
i  * 
* 
j ( x, t ) 



2m 
x
x 
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Flux entering
at x=a
minus
a
Flux leaving
at x=b
b
x
Note: a wavefunction that is real
carries no current
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21
Particle Flux (4)
i  * 
* 
j ( x, t ) 



2m 
x
x 
Check: apply to free-particle plane wave.
Makes sense:
# particles passing x per unit time = # particles per unit length ×
velocity
So plane wave wavefunction describes a “beam” of particles.
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22
Particle Flux (5): Notes
• Particle flux is nonlinear
i  * 
* 
j ( x, t ) 



2m 
x
x 
• Time-independent case: replace   x, t  with   x 
• 3D case,


x
• Can use this argument to prove CONSERVATION OF PROBABILITY.
Put a = -∞, b = ∞, then
Prab  N and
d Prab
dN
 j (a, t )  j (b, t ) 
 j (, t )  j (, t )
dt
dt
If      0, then
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dN
0
dt
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23
Potential Step
V(x)
Consider a potential which rises
suddenly at x = 0:
Case 1
V0
0
Case 1: E > V0 (above step)
x < 0, V = 0
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x
Boundary condition: particles only
incident from left
x > 0, V = V0
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24
Potential Step (2)
Continuity of ψ at x = 0:
Continuity of
d
at x  0 :
dx
Solve for reflection and
transmission amplitudes:
r
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kK
2k
, t
kK
kK
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Potential Step (3)
Transmission and Reflection Fluxes
Calculate transmitted and reflected fluxes
x<0
i  * 
 * 
j ( x) 



2m  x
x 
x>0
1 ( x)  eikx  r eikx
 2 ( x)  teiKx
(cf classical case: no reflected flux)
Check: conservation of particles
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

k
q 2
2
1 r 
t
m
m
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26
Potential Step (4)
Case 2: E < V0 (below step)
Solution for x < 0 same as before
V(x)
V0
0
Solution for x > 0 is now
evanescent wave
Matching boundary conditions:
Transmission and reflection amplitudes:
Transmission and reflection fluxes:
This time we have total reflected flux.
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27
Potential Step (5): Notes
• Some tunnelling of particles into classically forbidden region even for
energies below step height (case 2, E < V0).
• Tunnelling depth depends on energy difference V0  E ,
1

q
2m V0  E 
• But no transmitted particle flux, 100% reflection, like classical case.
• Relection probability is not zero for E > V0 (case 1). Only tends to zero
in high energy limit, E >> V (correspondence principle again).
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28
Rectangular Potential Barrier
V(x)
Now consider a potential barrier
of finite thickness:
I
II
III
V0
x
0
Assume 0  E  V0
Region I:
b
Boundary condition: particles only
incident from left
Region II:
u = exp(ikx) + B exp(−ikx) u = C exp(Kx) + D exp(−Kx)
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Region III:
u = F exp(ikx)
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Rectangular Barrier (2)
Match value and derivative of
wavefunction at boundaries:
x  b
x0
Match ψ:
1+B=C+D
C exp(Kb) + D exp(−Kb) = F exp(ikb)
Match dψ/dx:
1 − B = K/(ik)(C − D)
C exp(Kb) − D exp(−Kb) = ik/K F exp(ikb)
Eliminate wavefunction
in central region:
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Rectangular Barrier (3)
Transmission and reflection amplitudes:
F
For very thick or high barrier:
|F|2 =
Non-zero transmission (“tunnelling”) through classically forbidden barrier region.
Exponentially sensitive to height and width of barrier.
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31
Examples of Tunnelling
Tunnelling occurs in many situations in physics and astronomy:
1. Nuclear fusion (in stars and fusion reactors)
V
( Ze) 2
Barrier height ~
~ MeV
4 0 rnucleus
Repulsive
Coulomb
interaction
thermal energies (~keV) at T
Incident
particles
107 K
Assume a Boltzmann distribution for the KE,
P  E  e E / kT
Nuclear separation x
Strong
nuclear force
(attractive)
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Probability of nuclei having MeV energy is e
1000
Fusion (and life) occurs because
nuclei tunnel through the barrier
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32
Examples of Tunnelling
2. Alpha-decay
V
Distance of α-particle
from nucleus
Initial α-particle
energy
α-particle must overcome Coulomb repulsion barrier.
Tunnelling rate depends sensitively on barrier width and height.
t
2
e 2qa
Explains enormous range of α-decay rates, e.g. 232Th, t1/2 = 1010 yrs, 218Th, t1/2 = 10-7s.
Difference of 24 orders of magnitude comes from factor of 2 change in α-particle energy!
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33
Examples of Tunnelling
3. Scanning tunnelling microscope
V
x
A conducting probe with a very sharp tip is
brought close to a metal. Electrons tunnel through
the empty space to the tip. Tunnelling current is so
sensitive to the metal/probe distance (barrier
width) that even individual atoms can be mapped.
Tunnelling current proportional to
q 
2
2m  E  V0 
2
so
and
q 1 A 
t
2
a
Vacuum
Probe
e 2qa
E  V0  4eV
1
If a changes by 0.01A (~1/100th of the atomic size)
then current changes by a factor of 0.98,
i.e. a 2% change, which is detectable
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Material
STM image of Iodine atoms on platinum.
The yellow pocket is a missing Iodine atom
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34
Summary of Flux and Tunnelling
The particle flux density is
i  * 
* 
j ( x, t ) 



2m 
x
x 
Particles can tunnel through classically forbidden regions.
Transmitted flux decreases exponentially with barrier height and width
t
2
e 2qa
q 
2
2m  E  V0 
2
We get transmission and reflection at potential steps.
There is reflection even when E > V0.
Only recover classical limit for E >> V0 (correspondence principle)
kK
2k
, t
kK
kK
k  iq
2k
, t
 E  V0   r 
k  iq
k  iq
 E  V0   r 
2
2
2 2
k2
K2
q
E
, E  V0 
( E  V0 ), V0  E 
( E  V0 )
2m
2m
2m
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35
Simple Harmonic Oscillator
Example: particle on a
spring, Hooke’s law
restoring force with
spring constant k:
Mass m
Force F  kx
k
m
Potential energy V ( x)  1 kx 2  1 m0 2 x 2
2
2
Angular frequency 0 =
x
V(x)
Time-independent
Schrödinger equation:
Problem: still a linear differential
equation but coefficients are not constant.
Simplify: change to
dimensionless variable
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 m0 
y



1/ 2
x,  =2E/( 0 )
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x
36
Simple Harmonic Oscillator (2)
Asymptotic solution in the
limit of very large y:
 ( y)  exp( y2 / 2)
Try it:
Suggests we substitute  ( y)  H ( y)exp( y 2 / 2)
Equation for H(y):
d 2H
dH

2
y
   1 H  0
2
dy
dy
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37
Simple Harmonic Oscillator (3)
Solve this ODE by the power-series
method (Frobenius method):

H ( y)   a p y
p 0
p
d 2H
dH

2
y
    1 H  0
2
dy
dy
Find that series for H(y) must terminate for a normalizable solution
Can make this happen after n terms for either even or
odd terms in series (but not both) by choosing
Hence solutions are either even or odd functions (expected on parity considerations)
0
Label normalizable functions H by the
values of n (the quantum number)
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Hn is known as the nth
Hermite polynomial.
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38
Simple Harmonic Oscillator (4)
EXAMPLES OF HERMITE POLYNOMIALS
AND SHO WAVEFUNCTIONS
H0  y   1
  0  y   N 0e
H1  y   2 y
  1  y   N1 2 ye  y
 y2 / 2
2
/2
  2  y   N2  4 y  2 e
H2  y  4 y  2
2
2
 y2 / 2
H 3  y   8 y 3  12 y   3  y   N 3  8 y 3  12 y  e  y
N0 , N1 , N2
Hn  y 
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Nn
2
/2
are normalization constants
is a polynomial of degree
yn
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39
Simple Harmonic Oscillator (5)
Wavefunctions
High n state (n=30)
wavefunction
Probability density
Pcl ( x) 
1
 a2  x2
• Decaying wavefunction tunnels into classically forbidden region
• Spatial average for high energy wavefunction gives classical result:  E  1 m 2 a 2 
0


another example of the CORRESPONDENCE PRINCIPLE
2


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40
Summary of Harmonic Oscillator
1) The quantum SHO has discrete energy levels because of the normalization
requirement
1

En   n   0 , n  0,1, 2,3
2


2) There is ‘zero-point’ energy because of the uncertainty principle. E0 
3) Eigenstates are Hermite polynomials times a Gaussian
0
2
 ( y)  H ( y)exp( y 2 / 2)
4) Eigenstates have definite parity because V(x) = V(-x). They can tunnel into the
classically forbidden region.
5) For large n (high energy) the quantum probability distribution tends to the classical
result. Example of the correspondence principle.
6) Applies to any SHO, eg: molecular vibrations, vibrations in a solid (phonons),
electromagnetic field modes (photons), etc
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41
Example of SHOs in Atomic Physics:
Bose-Einstein Condensation
87Rb
atoms are cooled to nanokelvin temperatures in a harmonic trap. de Broglie
waves of atoms overlap and form a giant matter wave known as a BEC. All the atoms go
into the ground state of the trap and there is only zero point energy (at T=0). This is a
superfluid gas with macroscopic coherence and interference properties.
Signature of BEC phase transition:
The velocity distribution goes from classical Maxwell-Boltzmann form to the distribution
of the quantum mechanical SHO ground state.
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42
Example of SHOs: Molecular vibrations
VIBRATIONAL SPECTRA OF MOLECULES
Useful in chemical analysis and in astronomy
(studies of atmospheres of cool stars and interstellar clouds).
V(x)
H2 molecule
H
H
x
Nuclear
separation x
SHO levels
SHO very useful because any potential is approximately parabolic near a minimum
2015/7/17
Chang-Kui Duan, Institute of Modern Physics, CUPT
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