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CHAPTER 6
Quantum Mechanics II
6.1
6.2
6.3
6.4
6.5
The Schrödinger Wave Equation
Expectation Values
Infinite Square-Well Potential
Finite Square-Well Potential
Three-Dimensional InfinitePotential Well
6.6 Simple Harmonic Oscillator
6.7 Barriers and Tunneling
Erwin Schrödinger (1887-1961)
A careful analysis of the process of observation in atomic physics has
shown that the subatomic particles have no meaning as isolated
entities, but can only be understood as interconnections between the
preparation of an experiment and the subsequent measurement.
- Erwin Schrödinger
Prof. Rick Trebino, Georgia Tech, www.frog.gatech.edu
Opinions on quantum mechanics
I think it is safe to say that no
one understands quantum
mechanics. Do not keep saying
to yourself, if you can possibly
avoid it, “But how can it be like
that?” because you will get
“down the drain” into a blind
alley from which nobody has yet
escaped. Nobody knows how it
can be like that.
- Richard Feynman
Those who are not shocked
when they first come across
quantum mechanics cannot
possibly have understood it.
Richard Feynman (1918-1988)
- Niels Bohr
A superb book
on the historical
debates, issues,
and
personalities of
the scientists
who invented
modern physics
Properties of Valid Wave Functions
Conditions on the wave function:
1. In order to avoid infinite probabilities, the wave function must be
finite everywhere.
2. The wave function must be single valued.
3. The wave function must be twice differentiable. This means that it
and its derivative must be continuous. (An exception to this rule
occurs when V is infinite.)
4. In order to normalize a wave function, it must approach zero as x
approaches infinity.
Solutions that do not satisfy these properties do not generally
correspond to physically realizable circumstances.
Normalization and Probability
The probability P(x) dx of a particle being between x and x + dx is
given in the equation
P( x)dx   ( x, t )( x, t )dx
The probability of the particle being between x1 and x2 is given by
x2
P    dx
x1
The wave function must also be normalized so that the probability
of the particle being somewhere on the x axis is 1.



 ( x, t )( x, t )dx  1
6.2: Expectation Values
In quantum mechanics, we’ll compute expectation values.
The expectation value, x , is the weighted average of a
given quantity. In general, the expected value of x is:
x  P1 x1  P2 x2 
 PN xN 
P x
i
i
i
If there are an infinite number of possibilities, and x is continuous:

x  P( x) x dx
Quantum-mechanically:
x 



 ( x) x dx   ( x)  * ( x) x dx   * ( x) x  ( x) dx
2
And the expectation of some function of x, g(x):
g ( x) 

 * ( x) g ( x)  ( x) dx
Bra-Ket Notation
This expression is so important that physicists have a special
notation for it.

g ( x)   * ( x) g ( x)  ( x) dx 
The entire expression is called a bracket.
And  | is called the bra with |  the ket.
The normalization condition is then:
|   1
|g|
6.1: The Schrödinger Wave Equation
The Schrödinger Wave Equation for the wave function (x,t) for a
particle in a potential V(x,t) in one dimension is:
2

2
i

V 
2
t
2m x
where
i  1
The Schrodinger Equation is the fundamental equation of
Quantum Mechanics.
Note that it’s very different from the classical wave equation.
But, except for its inherent complexity (the i), it will have similar
solutions.
General Solution of the Schrödinger Wave
Equation when V = 0
2
2
i
Try the usual solution:



t
2m x 2
( x, t )  Aei ( kx t )  A[cos(kx  t )  i sin(kx  t )]

 i Aei ( kx t )  i 
t

i
 (i )(i )   
t
This works as
long as:
2
k 2 p2


2m 2m
2
2


k

2
x
2 2
 2 2
k


2
2m x
2m
which says that the total
energy is the kinetic energy.
General Solution of the Schrödinger
Wave Equation when V = 0
In free space (with V = 0), the wave function is:
( x, t )  Aei ( kxt )  A[cos(kx  t )  i sin(kx  t )]
which is a sine wave moving in the x direction.
Notice that, unlike classical waves, we are not taking the real part
of this function.  is, in fact, complex.
In general, the wave function is complex.
But the physically measurable quantities must be real.
These include the probability, position, momentum, and energy.
Time-Independent Schrödinger Wave Equation
The potential in many cases will not depend explicitly on time: V = V(x).
The Schrödinger equation’s dependence on time and position can then
be separated. Let:
 ( x, t )   ( x) f (t )
2

2
And substitute into: i

V 
2
t
2m x
which yields:
f (t )
 2 f (t )  2 ( x)
i ( x)

 V ( x) ( x) f (t )
2
t
2m
x
Now divide by (x) f(t):
1 df (t )
 2 1  2 ( x)
i

 V ( x)
2
f (t ) t
2m  ( x) x
The left side depends only on t, and the right side
depends only on x. So each side must be equal to
a constant. The time-dependent side is:
i
1 df
B
f t
1 df
i
B
f t
Time-Independent Schrödinger
Wave Equation
f
Multiply both sides by f /iħ:
t
 B f /i
which is an easy differential
equation to solve:
f (t )  e Bt / i  eiBt /
But recall our solution for the free particle:  ( x, t )  e
i  kx t 
in which f(t) = exp(-it), so:  = B / ħ or B = ħ, which means that: B = E !
f (t )  eiEt /
So multiplying by (x), the spatial Schrödinger equation becomes:
d 2 ( x)

 V ( x) ( x)  E ( x)
2
2m dx
2
Stationary States
The wave function can now be written as:
( x, t )   ( x)eiEt /   ( x)eit
The probability density becomes:
*   * ( x) eit  ( x) eit
  ( x)
2
The probability distribution is constant in time.
This is a standing-wave phenomenon and is called a stationary state.
Most important quantum-mechanical problems will have stationary-state
solutions. Always look for them first.
Operators
d 2

 V  E
2
2m dx
2
The time-independent Schrödinger wave equation is as
fundamental an equation in quantum mechanics as the timedependent Schrödinger equation.
So physicists often write simply:
Ĥ  E
where:
2

Hˆ  
V
2
2m x
2
Ĥ is an operator yielding
the total energy (kinetic
plus potential energies).
Operators
 2 d 2 ( x)

 V ( x) ( x)  E ( x)
2
2m dx
Operators are important in quantum mechanics.
All observables (e.g., energy, momentum, etc.) have
corresponding operators.
The kinetic energy operator is:
2
K 
2m x 2
2
Other operators are simpler, and some just involve multiplication.
The potential energy operator is just multiplication by V(x).
Momentum Operator
To find the operator for p, consider the derivative of the wave function
of a free particle with respect to x:

 i ( kx t )

[e
]  ikei ( kx t )  ik
x x

 p
With k = p / ħ we have:
 i 
x
 
This yields:

p   i
x
This suggests we define the momentum operator as: pˆ  i
The expectation value of the momentum is:

p  i   * ( x, t )

 ( x, t )
dx
x

.
x
Position and Energy Operators
The position x is its own operator. Done.
Energy operator: Note that the time derivative of the free-particle
wave function is:
  i ( kx t )
 [e
]  iei ( kx t )  i
t
t
Substituting   E / ħ yields: E   i

t
This suggests defining the energy operator as:

Eˆ  i
t
The expectation value of the energy is:
E i
( x, t )
 ( x, t )
dx
t



*
Deriving the Schrödinger Equation
using operators
The energy is:
p2
 E 
 V 
2m
p2
E  K V 
V
2m
Substituting operators:
Ê   i
E:

t
pˆ 2
1 
 
  Vˆ  

i

  V 
2m
2m 
x 
2
K+V :
2

V 
2
2m x
2
E  K V :
2

 2
i

V 
2
t
2m x
Operators and Measured Values
In any measurement of the observable associated with an
operator A,
ˆ the only values that can ever be observed are the
eigenvalues. Eigenvalues are the possible values of a in the
Eigenvalue Equation:
Â  a
where a is a constant and the value that is measured.
For operators that involve only multiplication, like position and
potential energy, all values are possible.
But for others, like energy and momentum, which involve
operators like differentiation, only certain values can be the
results of measurements. In this case, the function  is often a
sum of the various wave function solutions of Schrödinger’s
Equation, which is in fact the eigenvalue equation for the energy
operator.
Solving the
Schrödinger
Equation when
V is constant.
When V0 > E:
d 2

 V0  E
2
2m dx
2
d 2 2m
Rearranging:
 2 V0  E 
2
dx
d 2
2

a

2
dx
where:
a  2m V0  E  /
2
Because the sign of the constant a2 is positive, the solution is:
kx
 kx
1
cosh(
kx
)

(
e

e
)
2
ax
a x
Sometimes
 ( x)  Ae  Be
When E > V0:
d 2
2


k

2
dx
people use: sinh( kx)  1 (e kx  e  kx )
2
where:
k  2m( E  V0 ) /
2
Because the sign of the constant k2 is negative, the solution is:
 ( x)  Aeikx  Beikx
or
A sin(kx)  B cos(kx)
6.3: Infinite Square-Well Potential
Consider a particle trapped in a box with
infinitely hard walls that the particle cannot
penetrate. This potential is called an infinite
square well and is given by:

V ( x)  
0
x  0, x  L
0  x  L
0
L
x
Outside the box, where the potential is infinite, the wave function must
be zero.
Inside the box, where the potential is zero, the energy is entirely kinetic,
so E > 0 = V0:
So, inside the box, the solution is:
 ( x)  A sin(kx)  B cos(kx)
Taking A and B to be real.
where
k  2mE /  2
Quantization and Normalization
Boundary conditions dictate that the wave
function must be zero at x = 0 and x = L.
This yields solutions for integer values of n
such that kL = np.
The wave functions  ( x)  A sin  np x 
n


are:
 L 
0
L
x
The same functions as those for a vibrating string with fixed ends!
In QM, we must normalize the wave functions:


 ( x)  n ( x) dx  1  A

*
n
The normalized wave
functions become:
2

L
0
½  ½ cos(2npx/L)
 np x 
sin 
 dx  1
 L 
 n ( x) 
2
2
 npx 
sin 

L
L


 A  2/ L
Quantized Energy
We say that k is quantized:
Solving for the energy yields:
kn 
2mEn
np

L
2
En  n
2
p2
2
2
2mL
(n  1, 2, 3,...)
The energy also depends on n. So the energy is also quantized.
The special case of
n = 1 is called the
ground state.
E1 
p 2 2
2mL2
6.4: Finite SquareWell Potential
The finite square-well potential is:
V0

V ( x)   0
V
 0
x  0
Region I
0  x  L Region II
x  L
Region III
Assume:
E < V0
The solution outside the finite well in regions I and III, where E < V0, is:
Region I, x  0
 I ( x)  Aea x
 III ( x)  Bea x Region III, x  L
Realizing that the wave function must be zero at x = ±∞.
Finite Square-Well Solution (continued)
Inside the square well, where the potential V is zero, the solution is:
 II ( x)  C sin kx  D cos kx Region II, 0  x  L
Now, the boundary conditions
require that:
 I   II at x  0
 II   III at x  L
So the wave function is smooth
where the regions meet.
Note that the wave function is
nonzero outside of the box!
The particle penetrates the walls!
This violates classical
physics!
The penetration depth is
the distance outside the
potential well where the
probability decreases to
about 1/e. It’s given by:
x
1
a

2m(V0  E )
Note that the penetration
distance is proportional to
Planck’s constant.
Tunneling
Consider a particle of energy
E approaching a potential
barrier of height V0, and the
potential everywhere else is
zero. Also E < V0.
The solutions in Regions I and III are sinusoids.
The wave function in region II becomes:
Enforcing
continuity at the
boundaries:
 II  Ce a x  Dea x
Tunneling wave function
The particle can
tunnel through the
barrier—even
though classically it
doesn’t have
enough energy to
do so!
The transmission probability
for tunneling is:
 V0 sinh (L) 
T  1 

4
E
(
V

E
)
0


2
2
1
The uncertainty principle allows this violation of classical physics.
The particle can violate classical physics by DE for a short time,
Dt ~ ħ / DE.
Analogy with Wave Optics
If light passing through a glass prism
reflects from an internal surface with an
angle greater than the critical angle,
total internal reflection occurs.
However, the electromagnetic field isn’t
exactly zero just outside the prism.
If we bring another prism very close to
the first one, light passes into the
second prism.
The is analogous to quantummechanical tunneling.
The Scanning-Tunneling Microscope
Electrons must tunnel through the
vacuum (barrier) from the surface
to the tip. The probability is
exponentially related to the
distance, hence the ultrahigh
resolution.
Image of a molecule
Alpha-Particle Decay
Tunneling explains alpha-particle decay of heavy, radioactive nuclei.
Outside the nucleus, the Coulomb force dominates.
Inside the nucleus, the strong,
short-range attractive nuclear
force dominates the repulsive
Coulomb force. The potential
is ~ a square well.
The potential barrier at the nuclear
radius is several times greater than
the energy of an alpha particle.
In quantum mechanics, however,
the alpha particle can tunnel
through the barrier.
This is radioactive decay!
Radioactive
Decay
The number of
radioactive nuclei as
a function of time.
The time for the
number of nuclei to
drop to one half its
original value is the
well-known half life.
Potassium–
argon
dating
Potassium–argon dating is
used in archeology. It’s based
on the radioactive decay
(positron emission) of an
isotope of potassium 40K
(half-life = 1.248×109 yr) into argon (Ar). 40Ar is able to escape liquid
(molten) rock, but starts to accumulate when the rock solidifies (recrystallizes).
Time since re-crystallization is determined by measuring the ratio of
the amount of 40Ar accumulated to the amount of 40K remaining.
The long half-life of 40K allows the method to be used to calculate
the absolute age of samples older than a few thousand years.
Long-Time Dating Using Lead Isotopes
Nothing decays to or from
204Pb.
A plot of the abundance
ratio of 206Pb / 204Pb
versus 207Pb / 204Pb can
be a sensitive indicator of
the age of lead ores.
Such techniques have
been used to show that
moon rocks and
meteorites, believed to be
left over from the
formation of the solar
system, are 4.55 billion
years old.
6.5: Three-Dimensional
Infinite-Potential Well
y
The wave function must be a function of all
three spatial coordinates.
So consider momentum as an
operator in three
dimensions:
p2  p2 
x

pˆ x  i
x
x
z
py2  pz2

pˆ y  i
y

pˆ z  i
z
The three-dimensional Schrödinger wave equation is:
 2   2  2  2 
  2  2  2   V  E
2m  x
y
z 
or:

2
2m
2  V  E
The 3D infinite potential well
It’s easy to show that:
 ( x, y, z )  A sin(k x x) sin(k y y ) sin(k z z )
where:
k x  p nx / Lx
 nx2 n y2 nz2 
E
 2  2  2 
2m  Lx Ly Lz 
p2
and:
2
When the box is a cube:
E
p2
2
2
2mL
kz  p nz / Lz
k y  p n y / Ly
n
2
x
Lz
Ly
n n
2
y
y
2
z

z
x
Lx
Degeneracy
E
p2
2
2mL2
n
2
x
 ny2  nz2 
Try 10, 4, 3 and 8, 6, 5
Note that more than one wave function can have the same energy.
When more than one wave function has the same energy,
those quantum states are said to be degenerate.
Degeneracy results from symmetries of the potential energy
function that describes the system.
A perturbation of the potential energy can remove the degeneracy.
Examples of perturbations include external electric or magnetic
fields or various internal effects, like the magnetic fields due to the
spins of the various particles.
6.6: Simple Harmonic Oscillator
Simple harmonic
oscillators describe
many physical
situations: springs,
diatomic molecules
and atomic lattices.
Consider the Taylor expansion of an
arbitrary potential function:
1
V ( x)  V0  V1  [ x  x0 ]  V2  [ x  x0 ]2  ...
2
Near a minimum, V1[xx0] ≈ 0.
Simple Harmonic
Oscillator
Consider the second-order term
of the Taylor expansion of a
potential function:
Letting x0 = 0.
V ( x)  12  ( x  x0 ) 2  12  x 2
Substituting this into
Schrödinger’s equation:
 2 d 2 ( x)

 V ( x) ( x)  E ( x)
2
2m dx

 m x 2 2mE 
d 2
2m   x 2
  2 
 E    2  2 
We have:
2
dx
 2



2mE
m
Let a  2 and   2 , which yields:


2
d 2
2 2

a
x  
2
dx


The Parabolic
Potential Well
The wave function solutions
are:  n ( x)  H n ( x) exp(a x 2 / 2)
where Hn(x) are Hermite
polynomials of order n.
n
|n |2
The Parabolic
Potential Well

Classically, the
probability of finding the
mass is greatest at the
ends of motion (because
its speed there is the
slowest) and smallest at
the center.


Classical
result

Contrary to the classical
one, the largest
probability for the lowest
energy states is for the
particle to be at (or near)
the center.
Correspondence Principle for the
Parabolic Potential Well
As the quantum number (and the size scale of the motion) increase,
however, the solution approaches the classical result. This confirms the
Correspondence Principle for the quantum-mechanical simple
harmonic oscillator.
Classical
result
The Parabolic Potential Well
The energy levels are given by:
1
1
En  (n  )  / m  (n  )
2
2
The zero point
energy is
called the
Heisenberg
limit:
1
E 0  
2