Transcript I II III

tunnel effect
a famous cat
harmonic oscillator
“I wish you wouldn’t keep appearing and vanishing so suddenly; you make
one quite giddy!”
“All right,” said the Cat; and this time it vanished quite slowly, beginning with
the end of the tail, and ending with the grin, which remained some time
after the rest of it had gone.
“Well! I’ve often seen a cat without a grin,” thought Alice; “but a grin
without a cat! It’s the most curious thing I ever saw in all my life!”
5.10 Tunnel Effect
We found some very strange things going on with our particle
in a finite well.
We calculated a finite probability for a particle to exist inside
the wall (even though the wall was supposedly impenetrable).
There are plenty of examples of this in everyday
life. Have you ever had a brother, sister, or parent
show up somewhere they weren’t supposed to be;
where they couldn’t possibly be?
Just kidding, of course. But for quantum systems, we do
calculate finite probabilities for particles to exist where they are
not “supposed” to be.
Here’s a thought experiment. Throw a ball at an 8 foot high
brick wall.
Too low and it
reflects back.
High enough
and it goes
over.
How about a classical particle of energy E incident on a barrier
of height U?
E
E
If E < U it
reflects back.
U
If E > U it
goes over.
U
Actually, the particle “breaks through,” but this is just a conceptual picture.
How about a quantum mechanical particle of energy E incident
on a barrier of height U?
E
If E < U it
reflects back.
Or maybe not.
E
U
U
You should imagine the particle as a wave when we do this thought experiment.
E > U?
E
If E > U it
goes over.
E
Or maybe not
U
U
You should imagine the particle as a wave when we do this thought experiment.
Something to think about…
1
2 = 1
inside
E
E
U
The particle’s (kinetic) energy is the same before and after the
barrier (it loses no energy getting through the barrier!), so its
momentum and wavelength are the same before and after.
What about its wavelength inside the barrier?
KEinside = (E – U) < E = KEoutside so KEinside must be less.
Therefore pinside < poutside and inside > outside.
If E < U then inside is imaginary?!
“This is complete, utter
rubbish!”
No, this is real!
If you are wearing glasses, take the off, hold them six inches or
a foot away from your face, and look at them.
You can see through your glasses. If there is a bright object
behind you, you see its reflection “in” your glasses.
A person standing on the other side of your glasses would see
the bright object through your glasses.
Some of the photons go through your glasses, others are
reflected off them. There is a probability of transmission, and a
probability of reflection.
Another “everyday” example is the physics 24 lab on reflection
and refraction of light.
“Ahh,” but he says “that’s because photons are waves.”
BINGO! Gotcha! Particles are waves too! They reflect and
transmit.
“Prove it!”
I was hoping you would ask that…
The mathematics that follows is for your personal edification
and enjoyment, and is not specifically* testable material this
year (although problems based specifically* on it have
appeared on a few prior exams). Cultural gratification material
starts here.
In the following slides we solve Scrödinger's equation to find
that a quantum particle can tunnel through a high barrier, or be
reflected off a low barrier, quite unlike a classical particle.
*In other words, I will not ask you to derive the tunneling equation, which I do next.
E
Here's a diagram of the potential
for the problem.
The particle has mass m and
energy E,* and the barrier has
height U and length L.
There are three regions: I, II,
and III.
U
I
II
III
x
0 L
We must solve Schrödinger's equation for a particle of energy E
in the three different regions in the figure above.
Schrödinger's equation in regions I, II, and III becomes…
*Potentially confusing, because E has a double use: as a label for the
energy axis, and as the specific value of the particle’s energy—plus E
sometimes means “kinetic energy” in this course.
E
 2 I 2m
+ 2 EI = 0
2
x
 2II 2m
+ 2 E - U II = 0
2
x
 2  III 2m
+ 2 E III = 0
2
x
U
I
II
III
x
0 L
Our job is to solve these equations, subject to appropriate
boundary conditions on , and subject to the conditions that 
and its first derivative must be continuous and finite.
Our three equations are valid for both E > U and for E < U. We
will assume E < U when we solve Schrödinger's equation in
region II.
We begin by assuming wave-like
solutions for  in regions I and III,
where the potential U is zero.
E
U
I
II
III
Appropriate wave-like solutions are*
ψI = A e j k1x + B e- j k1x
x
0 L
and
ψIII = F e j k1x + G e- j k1x
where
k1 =
2
2mE
=
.

Why is k1 the same in regions I and
III? Because the kinetic energy is
the same in regions I and III. The
particle loses no energy getting past
the barrier.
*Why not use sines and cosines? The later math would prove
extremely difficult. Complex exponentials simplify the math.
E
ψI = A e j k1x + B e- j k1x
ψIII = F e
j k1 x
+Ge
Note that I and III are of the form
ψI = A e j k1x + B e- j k1x
+
I
= ψ
+
ψ
U
- j k1 x
I
I
II
III
x
0 L
and ψIII = F e j k1x + G e- j k1x
= ψ+III + ψIII
where I+ represents a wave in region I traveling to the right,
I- represents a wave in region I traveling to the left,
and III+ represents a wave in region III traveling to the right,
and III- represents a wave in region III traveling to the left.
E
We assume our incident wave
travels from left to right.
If our wave gets through the
barrier, there is nothing in region
III to reflect it back to the left.
ψIII = F e j k1x + G e- j k1x
U
I
II
III
x
0 L
This tells us III- = 0 so that G = 0.
A “full” solution of this problem requires calculation of 
everywhere, so we would need I, II, and III. However, often
in QM calculations, a full solution is not needed. Today I am
just calculating the probability of transmission through the
barrier.
Wave function particle wave incident on the barrier:
ψI = A e j k1x + B e- j k1x
=
ψ+I + ψI-
Are you trying to tell me the
particle is made up—all at
once—of parts going right
and parts going left???
I+*I+ is the probability density of the incoming wave. In one
dimension, it is a linear probability density having units of
particles/meter.
If v is the group velocity of the I+ wave, then the number of
particles per square meter per second incident at the barrier
from the left is
SI+ = I+* I+ v.
Similarly, the number of particles per square meter per second
emerging at the right of the barrier SIII+ = III+* III+ v.
Real-life transmission problem:
(http://www.nearingzero.net)
The transmission probability is then

S
T=
=
+ *
S
 I   +I  v
+
III
+
I
T =
F e
A e

+

 III  v

=
 A e  A
j k1 x *
j k1 x *
j k1 x
F
e


+ *
III
- j k1 x
j k1 x
*
F
e
F
e



j k1 x
*
e - j k1x  A e j k1x 
F*F
T = *
A A
To get a useful expression for T, we need to apply the
continuity of  and its derivative across the barrier to eliminate
A and F in the equation for T.
This means we need to look at Schrödinger's equation in region
II.
Recall that in region II, Schrödinger's equation is
 2II 2m
+ 2 E - U II = 0
2
x
Solutions in region II are:
where
k =
ψII = C e j k  x + D e- j k  x
2m E - U 
.
Do you see anything interesting about these solutions?
The wavenumber is k'. But E < U, so k' is pure imaginary.
What implications does that have? Answer: damped or
blowing up exponentials, not waves, in region II.
Does that make sense? Yes, if you think about it.
Since k' is imaginary, it makes sense to define k2 = - jk'. Now
we see that
ψII = C e - k 2 x + D e k 2 x
and it is clear that II does not represent a “wiggly” wave, but
rather a damped or blowing up exponential.
Philosophically, one might argue that since II does not
represent a wave, it does not represent a moving particle, and
therefore no particle exists in region II. But II*II does
represent a probability density, so there is a probability of
finding the particle inside the barrier.
Come on, commit yourself—does the particle exist inside the
barrier or not? Look, I’m just a simple-minded
experimentalist. Let’s move on to something we could
actually measure.
Let’s think about our calculations so far.
We have an expression for transmission, containing the
constants A and F. We need to determine them in order to
calculate the transmission.
We also wrote down solutions for the wavefunction in region II,
but this just introduced two new constants, C and D. That
doesn't seem to help. Seems to have made it worse
because we just have more “stuff” to calculate.
What we can do, now that we have solutions in regions I, II,
and III, is apply boundary conditions at the two boundaries.
Let's see... two boundaries, two conditions at each boundary
(wave function and its derivative continuous) gives a total of
four conditions, to go along with our 4 unknowns A, C, D, and
F. 4 unknowns, 4 conditions, and we can solve the problem.
By "problem" here, I mean
calculating the transmission.
E
The full problem is more
general. We actually have 5
coefficients on the wave
functions (A, B, C, D, and F)
to go along with four
boundary conditions.
I
II III
A,B C,D F
U
 &  cont.
0 L
x
 &  cont.
Getting T requires only four of those coefficients. If we want to
solve for the wave functions, we’ll need another condition.
If we want to solve for the wave functions, we need to apply
normalization, which gives a fifth condition to go along with the
five coefficients; the problem is then solvable.
E
Here are the boundary
conditions.
At the left-hand wall (x = 0),
ψI = ψII
 I  II
=
x
x
(1)
(2)
At the right-hand wall (x = L),
ψII = ψIII
(3)
 II  III
=
x
x
(4)
U
I
II III
A,B C,D F
 &  cont.
0 L
x
 &  cont.
Plugging I and II into (1) gives
Ae j k1 0 + Be -j k1 0 = Ce-k2 0 + De k2 0
A +B =C +D
Plugging I and II into (2) gives
jk1 Ae j k1 0 - jk1 Be -j k1 0 = -k 2 Ce-k2 0 + k 2 De k2 0
jk1 - jk1 = - k 2 + k 2
Next, we plug II and III into (3) and (4) to get (after a bit of
math)
Ce - k 2L + De k 2L = Fe j k1L
-k 2 Ce - k 2L + k 2 De k 2L = jk1 Fe j k1L
Here’s a little pop quiz. Sort the following list in order of least fun to most
fun (or least liked to most liked, if these things are not “fun”).
Solving an algebraic equation in one unknown.
Solving a system of four algebraic equations in four unknowns.
Solving a system of two algebraic equations in two unknowns.
Solving a system of three coupled linear differential equations.
Solving a single linear differential equation.
Here’s my solution:
1
2
3
4
5
Solving a system of three coupled linear differential equations.
Solving a single linear differential equation.
Solving a system of four algebraic equations in four unknowns.
Solving a system of two algebraic equations in two unknowns.
Solving an algebraic equation in one unknown.
Notice what we have done. We replaced this
 2 I 2m
+ 2 EI = 0
2
x
 2II 2m
+ 2 E - U II = 0
2
x
 2  III 2m
+ 2 E III = 0
2
x
by this
A +B =C +D
jk1 - jk1 = - k 2 + k 2
Ce - k 2L + De k 2L = Fe j k1L
-k 2 Ce - k 2L + k 2 De k 2L = jk1 Fe j k1L
I like this a lot!
A +B =C +D
jk1 - jk1 = - k 2 + k 2
Ce - k 2L + De k 2L = Fe j k1L
-k 2 Ce - k 2L + k 2 De k 2L = jk1 Fe j k1L
Whaddya mean you don’t like this? This is just a system of 4
linear equations. You should be able to solve it in your sleep.
Oops, there is one problem: there are five unknown
coefficients in the four equations. We need another equation.
Remember, however, that we can apply normalization if we
need to.
We are only interested in finding the ratio F*F / A*A. We can
always solve a system of N equations in N+1 unknowns for a
ratio of any two of the coefficients.
How would
you solve this
system for
A/F?
Once I get A/F, I just
multiply by the
complex conjugates
and invert the result.
A +B =C +D
jk1 - jk1 = - k 2 + k 2
Ce - k 2 L + De k 2 L = Fe j k1L
-k 2 Ce - k 2 L + k 2 De k 2 L = jk1 Fe j k1L
You would probably use your calculator.
I’m too old to learn newfangled calculators. I’d do it five
different times and if any two of the five trials produced
identical results, I’d figure those two (actually one, of course)
were the correct answer. Here’s the result:
A 1
j  k 2 k1    jk1 + k 2 L  1 j  k 2 k1    jk1 -k 2 L
= +  + -   e
 e
F  2 4  k 1 k 2 
 2 4  k 1 k 2 
Anybody claiming to find a typo must prove it by solving the
system by hand, all by him(her)self!
The ratio for A/F is "only" algebra, and if you correctly type it
into Mathcad once, you can copy and calculate and modify as
much as you want, so it's really not so bad.
We simplify the problem by assuming a high and wide barrier.
"High" means the barrier potential is high relative to the
incident kinetic energy. In that case k2/k1>k1/k2. (Look back at
the definitions of k1 and k2 to see this.)
"Wide" means that the wave function is severely attenuated in
the barrier region between x=0 and x=L. That means k2L >> 1.
With these approximations, the equation for A/F is considerably
simplified, and can be written
A 1
j  k 2    jk1 + k 2  L
=  +   e
F  2 4  k 1 
hey, that’s
not so bad
Of course, you have to find the complex conjugate A*/F* (just
replace j by -j everywhere). The transmission is then



F*F 
16
-2k 2L

T= * =
e
2
A A 
 k2  
4
+
 k  


1 
The quantity in the square brackets varies slowly compared to
the exponential, and is of order of magnitude 1, so we can
further simplify the transmission, and write (cultural gratification
material ends here)
T = e -2k 2L
where (remember?)
k2 =
2m  U - E 
.
That was fun enough to be worth writing again.
T=e
-2k 2L
k2 =
2m  U - E 
Keep in mind that this applies for a "high" and "wide" barrier. If
the barrier is not both "high" and "wide," you must use the full
expression for A/F (and then multiply by its complex
conjugate).
“Remember that stuff in the square brackets a page
ago. You said it was of order magnitude 1. Looked
more like 4 to me. Doesn’t it bother you to throw it
away?”
Well, yes. But 4 is closer to 1 than 10, so it really is of order
magnitude 1.
“An order of magnitude calculation means we accept
errors of up to a factor of 10. What kind of a theory is
that?”
A great theory, if that’s the best you can do! Or if you just
want to develop a feel for the physics. Obviously, in “real life”
we would use the full expression, with no approximations.
Let’s do an example. Consider a 1 eV electron incident on a
barrier 5 eV high and 0.5 nm wide.
5 is “sort of considerably" bigger than 1, and 0.5 nm is long
compared to a typical low-energy electron wavelength. Thus
we can probably get away with using the simplified
transmission equation.
First, we need to get k2. There are a couple of ideas important
enough to make me put the calculation on the next (full) slide…
k2 =
What mass do I use here?
What is the object?
Electron!
2m  U - E 
Can I keep energies
in eV and use the
eV·s version of ħ?
Absolutely not! The
square root “messes
up” your units. You
will be wrong every
time!
k2 =
2  9.11×10-31 kg  5 eV - 1 eV  1.6×10-19 J/eV 
-34
1.055×10
Js

k 2 = 1.00×10-10 m-1
The transmission is
T = e -2k 2L
T=e


-2 1.0×1010 5.0×10-10
Because of the exponential,
small differences in how you
round in calculating k2 can make
large apparent differences here.
 = 4.5×10-5
T had better be between 0 and 1. For this approximation it had
better be “small.” Looks like we are OK.
Could you calculate the reflection probability?
Remember this statement from earlier in this lecture?
Are you trying to tell me
the particle is made up—
all at once—of parts
going right and parts
going left???
Schrödinger’s cat! Argh, I don’t like this “many-world” stuff.
Another Schrödinger’s cat.
An engineer on Schrödinger’s cat: “Physicists agonize while that
Cheshire cat sits and smiles. They try to write wave functions
for cats and gamma radiation. They conclude goofy things:
maybe the cat in the unopened box is both alive and dead at
the same time.”
Ramblings:
What would happen if you threw a dead cat from the top of a
50 story building. Dead Cat Bounce.
The return of the dead cat.
101 uses for a dead cat (OK, so I exaggerated by 97).
A few more of the 101 uses. (Link seems to be dead, FS2003.)
What happens after a cat dies. (Caution—pop-up ads.)
A real-life Schrödinger’s cat: an ion living in two states and two
different places at the same time.
"When I hear of Schrödinger's cat, I reach for my gun." –Steven Hawking.
5.11 Harmonic Oscillator
Recall from math how functions can be written in the form of a
Maclaurin’s series (a Taylor series about the origin):
2
2
3
3



dF
x
d
F
x
d
F
 
F(x) = F(0) + x   +  2  +  3  + . . .
 dx 0 2!  dx 0 3!  dx 0
If F represents a restoring force (a force that “pulls the system
back to the origin”) then F(0) = 0.
For small displacements x, all the higher order terms (involving
x2, x3, etc.) are small, so
 dF 
F(x)  x   = - k x .
 dx 0
The – sign enters because F is a restoring force, so the derivative is
negative.
So in the limit of small displacements, any restoring force
obeys Hooke’s Law:
F(x) = - k x .
If any restoring force obeys Hooke’s Law, it must be worth
studying!
Classically, a harmonic oscillator is subject to Hooke's law.
Newton's second law says F = ma. Therefore
d2 x
- k x =m 2 .
dt
d2 x
m 2 +k x=0.
dt
Another differential equation to solve!
The solution to this differential equation is of the form
x = A cos (ωt + φ)
where the frequency of oscillation is f, and
k
ω = 2f =
.
m
Recall from your first semester (mechanics) physics course, that
the harmonic oscillator potential is
1
U(x) = k x 2 .
2
So what?
Many systems are described by harmonic oscillators. We had
better see what quantum mechanics has to say about them!
A truly classic example is the swinging bowling ball demo.
Before we continue, let’s think about harmonic oscillators…
Classically, all energies are allowed. What will QM say?
Only quantized energies?
Classically, an energy of zero is allowed. What will QM say?
Nonzero, like particle in box?
Classically, the oscillator can't exist in a state in "forbidden"
regions. For example, a pendulum oscillating with an amplitude
A cannot have a displacement greater than A.
Could there be a nonzero probability of finding the system in
"forbidden" regions. I wonder what that means for our
swinging bowling ball…
Now, let's solve Schrödinger's equation for the harmonic
oscillator potential.
2 2m  1 2 
+ 2  E - kx   = 0 .
2
x
2


Why  instead of ?
If we let y =
all you do is plug in the correct
potential and turn the math crank
2 m f
2E
x and  =
h
hf
then Schrödinger's equation becomes
2ψ
2
+
α
+
y
ψ =0.


2
y
Solutions to this equation must satisfy all the requirements we
have previously discussed, and  must be normalized.
The solution is not particularly difficult, but is not really worth a
day's lecture. Instead, I will quote the results.
The equation can be solved only for particular values of ,
namely =2n+1 where n = 0, 1, 2, 3, ...
For those values of , the wave function has the form
 2mf 
ψn = 

h


1
4
2 n!
n
-1
2
Hn (y) e
 y2 
- 2 


.
A normal human would say this looks nasty, but a mathematician would say it is simple. Just a bunch of numbers, an
exponential function, and the Hermite Polynomials Hn.
Polynomials are simple. H0(y) = 1, H1(y) = 2y, and other
polynomials are given in Table 5.2 of Beiser.
More important, we find that the wave equation is solvable only
for certain values of E (remember,  = 2E/hf = 2n+1), given by
1

En =  n +  hf ,
2

n = 0,1,2, ...
The energies of the quantum mechanical harmonic oscillator
are quantized in steps of hf, and the zero point energy is E0 =
½hf.
Here is a Mathcad document illustrating QM harmonic oscillator
energy levels, probabilities, and expectation values.
Because of the scaling we did in re-writing Schrödinger’s
equation, it is difficult to identify the classically forbidden
regions in the graphs in the Mathcad document. See Figures
5.12 and 5.13, page 191 of Beiser, for an illustration of how the
amount of wavefunction “tails” in the forbidden region shrinks
as n increases.
Here are a couple of plots.
Wave Functions
Probability Densities
2
n=1
Things you ought* to study in relation to harmonic oscillators:
Figure 5.13, to see how the quantum mechanical harmonic
oscillator "reduces" to the classical harmonic oscillator in limit of
large quantum numbers.
Figure 5.11, to see how the different potentials for different
systems lead to different energy levels (we will do the hydrogen
atom, figure 5.11a, in the next chapter).
Example 5.7, page 192, expectation values.
*Like, before exam 2!
The BIG PICTURE.
It took us forever to get through chapter 5. What are some big
ideas?
Wave functions – probability densities – normalization –
expectation values – “good” and “evil,” uhh I mean
“possible” and “not possible” wave functions – calculating
probabilities.
Particle in box – how to solve the SE – energy levels –
quantization – expectation values – effect of box length –
calculating probabilities.
Particle in well – how to solve the SE – energy levels –
quantization – expectation values – effect of well length –
effect of well height – calculating probabilities – compare
and contrast with infinite well – classically forbidden
regions.
Tunneling – (how to solve the SE) – transmission
probability – reflection probability – effect of particle mass
and energy on tunneling probability – effect of barrier
height on tunneling probability.
We didn’t discuss applications, but there are many.
Scanning tunneling microscope. Quantum effects as
integrated circuits shrink towards the quantum world!
Your life will be impacted by quantum effects in a big
way!
Harmonic Oscillator – (how to solve the SE) – energy
levels –zero point energy – quantization – expectation
values – classically forbidden regions – classical limit.
This is not guaranteed to be an all-inclusive list!