“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!”

tunnel effect

a famous cat

harmonic oscillator

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?

If E < U it reflects back.

If E > U it goes over.

U

U

E

E

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?

If E < U it reflects back.

Or maybe not.

U

U

E

E

You should imagine the particle as a wave when we do this thought experiment.

If E > U it goes over.

Or maybe not

U

U

E

E

E > U?

You should imagine the particle as a wave when we do this thought experiment.

UE E

Something to think about…

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?!

1 2 = 1

inside

“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 other words, I will not ask you to derive the tunneling equation, which I do next.

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.

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.

x

E

U

0 LThere are three regions: There are three regions: I, There are three regions: I, II, There are three regions: I, II, and III.

I II III

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.

x

E

U

0 L

I II III

2I

I2 2

2m + E = 0

x

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.

2III

III2 2

2m + E = 0

x

2

IIII2 2

2m + E-U = 0

x

We begin by assuming wave-like solutions for in regions I and III, where the potential U is zero.

x

E

U

0 L

I II IIIAppropriate wave-like solutions are*

ψ 1 1 j k x - j k xI = A e + B e

and

ψ 1 1 j k x - j k xIII = F e + G e

where

1

2 2mEk = = .

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.

ψ 1 1 j k x - j k xI = A e + B e

x

E

U

0 L

I II IIIψ 1 1 j k x - j k x

III = F e + G e

Note that I and III are of the form

ψ

ψ ψ

1 1 j k x - j k xI

+ -I I

= A e + B e

= +

and ψ

ψ ψ

1 1 j k x - j k xIII

+ -III III

= F e + G e

= +

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.

We assume our incident wave travels from left to right.

x

E

U

0 L

I II IIIIf our wave gets through the barrier, there is nothing in region III to reflect it back to the left.

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.

ψ 1 1 j k x - j k xIII = F e + G e

Wave function particle wave incident on the barrier:

ψ

ψ ψ

1 1 j k x - j k xI

+ -I I

= A e + B e

= +

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.

Are you trying to tell me the particle is made up—all at once—of parts going right and parts going left???

Real-life transmission problem:

(http://www.nearingzero.net)

The transmission probability is then

*+ ++III IIIIII

*+ + +I I I

vST = =

S v

1 1 1 1

1 11 1

* j k x j k x - j k x j k x*

* - j k x j k x* j k x j k x

F e F e F e F eT = =

A e A eA e A e

*

*

F FT =

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

Solutions in region II are: ψ j k x - j k xII = C e + D e

where

.2m E - U

k =

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? 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.

2

IIII2 2

2m + E-U = 0

x

Since k' is imaginary, it makes sense to define k2 = - jk'. Now we see that

ψ 2 2 - k x k xII = C e + D e

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? 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.

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.

x

E

U

0 L

I II IIIThe 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.

Getting T requires only four of those coefficients. If we want to solve for the wave functions, we’ll need another condition.

A,B C,D F

& cont. & cont.

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.

Here are the boundary conditions.

At the left-hand wall (x = 0),

ψ ψI II = (1)

I II = (2)x x

At the right-hand wall (x = L),

x

E

U

0 L

I II IIIA,B C,D F

& cont. & cont.

II III = (4)x x

ψ ψII III = (3)

Plugging I and II into (1) gives

1 1 2 2 j k 0 - j k 0 -k 0 k 0Ae + Be = Ce + De

A + B = C + D

Plugging I and II into (2) gives

1 1 2 2 j k 0 - j k 0 -k 0 k 01 1 2 2jk Ae - jk Be = -k Ce + k De

1 1 2 2jk - jk = -k + k

Next, we plug II and III into (3) and (4) to get (after a bit of math)

2 2 1 - k L k L j k LCe + De = Fe

2 2 1 - k L k L j k L2 2 1-k Ce + k De = jk Fe

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 two algebraic equations in two unknowns.

Solving a system of four algebraic equations in four unknowns.

Solving a single linear differential equation.

Solving a system of three coupled linear differential equations.

Here’s my solution:

5 Solving an algebraic equation in one unknown.

4 Solving a system of two algebraic equations in two unknowns.

3 Solving a system of four algebraic equations in four unknowns.

2 Solving a single linear differential equation.

1 Solving a system of three coupled linear differential equations.

Notice what we have done. We replaced this

by this

2I

I2 2

2m + E = 0

x

2III

III2 2

2m + E = 0

x

2

IIII2 2

2m + E-U = 0

x

A + B = C + D

1 1 2 2jk - jk = -k + k

2 2 1 - k L k L j k LCe + De = Fe

2 2 1 - k L k L j k L2 2 1-k Ce + k De = jk Fe

I like this a lot!

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.

A + B = C + D

1 1 2 2jk - jk = -k + k

2 2 1 - k L k L j k LCe + De = Fe

2 2 1 - k L k L j k L2 2 1-k Ce + k De = jk Fe

How would you solve this system for A/F?

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:

1 2 1 2jk + k L jk -k L2 1 2 1

1 2 1 2

k k k kA 1 j 1 j = + - e + - - e

F 2 4 k k 2 4 k k

Anybody claiming to find a typo must prove it by solving the system by hand, all by him(her)self!

A + B = C + D

1 1 2 2jk - jk = -k + k

2 2 1 - k L k L j k LCe + De = Fe

2 2 1 - k L k L j k L2 2 1-k Ce + k De = jk Fe

Once I get A/F, I just multiply by the complex conjugates and invert the result.

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

1 2jk + k L2

1

kA 1 j = + e

F 2 4 k

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

2

*-2k L

2*2

1

F F 16T = = e

A A k4 + k

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)

2-2k LT = e

where (remember?)

2

2m U - Ek = .

That was fun enough to be worth writing again.

2-2k LT = e

2

2m U - Ek =

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…

2

2m U - Ek =

-31 -19

2 -34

2 9.11×10 kg 5 eV - 1 eV 1.6×10 J /eVk =

1.055×10 J s

Absolutely not! The square root “messes up” your units. You will be wrong every time!

-10 -12k = 1.00×10 m

What mass do I use here?What is the object?

Electron!

Can I keep energies in eV and use the eV·s version of ħ?

The transmission is

2-2k LT = e

10 -10-2 1.0×10 5.0×10 -5T = e = 4.5×10

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?

Are you trying to tell me the particle is made up—all at once—of parts going right and parts going left???

Remember this statement from earlier in this lecture?

Schrödinger’s cat! Argh, I don’t like this “many-world” stuff.Another Schrödinger’s cat.

Because of the exponential, small differences in how you round in calculating k2 can make large apparent differences here.

Ramblings:

What would happen if you threw a dead cat from the top of a 50 story building. 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.)

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.”

"When I hear of Schrödinger's cat, I reach for my gun." –Steven Hawking.

A real-life Schrödinger’s cat: an ion living in two states and two different places at the same time.

What would happen if you threw a dead cat from the top of a 50 story building. Dead Cat Bounce.

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

2 30 0 0

dF x d F x d FF(x) = F(0) + x + + + . . .

dx 2! dx 3! dx

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

0

dFF(x) x = - k x .

dx

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.

Another differential equation to solve!

Newton's second law says F = ma. Therefore

2

2

d x- k x = m .

dt

2

2

d xm + k x = 0 .

dt

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

21U(x) = k x .

2

So what?

Recall from your first semester (mechanics) physics course, that the harmonic oscillator potential is

Many systems are described by harmonic oscillators. We had better see what quantum mechanics has to say about them!

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…

A truly classic example is the swinging bowling ball demo.

Now, let's solve Schrödinger's equation for the harmonic oscillator potential.

22

2 2

2m 1 + E - kx = 0 .

x 2

all you do is plug in the correct potential and turn the math crank

If we let2 m f 2E

y= x and = h hf

then Schrödinger's equation becomes

2ψ

α ψ .22 + + y = 0

y

Solutions to this equation must satisfy all the requirements we have previously discussed, and must be normalized.

Why instead of ?

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

ψ

21y- 14 - 2n 2

n n

2mf= 2 n! H (y) e .

h

A normal human would say this looks nasty, but a math-ematician 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

n

1E = n + hf , n = 0,1,2, ...

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

n = 1n = 2

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!