of All modern physics is governed by that magnificent and thoroughly confusing discipline called quantum mechanics...It has survived all tests and there is no reason to believe that there is any flaw in it.We all know how to use it and how to apply it to problems; and so we have learned to live with the fact that nobody can understand it. --Murray Gell-Mann
Lecture 11, p 1
Particles in Finite Potential Wells
U(x)
Lecture 11:
y(x) U0
n=2 n=1 n=4 n=3
AlGaAs
GaAs
AlGaAs
I
II
III
U(x)
0
L
x
Lecture 11, p 2
This week and last week are critical for the course:
Week 3, Lectures 7-9: Light as Particles Particles as waves Probability Uncertainty Principle Week 4, Lectures 10-12: Schrdinger Equation Particles in infinite wells, finite wells
Midterm Exam Monday, Feb. 14. It will cover lectures 1-11 and some aspects of lectures 11-12. Practice exams: Old exams are linked from the course web page. Review Sunday, Feb. 13, 3-5 PM in 141 Loomis. Office hours: Feb. 13 and 14
Next week: Homework 4 covers material in lecture 10 due on Thur. Feb. 17. We strongly encourage you to look at the homework before the midterm! Discussion: Covers material in lectures 10-12. There will be a quiz. Lab: Go to 257 Loomis (a computer room). You can save a lot of time by reading the lab ahead of time Its a tutorial on how to draw wave functions.
Lecture 11, p 3
Last Time
Schrodingers Equation (SEQ)
A wave equation that describes spatial and time dependence of Y(x,t). Expresses KE +PE = Etot Second derivative extracts -k2 from wave function.
Constraints that y(x) must satisfy
Existence of derivatives (implies continuity). Boundary conditions at interfaces.
Infinitely deep 1D square well (box)
Boundary conditions y(x) = Nsin(kx), where k = np/L. Discrete energy spectrum: En = n2E1, where E1 = h2/8mL2. Normalization: N = (2/L).
Lecture 11, p 4
Today
Normalizing the wave function
General properties of bound-state wave functions Particle in a finite square well potential
Solving boundary conditions
Comparison with infinite-well potential
Midterm material ends here.
Lecture 11, p 5
Particle in Infinite Square Well Potential
2p y n ( x) sin kn x sin n y(x)
n=2 n=1 n=3
np x sin L
x
for 0 x L
nn 2L
Lx
0
d 2y n (x) U ( x )y n ( x ) Eny n ( x ) 2 2m dx
2
The discrete En are known as energy eigenvalues:
electron
U=
En
U=
n=3 n=2 n=1
p2 h2 1.505 eV nm 2 En 2 2m 2mn n2 En E1n 2 where E1 h 8mL2
2
0
L
x
Lecture 11, p 6
Often what we measure in an experiment is the probability density, |y(x)|2.
y n ( x) B1 sin
U= y
Probabilities
np L
Wavefunction = x Probability amplitude
U= n=1
np y n ( x) B sin x L
2 2 1 2
Probability per unit length (in 1-dimension)
|y|2
0 y
L
x
0 |y|2 n=2
L
x
0 y
L
x
0 |y|2 n=3
L
x
0
L
x
0
L
x
Lecture 11, p 7
Probability and Normalization
We now know that y n ( x ) B1 sin
np x . How can we determine B1? L
|y|
We need another constraint. It is the requirement that total probability equals 1. 2 The probability density at x is |y (x)|2:
n=3 0
Integral under the curve = 1
|B1|2
L
x
Therefore, the total probability is the integral:
In our square well problem, the integral is simpler, because y = 0 for x < 0 and x > L:
Ptot
Ptot B1 B1
2
y x
0 L
2
dx
2
np sin x dx L
2
2 Requiring that Ptot = 1 gives us: B1 L
L 2
Lecture 11, p 8
Probability Density
In the infinite well: P x N 2 sin2
np x . (Units are m-1, in 1D) L
Notation: The constant is typically written as N, and is called the normalization constant. For the square well:
N
2 L
One important difference with the classical result: For a classical particle bouncing back and forth in a well, the probability of finding the particle is equally likely throughout the well. For a quantum particle in a stationary state, the probability distribution is not uniform. There are nodes where the probability is zero!
|y|2
N2
n=3 0
L
x
Lecture 11, p 9
Properties of Bound States
Several trends exhibited by the particle-in-box states are generic to bound state wave functions in any 1D potential (even complicated ones).
1: The overall curvature of the wave function increases with increasing kinetic energy. 2 d 2 y ( x ) p2
2m
dx
2
2m
for a sine wave
y(x)
2: The lowest energy bound state always has finite kinetic energy -- called zero-point energy. Even the lowest energy bound state requires some wave function curvature (kinetic energy) to satisfy boundary conditions.
n=2
n=1
n=3
0
Lx
3: The nth wave function (eigenstate) has (n-1) zero-crossings. Larger n means larger E (and p), which means more wiggles.
4: If the potential U(x) has a center of symmetry (such as the center of the well above), the eigenstates will be, alternately, even and odd functions about that center of symmetry.
Lecture 11, p 10
Act 1
The wave function below describes a quantum particle in a range Dx:
yx 1. In what energy level is the particle? n=
x
(a) 7
(b) 8
(c) 9
Dx
2. What is the approximate shape of the potential U(x) in which this particle is confined?
(a) U(x) E
(b)
U(x)
(c)
E
U(x)
E
Dx
Dx
Dxecture 11, p 11 L
Solution
The wave function below describes a quantum particle in a range Dx:
yx 1. In what energy level is the particle? n=
x
(a) 7
(b) 8
(c) 9
Eight nodes.
Dont count the boundary conditions.
Dx
2. What is the approximate shape of the potential U(x) in which this particle is confined?
(a) U(x) E
(b)
U(x)
(c)
E
U(x)
E
Dx
Dx
Dxecture 11, p 12 L
Solution
The wave function below describes a quantum particle in a range Dx:
yx 1. In what energy level is the particle? n=
x
(a) 7
(b) 8
(c) 9
Eight nodes.
Dont count the boundary conditions.
Dx
2. What is the approximate shape of the potential U(x) in which this particle is confined? Wave function is symmetric.
Wavelength is shorter in the middle.
U(x) U(x)
(a) U(x) E
Not symmetric
(b)
(c)
E
KE smaller in middle
E
Dx
Dx
Dxecture 11, p 13 L
Particle in a Finite Well (1)
What if the walls of our box arent infinitely high? We will consider finite U0, with E < U0, so the particle is still trapped. This situation introduces the very important concept of barrier penetration. As before, solve the SEQ in the three regions. Region II: U = 0, so the solution is the same as before:
U(x) U0
y II ( x ) B1 sin kx B2 cos kx
We do not impose the infinite well boundary conditions, because they are not the same here. We will find that B2 is no longer zero. I
E
II y L III
0
Before we consider boundary conditions, we must first determine the solutions in regions I and III.
Lecture 11, p 14
Particle in a Finite Well (2)
Regions I and III: U(x) = Uo, and E < U0 Because E < U0, these regions are forbidden in classical particles.
d 2 y ( x ) 2m The SEQ 2 (E U )y ( x ) 0 can be written: dx 2
d 2y (x) K 2y ( x ) 0 2 dx
In region II this was a + sign.
U0 > E: K is real.
where:
K
2m
2
U 0 E
U(x)
U0
The general to solution this equation is: Region I: y ( x ) C eKx C eKx
I 1 2
E
I
Region III:
y III ( x ) D1eKx D2eKx
y
0
II
y
L
III
C1, C2, D1, and D2, will be determined by the boundary conditions.
Lecture 11, p 15
Particle in a Finite Well (3)
Important new result! (worth putting on its own slide)
For quantum entities, there is a finite probability amplitude, y, to find the particle inside a classically-forbidden region, i.e., inside a barrier.
y I ( x ) C1e C2e
Kx
Kx
U(x)
U0 E
I
y
0
II
y
L
III
Lecture 11, p 16
ACT 2
In region III, the wave function has the form
U(x)
y III ( x ) D1e D2e
Kx
Kx
U0 E
1. As x , the wave function must vanish. (why?) What does this imply for D1 and D2? a. D1 = 0 b. D2 = 0
I
y
0
II
y
L
III
c. D1 and D2 are both nonzero.
2. What can we say about the coefficients C1 and C2 for the wave function in region I? Kx Kx
y I ( x ) C1e C2e
a. C1 = 0
b. C2 = 0
c. C1 and C2 are both nonzero.
Lecture 11, p 17
Solution
In region III, the wave function has the form
U(x)
y III ( x ) D1e D2e
Kx
Kx
U0 E
1. As x , the wave function must vanish (why?). What does this imply for D1 and D2? a. D1 = 0 b. D2 = 0
I
y
0
II
y
L
III
c. D1 and D2 are both nonzero.
Since eKx as x , D1 must be 0.
2. What can we say about the coefficients C1 and C2 for the wave function in region I? Kx Kx
y I ( x ) C1e C2e
a. C1 = 0
b. C2 = 0
c. C1 and C2 are both nonzero.
Lecture 11, p 18
Solution
In region III, the wave function has the form
U(x)
y III ( x ) D1e D2e
Kx
Kx
U0 E
1. As x , the wave function must vanish (why?). What does this imply for D1 and D2? a. D1 = 0 b. D2 = 0
I
y
0
II
y
L
III
c. D1 and D2 are both nonzero.
Since eKx as x , D1 must be 0.
2. What can we say about the coefficients C1 and C2 for the wave function in region I? Kx Kx
y I ( x ) C1e C2e
a. C1 = 0
b. C2 = 0
c. C1 and C2 are both nonzero.
Kx is negative for x < 0. e-Kx as x - . So, C2 must be 0.
Lecture 11, p 19
Particle in a Finite Well (4)
Summarizing the solutions in the 3 regions: Region I: Region II: Region III:
U(x) U0
y I ( x ) C1e
Kx
y II ( x ) B1 sin(kx ) B2 cos(kx )
I 0
II
y
L
E
III
y III ( x ) D2eKx
As with the infinite square well, to determine parameters (K, k, B1, B2, C1, and D2) we must apply boundary conditions.
Useful to know: In an allowed region, y curves toward 0. In a forbidden region, y curves away from 0.
Lecture 11, p 20
Particle in a Finite Well (5)
U(x)
The boundary conditions are not the same as for the finite well. We no longer require that y = 0 at x = 0 and x = L. Instead, we require that y(x) and dy/dx be continuous across the boundaries:
I
U0
II
y
E
III
0
L
y is continuous
At x = 0:
dy/dx is continuous
y I y II
y II y III
dy I dy II dx dx
dy II dy III dx dx
At x = L:
Unfortunately, this gives us a set of four transcendental equations. They can only be solved numerically (on a computer). We will discuss the qualitative features of the solutions.
Lecture 11, p 21
Particle in a Finite Well (6)
What do the wave functions for a particle in the finite square well potential look like? They look very similar to those for the infinite well, except
n=4 n=2 n=1 n=3
U(x) U0
The particle has a finite probability to leak out of the well !!
0
L
Some general features of finite wells: Due to leakage, the wavelength of yn is longer for the finite well. Therefore En is lower than for the infinite well. K depends on U0 - E. For higher E states, e-Kx decreases more slowly. Therefore, their y penetrates farther into the forbidden region.
A finite well has only a finite number of bound states. If E > U0, the particle is no longer bound.
Very nice Java applet:
http://www.falstad.com/qm1d/
Lecture 11, p 22
ACT 3
1. Which has more bound states?
a. particle in a finite well b. particle in an infinite well c. both have the same number of bound states.
2. For a particle in a finite square well, which of the following will decrease the number of bound states?
a. decrease well depth U0 b. decrease well width L c. decrease m, mass of particle
3. Compare the energy E1,finite of the lowest state of a finite well with the energy E1,infinite of the lowest state of an infinite well of the same width L.
a. E1,finite < E1,infinite b. E1,finite > E1,infinite c. E1,finite = E1,infinite
Solution
1. Which has more bound states?
a. particle in a finite well b. particle in an infinite well c. both have the same number of bound states.
A particle in an infinite well has an infinite number of states.
2. For a particle in a finite square well, which of the following will decrease the number of bound states?
a. decrease well depth U0 b. decrease well width L c. decrease m, mass of particle
3. Compare the energy E1,finite of the lowest state of a finite well with the energy E1,infinite of the lowest state of an infinite well of the same width L.
a. E1,finite < E1,infinite b. E1,finite > E1,infinite c. E1,finite = E1,infinite
Solution
1. Which has more bound states?
a. particle in a finite well b. particle in an infinite well c. both have the same number of bound states.
A particle in an infinite well has an infinite number of states.
2. For a particle in a finite square well, which of the following will decrease the number of bound states?
All three choices are correct: a. decrease well depth U0 a makes fewer energy levels have E < U0. b. decrease well width L c. decrease m, mass of particle b and c raise the energy of each energy level.
NOTE: For a particle in a 1-dimensional potential well, there is always at least one bound state.
Solution
3. Compare the energy E1,finite of the lowest state of a finite well with the energy E1,infinite of the lowest state of an infinite well of the same width L.
a. E1,finite < E1,infinite b. E1,finite > E1,infinite c. E1,finite = E1,infinite
www.falstad.com/qm1d U0 III
Look at the wavefunctions for the two situations: y(x) n=1 U= U0 I y(x)
U=
n=1
0
II
0
L
x
L
x
The wavelength in the finite well is longer, because it is not required to go to zero at x = 0 and x = L (it leaks out a little). Thus, the momentum p = h/ is smaller, and so is the energy. Thats true in general; the less one confines an object, the lower its energy can be - a consequence of the Heisenberg Kruse Demo Uncertainty Principle. (wvfn)
Summary
Particle in a finite square well potential
Solving boundary conditions: Youll do it with a computer in lab. We described it qualitatively here.
Particle can leak into forbidden region. Well discuss this more later (tunneling). Comparison with infinite-well potential: The energy of state n is lower in the finite square well potential of the same width. We can understand this from the uncertainty principle.
Lecture 11, p 27
Next Week
Superposition of states and particle motion
Measurement in quantum physics Schrdingers Cat Time-Energy Uncertainty Principle
Lecture 11, p 28