Physics 214 Lecture 11

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It was almost as incredible as if you fired a
15-inch shell at a piece of tissue paper, and it
came back to hit you! --E.
Rutherford (on the discovery of the
nucleus)
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Special (Optional) Lecture

• Quantum Information
• One of the most modern applications of QM
• quantum computing
• quantum communication cryptography,
  teleportation
• quantum metrology
• Prof. Kwiat will give a special 214-level lecture
  on this topic
• Sunday, March 1
• 3 pm, 151 Loomis
• Attendance is optional, but encouraged.

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Overview of the rest of the course Up to now
general properties and equations of quantum
mechanics Time-dependent and time-independent
Schr. Eqs. Eigenstates, superposition of
eignestates and time- dependence, tunneling,
Schrodingers cat, . . . This week quantum
states in 3 dimensions, electron spin, H atom,
exclusion principle, periodic table of atoms Next
week molecules and solids, consequences of Q.
M. Metals, insulators, semiconductors,
superconductors, lasers, . .
Final Exam Monday, March 9!
HW 6 due next Saturday (Mar. 7), 8 am
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Particles in 3D Potentials and the Hydrogen Atom
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Overview

• 3-Dimensional Potential Well
• Product Wavefunctions
• Concept of degeneracy
• Early (Incorrect!) Models of the Hydrogen Atom
• Planetary Model
• Schrödingers Equation for the Hydrogen Atom
• Semi-quantitative picture from uncertainty
  principle
• Ground state solution
• Spherically-symmetric excited states
  (s-states)
• contain details beyond what we expect you to
  learn here

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Quantum Particles in 3D Potentials
A real (3D) quantum dot

• So far, we have considered quantum particles
  bound in one-dimensional potentials. This
  situation can be applicable to certain physical
  systems but it lacks some of the features of many
  real 3D quantum systems, such as atoms and
  artificial quantum structures

• One consequence of confining a quantum particle
  in two or three dimensions is degeneracy -- the
  occurrence of several quantum states at the same
  energy level.

• To illustrate this important point in a simple
  system, we extend our favorite potential -- the
  infinite square well -- to three dimensions.

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Particle in a 3D Box (1)

• The extension of the Schrödinger Equation (SEQ)
  to 3D is straightforward in cartesian (x,y,z)
  coordinates

Kinetic energy term in the Schrödinger Equation

• Lets solve this SEQ for the particle in a 3D box

This simple U(x,y,z) can be separated A
special feature
U(x,y,z) U(x) U(y) U(z)
y
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Particle in a 3D Box (2)
and the wavefunction can be
separated into the product of three functions
Note that the functions f canbe different for x,
y, and z

• So, the whole problem simplifies into three
  one-dimensional equations that weve already
  solved in Lecture 7.

graphic
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Particle in a 3D Box (3)

• So, finally, the eigenstates and associated
  energies for a particle in a 3D box are

where nx,ny, and nz can each have values 1,2,3,.
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Lecture 11, Act 1
Consider a particle in a two-dimensional
(infinite) well, with Lx Ly. 1. Compare the
energies of the (2,2), (1,3), and (3,1) states?
a. E(2,2) gt E(1,3) E(3,1) b. E(2,2) E(1,3)
E(3,1) c. E(1,3) E(3,1) gt E(2,2) 2. If we
squeeze the box in the x-direction (i.e., Lx lt
Ly) compare E(1,3) with E(3,1) a. E(1,3) lt
E(3,1) b. E(1,3) E(3,1) c. E(1,3) gt E(3,1)
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Lecture 11, Act 1 - Solution
Consider a particle in a two-dimensional
(infinite) well, with Lx Ly. 1. Compare the
energies of the (2,2), (1,3), and (3,1) states?
a. E(2,2) gt E(1,3) E(3,1) b. E(2,2) E(1,3)
E(3,1) c. E(1,3) E(3,1) gt E(2,2) 2. If we
squeeze the box in the x-direction (i.e., Lx lt
Ly) compare E(1,3) with E(3,1) a. E(1,3) lt
E(3,1) b. E(1,3) E(3,1) c. E(1,3) gt E(3,1)
E(1,3) E(1,3) E0 (12 32) 10 E0
E(2,2) E0 (22 22) 8 E0
The tighter confinement along x will increase the
contribution to E. The effect will be greatest on
states with greatest nx
Example Lx Ly/2
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Energy levels (1)

• Now back to a 3D cubic box

Show energies and label (nx,ny,nz) for the first
11 states of the particle in the 3D box, and
write the degeneracy D for each allowed energy.
Use Eo h2/8mL2.
E
D3
D1
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Energy levels (1)

• Now back to a 3D cubic box

Show energies and label (nx,ny,nz) for the first
11 states of the particle in the 3D box, and
write the degeneracy D for each allowed energy.
Use Eo h2/8mL2.
E
12Eo
11Eo
9Eo
D3
D1
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Lecture 11, Act 2
For a symmetric cube infinite box, we just saw
that the 5th energy state has an energy of 12 E0
and a degeneracy of 1, with quantum numbers
(2,2,2). 1. What is the energy of the next
energy level? a. 13E0 b. 14E0 c. 15E0 2.
What is the degeneracy of this energy level?
a. 2 b. 4 c. 6
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Lecture 11, Act 2 - Solution
For a symmetric cube infinite box, we just saw
that the 5th energy state has an energy of 12 E0
and a degeneracy of 1, with quantum numbers
(2,2,2). 1. What is the energy of the next
energy level? a. 13E0 b. 14E0 c. 15E0 2.
What is the degeneracy of this energy level?
a. 2 b. 4 c. 6
E(1,2,3) E0 (12 22 32) 14 E0
Any ordering of the quantum numbers 1, 2, and 3
will give the same total energy. There are 6
possible ways to order them (1,2,3), (1,3,2),
(2,1,3), (3,1,2), (2,3,1), (3,2,1). Note For
this system, 6 is the maximum possible
degeneracy.
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Energy levels (2)

• Now consider a non-cubic box

Assume that the box is stretched only along the
y-direction. What do you think will happen to
the cubes energy levels below?
E
(nx,ny,nz)
11Eo
9Eo
6Eo
3Eo
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Energy levels (2)

• Now consider a non-cubic box

Assume that the box is stretched only along the
y-direction. What do you think will happen to
the cubes energy levels below?
E
Figure out next level.
(nx,ny,nz)
(1) The symmetry of U is broken for y, so the
three-fold degeneracy is lowereda two-fold
degeneracy remains due to 2 remaining equivalent
directions, x and z.
11Eo
9Eo
(2,1,1) (1,1,2)
6Eo
D2
(1,2,1)
D1
(2) There is an overall lowering of energies due
to decreased confinement along y.
3Eo
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Another 3D System The Atom-electrons confined
in Coulomb field of a nucleus
Early hints of the quantum nature of atoms

• Discrete Emission and Absorption spectra
• When excited in an electrical discharge, atoms
  emitted radiation only at discrete wavelengths
• Different emission spectra for different atoms

Atomic hydrogen
l (nm)

• Geiger-Marsden (Rutherford) Experiment (1911)
• Measured angular dependence of a particles
  (He ions) scattered from gold foil.
• Mostly scattering at small angles ? supported
  the plum pudding model. But
• Occasional scatterings at large angles ?Something
  massive in there!

• Conclusion Most of atomic mass is concentrated
  in a small region of the atom

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Rutherford Experiment
   
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Atoms Classical Planetary Model
(An early model of the atom)

• Classical picture negatively charged objects
  (electrons) orbit positively charged nucleus due
  to Coulomb force.
• There is a BIG PROBLEM with this
• As the electron moves in its circular orbit, it
  is ACCELERATING.
• As you learned in Physics 212, accelerating
  charges radiate electromagnetic energy.
• Consequently, an electron would continuously lose
  energy and spiral into the nucleus in about 10-9
  sec.

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Potential Energy for the Hydrogen Atom

• How do we describe the hydrogen atom quantum
  mechanically?
• We need to specify U, the potential energy of the
  electron
• We assume that the Coulomb force between the
  electron and the nucleus is the force responsible
  for binding the electron in the atom

This spherically symmetric problem can be solved
in spherical coordinates.
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But first Hydrogen atom qualitatively

• Why doesnt the quantum electron collapse into
  the nucleus, where its potential energy is
  lowest?
• The more confined it gets, the bigger p spread it
  has, from Heisenberg Uncertainty. More p2/2m
  means more KE.
• So theres a tradeoff between lowering PE and
  raising KE.
• About what spread a0 minimizes KEPE?
• Roughly,
• Take derivative,find a0 to minimize E

Doing the same thing carefully ends up with the
same result! So the general energy and size
scale of an atom are fixed by the Heisenberg
Uncertainty relation.
virial theorem Exact, generally true for atoms
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Lecture 11, Act 3
Consider an electron around a nucleus that has
two protons (and two neutrons, like an
ionized Helium atom). 1. Compare the
effective Bohr radius a0,He with the usual Bohr
radius for hydrogen, a0 a. a0,He gt a0 b.
a0,He a0 c. a0,He lt a0 2. What is the ratio
of ground state energies E0,He/E0,H? a.
E0,He/E0,H 1 b. E0,He/E0,H 2 c. E0,He/E0,H
4
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Lecture 11, Act 3 - Solution
Consider an electron around a nucleus that has
two protons (and two neutrons, like an
ionized Helium atom). 1. Compare the
effective Bohr radius a0,He with the usual Bohr
radius for hydrogen, a0 a. a0,He gt a0 b.
a0,He a0 c. a0,He lt a0 2. What is the ratio
of ground state energies E0,He/E0,H? a.
E0,He/E0,H 1 b. E0,He/E0,H 2 c. E0,He/E0,H
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Look at how a0 depends on the charge
This makes sense more charge ? stronger
attraction ? electron sits closer to the nucleus
Clearly the electron will be more tightly bound,
so E0,He gt E0,H . How much more tightly?
Look at E0
Depends on which state
In general, for a hydrogenic atom with Z
protons
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Product Wavefunction in Spherical Coordinates

• The potential is spherically symmetric, i.e., U(r
  ) depends only on the radius. Therefore the
  problem is separable and the solution to the SEQ
  in spherical coordinates is a product wave
  function of the form

Here Rnl is the radial part, and Ylm are
spherical harmonics. The Schrödinger equation
in spherical coordinates is complicated. An
appreciation of the problem can be gained by
considering only spherically symmetric states
(wavefunctions with no angular dependence). For
these s-states, l 0 and m 0, and the
radial SEQ takes the form
KE term PE term
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Radial Eigenstates of H

• The radial eigenstates, Rno(r) , for the electron
  in the Coulomb potential of the proton are
  plotted below

• s-state wavefunctions
• The zeros in the subscripts below are a reminder
  that these are states with zero angular momentum.

Plug these into radial SEQ (Appendix) ?
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Probability Density of Electrons
Probability density Probability per unit volume
?y?2 ? Rn02 for s-states. The density of dots
plotted below is proportional to Rn02.
1s state
2s state
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Radial Probability Densities for s-states

• Summary of wave functions and radial probability
  densities for some s-states

These P(r) include both y2(r) and a factor of
r2 because bigger r shells have more volume.
Question If the classical planetary model were
correct, then what would you expect for P(r)?
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Optical Transitions in H -- Example

• An electron, initially excited to the n 3
  energy level of the hydrogen atom, falls to the n
  2 level, emitting a photon in the process.
  What are the energy and wavelength of the photon
  emitted?

Discharge tubes
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Optical Transitions in H --Example

• An electron, initially excited to the n 3
  energy level of the hydrogen atom, falls to the
  n 2 level, emitting a photon in the process.
  What are the energy and wavelength of the photon
  emitted?

Therefore
You will experimentally measure several
transitions in Lab.
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Next week Laboratory 4
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Example Problems (1)
a) What is the energy of the second excited state
of a 3-D cubic well? b) How many states have
this energy?
Solution
Starting with E211 (the first excited state), we
must increase one of the quantum numbers. Which
choice adds the least energy? E221 or
E311? There are three distinct ways to arrange
the three numbers, 2, 2, and 1. The different
arrangements correspond to different x, y, and z
components of the momentum.
a) E311 (321212) Eo 11 Eo E221
(222212) Eo 9 Eo with Eo h2/8mL2 b)
E221 E212 E122
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Example Problems (2)
Consider the three lowest energy states of the
hydrogen atom. What wavelengths of light will be
emitted when the electron jumps from one state to
another?
Solution
E -13.6 eV/n2, so E1 -13.6 eV, E2 -3.4 eV,
and E3 -1.5 eV. There are three jumps to
consider, 2-to-1, 3-to-1, and 3-to-2. The photon
carries away the energy that the electron
loses. hc 1240 eVnm Two wavelengths are all
in the ultraviolet. Note that the 3-to-2
transition gives a visible (red) photon, l32
653 nm.
DE21 10.2 eV DE31 12.1 eV DE32 1.9 eV l
h/p hc/E l21 122 nm l31 102 nm l32 653
nm
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Example Problems (3)
What happens to the ground state radius for the
first electron bound to a helium nucleus (Z 2),
i.e., for a helium ion? What is the ratio of its
typical radius to that of the H atom?
Solution
The potential energy for a single electron and a
nucleus with Z protons involves the product of
those charges, i.e. -e ? Ze. The Bohr radius,
a0, also has the factor e2 for hydrogen it
becomes Ze2 for a nucleus with a single electron.
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Appendix Solving the SEQ for H --deriving ao and
E

• Substituting into
  , we get

• For this equation to hold for all r, we must have

AND

• Evaluating the ground state energy