Physics
2203,
Fall
2012
Modern
Physics
.
  Wednesday,
Oct.
10th
,
2012.
‐‐‐‐Finish
examples
from
Ch.
8
‐‐‐‐Chapter
9
with
examples
‐‐‐‐Take
home
average
85,
sum
average
63
.
  Announcements:
‐‐‐‐Quiz
on
Friday
on
Ch.
8
or
Ch.
9
‐‐‐‐Midterm
grades
posted
early
next
week‐‐conservaNve
‐‐‐Submit
a
detailed
outline
of
your
paper
by
Monday,
Oct.
29th
hVp://www.nobelprize.org/mediaplayer/
index.php?id=1829
David
Wineland
(US),
Serge
Horoche
(France)
hVps://lasers.llnl.gov/mulNmedia/virtual_tours/
NIF:
The
“Crown
Joule”
of
Laser
Science
This
could
be
worse
than
the
death
of
the
“Supercollider”
for
American
Science
The allowed magnitudes of the angular momentum are L= l (l +1) in QM.
They are l in the Bohr model.
(a) Find the ratio of correct/Bohr for l=1, 2, 3, 4, 10, and 100
(b) How large does l have to be to have only a 1% error?
l ( l + 1)
correct
The ratio of
=
Bohr
l
l ( l + 1)
correct
(b)
=
= 1.01
Bohr
l
l ≈ 50
l
Ratio
1
2
1.41
1.22
3
1.15
4
10
100
1.12
1.05
1.005
What
conclusion
do
you
draw?
Hydrogen
Atom
The
picture
we
have
developed
for
a
H
like
atom
fails
when
you
apply
a
magneNc
field.
What
was
a
single
emission
line
in
zero
field
become
a
mulNplet
when
a
field
is
applied
First
an
electron
moving
in
a
circular
orbit
has
a
magneNc
moment.
You
can
think
about
this
in
analogy
with
the
total
angular
momentum
of
the
earth
around
the
sun.
Second,
the
electron
has
an
inherent
magneNc
moment
called
spin
 + S
J = L
L is the angular momentum around the sun rxp
S is orbital angular momentum Iω
An
electron
moving
in
a
circular
orbit
has
a
magneNc
moment.
If
there
is
a
magneNc
field
B
there
will
be
a
torque.
B
If the current i flowing around a small plane loop of area A in
a magnetic field B, there will be a torque τ
 
τ = iAxB
A is a vector with magnitude of the area, direction according to right hand rule with i
 
But usual notation is µ = iA
µ is magnetic moment of the loop.
Work is done by torque moving through angle θ
W = − ∫ τ dθ = −µ B ∫ sin θdθ = µ B cosθ + constant
The potential U is - work
We can always define a zero for the potenital
•B

U = − µ B cosθ = − µ
Lets calculate the classical µ
The current is the charge divided by the period.
e
2π r
i = : but v=
T
T
v
µ = iA = e
πr2
2π r
µ = iA =
µ
e
=
Gyromagnetic Ratio
L 2m
ev
evr
πr2 =
2π r
2
But L=mvr

e 
µ=−
L
2m
Electrons
charged
negaNve
e 

µ=−
L
2m
µB =
MagneNc
moment
of
an
orbiNng
charge
e
= 5.79x10 −5 eV / T
2me
Bohr
Magneton
We
know
that
angular
momentum
will
be
quanNzed‐‐figure
µz = −
e
e
Lz = −
ml = µ B ml
2me
2me
  
τ = µxB
Classical
this
creates
a
torque
that
would
align
the
moment
with
the
B
field.
With
quanNzed
angular
moment,
we
get
precession
around
the
field
direcNon:
Larmor
precession

Look
at
the
boVom
of
the
figure:
L sin θ idφ =| dL |
 
e
Giving:
L sin θ idφ =| dL |=| τ | dt =
LBsin θ dt
2me
ωL =
dφ
1 | dL |
e
=
=
B
dt L sin θ dt
2me
Larmor
frequency
 
Work done dW=-µBsinθ dθ =d(µBcosθ )=d(µ iB)
 
U = −µ iB → look back.

e 
µ=−
L
2m
  
τ = µxB

e 
µ=−
L
2m
e
ωL =
B
2me
 
e   eB
U = −µ iB =
LiB =
Lz = ω L ml
2me
2me
The
Total
Energy
of
the
electron
is
energy
without
the
field
+
U
 
E = E0 +U = −µ iB = E0 + ω L ml
What
does
this
look
like
for
n=1
system?
Example
for
l=2

B = B0 k
ml=0
Δml = 0, ± 1
Δl = ±1
Calculate
the
magneNc
energy
and
Larmor
frequency
for
an
electron
in
the
n=2
state
of
hydrogen,
assuming
the
atom
is
in
a
magneNc
field
of
strength
B=1.00T.
Take
the
z
axis
to
be
along
B
ω L =
U = ω L ml
e
B=µb = (5.79x10 −5 eV / T )(1T )
2me
ω L = 5.79x10 −5 eV
n = 2, l=0, m l = 0 no U
n = 2, l=1, m l = ±1 no U=±ω L
allowed Energies, -ω L , 0,+ω L
ω L

ω L = 8.80x10+10 rad / s
Larmor frequency=
Red
allowed
Black
Forbidden
Pieter
Zeeman
1865‐1943
Dutch
A
magneNc
field
splits
emission
lines
in
the
Rydberg
series.
Proves
electrons
are
moving
around
the
nucleus.
Shared
1902
Nobel
Prize
with
Lorentz
To
make
the
arguments
simple
we
need
a
system
where
we
can
focus
on
the
orbital
moNon
of
the
electron
without
having
to
worry
about
the
spin.
To accomplish this we need a system with two electrons where the spins are opposite.
This is called a singlet state.
One option would be two electron in the 1s state of He, but that will not work
because we need m l ≠ 0. What will work is having electrons in two different orbits of He
but still as a singlet.
Pieter
Zeeman
1865‐1943
Dutch
It is easy to calculate the energy ΔE
•B
= e L
•B

ΔE=- µ
 2m 
e
Pick the z axis to be aligned with B
L z = m
Shared
1902
Nobel
Prize
with
Lorentz
 e 
 e 
ΔE= 
L
B
=
ml B
z



 2me 
 2me 
µB =
e
= 9.27x10 −24 A • m 2
2me
µB =
e
= 5.79x10 −5 eV / T
2me
e
2me
BohrMagnetron
µB =
Pieter
Zeeman
1865‐1943
Dutch
e
2me
BohrMagnetron
µB =
µB =
e
= 5.79x10 −5 eV / T
2me
ΔE=mµ B B
Consider
a
He
atom
with
two
electrons,
one
is
in
the
1s
state
and
the
excited
one
is
in
the
2p
state,
singlet
state.
Shared
1902
Nobel
Prize
with
Lorentz
Lets
see
what
happens
in
a
5
T
magneNc
field.
The magnetic field spits the one line (B=0) into
three for m l = −1, 0, +1
Each line separated by ΔE=µ B B = 3.0x10 −4 eV
ΔE
= 1x10 −5 eV
E
Pieter
Zeeman
1865‐1943
Dutch
e
2me
BohrMagnetron
µB =
µB =
e
= 5.79x10 −5 eV / T
2me
ΔE=mµ B B
Consider
a
He
atom
with
two
electrons,
one
is
in
the
1s
state
and
the
excited
one
is
in
the
d
state.
Nothing
happens
to
the
1s.
Shared
1902
Nobel
Prize
with
Lorentz
Now
what
happens?
How
many
lines
will
you
see?
Good
Quiz
quesNon!
Imagine
a
hydrogen
atom
in
which
the
electron
has
no
spin
(only
orbital
moment
counts).
The
atom
is
placed
in
a
magneNc
field
of
1.5T
along
the
z
axis
(a)  Describe
the
effect
of
the
B
field
on
the
1s
and
2p
states
of
the
hypotheNcal
H
atom.
Sketch
the
energy
levels.
(b)  When
B=0,
there
is
a
single
spectral
line
corresponding
to
the
2p
to
1s
transiNon.
How
many
lines
does
this
become
when
B≠0?
(c)  What
is
the
fracNonal
separaNon
between
the
adjacent
lines.
ΔE=µ B B
ΔE=µ B B = 0.9x10 −4 eV

E(2 p → 1s) = 13.6  1 −

1
 = 10.2eV
4
ΔE 0.9x10 −4
=
0.88x10 −5
E
10.2
Fact,
The
“Normal
Zeeman”
effect
is
frequently
not
observed.
Instead
you
see
four,
six
or
even
more
unequally
spaced
lines
in
the
emission
spectra.
This
is
known
as
the
“Anomalous
Zeeman
Effect.”
There
is
also
another
effect
which
doubles
the
number
of
lines
call
Fine
structure.
Everything
is
a
result
of
the
electron
spinning
around
its
axis
and
having
a
magneNc
moment.
Stern‐Gerlach
experiment
Sz = ms  m s = ±1 / 2
If
we
create
a
beam
of
atoms
such
that
we
are
in
the
n=2,
l=1
state
there
are
2l+1=3
different
values
of
ml.
If
this
beam
is
directed
into
a
region
with
a
constant
magneNc
field
all
that
happens
is
the
vector
precesses
around
the
field.
But
in
an
a
nonuniform
magneNc
field
there
is
a
force
and
we
should
see
three
images
of
the
slit
as
shown.
–2(l+1)
Stern
and
Gerlach
carried
out
this
experiment
on
Ag
atoms
where
l=0
so
they
should
have
seen
no
splipng
but
they
saw
two
lines,
caused
by
the
intrinsic
spin
of
the
electron
OVo
Stern
Nobel
Prize
1943
Spin
Quantum
Number
s
for
the
electron
is
½!
Stern‐Gerlach
magnet
experiment
split
the
beam
into
to
lines
=(2s+1);
Spin
up—spin
down.
According
to
quanNzaNon
rules
for
angular
momentum
–applied
to
spin.
1
Sz = m s  m s = ±
2
By
definiNon
the
magnitude
of
the
spin
angular
momentum
is

S = s(s +1) =
3

2
The
angular
momentum
of
rotaNon
cannot
be
changed
in
any
way,
but
is
an
intrinsic
property
of
the
electron
Spin,
as
angular
momentum
and
torque,
is
known
as
a
pseudovector
(or
an
axial
vector),
Opposed
to
a
true
or
polar
vector
such
as
velocity.
The
property
of
a
pseudovector
is
that
its
mirror
image
is
equal
in
magnitude
but
flipped
in
direcNon
because
a
pseudovector
has
a
chirality.
An
ordered
array
of
spin
in
a
solid
breaks
the
symmetry
of
the
crystal.
How
can
you
see
this?
PHYSICAL
REVIEW
B
86,
060512(R)
(2012)
An
LSU
team,
first
authored
by
graduate
student
Guorong
Li,
has
found
a
way
to
see
such
a
broken
symmetry.
They
reported
in
late
August
(Phys.
Rev.
B
86,
060512(R)
(2012))
that,
in
one
of
the
parent
compounds
of
new
Fe‐based
superconductors
BaFe2As2,
the
surface
dramaNcally
enhances
the
spin‐lapce
coupling
so
that
dynamic
anNphase
spin
domains
in
the
bulk
are
frozen
out
at
the
surface.
This
leads
to
the
coexistence
of
structural
and
spin
anNphase
domain
walls.
Due
to
the
large
spin‐electron‐lapce
coupling
at
the
surface,
these
domain
walls
have
the
broken
mirror‐plane
symmetry
associated
with
spin,
resulNng
in
a
leu‐
and
right‐handed
chirality,
which
can
be
described
as
a
spin
toroidal
moment
chirality
(Fig.
1b).
While
layer
ruthenate
materials
are
well
known
to
display
an
array
of
exciNng
phenomena
such
as
metal‐insulator
transiNons
(MIT),
spin‐orbital
ordering,
exoNc
superconducNvity,
and
quantum
criNcality,
this
neutron
scaVering
work
has
revealed
an
unusual
E‐type
anNferromagneNc
(AFM)
structure
in
Mn‐subsNtuted
Sr3Ru2O7
(x
=
0.16)
[See
Fig.].
They
found
that
this
layered
ruthenate
behaves
as
a
quasi‐two‐dimensional
(2D)
anNferromagnet
with
in‐plane
(ab)
long‐range
ordering
but
only
single
bilayer
(5‐6Å)
ferromagneNc
correlaNons
along
the
c
direcNon
bellow
TN
=
78
K.
Such
Mn‐induced
magneNc
structure
is
unusual
because
the
criNcal
behavior
of
the
staggered
magneNzaNon
(i.e.
the
AFM
order
parameter)
does
not
reflect
the
expected
behavior
of
a
2D
magneNc
phase
transiNon.
‐PRB
85,
180410(R)
(2012)].
We
will
do
many
examples
of
this
on
Friday.
We now have four quantum numbers What is the degeneracy
n = 1, 2, 3,....
for n=1?
l = 0,1, 2, 3,....,(n −1)
n =1
ml = l,l −1,., 0,...,−l
l=0
1
ml = 0
ms = ±
2
1
ms = ±
What is the degeneracy?
2
n = 2,
l = 0 and l=1
ml = 0 and 1,0,-1
1
ms = ±
2
Degeneracy is 2n 2
Make
a
table
showing
the
values
of
the
four
quantum
numbers,
n,
l,
m,
and
ms
for
10
lowest
lying
states
of
the
H
atom.
n=1, we have l=0, m=0, m s =±
1
2
1
n=2, l=1, m=1, m s =±
2
1
n=2, l=1, m=0, m s =±
2
1
n=2, l=1, m=-1, m s =±
2
n=2, l=0, m=0, m s =±
1
2
Compute
the
raNon
S/L
of
the
magnitudes
of
the
spin
angular
momentum
to
the
orbital
angular
momentum
for
(a)
an
s
electron,
(b)
a
p
electron,
(c)
a
d
electron,
and
(d)
an
f
electron.
s ( s + 1)
S = s ( s + 1) : L = l ( l + 1) : S/L=
l ( l + 1)
l
S/L
0
∞
1
0.61
When
will
this
raNo
be
10%?
2
0.35
3
0.25