Quantum Mechanics
Model Solutions for Sample Exam for Final Examination
1. (a) Expectation value. In terms of ladder operators,
1
J x = 2 (J + + J ).
Thus
1
J x = 2 ( j, m|J + |j, m + j, m|J |j, m ) = 0.
(b) Uncertainty. We have
2
(Jx )2 = J x J x
2
2
=
Notes on
The Physics of Quantum Mechanics
Daniel F. Styer
Professor of Physics, Oberlin College
[email protected]
Copyright c 15 August 2011
Everything in this book is a sketch it is not polished or complete.
Contents
Preface
7
1 What is Quantum Mecha
Matrix Calisthenics
The trace
If S = AB and T = BA, then
N
sij =
N
aik bkj
tij =
k=1
whence
bik akj
k=1
N
tr(S) =
N
aik bki
tr(T ) =
i,k=1
bik aki
i,k=1
so
tr(S) = tr(T ).
Now dene X = ABC, so tr(ABCD) = tr(XD) = tr(DX) = tr(DABC), etc.
Finally, try
A=
1
Degenerate Pertubation Theory in a Two-State System
a. Using the initial basis (call it cfw_|1init , |2init ):
1init |H |1init
=
1
0
a3
a1
a1
a3
1
0
= a3
2init |H |2init
=
0
1
a3
a1
a1
a3
0
1
= a3
Using the second basis (call it cfw_|1second , |2second ):
Quantum Mechanics
Model Solutions for Sample Exam for Second Examination
1. The state evolves in time to
h
h
h
|(t) = 2 e(i/ )E2 t |2 + 6 e(i/ )E6 t |6 + 8 e(i/ )E8 t |8 ,
so it remains always a linear combination of even energy eigenstates. Now the momen
Anharmonic Oscillator
a. Using the results from the problem Ladder Operators for the Simple Harmonic Oscillator,
m|3 |n
x
=
=
m|
x
h
2m
|2 |n
x
3/2
m,
=
h
2m
3/2
+
+ 1 m,
n(n 1)
,n2
+1
+ (2n + 1)
,n
+
(n + 1)(n + 2)
,n+2
m,
+
h
2m
1
3/2
1
n(n 1)
,n2
+
Angular Momentum
Angular momentum trivia
a. Assume that A commutes with Lx and Ly :
[A, Lx ] = 0,
[A, Ly ] = 0.
(1)
Then it follows from
[Lx , Ly ] = i Lz ,
h
that
[A, Lz ]
1
[A, [Lx , Ly ]
[Use the Jacobi idenity. . . ]
i
h
1
=
[Ly , [A, Lx
Square Well with a Bump
V (x) =
0
2
p
H0 =
+ V ()
x
2M
H = V ()
x
where
0
V (x) =
V
0
where
L < x
0 < x < L
x < 0
(L + a)/2
(L a)/2
< x
< x <
x <
(L + a)/2
(L a)/2
The rst-order energy corrections are
(1)
En
=
n |H |n
(L+a)/2
=
=
=
=
=
=
=
n (x) V n (x
Ladder Operators for the Simple Harmonic Oscillator
a. Simple algeba shows that
h
( + a )
a
2m
m
h
( a )
a
p = i
2
x
=
b. Matrix elements.
m|n
a
=
=
m| |n
a
n m,n1
n + 1 m,n+1
h
=
n m,n1 + n + 1 m,n+1
2m
m
h
n m,n1 n + 1 m,n+1
= i
2
m|n
x
m|n
p
[Ques
Expressions for SHO Ladder Operators
The lowering operator a acts upon energy eigenstate |n as
a|n =
n |n 1 .
Since we know how a acts upon every element of a basis, we know how it acts upon any state.
The outer product expression
m |m 1 m|
m=0
similarly
Quantum Mechanics
Sample Exam for Second Examination
1 A one-dimensional simple harmonic oscillator has initial state
| = 2 |2 + 6 |6 + 8 |8 .
What is the expected momentum p , and how does it change with time?
2 Evaluate mcfw_ xp for any state | . ( mcf
Stern-Gerlach Analyzers
Exit probabilities
measure projection onto this axis
orientation of definite magnetic
moment for atoms with z = B
The second analyzer is rotated by angle = relative to the orientation axis of the atoms with
z = B . Thus the probabi
Quantum Mechanics
Model Solutions for Sample Exam for First Examination
1. State representations
The representation of |z in the basis cfw_|+ , | is
+ |z
|z
sin /2
cos /2
=
.
2. Photon polarization
Using the amplitudes determined in the third problem o
Corrections to the 7th printing, September 14, 2009
Introduction to Quantum Mechanics, 2nd ed.
by David Griths
page vii, last line of second paragraph: change following to after nishing.
page 8, line 4: change x to j (twice).
page 23, Problem 1.18(b):
Ground State of the Simple Harmonic Oscillator
From the discussion in the problem statement,
P.E. =
m 2 2
d ,
8
and the total energy is
E(d) =
Energy
K.E. =
h
2 1
,
32 m d2
m 2 2
h
2 1
.
d +
8
32 m d2
total energy
P.E.
K.E.
0
0
d
The minimum falls at d0 w
Quantum Mechanics
Sample Exam for Final Examination
1 A one-particle system is in angular momentum eigenstate |j, m so that
2
J |j, m
J z |j, m
=
2 j(j + 1)|j, m
h
=
m|j, m .
h
What is the expectation value and uncertainty of J x in this state?
2 For a
Quantum Mechanics
Sample Exam for First Examination
1 State representations
Write the column matrix that represents the state |z in the basis cfw_|+ , | , as a function of the
angle .
2 Photon polarization
A photon linearly polarized at an angle to the ve
Quantum Mechanics 2011
Model Solutions for Second Exam
1. Commutator
Apply
[A, B] = AB B A = cB A
to the ket |a which has A|a = a|a . We nd
AB|a B A|a = cB A|a
AB|a B(a|a ) = cB(a|a )
A(B|a ) a(B|a ) = ca(B|a )
A(B|a ) = a(B|a ) + ca(B|a )
A(B|a )