PHY4604 Fall 2007
Problem Set 5 Solutions
PHY 4604 Problem Set #5 Solutions
Problem 1 (20 points): Use separation of variables in Cartesian coordinates to solve the infinite
cubical well (or particle in a box):
0 0 < x < L,0 < y < L,0 < z < L
V ( x, y ,
PHY4604
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The Infinite Square Well (5)
Case II: Another set of solutions comes from taking A' = 0 and sin(kL/2) =
0, which implies that kL/2 = n- with n- = 1, 2, 3, . Thus,
n ( x ) = 2 B ' sin( 2nx / L )
and
h 2 k 2 2h 2 2 ( n ) 2
2h 2 2
2
En =
PHY4604
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Useful Math
Trigonometric Relations:
sin( A B ) = sin A cos B cos A sin B
cos( A B ) = cos A cos B m sin A sin B
2 cos A cos B = cos( A + B ) + cos( A B )
2 sin A sin B = cos( A + B ) cos( A B )
2 sin A cos B = sin( A + B ) + sin( A B
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Hadrons PseudoScalar Meson Nonet
(JP = 0- bosons, B = 0, Ch = 0, Bo = 0, To = 0)
Y = B + S +Ch +Bo + To
Symbol
Name
+
0
pion
Mass
MeV
140
pion
135
K+
K0
K0bar
K
pion
140
kaon
494
kaon
478
kaon
478
kaon
494
eta
549
-1
+1
0
0
-1
0
eta-pr
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Hermitian Operators
Operators: Operators acting on a function maps it into another function.
The following are examples of operators:
Oop f ( x ) = f ( x ) + x 2
Oop f ( x ) = [ f ( x )]
2
Oop f ( x ) = f (3 x 2 + 1)
df ( x )
2 f ( x)
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Hadrons + Baryon Octet
(JP = + fermions, B = 1, Ch = 0, Bo = 0, To = 0)
Symbol
Name
+
0
p
n
0
Sigma
Mass
MeV
1189
Sigma
1193
Sigma
1189
Proton
Neutron
Cascade
938
940
1315
Cascade
1321
Lambda
1116
Qem/e
Net
Quarks
I
Iz
Y
S
Qcolor
+1
0
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Dirac Bracket Notation (1)
It is very convenient to make the following definitions
< 2 | 1 >
*
2
( x, t ) 1 ( x, t ) dx ,
and
< 2 | O | 1 >
Like (x,t)!
Like (x,t)!
*
2
( x, t )Oop 1 ( x, t ) dx .
Note that |1> is called the Ket and <2|
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The Simple Structure of our Universe
Elementary Particle: Indivisible piece of matter without internal
structure and without detectable size or shape .
.
Mass and chage located
inside sphere of radius zero!
Four Forces:
Gravity (Solar
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The Harmonic Oscillator (1)
One Dimensional Simple Harmonic Oscillator: The simple harmonic
oscillator has a linear restoring force (i.e. Hookes Law spring) and
dV ( x )
Fx =
= Kx and V ( x) = 1 Kx 2
2
dx
where K is the spring constan
PHY4604
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The Harmonic Oscillator (2)
Hamiltonian Operator: The Hamiltonian operator for the simple harmonic
oscillator is given by
1
1
2
H op =
( p x ) op + m 2 ( x 2 ) op since V ( x ) = 1 Kx 2 = 1 m 2 x 2
2
2
2m
2
where K is the spring consta
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The Harmonic Oscillator (3)
Energy Eigenvalue Equation: We are looking for solutions of the equation
Hop|En> = En|En>
where |En> are the eigenkets and En are the allowed energies (i.e.
eigenvalues). The state (a+)op|En> = |a+En> is an
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The Harmonic Oscillator (4)
Properties of the Ground State: We can use
hm
(a+ ) op (a ) op ) and ( x)op = h (a+ )op + (a )op )
2m
2
to calculate (x)(px) for the ground state. We see that
h
< x >=< E 0 | xop | E 0 >=
< E 0 | ( a + ) op
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The Dirac Delta Function
Dirac Delta Function: The Dirac delta function is not really
a function (mathematically it is called a distribution). It
corresponds to an infinitely high, infinitesimally narrow spike
at the point x = a.
+
0 x
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The Free Particle
A stationary state free particle (i.e. V(x) = 0) with energy E must satisfy
h 2 d 2 ( x )
= E ( x ) and hence ( x ) = Ae ikx
2
2m dx
where A is a constant and h 2 k 2 /(2m) = E . But (x) is also an eigenstate of
(px)o
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Momentum-Space Wavefunctions
Change Variable: Since px = hk we can express the position-space wave
function as follows:
+
+
1
1
i ( p x x E ( p x )t ) / h
ip x / h
( x, t ) =
dp x =
( p x )e
( p x , t )e x dp x ,
2h
2h
where E ( p
PHY4604
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The Infinite Square Well (4)
One Dimensional Box
Particle in a One-Dimensional Box: Consider the
solution of
h 2 d 2 ( x )
+ V ( x ) ( x ) = E ( x ) ,
2m dx 2
where
-L/2
( x, t ) = ( x )e iEt / h ,
for the case V(x) = if x -L/2 and V
PHY4604
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The Infinite Square Well (3)
Average Value of px: The average value of px for the nth state is
dn ( x )
< p x > n = n ( x )( p x ) op n ( x ) dx = ih n ( x )
dx
dx
n
2i h n
2i h
=
sin( nx / L ) cos( nx / L ) dx =
sin( ) cos( ) d = 0
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Free-Particle Spinors (p 0)
Free-Particle Solutions: In the Dirac-Pauli representation and p 0 the free
particle equation
rr
r
r
r
H op u ( p ) = c p + mc 2 u ( p ) = Eu ( p ) .
becomes
rr
mc 2 c p u A
u
= E A ,
rr
u
c p mc 2 u
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Expectation Values and Differential Operators (2)
Dynamical Variables become Differential Operators:
2
2
E op = ih
( p x ) op = ih
( p x ) op = h 2 2
x
t
x
Expectation Values of Dynamical Quantities: The average momentum is
given by
(
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The Helicity Operator
The bonus embodied in the Dirac equation is the extra two-fold degeneracy.
For example, the two positive energy solutions
+
r
r
rr
rr
(1)
( 2)
u E > 0 ( p ) = N c p + and u E > 0 ( p ) = N c p
E + mc 2
E + mc
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Schrdingers Equation
The Classical Hamiltonian: Classically the energy is the sum of the kinetic
energy plus the potential energy as follows (in one dimension):
p2
E = x + V ( x) ,
2m
and hence corresponding Quantum Mechanical Hamilton
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Angular Momentum and Spin
Orbital Angular Momentum: The orbital angular momentum operator is
defined as follows:
r
r
r
Lop = rop p op
It is easy to show that the orbital angular momentum operator does not
r
commute with the Dirac Hamil
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Probability Flux
Probability Flux: Look at the time dependence of the
probability that the particle lies in the region x1 x x2
j(x1,t)
j(x2,t)
P(x1,x2,t)
x2
P ( x1 , x 2 , t ) = * ( x, t ) ( x, t ) dx .
x1
x2
x1
We see that
x
dP ( x1 ,
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Total Angular Momentum J
We see that
r
r
rr
rr
[ H op , Lop ] = ihc ( p op ) and [ H op , S op ] = + ihc ( p op )
r
L or the spin angular
and hence neither the orbital angular momentum
r
S is conserved.
momentum
Total Angular Momentum:
PHY4604
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Classical Mechanics vs Quantum Mechanics
Classical Mechanics: The goal of classical mechanics is to determine the
position of a particle at any given time, x(t). Once we know x(t) then we
can compute the velocity vx = dx/dt, the moment
PHY4604
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Relativistic Energy and Momentum (Summary)
Relativistic Energy:
The total relativistic energy is the sum of the kinetic energy (energy of
motion) plus the rest mass energy (RME = m0c2).
2
0
Also, the relativistic energy is equal to the
Quantum Mechanics
Solution Set 1
Due: 31 August 2016
Reading: Shankar Sections 1.1-1.4, and 4.1; read and study (i.e. learn) appendix sections
A.1-A.3; Lecture notes, Chapter 1 and Sections 2.1-2.4.
In this first solution set I give each problem a typical
Quantum Mechanics
Solution Set 5
Due: 28 September 2016
Reading: Shankar, Chapter 4 and Sections 2.1-2.6; Lecture notes, Sections 3.1-3.6.
21. a) S, Exercise 4.2.2; b) S, Exercise 4.2.3.
Solution:
a) The momentum space wave function
hp|i =
Z
dx ipx/~
e
(x
Quantum Mechanics
Solution Set 4
Due: 21 September 2016
Reading: Shankar Sections 1.9-1.10, and Chapter 9; Lecture notes, Sections 2.11-2.12.
16. The matrix
1 1 1
=
2 1 1
is Hermitian and hence is an observable. Using an eigenbasis of to define any functi
Quantum Mechanics
Problem Set 1
Due: 31 August 2016
This and all future homework will be posted on the course webpage:
http:/www.phys.ufl.edu/thorn/homepage/qminfo.html
I will refer to problems in Shankar by prefixing the problem number with S.
Reading: S
Quantum Mechanics
Solution Set 3
Due: 14 September 2016
Reading: Shankar Sections 1.8-1.9; Lecture notes, Sections 2.7-2.10. Try to do as many of
the exercises in the text as you can. However, turn in only the ones that I assign!
11. (a) S, Exercise 1.8.2