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5531' A
FOR EACH OF THE FOLLOWING QUEsTIONs IN PART A, ENTER THE MOST APPROPRIATE RESPONSE ON
THE OMR SHEET.
A1.
A
A3.
A4.
In gravityfree space a rock, which is attached to the end of a string. is swung in a circular
PHYS 455
Short Answers to Homework I
1.
(a) Eigenvalues of A are 1 = 2 = 1 with degeneracy 2 and 3 = 1 with degeneracy
1.
(b) Two eigenvalues and therefore two projectors to these subspaces. For the case
where A has only eigenvalues 1, these projectors ca
PHYS 455  Introduction to Quantum Information Theory
Answers to the First Midterm
1. In this problem, we will look at the spectral problem of any component of the spin of
a spin1/2 particle. Let n = nx + ny + nz k be any unit vector (n2 + n2 + n2 = 1).
Answers to First Midterm Examination
1. Two spin 1/2 particles possessed by Alice and Bob are in the following entangled state
1
= ( +  + i  )
3
1
= ( A  B +  A 
3

B
+ i 
A
 B )
(a) What is the reduced density matrix A of Alices particle?
Let
PHYS 455  Answers to Homework II
1. The observable B is measured in state where
1
2 0 0
1
B= 0 5 0 , = 2
3
0 0 2
2i
.
(i) the values measured are the eigenvalues of B, namely 1 = 2 and 2 = 5.
(ii) the probabilities can be computed by the lengthsquares
PHYS 455  Answers to Homework I
1. The eigenvalue equation. Consider the following 3 3 matrix.
0 1 1
A= 1 0 1
1 1 0
We will solve the eigenvalue equation for this matrix.
(a) p() = det(I A) = 3 3 2 = ( + 1)2 ( 2). The eigenvalues are
1 = 2 = 1 (double d
PHYS 455
Short Answers to Homework II
1.
(a)
2
6
=uv =
i
3i
2
i
=vu=
6
3i
(b)  = 50,  = 13.
2.
(a)
0 0 0 i
0 0 i 0
x y =
0 i 0 0
i 0 0 0
(b)
0
0
y x =
0
i
0 0 i
0 i 0
i 0 0
0 0 0
3
1
(x y )(u v) = (x u) (y v) =
6i
2i
(c)
1x
0
0
PHYS 455
Short Answers to Homework III
1. The GHZ state:GHZ = ( +  ) / 2.
(a) Eigenstates of x .
x =
 + 
2
x =
 
2
Therefore, the eigenstates of z can be expressed as
 =
x + x
2
 =
x x
2
Inserting these into GHZ we nd
GHZ =
1
(x x x + x
Solutions to Phys455 Midterm exam (2005)
a)
L2
p2
dp2p exp(
Z1 =
).
(2 )2 0
h
2m
0
Dene p2 /2m = x and take into account
dx exp(x) = 1, one obtains
Z1 indicates that
L2
2
h
, T =
.
2
T
2mkT
(2)
N
L2
Z1 =
N
(1)
(3)
1
.
2
T
This leads to a characteristic