Download Problem Set 2

OPTI 646
Jessen
Homework Set #2
Distributed 18 October 2012
Due 1 November 2012
I
Quantum Bit Commitment
The Yankees are going to play the Diamondbacks in the World Series. Alice is sure that she
knows who will win. Alice doesn’t like Bob, so she doesn’t want to tell him who the winner will
be. But after the Series is over, Alice wants to be able to convince Bob that she knew the
outcome all along. What to do?
Bob suggests that Alice write down her choice (Yankees or Diamondbacks) on a piece of paper,
and lock the paper in a box. She is to give the box to Bob, but she will keep the key for herself.
Then, when she is ready to reveal her choice, she will send the key to Bob, allowing him to open
the box and read the paper.
Alice rejects this proposal. She doesn’t trust Bob, and she knows he is a notorious safe cracker.
Who’s to say whether he will be able to open the box and sneak a look, even if he doesn’t have
the key?
Instead, Alice proposes to certify her honesty in another way, using quantum information. She
will prepare one of two orthogonal quantum states of the bipartite system AB : "Yanks AB if she
picks the Yankees, or " DBs AB if she picks the Diamondbacks. Then she will send system B to
Bob, keeping system A for herself. The two states are to be chosen so that the density matrix for
system B is the same in either case,
! !
! " (Yanks) = tr #
#
= " ( DBs) = tr #
#
B
A
(
Yanks AB AB
Yanks
)
B
A
(
DB' s AB AB
DBs
)
That way, no measurement Bob could conceivably make on system B alone will uncover any
information about Alice’s choice. Finally, after the series is over, Alice can send system A to
Bob, and! he can perform a suitable orthogonal measurement to determine whether the state now
in his possession is "Yanks AB or " DBs AB . Bob thinks this sounds fair. Is it?
(Hint: Read about Schmidt decomposition in Preskill, chapter 2.4. Alice can apply a unitary
transformation of choice to system A before she ships it to Bob. Use the Schmidt decomposition
! how the!states "Yanks
to investigate
AB and " DBs AB are related, then think about how Alice might
cheat. Note: the systems A and B are not necessarily qubits so assume their respective Hilbert
spaces can have arbitrary dimension " 2 .)
!
!
!
III
Approximate Cloning
The no-cloning theorem shows that we cannot build a unitary machine that will make a perfect
copy of an unknown quantum state. But suppose we are willing to settle for an imperfect copy –
what fidelity might we achieve?
Consider a machine that acts on three qubit states according to
(a)
000
ABC
100
ABC
2
1 +
00 AB 0 C +
# AB 1 C
3
3
2
1 +
"
11 AB 1 C +
# AB 0 C .
3
3
"
!
Is such a device physically realizable, in principle?
!
If the machine operates
on the initial state " A 00 BC , it produces an entangled pure state " ABC
of the three qubits. But if we observe qubit A alone, its final state is the density operator
# "A = TrBC ( $ ABC ABC $ ) . Similarly, the qubit B, observed in isolation, has the final state # "B . It
is easy to see that # "A = # "B - these are the identical, but imperfect copies of the input pure state
!
!
" A.
!
(b)
!
!
!
For " A = a 0 A + b 1 A , find # A" , and compute the copier fidelity F= A " $ A# " A . Compare
!
this fidelity to a strategy based on the procedure outlined in HW Set 1, Problem 3d, where
Bob first measures " z for Alices qubit and then prepares Alices and his own qubit in the
most likely state given the measurement outcome.
!
!
!
IV
Selective Absorption
Suppose that we observe a qubit by directing a probe particle at the qubit, and detecting whether
the particle is scattered. If the qubit is in its “excited” state 1 , it is invisible to the probe, but if
it is in the “ground state” 0 , then there is a probability p that the qubit will scatter the probe.
This observation can be described as a two-outcome POVM, with operators
!
Fscatt = p 0 0 ,
! = (1" p) 0 0 + 1 1 .
Fnoscatt
Explain how this POVM can be realized by introducing a second (“ancilla”) qubit, performing a
suitable unitary transformation on the two-qubit system, and then performing an orthogonal
!
measurement !
on the ancilla.
V
Quantum Key Distribution
Alice and Bob want to execute a quantum key distribution protocol. Alice is equipped to prepare
either one of the two states u or v . These two states, in a suitable basis, can be expressed as
" , v = sin " ,
u = cos
sin "
cos"
( )
( )
!
!
where 0 < " < # 4 . Alice decides at random to send either u or v to Bob, and Bob is to make
a measurement to determine what she sent. Since the two states are not orthogonal, Bob cannot
!
!
distinguish the states perfectly.
! (a)
!
! to !
Bob realizes that he cannot expect to be able
identify Alice’s qubit every time, so he
settles for a procedure that is successful only some of the time. He performs a POVM with
three possible outcomes: ¬u , ¬v , or DON’T KNOW. If he obtains the result ¬u , he is
certain that v was sent, and if he obtains ¬v , he is certain that u was sent. If the result
is DON’T KNOW, then his measurement is inconclusive. This POVM is defined by the
operators
! !
!
!F¬u = A(1 " u u ) , F¬v = A(1!" v v ) , FDK = (1" 2A
! )1 + A( u u + v v ) , (i)
where A is a positive real number. How should Bob choose A to minimize the probability
of the outcome DK, and what is this minimal DK probability (assuming that Alice chooses
!
from { u , v }!equiprobably)?
Hint: if A is too large, FDK will have negative eigenvalues, and (i) is not a POVM.
!
(b)
Design a quantum key distribution protocol using Alice’s source and Bobs POVM.
(c)
! also wants to know what Alice is sending to Bob. Hoping that Alice and
Of course, Eve
Bob won’t notice, she intercepts each qubit that Alice sends, by performing an orthogonal
measurement that projects onto the basis
{(10), (01)}.
If she obtains the outcome
sends the state u on to Bob, and if she obtains the outcome
( ), she sends
0
1
( ) she
1
0
v on to Bob.
Therefore, each time Bob’s POVM has a conclusive outcome, Eve knows with certainty
! tampering causes detectable errors;!sometimes Bob
what that outcome is. But Eve’s
obtains
sent. What is the
! a “conclusive” outcome that actually differs from what Alice
!
!
probability of such an error?