Quantum Teleportation and Bit
Commitment
Chi-Yee Cheung
Chung Yuan Christian University
June 9, 2009
What are teleportation and bit commitment?
(1) Teleportation: Arguably the most novel
of all quantum protocols (procedures). It
allows one to recreate an unknown
quantum state at a remote site without
physically transferring the quantum state
itself.
(2) Bit commitment is a primitive (basic)
two-party quantum protocol which can be
use as the building block of other more
complicated two-party protocols.
Two parties Alice and Bob…
What do these two protocols have in
common?
Before answering this question, I will have
to tell you something about density matrix
It is well know that for a pure state
the density matrix is simply given by
For a mixed state denoted by
then
Given a density matrix, there are always
Infinitely many different representations,
e.g.,
and etc.
Purification:
A mixed state
in a Hilbert space
can always be written as a pure state
in a higher dimension Hilbert space
Such that
e.g., for the mixed state
If party-A applies an unitary transformation
on her states orthrononmal set
namely,
It is easy to show that the reduced density
matrix
on B’s side remain unchanged.
*** It must be so, otherwise one could
transfer information instantaneously!
Theorem:
All representations of
can be
generated in this manner.
Hughston, Jozsa, and Wooters
(Phys. Lett. A 183 (1993) 14.)
This property of mixed quantum states has been
found to be useful in discussions related to
teleportation, bit commitment.
Note that, although it is impossible for
party-A (Alice) to remotely change the
density matrix
of party-B (Bob), it is
nevertheless possible for her to change the
representation of
instantaneously at will!
All she has to do is apply an appropriate
transformation on the basis
in her hand.
(1) Teleportation
Suppose Alice has an unknown quantum
state
, and in addition she shares a
Bell (entangled) state with Bob:
Then she can use this quantum channel
to recreate
at Bob’s site without
sending the particle itself.
It goes as follows:
where
and
Therefore If Alice makes a “Bell measurement”,
and tells Bob what she gets (say i=3 ), then Bob
will be able to recreate
in his lab. by making
a compensating unitary transformation
Note that:
(1) The protocol is linear, that is,
could be part of a N-particle state, so that one can
teleport the whole N-particle state by repeating the
above process N times. Provided, of course, Alice
shares N pairs of Bell states with Bob,
(2)
could be a mixed state. Since as
we have seen, any mixed state can be regarded
as a pure state if one enlarges the Hilbert space.
Question: What if the given quantum channel
cannot be written as a product of Bell states?
(i) How do we know if it could be used for
teleportation?
(ii) If so, how does one proceed?
There exists no general result in the
literature, and one just have to treat the
problem on a case by case basis.
For example, Zha and Song (Phys. Lett. A, 2007)
considered the teleportation of a 2-particle
state
when the given channel is an
arbitrary state
(Alice: 1234; Bob: 56)
They found that the combined state can
always be recast in the form
From which they conclude that if
is
unitary then Alice can teleport faithfully the
2-particle state. Otherwise, no.
Also, Yeo and Chua (Phys. Rev. Lett. 96 (2006) 060502)
found a class of so-called 4-particle genuinely
entangled states which cannot be recast into a
product of Bell states, but they can be used to
teleport faithfully arbitrary 2-particle state.
And there are others …
1.
2.
3.
4.
5.
6.
P.X. Chen et al, PRA 74 (2006) 032324
J. Lee et al, PRA 66 (2002) 052318
G. Rigolin, PRA 71 (2005) 032303
P. Agrawal, PRA 74 (2006) 06232
S. Muralidharan et al, PRA 77 (2008) 032321
…
However, all of the results are not general,
in the sense that they work for some
specifically constructed quantum channels
only.
.
Note: (1) d is the number of unpolarized qubits in H_B
(2) Local operations on H_B does not change the
degree of entanglement between Alice and Bob.
Conclusion:
We have found a criterion which allows
one to judge if any given quantum
channel is good for faithful teleportation,
and if so, how many qubits can it teleport.
(2). Quantum bit commitment
QBC is a quantum cryptographic
protocol involving two parties:
Alice and Bob
They do not trust each other
and they will do whatever it
takes to gain an advantage!
The security of QBC is an important
issue because it can be used as the
building block of various other twoparty quantum protocols, such as
quantum coin tossing, etc.
QBC:
Commitment + Unveiling
Commitment phase:
• Alice secretly commits to a bit b = 0 or
1, which is to be revealed to Bob at a
later time.
• To ensure that Alice will not change her
mind before unveiling, Alice and Bob
execute a series of quantum and
classical exchanges such that in the end,
Bob has a quantum state
in his
hand.
Unveiling phase:
1. Alice reveals the value of b.
2. With some additional information from
Alice, Bob uses
to check whether
Alice is honest.
Security issues:
A QBC protocol is secure if it
is
1. Binding: Alice cannot change her
commitment without Bob’s
knowledge.
2. Concealing: Bob cannot find out the
value of b before Alice unveils it.
Concealing condition implies:
Ideal case
Non-ideal case
• Unconditional Security:
If the protocol remains secure even
when A and B had capabilities
limited only by the laws of nature.
(Security unaffected by any possible
technological advances.)
Example (1):
1. Alice writes b on a piece of paper and
locks it in a box.
2. She gives the box (but not the key) to
Bob as evidence of her commitment.
Concealing: Bob cannot read the paper
Binding: Paper is in Bob’s hand
However classical BC cannot be
unconditionally secure, because it’s
security is always dependent on
some unproven assumptions:
(1) The box is hard to break
(2) Bob cannot pick the lock
(3) Etc….
How about QBC?
Example (2):
To commit, Alice sends Bob a sequence of qubits
Where
Concealing:
Binding:
If Alice commits to b=0 initially, she
cannot open as b=1, for if she did, her
chance of success on each qubit is ½,
therefore the overall chance of cheating
successfully is exponentially small.
So naively, it seems that such a
protocol is unconditionally
secure!
•No-Go Theorem
Lo and Chau (1997), Mayers (1997)
If a protocol is concealing, it cannot be
binding at the same time.
Unconditionally secure QBC is
impossible as a matter of principle.
Main Idea of the Proof:
Purification – Alice leaves all undisclosed
classical information undetermined at the
quantum level by entangling with ancillas.
• Any action taken on a quantum system can be
represented by an unitary transformation on
system + environment (ancillas)
• Needs quantum computers.
Purification
At the end of commitment phase:
• Instead of a mixed state
in
• There exists a pure state
in
• HA is used to store Alice’s undisclosed info.
As long as
Bob cannot tell whether Alice purifies or not!
Then Alice has a perfect
cheating strategy (EPR attack):
Because
Hence,
Notice: UA acts on HA only
Alice can execute it without Bob’s
knowledge
Therefore Alice can unveil b=0 or
b=1 at will.
Cheating!
(That is, the protocol is not binding!)
How does it work in example (2)?
Instead of honestly producing
She generates
That is, instead of fixing undisclosed information
at the beginning, Alice leaves it undetermined at
the quantum level. That is, recorded in ancillary
qubit states.
Then she can cheat perfectly:
So it seems that
unconditionally secure
QBC is impossible!
Question: Does the “no-go theorem” cover
all possible cases?
Note that QBC has a definite objective, but
the corresponding procedure is not precisely
defined.
There are infinitely many ways to do QBC.
So how could one be sure the no-go result
is universally valid?
* Secret Parameters
In the impossibility proof, in order that
Alice knows UA, it is assumed that Alice
knows every details of the protocol, so
that no secret parameters exist.
What if Bob is allowed to generate secret
parameters unknown to Alice?
The point is:
If the pure state
depends on some
parameter
unknown to Alice?
Then
1. In general,
depends on
2. But
is not known to Alice, therefore
she cannot calculate
3. If so, then unconditionally secure QBC
might be possible (?)
The point is, the no-go theorem only proves the
existence of
, but there is no guarantee
that it is known to Alice!
However, we find that for a concealing QBC
protocol, the cheating unitary transformation
is independent of any secret parameter
(probability distribution), , chosen by Bob.
(Purification is assumed to hold. In a purified
approach, probability distributions are the only
possible unknowns left.)
Proof:
If Bob is allowed to choose
from
in secret, then he can purify his
choices with an associated probability
distribution
That is, instead of picking a particular
,
Bob entangles all his possible choices
Where
are orthonormal ancilla states.
Now the protocol is concealing. Therefore no
matter what
Bob uses, the resulting
reduced density matrix must be independent of b
Then there exists an
that
Since
Whereas
such
`s are orthogonal states in
acts on HA only.
Then it is easy to show that
Hence
Is independent of
!
Hence we have proved that Alice can cheat
perfectly whether Bob uses secret parameters
or not.
*Conclusion
For a concealing QBC protocol, i.e.,
is independent of any of Bob’s secret
choices, and Alice can calculate it without
knowing them.
Therefore, even if Bob is allowed to
generate secret parameters unknown to
Alice, she can still cheat perfectly.
Our result enlarges the applicability of
the no-go theorem to include cases
where the parties are allowed to use
parameters not known to each other.
Thank you!
• Lo and Chau: “In order that Alice and
Bob can follow the procedures, they
must know the exact forms of all unitary
transformations involved“
• Mayers: “It is a principle that we must
assume that every participant knows
every detail of the protocol, including
the distribution of probability of a
random variable generated by another
participant"