Transcript Document
Quantum Information
Stephen M. Barnett
University of Strathclyde
[email protected]
The Wolfson Foundation
1. Probability and Information
2. Elements of Quantum Theory
3. Quantum Cryptography
4. Generalized Measurements
5. Entanglement
4.1
4.2
4.3
4.4
4.5
6. Quantum Information Processing
7. Quantum Computation
8. Quantum Information Theory
Ideal von Neumann measurements
Non-ideal measurements
Probability operator measures
Optimised measurements
Operations
4.1 Ideal von Neumann measurements
Measurement model
What are the probabilities for the measurement outcomes?
How does the measurement change the quantum state?
Red
‘Black box’
von Neumann measurements
Mathematical Foundations of Quantum Mechanics
(i) An observable A is represented by an Hermitian operator:
Aˆ n n n
(ii) A measurement of A will give one of its eigenvalues as a
result. The probabilities are
P(n ) n
or
2
n n
P(n ) n ˆ n Tr n n ˆ
(iii) Immediately following the measurement, the system is left
in the associated eigenstate:
ˆ n n
More generally for observables with degenerate spectra:
(i’) Let Pˆn be the projector onto eigenstates with eigenvalue n :
Aˆ Pˆn n Pˆn
(ii’) The probability that the measurement will give the result n
is
P(n ) Tr Pˆn ˆ
(iii’) The state following the measurement is
Pˆn ˆPˆn
ˆ
Tr( Pˆn ˆ )
b
PˆDetect x x dx
a
PˆNo detect ˆI - PˆDetect
a
b
Detector
Properties of projectors
†
ˆ
Pn Pˆn
I. They are Hemitian
II. They are positive
Pˆn 0
III. They are complete
ˆ
ˆ
P
n I
Observable
Probabilities
Probabilities
n
IV. They are orthonormal
Pˆi Pˆj ij Pˆi
??
4.2 Non-ideal measurements
Real measurements are ‘noisy’ and this leads to errors
P(0) 1 p
0
‘Black box’
P(1) p
P(0) p
1
‘Black box’
P(1) 1 p
For a more general state:
P(0) (1 p) 0 ˆ 0 p 1 ˆ 1
P(1) (1 p) 1 ˆ 1 p 0 ˆ 0
We can write these in the form
P(0) T r(ˆ 0 ˆ )
P(1) T r(ˆ1 ˆ )
where we have introduced the probability operators
ˆ 0 (1 p) 0 0 p 1 1
ˆ1 (1 p) 1 1 p 0 0
These probability operators are
†
ˆ
0,1 ˆ0,1
I. Hermitian
II. positive
ˆ 0 (1 p) 0
ˆ1 (1 p ) 1
III. complete
2
2
p 1
2
0
p 0
2
0
ˆ0 ˆ1 0 0 1 1 ˆI
But
IV. they are not orthonormal
ˆ0ˆ1 p(1 p)ˆI 0
We seem to need a generalised description of measurements
4.3 Probability operator measures
Our generalised formula for measurement probabilities is
P(i) T rˆi ˆ
The set probability operators describing a measurement is
called a probability operator measure (POM) or a positive
operator-valued measure (POVM).
The probability operators can be defined by the properties
that they satisfy:
Properties of probability operators
I. They are Hermitian
II. They are positive
III. They are complete
ˆ ˆn
†
n
ˆn 0
ˆ
ˆ
n I
Observable
Probabilities
Probabilities
n
IV. Orthonormal
ˆiˆ j ijˆi
??
Generalised measurements as comparisons
Prepare an ancillary system in
a known state:
A
S
Perform a selected unitary
transformation to couple the system
and ancilla:
S+A
S
S A
Uˆ S A
A
Perform a von Neumann measurement
on both the system and ancilla:
i Si Ai
The probability for outcome i is
P(i) i Uˆ A S S A Uˆ † i
S
A Uˆ † i
i Uˆ A S
ˆi
The probability operators ˆi
act only on the system
state-space.
POM rules:
I. Hermiticity:
A Uˆ † i
i Uˆ A
A Uˆ † i
†
i Uˆ A
II. Positivity:
2
ˆi i Uˆ S A 0
III. Completeness follows
from:
i
i
i ˆI A,S
Generalised measurements as comparisons
We can rewrite the detection probability as
P(i) A S Pˆi S A
†
ˆ
ˆ
Pi U i
i Uˆ
is a projector onto correlated (entangled) states of the system
and ancilla. The generalised measurement is a von Neumann
measurement in which the system and ancilla are compared.
ˆi A Pˆi A
ˆ nˆ m A Pˆn A A Pˆm A 0
Simultaneous measurement of position and momentum
The simultaneous perfect measurement of x and p would violate
complementarity.
p
Position measurement gives no
momentum information and
depends on the position probability
distribution.
x
Simultaneous measurement of position and momentum
The simultaneous perfect measurement of x and p would violate
complementarity.
Momentum measurement gives no
position information and
p
depends on the momentum probability
distribution.
x
Simultaneous measurement of position and momentum
The simultaneous perfect measurement of x and p would violate
complementarity.
p
Joint position and measurement
gives partial information on both
the position and the momentum.
x
Position-momentum minimum
uncertainty state.
POM description of joint measurements
Probability density:
( xm , pm ) Trˆˆ ( xm , pm )
Minimum uncertainty states:
xm , pm 2
2 1/ 4
( x xm ) 2
pm x
dx exp 4 2 i x
1
ˆI
dx
dp
x
,
p
x
,
p
m
m
m
m
m
m
2
This leads us to the POM elements:
1
ˆ ( xm , pm )
xm , p m xm , p m
2
The associated position probability distribution is
( x xm ) 2
( xm ) dx x ˆ x exp
2
2
Var ( xm ) x 2 2
2
& Var ( pm ) p 2
4 2
Increased uncertainty is the price we pay for measuring x and p.
The communications problem
‘Alice’ prepares a quantum system in one of a set of N possible
signal states and sends it to ‘Bob’
i selected.
prob. pi
Preparation
device
ˆ i
Measurement
device
P( j | i) Trˆ j ˆi
Bob is more interested in
T rˆ j ˆ i pi
P(i | j )
T r(ˆ j ˆ )
Measurement
result j
In general, signal states will be non-orthogonal. No measurement
can distinguish perfectly between such states.
Were it possible then there would exist a POM with
1 ˆ1 1 1 2 ˆ 2 2
2 ˆ1 2 0 1 ˆ 2 1
Completeness, positivity and 1 ˆ1 1 1
ˆ1 1 1 Aˆ
2 ˆ1 2 1 2
Aˆ positiveand Aˆ 1 0
2
2
Aˆ 2 1 2
2
0
What is the best we can do? Depends on what we mean by ‘best’.
Minimum-error discrimination
We can associate each measurement operator ˆi with a signal
state ˆ i . This leads to an error probability
Pe 1 p jTr ˆ j ˆ j
N
j 1
Any POM that satisfies the conditions
ˆ j ( p j ˆ j pk ˆ k )ˆ k 0
N
p ˆ ˆ
k 1
k
k
k
p j ˆ j 0
will minimise the probability of error.
j, k
j
For just two states, we require a von Neumann measurement with
projectors onto the eigenstates of p1 ˆ1 p2 ˆ 2 with positive (1)
and negative (2) eigenvalues:
Pemin 12 1 Tr p1ˆ1 p2 ˆ 2
Consider for example the two pure qubit-states
1 cos 0 sin 1
1 2 cos(2 )
2 cos 0 sin 1
The minimum error is achieved by measuring in
the orthonormal basis spanned by the states 1
and 2 .
1
1
2
2
We associate 1 with 1 and 2 with 2 :
Pe p1 1 2
2
p2 2 1
2
The minimum error is the Helstrom bound
Pemin
1
2 1 1 4 p1 p2 1 2
1/ 2
2
A single photon only gives one “click”
P = |a|2
a
+b
P = | b| 2
But this is all we need to discriminate between our two states
with minimum error.
A more challenging example is the ‘trine ensemble’ of three
equiprobable states:
0
31
1 12 0 3 1
p1
2 12
p2
3 0
1
3
p3
1
3
1
3
It is straightforward to confirm that the minimum-error conditions
are satisfied by the three probability operators
ˆi 23 i i
Simple example - the trine states
Three symmetric states of photon polarisation
3
2
3
1
2
3
2
Minimum error probability
is 1/3.
This corresponds to a POM
with elements
2
ˆ
j j j
3
How can we do a polarisation
measurement with these three
possible results?
Polarisation interferometer - Sasaki et al, Clarke et al.
1 / 6 2 / 3 1 / 6
0 0 0
0 0 0
1 / 6 1 / 6 2 / 3
PBS
2
0 0 0
0 3 / 2 3 / 2
PBS
1 1 / 2 1 / 2
0 3 / 2 3 / 2
2
11/ 2 1/ 2
0 0 0
PBS
0 0 0
2 / 3 1 / 6 1 / 6
1 / 3 1 / 12 1 / 12
2 / 3 1 / 6 1 / 6
Unambiguous discrimination
The existence of a minimum error does not mean that error-free
or unambiguous state discrimination is impossible. A von Neumann
measurement with
Pˆ1 1 1
ˆ
P1 1 1
will give unambiguous identification of 2 :
result
1 2
error-free
result
1 ?
inconclusive
There is a more symmetrical approach with
ˆ1
ˆ 2
1
1 1 2
1
1 1 2
2 2
1 1
ˆ? ˆI ˆ1 ˆ 2
Result 1
Result 2
Result ?
State 1
1 1 2
0
1 2
State 2
0
1 1 2
1 2
How can we understand the IDP measurement?
Consider an extension into a 3D state-space
b
a
a
b
Unambiguous state discrimination - Huttner et al, Clarke et al.
a
?
b
a
b
A similar device
allows minimum
error discrimination
for the trine states.
Measurement model
What are the probabilities for the measurement outcomes?
How does the measurement change the quantum state?
Red
‘Black box’
4.5 Operations
How does the state of the system change after a measurement is
performed?
Problems with von Neumann’s description:
1) Most measurements are more destructive than von Neumann’s
ideal.
2) How should we describe the state of the system after a
generalised measurement?
Physical and mathematical constraints
What is the most general way in which we can change a
density operator?
Quantum theory is linear so …
ˆ Aˆ ˆBˆ
or more generally
ˆ Aˆi ˆBˆi
i
BUT this must also be a density operator.
Properties of density operators
I. They are Hermitian
†
ˆ
ˆ
II. They are positive
ˆ 0
III. They have unit trace
T rˆ 1
The first of these tells us that
the second is satisfied.
†
ˆ
ˆ
Bi Ai and this ensures that
The final condition tells us that
i
† ˆ
ˆ
Ai Ai ˆI
The operator
† ˆ
ˆ
Ai Ai
is positive and this leads us to associate
† ˆ
ˆ
ˆ
i Ai Ai
ˆ .)
(Knowing ˆi does not give us A
i
If the measurement result is i then the density operator changes
as
Aˆi ˆAˆi†
Aˆi ˆAˆi†
ˆ
†
T r( Aˆi Aˆi ˆ ) T r(ˆi ˆ )
This replaces the von Neumann transformation
Pˆn ˆPˆn
ˆ
Tr( Pˆn ˆ )
If the measurement result is not known then the transformed
density operator is the probability-weighted sum
†
ˆ
ˆ
ˆ
A
A
i
i
ˆ ˆAˆ †
ˆ T r( Aˆi† Aˆi ˆ )
A
i
i
† ˆ
ˆ
T r( Ai Ai ˆ ) i
i
which has the form required by linearity.
We refer to the operators
or an an ‘effect’.
Aˆi
and
ˆA†
i
as Krauss operators
Repeated measurements
Suppose we perform a first measurement with results i and effect
ˆ and then a second with outcomes j and effects Bˆ .
operators A
j
i
The probability that the second result is j given that the first was i is
T r( Bˆ †j Bˆ j Aˆi ˆAˆi† )
P( j | i )
T r( Aˆ † Aˆ ˆ )
i
i
P(i, j ) P( j | i) P(i) T r( Aˆi† Bˆ †j Bˆ j Aˆi ˆ )
Hence the combined probability operator for the two measurements
is
ˆij Aˆi† Bˆ †j Bˆ j Aˆi
If the results i and j are recorded then
ˆ
† ˆ†
ˆ
ˆ
ˆ
ˆ
B j Ai Ai B j
P(i, j )
If they are not known then
† ˆ†
ˆ
ˆ
ˆ
ˆ
ˆ
B j Ai Ai B j
i, j
Unitary and non-unitary evolution
The effects formalism is not restricted to describing measurements,
e.g. Schroedinger evolution
ˆ (0) ˆ (t ) exp iHˆ t / ˆ (0) exp iHˆ t /
ˆ exp iHˆ t / .
which has a single Krauss operator A
We can also use it to describe dissipative dynamics:
e
Spontaneous decay rate 2g
g
We can write the evolved density operator in terms of two effects:
†
†
ˆ
ˆ
ˆ
ˆ
ˆ (t ) AN (t ) ˆ (0) AN AY (t ) ˆ (0) AY
Aˆ (t ) e gt e e g g
N
ˆA (t ) 1 e 2gt g e
Y
Measurement interpretation?
ˆ N Aˆ N† Aˆ N e 2gt e e g g
† ˆ
2gt
ˆ
ˆ
Y AY AY (1 e ) e e
Has the atom decayed?
e
g
e
g
No detection
ee
ˆ
ge
eg eee 2gt
gg ge e gt
eg e gt
( eee 2gt gg ) 1
gg
e
g
Detection
ˆ g g
Optimal operations: an example
What processes are allowed? Those that can be described by
effects.
State separation
Suppose that we have a system known to have been prepared
1
1
in one of two non-orthgonal states and . Our task
is to separate the states, i.e. to transform them into 2 and
2 with
2 2 1 1
so that the states are more orthogonal and hence more
distinguishable. This process cannot be guaranteed but can
succeed with some probability PS . How large can this be?
We introduce an effect associated with successful state separation
Aˆ S 1 2 PS | |2
We can bound the success probability by considering the action
1
ˆ
of AS on a superposition of the states and noting that the
result must be an allowed state:
PS | |2
1 1 1
1 2 2
There is a natural interpretation of this in terms of unambiguous
state discrimination:
1( 2 )
Conc
P
1 P
1( 2 )
?
1
1( 2 )
1( 2 )
Because this is the maximum allowed, state separation cannot
better it so
2
1
PS PConc
PConc
PS
1 1 1
1
2
2
No cloning theorem - Wootters & Zurek, Dieks
Can we copy an unknown state s ?
Suppose it is possible:
s A
s s
A
A
But the superposition principle then gives:
a b A
a A b A
clone
a b a b a b
Exact cloning? - Duan & Guo, Chefles
Can we clone exactly a quantum
system known to be in the state |a
or |b?
Pclone
a A
b b
b A
1 Pc lone
Error-free discrimination probability
PIDP 1 a b
For the cloned states
a a
?
Cloning cannot increase the
discrimination probability
cloned
PIDP Pclone PIDP
cloned
PIDP
1 a a b b
1
Pclone
1 a b
1 a b
separation bound
Summary
• The projective von Neumann measurements do not provide the
most general description of a measurement.
• The use of probability operators (POMs) allows us to seek
optimal measurements for any given detection problem.
• The language of operations and effects allows us to describe, in
great generality, the post-measurement state.
Naimark’s theorem
All POMs correspond to measurements. Consider a POM for a
qubit with N probability operators:
ˆ j j j ,
j j 0 0 j1 1
N
POM conditions:
j 1
2
j0
N
1 j1
2
j 1
N
N
j 1
j 1
*
*
0
j1 j 0
j 0 j1
Can we treat this as a von Neumann measurement in a enlarged
state-space?
Our task is to represent the vectors j as the components in the
qubit space of a set of orthonormal states j in an enlarged space.
Enlarged space spanned by N orthornormal states j .
The extra states can be other states of the system or by introducing
an ancilla:
0 A0 , 1 A0 , 0 A1 , 0 A2 ,, 0 AN 2
An alternative orthonormal basis is
N
i *ji j Uˆ j
j 1
N
m n jm *jn mn
j 1
Because Uˆ is unitary, we can also form the orthonormal basis:
N 1
j Uˆ † j ji i
i 0
These are the required orthonormal states for our von Neumann
measurement:
P( j ) j ˆ j
j ˆ j
T rˆ j ˆ
All measurements can be described by a POM and all POMs
describe possible measurements.
Summary
The probabilities for the possible outcomes of any given measurement can be
written in the form:
P(i ) Tr ˆˆ i
The probability operators satisfy the three conditions
ˆ i† ˆi ,
ˆ i 0,
ˆ i ˆI
i
All measurements can be described in this way
and all sets of operators with these properties represent possible measurements.
Spin and polarisation Qubits
Poincaré and Bloch Spheres
Two state quantum system
Bloch Sphere
Electron spin
Poincaré Sphere
Optical polarization
States of photon polarisation
Horizontal
0
Vertical
1
Diagonal up
1
2
0 1
Diagonal down
1
2
Left circular
1
2
Right circular
1
2
0 1
0
i 1
0
i 1
Example from quantum optics:
How can we best discriminate (without error) between the coherent states
a and a ?
Coherent states:
a
n 0
an
n!
n
Interfere like classical amplitudes.
ra tb
rb t a
a
b
Symmetric beamsplitter
Unambiguous discrimination between coherent states - Hutner et al
2 1/ 2 (i i )a 21/ 2 ia or 0
a
2 1/ 2 (1 1)a 0 or 21/ 2 a
ia
This interference experiment puts all the light into a single output.
Inconclusive results occur when no photons are detected. This
happens with probability
P? 2 a 0
1/ 2
2
exp(2 | a |2 ) a a