Transcript Density matrices, traces, partial traces. Operators and

DENSITY
MATRICES, traces,
Operators and
Measurements
Lectures 10 ,11 and 12
Richard Cleve
Michael A. Nielsen
Sources:
Michele Mosca
1
Density matrices
of pure states
Review:
We have represented quantum states as vectors (e.g. ψ,
and all such states are called pure states)
An alternative way of representing quantum states is in terms
of density matrices (a.k.a. density operators)
The density matrix of a pure state ψ is the matrix  = ψψ
Example: the density matrix of 0 + 1 is

α  
ρ α
β 
2


α
αβ
β   
2 
α β β 

2
Reminder:
Trace of a matrix
The trace of a matrix is the sum of its diagonal elements
e.g.
a00
Tr  a10
a20
a01 a02 
a11 a12   a00  a11  a22
a21 a22 
Some properties: Tr xA  yB   xTrA  yTr B 
Tr AB  Tr BA 
Tr[ ABC ]  Tr[CAB]


Tr UAU t  Tr A
Orthonormal basis { φi } Tr  A   φi A φi
3
Example: Notation of Density
Matrices and traces
φ  α0 0  α1 1
Notice that 0=0|, and 1=1|.
So the probability of getting 0 when measuring | is:
p(0)  0  0 
2
 0φ
2
0φ 

 0φ φ 0
 0 φ φ 0  Tr  0 φ φ 0
 Tr  0 0 φ φ   Tr  0 0 ρ 

where  = || is called
the density matrix for the
4
state |
Review: Mixture of pure states
A state described by a state vector | is called a
pure state.
What if we have a qubit which is known to be in the
pure state |1 with probability p1, and in |2 with
probability p2 ?
More generally, consider probabilistic mixtures of
pure states (called mixed states):
φ    φ1 , p1  ,  φ2 , p2  , ... 
5
Density matrices of mixed states
A probability distribution on pure states is called a mixed state:
( (ψ1, p1), (ψ2, p2), …, (ψd, pd))
The density matrix associated with such a mixed state is:
d
ρ   pk ψ k ψ k
k 1
Example: the density matrix for ((0, ½ ), (1, ½ )) is:
1 1 0 1 0 0 1 1 0
 
 



2 0 0 2 0 1 2 0 1
Question: what is the density matrix of
((0 + 1, ½ ), (0 − 1, ½ )) ?
6
Density matrix of a mixed
state (use of trace)
…then the probability of measuring 0 is given by
conditional probability:
p(0)   pi  prob. of measuring 0 given pure state i
  pi  Tr  0 0 φi φi
i


i
 Tr  pi 0 0 φi φi
i
 Tr  0 0 ρ 
where  
p
i i
is the density matrix for the mixed
i
state 
Density matrices contain all the useful information about an7
arbitrary quantum state.
i
Recap: operationally
indistinguishable states
Since these are expressible in
terms of density matrices alone
(independent of any specific
probabilistic mixtures), states with
identical density matrices are
operationally indistinguishable
8
Applying Unitary Operator to a
Density Matrix of a pure state
If we apply the unitary operation U to 
the resulting state is U 
with density matrix
U  U 

t
U   U
t
9
Applying Unitary Operator to a
Density Density
Matrix ofMatrix
a mixed state
How do quantum operations work for these mixed states?
If we apply the unitary operation U to qk , ψk
the resulting state is qk ,U ψk 
with density matrix
t
q
U
ψ
ψ
U
 k k k
k

 t
 U   qk ψk ψk U
 k

t
 UρU
10

Operators on Density matrices of
mixed states.
Effect of a unitary operation on a density matrix:
†
applying U to  still yields U U
Thus this
is true
always
Effect of a measurement on a density matrix:
measuring state  with respect to the basis 1, 2,..., d,
still yields the k th outcome with probability k k
Why?
11
How do quantum operations
work using density matrices?
Effect of a measurement on a density matrix:
measuring state  with respect to the basis 1,
2,..., d, yields the k th outcome with probability
k k
(this is because k k = kψψk = kψ2 )
—and the state collapses to kk
12
More examples of density matrices
The density matrix of the mixed state
d
((ψ1, p1), (ψ2, p2), …,(ψd, pd)) is: ρ   pk ψ k ψ k
Examples (from previous lecture):
k 1
1 1 1
1. & 2. 0 + 1 and −0 − 1 both have ρ  
2 1 1
3. 0 with prob. ½
1 with prob. ½
4. 0 + 1 with prob. ½
0 − 1 with prob. ½
6. 0
1
0 + 1
0 − 1
with prob. ¼
with prob. ¼
with prob. ¼
with prob. ¼
1 1 0
ρ 
2 0 1
13
More examples of density matrices
Examples (continued):
5. 0
with prob. ½
0 + 1 with prob. ½
has:
1 1 0 1 1 / 2 1 / 2 3 / 4 1 / 2
 
 



2 0 0 2 1 / 2 1 / 2 1 / 2 1 / 4
7. The first qubit of 01 − 10 ...? (later)
14
To Remember: Three Properties of Density
Matrices
d
ρ   pk ψ k ψ k
k 1
Three properties of  :
• Tr = 1 (Tr M = M11 + M22 + ... + Mdd )
•  =† (i.e.
 is Hermitian)
•    0, for all states 
Moreover, for any matrix  satisfying the above properties,
there exists a probabilistic mixture whose density matrix is 
Exercise: show this
15
Use of Density Matrix and Trace to
Calculate the probability of obtaining
state in measurement
If we perform a Von Neumann measurement
of the state     wrt a basis
containing  , the probability of
obtaining  is
 
2
Tr   

This is for a pure
state.
How it would be for
a
16
mixed state?
Use of Density Matrix and Trace to Calculate the
probability of obtaining state in measurement (now
DensityaMatrix
for measuring
mixed state)
If we perform a Von Neumann measurement
of the state qk , ψk 
wrt a basis containing  the probability
of obtaining 
is
q
k
k
ψk φ
2
  qk Tr  ψ k ψ k φ φ

k


 Tr   qk ψ k ψ k φ φ 
 k

 Tr ρ φ φ 
The same
state
17
Conclusion: Density Matrix Has
Complete Information
In other words, the density matrix contains
all the information necessary to compute
the probability of any outcome in any
future measurement.
18
Spectral decomposition can be used to
represent a useful form of density matrix


Often it is convenient to rewrite the
density matrix as a mixture of its
eigenvectors
Recall that eigenvectors with distinct
eigenvalues are orthogonal;
 for the subspace of eigenvectors with a
common eigenvalue (“degeneracies”), we
can select an orthonormal basis
19
Continue - Spectral decomposition used
to diagonalize the density matrix

In other words, we can always
“diagonalize” a density matrix so that it
is written as
ρ   pk φ k φ k
k
where φk is an eigenvector with
eigenvalue pk and φ forms an
k
orthonormal basis
 
20
Taxonomy of
various normal
matrices
21
Normal matrices
Definition: A matrix M is normal if M†M = MM†
Theorem: M is normal iff there exists a unitary U such that
M = U†DU, where D is diagonal (i.e. unitarily diagonalizable)
 λ1
0
D


0
0
0 
 

 λd 
0 
λ2 

0
Examples of abnormal matrices:
1
0

1 is not even
1 diagonalizable
1 1 is diagonalizable,
0 2 but not unitarily


eigenvectors:
22
Unitary and Hermitian matrices
Normal:
 λ1
0
M 


0
0
0 
 

 λd 
0 
λ2 

0
with respect to some
orthonormal basis
Unitary: M†M = I which implies |k |2 = 1, for all k
Hermitian: M = M† which implies k  R, for all k
Question: which matrices are both unitary and Hermitian?
Answer: reflections (k  {+1,1}, for all k)
23
Positive semidefinite matrices
Positive semidefinite: Hermitian and k  0, for all k
Theorem: M is positive semidefinite iff M is Hermitian and,
for all , M  0
(Positive definite: k > 0, for all k)
24
Projectors and density matrices
Projector: Hermitian and M 2 = M, which implies that M is
positive semidefinite and k  {0,1}, for all k
d
Density matrix: positive semidefinite and Tr M =1, so
λ
k 1
k
1
Question: which matrices are both projectors and density
matrices?
Answer: rank-one projectors (k = 1 if k = k0 and k = 0 if k  k0 )
25
Taxonomy of normal matrices
normal
unitary
If Hermitian then
normal
Hermitian
reflection
positive
semidefinite
density
matrix
projector
rank one
projector
26
Review: Bloch sphere for qubits
Consider the set of all 2x2 density matrices 
They have a nice representation in terms of the Pauli matrices:
0 1 
σx  X  

1
0


1 0
σz  Z  

0

1


0  i 
σy  Y  

i
0


Note that these matrices—combined with I—form a basis for
the vector space of all 2x2 matrices
We will express density matrices
 in this basis
27
Note that the coefficient of I is ½, since X, Y, Z have trace zero
Bloch sphere for qubits: polar
coordinates
We will express ρ 
I  c x X  c yY  c z Z
2
First consider the case of pure states , where, without
loss of generality,  = cos()0 + e2isin()1 (,   R)
 cos 2θ
e i 2φ cosθsin θ
1 1  cos2θ  e i 2φsin 2θ 
ρ   i 2φ
   i 2φ

2
2 e sin 2θ  1  cos2θ  
sin θ
e cosθsin θ

Therefore cz = cos(2), cx = cos(2)sin(2), cy = sin(2)sin(2)
These are polar coordinates of a unit vector (cx , cy , cz)  R3
28
Bloch sphere for qubits: location of
pure and mixed states
0
+ = 0 +1
– = 0 – 1
–i
+
–
+i = 0 + i1
–i = 0 – i1
+i
1
Note that orthogonal corresponds to antipodal here
Pure states are on the surface, and mixed states are inside
(being weighted averages of pure states)
29
General
quantum
operations
Decoherence, partial traces,
measurements.
30
General quantum operations (I)
General quantum operations are also called
“completely positive trace preserving maps”, or
“admissible operations”
condition
Let A1, A2 , …, Am be matrices satisfying
m
t
A
 j Aj  I
j 1
m
t


A

A
Then the mapping
 j j is a general quantum
j 1
operator
Example 1 (unitary op): applying U to
 yields U U†
31
General quantum operations:
Decoherence Operations
Example 2 (decoherence): let A0 = 00 and A1 = 11
This quantum op maps  to 0000 + 1111
For ψ =
0 + 1,
 α 2 αβ    α 2
 
2  
α β β   0
0 
2
β 
Corresponds to measuring  “without looking at the outcome”
After looking at the outcome,  becomes 00 with prob. ||2
11 with prob. ||2
32
General quantum operations: measurement
operations
Example 3 (trine state “measurement”):
Let 0 = 0, 1 = 1/20 + 3/21, 2 = 1/20  3/21
Define A0 =2/300
2 1 0

3 0 0
1  2 3  2
A1=2/311  

4  2
6
Then
t
t
1
1  2 3  2
A2=2/322  

4  2
6
t
A0 A0  A A1  A2 A2  I
Condition satisfied
m
We apply the general quantum mapping operator
   A j  Atj
j 1
•The probability that state k results in “outcome” state Ak is 2/3.
•This can be adapted to actually yield the value of k with this success
probability
33
General quantum operations: Partial trace discards
the second of two qubits
Example 4 (discarding the second of two qubits):
 1 0 0 0
0 1 0 0
Let A0 = I0  
and A1 = I1  


0
0
1
0
0
0
0
1




m
   A j  Atj
We apply the general quantum mapping operator
j 1
State
  becomes 
State

1
2
00 
1
2
 
11 
1
2
00 
1
2
11

1 1 0
becomes 
2 0 1
Note 1: it’s the same density matrix as for ((0, ½), (1, ½))
Note 2: the operation is the partial trace Tr2 
34
Distinguishing
mixed states
Several mixed states can have the same
density matrix – we cannot distinguish
between them.
How to distinguish by two different density
matrices?
Try to find an orthonormal basis 0, 1 in which both
density matrices are diagonal:
35
Distinguishing mixed states (I)
What’s the best distinguishing strategy between these two
mixed states?
0
with prob. ½
0 + 1 with prob. ½
3 / 4 1 / 2
ρ1  

1
/
2
1
/
4


0 with prob. ½
1 with prob. ½
1
1 also arises from this
orthogonal mixture:
1 1 0
ρ2  
2 0 1
+
0
… as does 2 from:
0
0 with prob. cos2(/8)
1 with prob. sin2(/8)
/8=180/8=22.5
0 with prob. ½
1 with prob. ½
36
Distinguishing mixed states (II)
Density matrices 1 and 2 are simultaneously diagonalizable
We’ve effectively found an orthonormal basis 0, 1 in
which both density matrices are diagonal:
cos 2 π / 8

0
ρ2  

2


0
sin
π
/
8


1 1 0
ρ1  
2 0 1
1
1
Rotating 0, 1 to 0, 1 the scenario can now
be examined using classical probability theory:
+
0
0
Distinguish between two classical coins, whose probabilities
of “heads” are cos2(/8) and ½ respectively (details: exercise)
Question: what do we do if we aren’t so lucky to get two
density matrices that are simultaneously diagonalizable?
37
Reminder:
Basic properties of
the trace
The trace of a square matrix is defined as Tr M 
d
M
k 1
It is easy to check that
k ,k
Tr M  N   Tr M  Tr N and Tr M N   Tr N M 

 λ
The second property implies Tr M   Tr U MU 
1
d
k 1
k
Calculation maneuvers worth remembering are:
Tr  a b M   b M a and Tr  a b c d   b c d a
Also, keep in mind that, in general, Tr M N   Tr M Tr N
38
Partial Trace
How can we compute probabilities for
a partial system?
 E.g.
 xy x y

x ,y


     xy x  y
y  x


y
  xy

py  
x y
 x py



Partial
measurement
39
Partial Trace
If the 2nd system is taken away and never
again (directly or indirectly) interacts with
the 1st system, then we can treat the first
system as the following mixture
From previous
 E.g.

slide


α
xy

 y ρ
p
x
y y  x p 
y






α
Trace2


xy

 p y , 
x   ρ2  Tr2 ρ


p
x
y


40
Partial Trace: we derived an important
formula to use partial trace


α
xy

 y ρ
p
x
y y  x p 
y






α
Trace2


xy

 p y , 
x   ρ2  Tr2 ρ

py
x


Tr2 ρ   p y Φ y Φ y
y
Φy  
x
Derived in
previous
slide
α xy
py
x
41
Why?

the probability of measuring e.g. w in
the first register depends only on Tr2 ρ
α
2
wy
  py
y
y

α wy
2
py
  p yTr w w Φ y Φ y

y



 Tr  w w   p y Φ y Φ y  


y



 Tr  w w Tr2 ρ 
42
Partial Trace can be calculated in
arbitrary basis

Notice that it doesn’t matter in which
orthonormal basis we “trace out” the
2nd system, e.g.
α 00  β 11  α 0 0  β 1 1
Tr2

2
2
In a different basis
1
1
 1

α 0  β 1  0  1 
α 00  β 11 
2
2 
 2
1
1
 1

α 0  β 1  0  1 

2
2 
 2
43
(cont) Partial Trace can be calculated in
Partial Trace
arbitrary
basis
1

α 0  β 1 
2

1

α 0  β 1 

2

1
Tr2
 α 0
2
1
 α 0
2
1
1

0 
1
2
2 
1
1

0 
1
2
2 




 β 1  α * 0  β* 1
 β 1  α * 0  β* 1
α 0 0 β 1 1
2
2
Which is the same as in
previous slide for other
base
44
Methods to calculate the Partial Trace



Partial Trace is a linear map that takes
bipartite states to single system states.
We can also trace out the first system
We can compute the partial trace directly
from the density matrix description
Tr2  i k  j l
 i
k Tr  j l
 i k  l j  l j i k
45

Partial Trace using matrices

a00
a
 10
a20

 a30
Tracing out the 2nd system
a01 a02
a11 a12
a21 a22
a31 a32
Tr 2
a03 
 a00
Tr 


a13  Tr2   a10

 a20
a23 
Tr 

a33 
  a30
a01 
a02
Tr 

a11 
 a12
a21 
a22
Tr 

a31 
 a32
a03  


a13  
a23  


a33  
 a00  a11 a02  a13 


a20  a31 a22  a33 
46
Examples: Partial trace (I)
Two quantum registers (e.g. two qubits) in states  and 
(respectively) are independent if then the combined system
is in state  =  
In such circumstances, if the second register (say) is discarded then the
state of the first register remains 
In general, the state of a two-register system may not be of the
form   (it may contain entanglement or correlations)
We can define the partial trace, Tr2  , as the unique linear
operator satisfying the identity Tr2( ) = 
index means
For example, it turns out that
Tr2( 
1
2
00 
1
2
 
11 
1
2
00 
1
2
11
)
1 1 0
= 
2 0 1
2nd system
traced out
47
Examples: Partial trace (II)
We’ve already seen this defined in the case of 2-qubit systems:
discarding the second of two qubits
 1 0 0 0
0 1 0 0
Let A0 = I0  
and
A
=
I

1

1


0
0
1
0
0
0
0
1




For the resulting quantum operation, state   becomes 
For d-dimensional registers, the operators are Ak = Ik ,
where 0, 1, …, d1 are an orthonormal basis
As we see in last slide, partial trace is a
matrix.
How to calculate this matrix of partial trace?
48
Examples: Partial trace (III): calculating
matrices of partial traces
For 2-qubit systems, the partial trace is explicitly
 ρ00,00
ρ
Tr2  01,00
 ρ10,00

 ρ11,00
ρ00,01
ρ01,01
ρ10,01
ρ11,01
ρ00,10
ρ01,10
ρ10,10
ρ11,10
ρ00,11 
ρ01,11   ρ00,00  ρ01,01

ρ10,11   ρ10,00  ρ11,01

ρ11,11 
ρ00,10  ρ01,11 
ρ10,10  ρ11,11 
ρ00,01
ρ01,01
ρ10,01
ρ11,01
ρ00,10
ρ01,10
ρ10,10
ρ11,10
ρ00,11 
ρ01,11   ρ00,00  ρ10,10

ρ10,11   ρ01,00  ρ11,10

ρ11,11 
ρ00,01  ρ10,11 
ρ01,01  ρ11,11 
and
 ρ00,00
ρ
Tr1  01,00
 ρ10,00

 ρ11,00
49
Unitary transformations don’t
change the local density matrix
A unitary transformation on the system
that is traced out does not affect the
result of the partial trace
 I.e.


p y Φ y U y  I  U ρ
y

 p y , Φ y
Trace2
 ρ
2
 Tr2 ρ
50
Distant transformations don’t
change the local density matrix
In fact, any legal quantum transformation
on the traced out system, including
measurement (without communicating
back the answer) does not affect the
partial trace
 I.e.
py , Φ y y




 p y , Φ y
Trace2
 ρ
2
 Tr2 ρ
51
Why??


Operations on the 2nd system should not
affect the statistics of any outcomes of
measurements on the first system
Otherwise a party in control of the 2nd
system could instantaneously
communicate information to a party
controlling the 1st system.
52
Principle of implicit
measurement


If some qubits in a computation are
never used again, you can assume (if
you like) that they have been
measured (and the result ignored)
The “reduced density matrix” of the
remaining qubits is the same
53
POVMs (I)
Positive operator valued measurement (POVM):
m
t
A
Let A1, A2 , …, Am be matrices satisfying  j A j  I
j 1
Then the corresponding POVM is a stochastic operation on
 that, with probability Tr Aj ρ Atj 
produces the outcome:
j (classical information)
A j ρ Atj

Tr A j ρ Atj

(the collapsed quantum state)
Example 1: Aj = jj (orthogonal projectors)
This reduces to our previously defined measurements …
54
POVMs (II): calculating the measurement
outcome and the collapsed quantum state
When Aj = jj are orthogonal projectors and  = ,


Tr Aj ρ Atj = Trjjjj
= jjjj
= j2
Moreover,
A j ρ Atj

Tr A j ρ A
t
j


probability
of the
outcome:
φj φj ψ ψ φj φj
φj ψ
2
 φj φj
(the collapsed quantum state)
55
The measurement postulate
formulated
in terms of “observables”
Our form: A measurement is described by a complete set of
projectors Pj onto orthogonal subspaces. Outcome j occurs
with probability
Pr(j )   Pj  .
The corresponding post-measurement state is
Pj 
.
 Pj 
This is a
projector
matrix
56
The measurement postulate formulated
in terms of “observables”
Our form: A measurement is described by a complete set of
projectors Pj onto orthogonal subspaces. Outcome j occurs
with probability
Pr(j )   Pj  .
The corresponding post-measurement state is
Pj 
.
 Pj 
Old form: A measurement is described by an observable,
a Hermitian operator M , with spectral decomposition
M   j j Pj .
The possible measurement outcomes correspond to the
eigenvalues j , and the outcome j occurs with probability
Pr(j )   Pj  .
The corresponding post-measurement state is
Pj 
.
The same
 Pj 
57
An example of observables in action
Example: Suppose we "measure Z".
Z has spectral decomposition Z  0 0 - 1 1 , so
this is just like measuring in the computational basis,
and calling the outcomes "1" and "-1", respectively, for
0 and 1.
Exercise: Find the spectral decomposition of Z  Z .
Show that measuring Z  Z corresponds to measuring
the parity of two qubits, with the result +1 corresponding
to even parity, and the result -1 corresponding to odd
parity.
Hint: Z  Z  00 00  11 11  10 10  01 01
58
An example of observables in action
Exercise: Suppose we measure the observable M for a
state  which is an eigenstate of that observable. Show
that, with certainty, the outcome of the measurement is
the corresponding eigenvalue of the observable.
59
What can be measured in
quantum mechanics?
Computer science can inspire fundamental questions about
physics.
We may take an “informatic” approach to physics.
(Compare the physical approach to information.)
Problem: What measurements can be performed in
quantum mechanics?
60
What can be measured in quantum mechanics?
•“Traditional” approach to quantum measurements:
A quantum measurement is described by an observable M
•M is a Hermitian operator acting on the state space
of the system.
Measuring a system prepared in an eigenstate of
M gives the corresponding eigenvalue of M as the
measurement outcome.
“The question now presents itself – Can every observable
be measured? The answer theoretically is yes. In practice
it may be very awkward, or perhaps even beyond the ingenuity
of the experimenter, to devise an apparatus which could
measure some particular observable, but the theory always
allows one to imagine that the measurement could be made.”
- Paul A. M. Dirac
61
“Von Neumann measurement
in the computational basis”
Suppose we have a universal set of quantum
gates, and the ability to measure each qubit
in the basis { 0 , 1 }
 If we measure   ( 0 0  1 1 ) we get
2
with
probability
b
α

b
62
In section 2.2.5, this is described as follows
We have the projection operators P0  0 0
and P  1 1 satisfying P0  P1  I
1
 We consider the projection operator or
“observable” M  0P  1P  P
0
1
1
 Note that 0 and 1 are the eigenvalues
 When we measure this observable M, the
probability of getting the eigenvalueb is
2
and we are in
Pr(b)  Φ Pb Φ  α b
that case left with the state Pb   b b 63 b

p(b)
b
What is an “Expected value”
of an observable
If we associate with outcome b the
eigenvalue b then the expected outcome is
 b Pr(b)
b


  b Φ Pb Φ  Φ   bPb  Φ
b
 b

 
 
 Tr  Φ   bPb  Φ   Tr M Φ Φ 
 
  b
64
“Von Neumann measurement in
the computational basis”
Suppose we have a universal set of quantum
gates, and the ability to measure each qubit
in the basis { 0 , 1 }

x
x
 Say we have the state
x{ 0,1 }n
 If we measure all n qubits, then we obtain
2
x with probability  x
 Notice that this means that probability of
measuring a 0 in the first qubit equals



x0 { 0,1 }
2
x
n 1
65
Partial measurements
If we only measure the first qubit and leave
the rest alone, then we2 still get 0 with
x
probability p0  x0
{ 0,1 }
 The remaining n-1 qubits are then in the
renormalized state
x

n 1

x0 { 0,1 }n 1

p0
x
(This is similar to Bayes Theorem)
66
Most general measurement

k
k
000
U
67
In section 2.2.5

This partial measurement corresponds to
measuring the observable
M  0 0 0 I
n 1
11 1 I
n 1
68
Von Neumann Measurements

A Von Neumann measurement is a type of
projective measurement. Given an
orthonormal basis { k } , if we perform a
Von Neumann measurement with respect to
{ k } of the state  
k k then
we measure k with probability

k
2
 k 
2
 k   k
 Tr  k    k
  Tr 
k
k  

69
Von Neumann Measurements
E.x. Consider Von Neumann measurement of
the state   ( 0   1 ) with respect to
the orthonormal basis  0  1 , 0  1 
2
2 

 Note that

  0  1

 
2 
2

   0  1
 

2 
2




We therefore get  0  1  with probability



2


2
2
70
Von Neumann Measurements

Note that  0  1 


  
2 
2

 0  1   *  *
 
 
2 
2

 0  1
 0  1 

   

2 
2 


  0  1  0  1 
   2

    
 Tr 

2
2
2




71
How do we implement
Von Neumann measurements?

If we have access to a universal set of
gates and bit-wise measurements in the
computational basis, we can implement Von
Neumann measurements with respect to an
arbitrary orthonormal basis { k } as
follows.
72
How do we implement
Von Neumann measurements?

Construct a quantum network that
implements the unitary transformation
U k  k

Then “conjugate” the measurement
operation with the operation U
 k k
U
k
prob  k
U
2
1
k
73
Another approach
 k k
U
U
1
k  k
prob  k
000
 
 
k
k
k
2
 000   k k  000
k  k   k  k  k
These two approaches74will
be illustrated in next slides
Example: Bell basis change


Consider the orthonormal basis consisting
of the “Bell” states
00  00  11
01  01  10
10  00  11
11  01  10
Note that
x
y
H
 xy
We discussed Bell basis in
lecture about superdense
75
coding and teleportation.
Bell measurements: destructive and
non-destructive

We can “destructively” measure
x
H
 x,y xy
y
x, y

prob  xy
2
Or non-destructively project

x, y
x, y
xy
00
H
H
 x, y xy
prob 76xy
2
Most general measurement

000
U
  Tr2   000 000 
77
Simulations among operations:
general quantum operations
Fact 1: any general quantum operation can be simulated
by applying a unitary operation on a larger quantum system:
input
zeros


0
0
0
Example: decoherence
0 + 1
0
output
U
discard
discard
α 2
ρ
 0
0 
2
β 
78
Simulations among operations:
simulations of POVM
Fact 2: any POVM can also be simulated by applying a
unitary operation on a larger quantum system and then
measuring:
input

0
0
0
U

quantum output
j
classical output
79
Separable states
A bipartite (i.e. two register) state  is a:
• product state if
 = 
• separable state if  
m
p
j 1
j
j
 j
( p1 ,…, pm  0)
(i.e. a probabilistic mixture
of product states)
Question: which of the following states are separable?
ρ1  12  00  11  00  11 
ρ2  12  00  11  00  11   12  00  11  00  11 
80
Continuous-time evolution
Although we’ve expressed quantum operations in discrete
terms, in real physical systems, the evolution is continuous
Let H be any Hermitian matrix and t  R
1
Then eiHt is unitary – why?
 λ1



†
D


H = U DU, where



λd 
e iλ1t
t
† iDt
iHt
Therefore e = U e U = U 


0


(unitary)

U
e iλd t 
81
Partially covered in 2007:
•
•
•
•
•
•
•
•
•
Density matrices and indistinguishable states
Taxonomy of normal operators
General Quantum Operations
Distinguishing states
Partial trace
POVM
Simulations of operators
Separable states
Continuous time evolution
82