1 richard cleve lectures 10,11 and 12 density matrices, traces, operators and measurements michael...
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Richard Cleve
Lectures 10 ,11 and 12
DENSITY DENSITY MATRICES, traces, MATRICES, traces,
Operators and Operators and MeasurementsMeasurements
Michael A. Nielsen
Michele Mosca
Sources:
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ReviewReview: : Density matrices Density matrices of pure statesof pure states
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
2
2
ββα
αβαβα
β
αρ
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Reminder:Reminder: Trace of a matrix Trace of a matrixThe trace of a matrix is the sum of its diagonal elementse.g.
221100
222120
121110
020100
aaa
aaa
aaa
aaa
Tr
Some properties:
i
ii AATr
ATrUAUTr
CABTrABCTr
BATrABTr
ByTrAxTryBxATr
φφ
t
][][
Orthonormal basis { }iφ
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Example:Example: Notation of Density Notation of Density Matrices and tracesMatrices and traces
Notice that 0=0|, and 1
=1|.So the probability of getting 0 when measuring | is:
22
0 0)0( p
ρφφ
φφφφ
φφφφ
0000
0000
0000
TrTr
Tr
where = || is called the density matrix for the state |
10 10 ααφ
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Review: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):
... , ,, , 2211 pp φφφ
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Density matrices of mixed statesDensity 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
kkkkp
1
ψψρ
Example: the density matrix for ((0, ½ ), (1, ½ )) is:
10
01
2
1
10
00
2
1
00
01
2
1
Question: what is the density matrix of
((0 + 1, ½ ), (0 − 1, ½ )) ?
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Density matrix of a mixed Density matrix of a mixed state (use of trace)state (use of trace)
…then the probability of measuring 0 is given by conditional probability:
i
iipp state pure given 0 measuring of prob.)0(
ρ
φφ
φφ
00
00
00
Tr
pTr
Trp
iiii
iiii
where
i
iiip is the density matrix for the mixed state
Density matrices contain all the useful information about an arbitrary quantum state.
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RecapRecap: operationally : operationally indistinguishable statesindistinguishable 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
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Applying Unitary Operator to a Applying Unitary Operator to a Density Matrix of a pure stateDensity Matrix of a pure state
If we apply the unitary operation U to the resulting state is
with density matrix
U
tt UUUU
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Density Matrix
If we apply the unitary operation U to the resulting state is with density matrix
kkq ψ,
kk Uq ψ,
t
t
t
UU
UqU
UUq
kk
kk
kk
kk
ρ
ψψ
ψψ
Applying Unitary Operator to a Applying Unitary Operator to a Density Matrix of a Density Matrix of a mixedmixed state state
How do quantum operations work for these mixed states?
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Operators on Density matrices of Operators on Density matrices of mixed mixed states.states.
Effect of a unitary operation on a density matrix:
applying U to still yields U U†
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 kWhy?
Thus this is true always
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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
How do quantum operations How do quantum operations work using density matrices?work using density matrices?
(this is because k k = kψ ψk = kψ2 )
—and the state collapses to k k
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More examples of density matricesMore examples of density matrices
The density matrix of the mixed state
((ψ1, p1), (ψ2, p2), …,(ψd, pd)) is:
d
kkkk ψψpρ
1
1. & 2. 0 + 1 and −0 − 1 both have
3. 0 with prob. ½ 1 with prob. ½
4. 0 + 1 with prob. ½ 0 − 1 with prob. ½
6. 0 with prob. ¼ 1 with prob. ¼ 0 + 1 with prob. ¼ 0 − 1 with prob. ¼
Examples (from previous lecture):
11
11
2
1ρ
10
01
2
1ρ
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5. 0 with prob. ½ 0 + 1 with prob. ½
7. The first qubit of 01 − 10
Examples (continued):
4/12/1
2/14/3
2/12/1
2/12/1
2
1
00
01
2
1has:
...? (later)
More examples of density matricesMore examples of density matrices
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To Remember:To Remember: Three Properties of Density Three Properties of Density MatricesMatrices
Three properties of :• Tr = 1 (Tr M = M11 + M22 + ... + Mdd )
• =† (i.e. is Hermitian)
• 0, for all states
d
kkkk ψψpρ
1
Moreover, for any matrix satisfying the above properties,
there exists a probabilistic mixture whose density matrix is
Exercise: show this
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Use of Density Matrix and Trace to Use of Density Matrix and Trace to Calculate the probability of obtaining Calculate the probability of obtaining
state in measurementstate in measurement
If we perform a Von Neumann measurement of the state wrt a basis containing , the probability of obtaining is
Tr2 This is for a pure
state.
How it would be for a mixed state?
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Density Matrix
If we perform a Von Neumann measurement of the state
wrt a basis containing the probability of obtaining is
kkq ψ,
φφρ
φφψψ
φφψψφψ
Tr
qTr
Trqq
kkkk
kkkk
kkk
2
Use of Density Matrix and Trace to Calculate the Use of Density Matrix and Trace to Calculate the probability of obtaining state in measurement (now probability of obtaining state in measurement (now
for measuring a mixed state)for measuring a mixed state)
The same state
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Conclusion: Conclusion: Density Matrix HasDensity Matrix Has Complete InformationComplete Information
In other words, the density matrix contains all the information necessary to compute the probability of any outcome in any future measurement.
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Spectral decomposition can be used to Spectral decomposition can be used to represent a represent a useful form of density matrixuseful 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
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Continue Continue - Spectral decomposition used - Spectral decomposition used to diagonalize the density matrixto diagonalize the density matrix
In other words, we can always “diagonalize” a density matrix so that it is written as
kk
kkp φφρ
where is an eigenvector with eigenvalue and forms an orthonormal basis
kφ
kp kφ
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Taxonomy of Taxonomy of various normal various normal
matricesmatrices
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Normal matricesNormal matricesDefinition: 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)
Examples of abnormal matrices:
10
11 is not even diagonalizable
20
11 is diagonalizable, but not unitarily
eigenvectors:
dλ
λ
λ
D
00
00
00
2
1
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Unitary and Hermitian matricesUnitary and Hermitian matrices
dλ
λ
λ
M
00
00
00
2
1 with respect to some orthonormal basis
Normal:
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)
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Positive semidefinite matricesPositive 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)
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Projectors and density matricesProjectors and density matrices
Projector: Hermitian and M 2 = M, which implies that M is
positive semidefinite and k {0,1}, for all k
Density matrix: positive semidefinite and Tr M = 1, so 11
d
kkλ
Question: which matrices are both projectors and density matrices?
Answer: rank-one projectors (k = 1 if k = k0 and k = 0 if k k0 )
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Taxonomy of normal matricesTaxonomy of normal matrices
normal
unitary Hermitian
reflectionpositive
semidefinite
projectordensitymatrix
rank oneprojector
If Hermitian then If Hermitian then normalnormal
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Review:Review: Bloch sphere for qubits Bloch sphere for qubits
Consider the set of all 2x2 density matrices
Note that the coefficient of I is ½, since X, Y, Z have trace zero
They have a nice representation in terms of the Pauli matrices:
01
10σ Xx
0
0σ
i
iYy
10
01σ Zz
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
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Bloch sphere for qubits: Bloch sphere for qubits: polar polar coordinatescoordinates
2
ZcYcXcIρ zyx We will express
First consider the case of pure states , where, without
loss of generality, = cos()0 + e2isin()1 (, R)
θ2cos1θ2sin
θ2sinθ2cos1
2
1
θsinθsinθcos
θsinθcosθcosρ
φ2
φ2
2φ2
φ22
i
i
i
i
e
e
e
e
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
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Bloch sphere for qubits: location of pure Bloch sphere for qubits: location of pure and mixed statesand mixed states
+
0
1
–
+i
–i
+i = 0 + i1
–i = 0 – i1
– = 0 – 1
+ = 0 +1
Pure states are on the surface, and mixed states are inside (being weighted averages of pure states)
Note that orthogonal corresponds to antipodal here
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General General quantum quantum
operationsoperationsDecoherence, partial traces, measurements.
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General quantum operations (I)General quantum operations (I)
Example 1 (unitary op): applying U to yields U U†
General quantum operationsGeneral quantum operations are also called “completely positive trace preserving maps”, or “admissible operations”
IAA j
m
jj
1
t
Then the mapping
m
jjj AA
1
t is a general quantum operator
Let A1, A2 , …, Am be matrices satisfying
conditioncondition
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General quantum operations: General quantum operations: Decoherence Operations Decoherence Operations
Example 2 (decoherence): let A0 = 00 and A1 = 11
This quantum op maps to 0000 + 1111
Corresponds to measuring “without looking at the outcome”
2
2
2
2
0
0
β
α
ββα
αβαFor ψ = 0 + 1,
After looking at the outcome, becomes 00 with prob. ||2
11 with prob. ||2
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General quantum operations: measurement General quantum operations: measurement operationsoperations
Example 3 (trine state “measurement”):
Let 0 = 0, 1 = 1/20 + 3/21, 2 = 1/20 3/21
Then IAAAAAA 221100ttt
•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
00
01
3
2Define A0 = 2/300
A1= 2/311 A2= 2/322
62
232
4
1
62
232
4
1
We apply the general quantum mapping operator
m
jjj AA
1
t
Condition Condition satisfiedsatisfied
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General quantum operations: General quantum operations: Partial tracePartial trace discards discards the second of two qubitsthe second of two qubits
Example 4 (discarding the second of two qubits):
Let A0 = I0 and A1 = I1
0100
0001
1000
0010
State becomes
State becomes 110011002
12
12
12
1
Note 1: it’s the same density matrix as for ((0, ½), (1, ½))
10
01
2
1
Note 2: the operation is the partial trace Tr22
We apply the general quantum mapping operator
m
jjj AA
1
t
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Distinguishing Distinguishing mixed statesmixed 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:
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Distinguishing mixed states (I)Distinguishing mixed states (I)
10
01
2
12ρ
0 with prob. ½ 0 + 1 with prob. ½
0 with prob. ½ 1 with prob. ½
4121
21431 //
//ρ
0 with prob. cos2(/8) 1 with prob. sin2(/8)
0
+
0
1
0 with prob. ½ 1 with prob. ½
What’s the best distinguishing strategy between these two mixed states?
1 also arises from this orthogonal mixture: … as does 2 from:
/8=180/8=22.5
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Distinguishing mixed states (II)Distinguishing mixed states (II)
8πsin0
08πcos2
2
2/
/ρ
0
+
0
1
10
01
2
11ρ
We’ve effectively found an orthonormal basis 0, 1 in which both density matrices are diagonal:
Rotating 0, 1 to 0, 1 the scenario can now be examined using classical probability theory:
Question: what do we do if we aren’t so lucky to get two density matrices that are simultaneously diagonalizable?
Distinguish between two classical coins, whose probabilities of “heads” are cos2(/8) and ½ respectively (details: exercise)
1
Density matrices 1 and 2 are simultaneously diagonalizable
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Reminder:Reminder: Basic properties of Basic properties of the tracethe trace
d
kk,kMM
1
Tr
NMNM TrTrTr
NMNM TrTrTr
MNNM TrTr
adcbdcba Tr
d
kkMUUM
1
1 λTrTr
The trace of a square matrix is defined as
It is easy to check that
The second property implies
and
Calculation maneuvers worth remembering are:
aMMa bb Tr and
Also, keep in mind that, in general,
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Partial Partial TraceTrace How can we compute probabilities
for a partial system? E.g.
yxp
p
yx
yx
y x y
xyy
y xxy
yxxy
,
Partial measurement
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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
E.g.
ρρα
ρα
22,2 Trx
pp
yxp
p
x y
xyy
Trace
y x y
xyy
From previous slide
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Partial Trace: we derived an important formula to use partial trace
ρρα
ρα
22,2 Trx
pp
yxp
p
x y
xyy
Trace
y x y
xyy
yyy
ypTr ΦΦ2 ρ x y
xyy x
p
αΦ
Derived in previous slide
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42
Why? the probability of measuring e.g.
in the first register depends only on
ρ
αα
2
2
2
ΦΦ
ΦΦ
TrwwTr
pwwTr
wwTrp
pp
yyy
y
yyy
y
y y y
wyywy
w
ρ2Tr
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43
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.
1100110022
2 βαβα Tr
In a different basis
1
2
10
2
110
2
11100 βαβα
1
2
10
2
110
2
1βα
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44
Partial Trace
1100
10102
1
10102
1
22
**
**2
βα
βαβα
βαβα
Tr
1
2
10
2
110
2
1βα
1
2
10
2
110
2
1βα
(cont) Partial Trace can be calculated in arbitrary basis
Which is the same as in Which is the same as in previous slide for other previous slide for other basebase
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45
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
kijljlki
ljTrkiljkiTr
2
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46
Partial Trace using matrices
Tracing out the 2nd system
33223120
13021100
3332
2322
3130
2120
1312
0302
1110
0100
33323130
23222120
13121110
03020100
2
aaaa
aaaa
aa
aaTr
aa
aaTr
aa
aaTr
aa
aaTr
aaaa
aaaa
aaaa
aaaa
Tr
Tr Tr 22
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47
ExamplesExamples: Partial trace (I): Partial trace (I)
In such circumstances, if the second register (say) is discarded then the state of the first register remains
Two quantum registers (e.g. two qubits) in states and (respectively) are independent if then the combined system is in state =
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( ) = For example, it turns out that
110011002
12
12
12
1
10
01
2
1Tr2( ) =
index means 2nd system traced out
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48
Examples:Examples: Partial trace (II) Partial trace (II)We’ve already seen this defined in the case of 2-qubit systems: discarding the second of two qubits
Let A0 = I0 and A1 = I1
0100
0001
1000
0010
For the resulting quantum operation, state becomes
For d-dimensional registers, the operators are Ak = Ik ,
where 0, 1, …, d1 are an orthonormal basis
As we see in last slide, partial trace is a matrix.
How to calculate this matrix of partial trace?
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49
Examples:Examples: Partial trace (III): Partial trace (III): calculating calculating matrices of partial tracesmatrices of partial traces
1111101001110010
1101100001010000
1111101101110011
1110101001100010
1101100101010001
1100100001000000
2Tr,,,,
,,,,
,,,,
,,,,
,,,,
,,,,
ρρρρ
ρρρρ
ρρρρ
ρρρρ
ρρρρ
ρρρρ
For 2-qubit systems, the partial trace is explicitly
1111010110110001
1110010010100000
1111101101110011
1110101001100010
1101100101010001
1100100001000000
1Tr,,,,
,,,,
,,,,
,,,,
,,,,
,,,,
ρρρρ
ρρρρ
ρρρρ
ρρρρ
ρρρρ
ρρρρ
and
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50
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.
ρρ
ρ
22,2 Φ
Φ
Trp
UIyUp
yy
Trace
yyy
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51
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. ρρ 22,
2 Φ
Φ,
Trp
yp
yy
Trace
yy
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52
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.
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53
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
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54
POVMs (I)POVMs (I)Positive operator valued measurementPositive operator valued measurement (POVM):
Let A1, A2 , …, Am be matrices satisfying IAA j
m
jj
1
t
Then the corresponding POVM is a stochastic operation on
that, with probability produces the outcome:
j (classical information)
tjj AρATr
t
t
jj
jj
AρA
AρA
Tr(the collapsed quantum state)
Example 1Example 1: Aj = jj (orthogonal projectors)
This reduces to our previously defined measurements …
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55
POVMs (II):POVMs (II): calculating the measurement calculating the measurement outcome and the collapsed quantum stateoutcome and the collapsed quantum state
Moreover,
tjj AρATr
jj
j
jjjj
jj
jj φφψφ
φφψψφφ
AρA
AρA 2Tr t
t
When Aj = jj are orthogonal projectors and = ,
= Trjjjj
= jjjj
= j2
(the collapsed quantum state)
probability of the outcome:
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56
The measurement postulate The measurement postulate formulatedformulated
in terms of “observables”in terms of “observables”
A measurement is described by a complete set of projectors onto orthogonal subspaces. Outcome occurs with probability Pr( ) .The corresponding post-measure
Our
ment state is
f orm
:j
j
P j
j P
P
.j
jP
This is a projector matrix
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57
The measurement postulate formulatedThe measurement postulate formulatedin terms of “observables”in terms of “observables”
A measurement is described by a complete set of projectors onto orthogonal subspaces. Outcome occurs with probability Pr( ) .The corresponding post-measure
Our
ment state is
f orm
:j
j
P j
j P
P
.j
jP
A measurement is described by an ,
a Hermitian operator , with spectral decomposiOld f orm: o
tion
bservable
.j jj
MM P
The possible measurement outcomes correspond to theeigenvalues , and the outcome occurs with probability Pr( ) .
j j
j jP
The corresponding post-measurement state is
.j
j
P
P
The same
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58
An example of observables in action
Suppose we "measE uxample: re ".Z has spectral decomposition 0 0 - 1 1, so
this is just like measuring in the computational basis,and calling the outcomes "1" and "-1", respectively, f or0 and 1.
Z Z
Find the spectral decomposition of .Show that measuring corresponds to measuringthe parity of two qubits, with the result +1 correspondingto even parity, and the result
Exercis
-1 correspon
:
i
e
d
Z ZZ Z
ng to oddparity.
00 00 11Hint: 11 10 10 01 01Z Z
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59
An example of observables in action
Suppose we measure the observable f or astate which is an eigenstate of that observable. Showthat, with certainty, the outcome of the measurement isthe corresponding eigenvalue
Exerci
of the ob
se: M
servable.
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60
What can be measured in What can be measured in quantum mechanics?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?
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61
•“Traditional” approach to quantum measurements: A quantum measurement is described by an observable M•M is a Hermitian operator acting on the state spaceof the system.Measuring a system prepared in an eigenstate of
M gives the corresponding eigenvalue of M as themeasurement outcome.
“The question now presents itself – Can every observablebe measured? The answer theoretically is yes. In practiceit may be very awkward, or perhaps even beyond the ingenuityof 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
What can be measured in quantum mechanics?
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62
““Von Neumann measurement Von Neumann measurement in the computational basis”in the computational basis”
Suppose we have a universal set of quantum gates, and the ability to measure each qubit in the basis
If we measure we get with probability
}1,0{
2
bαb)10( 10
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63
In section 2.2.5, this is described as follows
00P0 11P1
We have the projection operatorsand satisfying
We consider the projection operator or “observable”
Note that 0 and 1 are the eigenvalues When we measure this observable M, the
probability of getting the eigenvalue is and we
are in that case left with the state
IPP 10
110 PP1P0M
b2
ΦΦ)Pr( bbPb αbb
)b(p
P
b
bb
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64
What is an “Expected value” What is an “Expected value” of an observableof an observable
b If we associate with outcome the
eigenvalue then the expected outcome is
ΦΦΦΦ
ΦΦΦΦ
)Pr(
MTrbPTr
bPPb
bb
bb
bb
bb
b
b
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65
““Von Neumann measurement in Von Neumann measurement in the computational basis”the computational basis”
Suppose we have a universal set of quantum gates, and the ability to measure each qubit in the basis
Say we have the state If we measure all n qubits, then we
obtain with probability Notice that this means that probability of
measuring a in the first qubit equals
}1,0{x
n}1,0{xx
x 2x
0
1n}1,0{0x
2x
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66
Partial measurementsPartial measurements
(This is similar to Bayes Theorem)
xp1n}1,0{0x 0
x
0
1n}1,0{0x
2x0p
If we only measure the first qubit and leave the rest alone, then we still get with probability
The remaining n-1 qubits are then in the renormalized state
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Most general measurementMost general measurement
kk
000U
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In section 2.2.5
This partial measurement corresponds to measuring the observable
1n1n I111I000M
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69
Von Neumann MeasurementsVon Neumann Measurements
A Von Neumann measurement is a type of projective measurement. Given an orthonormal basis , if we perform a Von Neumann measurement with respect to of the state
then we measure with probability
}{ k
kk}{ k
k
kkkk
kk
2
k
2
k
TrTr
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70
Von Neumann Measurements
E.x. Consider Von Neumann measurement of the state with respect to the orthonormal basis
Note that
2
10,
2
10
2
10
22
10
2
)10(
We therefore get with probability
2
10
2
2
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71
Von Neumann Measurements
Note that22
10
22
10 **
22
10
2
10Tr
2
10
2
10
2
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72
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
as follows.}{ k
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73
How do we implement Von Neumann measurements?
Construct a quantum network that implements the unitary transformation
kU k Then “conjugate” the measurement
operation with the operation U
kk U k2
kprob
1Uk
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74
Another approach
kk U 1U
kk
000
kkk
000k000
kkk
kkk
2
kprob
These two approaches will These two approaches will be illustrated in next slidesbe illustrated in next slides
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75
Example:Example: Bell basis changeBell basis change
100101
Consider the orthonormal basis consisting of the “Bell” states
110000
110010 100111 Note that
xyx
y
H
We discussed Bell basis in lecture about superdense coding and teleportation.
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76
Bell measurements: destructivedestructive and non-destructive
We can “destructively” measure
Or non-destructively project
xyy,x
y,x x
y
H2
xyprob
xyy,x
y,x xyy,x
H
2
xyprob 00
H
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77
Most general measurement Most general measurement
000U
0000002 Tr
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78
SimulationsSimulations among operations: among operations: general quantumgeneral quantum operations operations
Fact 1:Fact 1: any general quantum operation can be simulated by applying a unitary operation on a larger quantum system:
U000
Example: decoherence
0
0 + 1
2
2
0
0
β
αρ
output
discard
input
zeros discard
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79
Simulations among operations: Simulations among operations: simulations of POVMsimulations of POVM
Fact 2:Fact 2: any POVM can also be simulated by applying a unitary operation on a larger quantum system and then measuring:
U000
quantum outputinput
classical outputj
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80
Separable statesSeparable states
m
jjjjp
1
• product state if =
• separable state if
A bipartite (i.e. two register) state is a:
Question: which of the following states are separable?
1100110011001100 21
21
2 ρ
(i.e. a probabilistic mixture of product states)
( p1 ,…, pm 0)
1100110021
1 ρ
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81
Continuous-time evolutionContinuous-time evolutionAlthough we’ve expressed quantum operations in discrete terms, in real physical systems, the evolution is continuous
0
1Let H be any Hermitian matrix and t R
Then eiHt is unitary – why?
H = U†DU, where
dλ
λ
D 1
Therefore eiHt = U† eiDt U = U
e
e
Utλi
tλi
d
1
t (unitary)
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82
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