Introduction to Quantum Information Processing
Lecture 4
Michele Mosca
Overview
Von Neumann measurements General measurements Traces and density matrices and
partial traces
“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
If we measure we get with probability
}1,0{
2
bαb)10( 10
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
“Expected value” of 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
“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
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
Partial 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
In section 2.2.5
This partial measurement corresponds to measuring the observable
1n1n I111I000M
Von 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
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
Von Neumann Measurements
Note that22
10
22
10 **
22
10
2
10Tr
2
10
2
10
2
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
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
Another approach
kk U 1U
kk
000
kkk
000k000
kkk
kkk
2
kprob
Ex. Bell basis change
100101
Consider the orthonormal basis consisting of the “Bell” states
110000
110010 100111 Note that
xyx
y
H
Bell measurement 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
Most general measurement
kk
000U
Trace of a matrix
The 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
Density Matrices
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
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
Density matrix of a mixed state
…then the probability of measuring 0 is given by conditional probability:
i
iipp state pure given 0 measuring of prob.)0(
ρφφ
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.
Density Matrix
If we apply the unitary operation U to the resulting state is
with density matrix
U
tt UUUU
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
Density Matrix
If we perform a Von Neumann measurement of the state wrt a basis containing , the probability of obtaining is
Tr2
Density MatrixIf 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
Density Matrix
In other words, the density matrix contains all the information necessary to compute the probability of any outcome in any future measurement.
Spectral decomposition
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
Spectral decomposition
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
Partial Trace
How can we compute probabilities for a partial system?
E.g.
yxp
p
yx
yx
y x y
xyy
y xxy
yxxy
,
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
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
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
Partial Trace
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
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
Distant transformations don’t change the local density
matrix
Notice that the previous observation implies that 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
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
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.
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
Partial Trace
This 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
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
Most general measurement
000U
0000002 Tr