Lecture 4
Fidelity and trace norm
In this lecture we will see two simple examples of semidefinite programs for quantities that are interesting in the theory of quantum information: the first is the
trace norm of a given operator and the second is the fidelity between two positive
semidefinite operators. When considering both of these semidefinite programming
formulations, we will make use of a lemma that will come up several more times
throughout the course; it gives a precise characterization of the relationship between the blocks of any 2-by-2 positive semidefinite block operator.
4.1 Trace norm
Suppose X and Y are complex Euclidean spaces and A ∈ L(Y, X) is an arbitrary
operator. By the singular value decomposition theorem, one may always write
r
A=
∑ sk xk y∗k
(4.1)
k =1
for r = rank( A) and for some choice of orthonormal sets { x1 , . . . , xr } ⊂ X and
{y1 , . . . , yr } ⊂ Y and positive real numbers s1 , . . . , sr (which we always assume are
ordered from largest to smallest, so s1 ≥ s2 ≥ · · · ≥ sr > 0). The values s1 , . . . , sr
are the singular values of A, and (up to their ordering) are uniquely determined
by A. The trace norm of A, which is denoted k A k1 , is defined as the sum of the
singular values of A:
k A k1 = s1 + · · · + sr .
(4.2)
There are other, equivalent expressions of the trace norm, such as
√
√
k A k1 = Tr
AA∗ = Tr
A∗ A
(4.3)
and
k A k1 = max |h B, Ai| : B ∈ L(Y, X), k B k ≤ 1 .
1
(4.4)
CS 867/QIC 890 Semidefinite Programming in Quantum Information
The second expression may alternatively be written as
k A k1 = max Re(h B, Ai) : B ∈ L(Y, X), k B k ≤ 1 ;
(4.5)
it always holds that Re(h B, Ai) ≤ |h B, Ai|, and for an arbitrary choice of B one can
choose a complex number γ with |γ| = 1, and therefore k γB k = k B k, such that
Re(hγB, Ai) = |h B, Ai|.
We’ll write down an SDP for the trace norm of a given operator A momentarily,
but first let us state and prove the lemma about block operators that was suggested
above.
Lemma 4.1. Let X and Y be complex Euclidean spaces, let P ∈ Pos(X) and Q ∈ Pos(Y)
be positive semidefinite operators, and let X ∈ L(Y, X) be an operator. It holds that
P X
X∗ Q
∈ Pos(X ⊕ Y)
(4.6)
√
√
PK Q for K ∈ L(Y, X) satisfying k K k ≤ 1.
√ √
Proof. Suppose first that X = PK Q for K ∈ L(Y, X) being an operator for which
k K k ≤ 1. It follows that KK ∗ ≤ 1X , and therefore
if and only if X =
√
0≤
PK
√
Q
!
K
√
∗
√
√
√ P X
PKK ∗ P X
≤
.
P
Q =
X∗ Q
X∗
Q
(4.7)
For the reverse implication, assume
P X
X∗ Q
and define
√
K=
∈ Pos(X ⊕ Y),
P+ X
p
(4.8)
Q+ .
(4.9)
Here P+ and Q+ refer to the Moore–Penrose pseudo-inverses of P and Q, respectively. For a case such as this, where P and Q are positive semidefinite, they may
be obtained by taking a spectral decomposition of either P or Q and replacing each
nonzero eigenvalue by its reciprocal.
√ √
It will be proved that X = PK Q and k K k ≤ 1. Observe first that, for every
Hermitian operator H ∈ Herm(X), the block operator
H 0
0 1
P X
X∗ Q
H 0
0 1
2
=
HPH HX
X∗ H Q
(4.10)
Lecture 4
is positive semidefinite. In particular, for H = Πker( P) being the projection onto the
kernel of P, one has that the operator
!
0
Πker( P) X
(4.11)
X ∗ Πker( P)
Q
is positive semidefinite, which implies that Πker( P) X = 0, and therefore Πim( P) X =
X. Through a similar argument, one finds that XΠim(Q) = X. It therefore follows
that
√ p
PK Q = Πim( P) XΠim(Q) = X.
(4.12)
Next, note that
x ∗ Px
x ∗ Xy
y∗ X ∗ x y∗ Qy
!
=
x∗
0
0
y∗
!
P
X
!
X∗ Q
x 0
0 y
!
≥0
for every choice of vectors x ∈ X and y ∈ Y. Setting
√
p
x = P+ u and y = Q+ v
for arbitrarily chosen unit vectors u ∈ X and v ∈ Y, one finds that
!
u∗ Πim( P) u
u∗ Kv
1
u∗ Kv
≥
≥0
v∗ K ∗ u
1
v∗ K ∗ u
v∗ Πim(Q) v
(4.13)
(4.14)
(4.15)
and therefore |u∗ Kv| ≤ 1. As this inequality holds for all unit vectors u and v, it
follows that k K k ≤ 1, as required.
Now we can devise an SDP whose optimal value is k A k1 for a given operator
A ∈ L(Y, X). The primal problem, in a slightly simplified form, will be the following:
Primal problem
maximize:
subject to:
1
1
hK, Ai + hK ∗ , A∗ i
2
2
1X K
≥ 0,
K ∗ 1Y
K ∈ L(Y, X).
The optimal value of this primal form is evidently k A k1 , because the objective
function can alternatively be expressed as Re(hK, Ai), and by the lemma above K
is free to range over all elements of L(Y, X) with spectral norm at most 1.
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CS 867/QIC 890 Semidefinite Programming in Quantum Information
The primal problem above can be matched to the rigid formal definition of
semidefinite programs we have adopted by first defining Φ ∈ T(X ⊕ Y) as
X ·
X 0
Φ
=
(4.16)
· Y
0 Y
for all X ∈ L(X) and Y ∈ L(Y) (and the dots representing unnamed elements of
L(Y, X) or L(X, Y) that get zeroed out by the map). The primal problem above is
then equivalent to the primal problem of the semidefinite program
1 0 A
1X 0
Φ,
,
.
(4.17)
0 1Y
2 A∗ 0
The dual problem is particularly easy to calculate for this semidefinite program
because Φ is its own adjoint. After just a bit of simplification we obtain the following problem:
Dual problem
minimize:
subject to:
Tr( X ) + Tr(Y )
1 0 A
X 0
≥
,
0 Y
2 A∗ 0
X ∈ Herm(X), Y ∈ Herm(Y).
We can further simplify this problem by first noting that we must have X ∈ Pos(X)
and Y ∈ Pos(Y) for every feasible X and Y; and by setting P = 2X and Q = 2Y we
get the following problem statement:
Dual problem
minimize:
subject to:
1
1
Tr( P) + Tr( Q)
2
2
P
−A
≥ 0,
∗
−A
Q
P ∈ Pos(X), Q ∈ Pos(Y).
If you think about a singular value decomposition
r
A=
∑ sk xk y∗k
(4.18)
k =1
of A, then the dual problem makes perfect sense: we can take
r
P=
∑
sk xk xk∗
r
and
Q=
∑ sk yk y∗k
(4.19)
k =1
k =1
to obtain a feasible solution whose objective value is k A k1 , and (by weak duality)
this is the best we can do.
4
Lecture 4
4.2 Fidelity
The fidelity between two positive semidefinite operators P, Q ∈ Pos(X) is defined
as
√ p F( P, Q) = P Q .
(4.20)
1
Sometimes the fidelity is defined as this quantity squared, but we’ll use the definition above, without the square. The alternative definition
q√ √ F( P, Q) = Tr
PQ P
(4.21)
is obtained by expanding the trace norm using the expression (4.3).
We will now describe a semidefinite program whose optimal value coincides
with F( P, Q). While this task is easily accomplished
√ √ by applying the semidefinite
program for the trace norm to the operator P Q, we will aim for something
simpler that does not require the computation of operator square roots.
The primal problem of our semidefinite program will be as follows:
Primal problem
maximize:
subject to:
1
1
Tr( X ) + Tr( X ∗ )
2
2
P X
≥ 0,
X∗ Q
X ∈ L(X).
Notice that the objective function can alternatively be expressed as Re(Tr( X )). The
lemma proved in the previous section reveals that the optimal value of the primal
problem is the fidelity between
√ √ P and Q: the operator X ranges precisely over
those operators equal to PK Q for K ∈ L(X) satisfying k K k ≤ 1, and taking the
maximum over all such K yields
o
n √ p max Re Tr
PK Q
: K ∈ L(X), k K k ≤ 1
n D√ p
E
o
= max Re
P Q, K
: K ∈ L(X), k K k ≤ 1
o
n D √ p E
(4.22)
= max Re K, P Q
: K ∈ L(X), k K k ≤ 1
√ p = P Q .
1
The primal problem above can be matched to the primal problem of a formally
specified semidefinite program in a way that is quite similar to the trace norm
5
CS 867/QIC 890 Semidefinite Programming in Quantum Information
semidefinite program from earlier in the lecture. More specifically, we may define
a map Φ ∈ T(X ⊕ X) exactly as in (4.16), taking Y = X so that each block is an
element of L(X), and then observe that the primal problem above is in agreement
with the primal problem for the semidefinite program
1
Φ,
2
0 1
P 0
,
.
1 0
0 Q
(4.23)
The dual problem, after just a bit of simplification (similar to the dual problem
in the trace norm semidefinite program), is as follows:
Dual problem
minimize:
subject to:
1
1
hY, Pi + h Z, Qi
2
2
Y −1
≥ 0,
−1 Z
Y, Z ∈ L(X).
The primal problem is feasible (although not strictly feasible in case P or Q
is singular) and the dual is strictly feasible, so strong duality holds by Slater’s
theorem.
This dual problem can simplified further using the following proposition.
Proposition 4.2. Let Y, Z ∈ Herm(X). It holds that
Y −1
∈ Pos(X ⊕ X)
−1 Z
(4.24)
if and only if Y, Z > 0 and Z ≥ Y −1 .
Proof. Suppose Y, Z > 0 and Z ≥ Y −1 . It holds that
and therefore
Y −1
−1 Z
=
1
0
−
1
−Y
1
Y −1
−1 Z
Y −1
−1 Z
Y
0
0 Z − Y −1
1 −Y − 1
0
1
(4.25)
∈ Pos(X ⊗ X).
(4.26)
∈ Pos(X ⊕ X).
(4.27)
Conversely, suppose that
6
Lecture 4
It holds that
0≤
u∗
v∗
Y −1
−1 Z
u
= u∗ Yu − u∗ v − v∗ u + v∗ Zv
v
(4.28)
for all u, v ∈ X. If Y were not positive definite, there would exist a unit vector v for
which v∗ Yv = 0, and one could then set
u=
1
(k Z k + 1)v
2
(4.29)
to obtain
k Z k ≥ v∗ Zv ≥ hu, vi + hv, ui = k Z k + 1,
(4.30)
which is absurd. Therefore Y > 0, and by inverting the expression above, we have
Y
0
1 0
Y −1
1 Y −1
=
≥ 0.
(4.31)
0 Z − Y −1
Y −1 1
−1 Z
0 1
This implies Z ≥ Y −1 (and therefore Z > 0) as required.
Now, given that Q is positive semidefinite, it holds that h Q, Z i ≥ h Q, Y −1 i
whenever Z ≥ Y −1 , so there would be no point in choosing any Z other than Y −1
when aiming to minimize the dual objective function subject to that constraint. The
dual problem above can therefore be phrased as follows:
Dual problem
1
1
hY, Pi + hY −1 , Qi
2
2
subject to: Y > 0,
minimize:
Y ∈ Pos(X).
Because we have strong duality for our semidefinite program, we obtain the following alternative characterization of the fidelity:
1
1 −1
F( P, Q) = inf
hY, Pi + hY , Qi .
(4.32)
2
Y >0 2
This characterization is equivalent to Alberti’s theorem, which states that
F( P, Q)2 = inf hY, PihY −1 , Qi.
Y >0
(4.33)
(The equivalence can be established through the use of the arithmetic-geometric
mean inequality.)
The following two theorems establish well-known properties of the fidelity
function, but through proofs based on the semidefinite programming characterization we have obtained.
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CS 867/QIC 890 Semidefinite Programming in Quantum Information
Theorem 4.3. Let P0 , P1 , Q0 , Q1 ∈ Pos(X) be positive semidefinite operators, for X a
complex Euclidean space. It holds that
F( P0 + P1 , Q0 + Q1 ) ≥ F( P0 , Q0 ) + F( P1 , Q1 ).
(4.34)
Proof. By considering the primal problem associated with our semidefinite program, and noting that Slater’s theorem implies that an optimal primal solution is
always obtained, we see that we may choose operators X0 , X1 ∈ L(X) such that
the block operators
P1 X1
P0 X0
(4.35)
and
X1∗ Q1
X0∗ Q0
are both positive semidefinite, and such that
Tr( X0 ) = F( P0 , Q0 )
and
Tr( X1 ) = F( P1 , Q1 ).
(4.36)
The sum of two positive semidefinite operators is positive semidefinite, and therefore
P0 + P1
X0 + X1
P0 X0
P1 X1
=
+
(4.37)
( X0 + X1 ) ∗ Q 0 + Q 1
X0∗ Q0
X1∗ Q1
is positive semidefinite. By again considering the primal problem of our semidefinite program, this time for F( P0 + P1 , Q0 + Q1 ), we find that
F( P0 + P1 , Q0 + Q1 ) ≥ Tr( X0 + X1 ) = F( P0 , Q0 ) + F( P1 , Q1 ),
(4.38)
as required.
Note that by combining this theorem with the fact that F(λP, λQ) = λ F( P, Q),
we find that the fidelity is jointly concave:
F(λP0 + (1 − λ) P1 , λY0 + (1 − λ)Y1 )
≥ λ F( P0 , Y0 ) + (1 − λ) F( P1 , Y1 ).
(4.39)
Theorem 4.4. Let X and Y be complex Euclidean spaces, let P, Q ∈ Pos(X), and let
Φ ∈ C(X, Y) be a channel. It holds that
F( P, Q) ≤ F(Φ( P), Φ( Q)).
(4.40)
Proof. One may choose X ∈ L(X) so that
P X
X∗ Q
8
(4.41)
Lecture 4
is positive semidefinite and satisfies Re(Tr( X )) = F( P, Q). By the complete positivity of Φ, the block operator
Φ( P) Φ( X )
Φ( P) Φ( X )
=
(4.42)
Φ( X )∗ Φ( Q)
Φ( X ∗ ) Φ( Q)
is positive semidefinite as well. By again considering the primal problem of the
semidefinite program, this time for F(Φ( P), Φ( Q)), along with the fact that Φ is
trace-preserving, we find that
F(Φ( P), Φ( Q)) ≥ Re(Tr(Φ( X ))) = Re(Tr( X )) = F( P, Q),
as required.
9
(4.43)