6 The Dualities mathcalC(mathcalH)ast = mathcalT(mathcalH) and mathcalT(mathcalH)ast = mathcalL(mathcalH)
Tải bản đầy đủ - 0trang
3.6 The Dualities C (H)∗ = T (H) and T (H)∗ = L(H)
53
so that ∞
k=1 ck < ∞, and by Theorem 3.10 (b) |T | ∈ T (H). Hence T ∈ T (H) and
T 1 = tr |T | ≤ f . The functional f and f T , which are continuous functions on
C(H), agree on the dense subspace F(H), and so f = f T . The mapping T → f T is
clearly linear, and when it is shown to be an injection, the theorem is proved. But if
T ∈ T (H) is such that f T = 0, then ϕ|T ψ = tr T |ψ ϕ| = f T (|ψ ϕ|) = 0 for
all ϕ, ψ ∈ H (see Theorems 3.3 (b), 3.11 (b)), so that T = 0.
Since C(H)∗ is a Banach space, we immediately obtain the following corollary:
Corollary 3.2 The space (T (H), ·
1)
is complete.
Theorem 3.13 For each S ∈ L(H) define the mapping f S : T (H) → C via the formula f S (T ) = tr ST for all T ∈ T (H). In this way we get an isometric linear bijection S → f s from the space L(H) onto the dual of the Banach space (T (H), · 1 ).
Proof From Theorem 3.8 it follows that each f S is defined on T (H) and linear.
By Theorem 3.9 | f S (T )| = |tr ST | ≤ S T 1 for all T ∈ T (H), so that f S is
continuous and f S ≤ S . The mapping S → f S is clearly linear. If ϕ, ψ ∈ H,
then by using Theorems 3.3 (b) and 3.11 (b) we see that
| ϕ|Sψ | = |tr[|Sψ ϕ|]| = | f S (|ψ ϕ|)|
≤ fS
|ψ ϕ| 1 = f S ψ
ϕ
so that
S = sup | ϕ|Sψ |
ϕ ≤ 1, ψ ≤ 1 ≤ f S .
Thus the mapping S → f S is isometric and in particular injective. We still show that it
is a surjection onto the dual of T (H). Let f ∈ T (H)∗ . We define a mapping B : H ×
H → C via the formula B(ϕ, ψ) = f (|ψ ϕ|). Then |B(ϕ, ψ)| ≤ f
|ψ ϕ| 1 =
f ϕ ψ by Theorem 3.11 (a). Since B is also conjugate linear with respect to
the first and linear with respect to the second argument, by Proposition 2.1 there is
S ∈ L(H) such that B(ϕ, ψ) = ϕ|Sψ for all ϕ, ψ ∈ H. We show that f = f S , i.e.
tr ST = f (T ) for all T ∈ T (H).
(3.11)
Both sides of (3.11) are linear functions of T that are continuous on T (H) with
respect to the norm · 1 (see Theorem 3.9). Since F(H) is dense in T (H) (see
Theorem 3.11 (e)), in view of Theorem 3.3 (c) it is enough to show that (3.11) holds
whenever T = |ψ ϕ| for some ϕ, ψ ∈ H. But according to Theorems 3.3 (b) and
3.11 (b) we get
tr S|ψ ϕ| = tr |Sψ ϕ| = ϕ|Sψ = B(ϕ, ψ) = f (|ψ ϕ|).
54
3 Classes of Compact Operators
Remark 3.3 Let us recall the notation Ls (H) = {T ∈ L(H)|T = T ∗ }. In the sequel
we write Ts (H) = {T ∈ T (H)|T = T ∗ }, and Ts (H)+
1 = {T ∈ T (H)|T ≥ 0, tr[T ] =
1}. Both Ls (H) and Ts (H) are in a natural way real Banach spaces, for they are
closed with respect to addition and multiplication by a real number, and in addition
norm closed because e.g. T ∗ 1 = T 1 for all T ∈ T (H) (a consequence of Theorems 3.3 (b) and 3.10 (b)). We now prove a result which shows that the dual Ts (H)∗
of the real Banach space Ts (H) (i.e. the space of the continuous R-linear functionals
on Ts (H)) and Ls (H) can be identified.
Theorem 3.14 Let S ∈ L(H) and f S be as in Theorem 3.13.
(a) S ≥ 0, if and only if f S (T ) ≥ 0 whenever T ∈ Ts (H), T ≥ 0.
(b) S = S ∗ , if and only if f S (T ) ∈ R whenever T ∈ Ts (H), and then
S = f S = sup{| f S (T )| |T ∈ Ts (H), T
1
≤ 1}.
(c) If g : Ts (H) → R is a continuous R-linear functional, there is a uniquely determined S ∈ Ls (H) such that g = f S |Ls (H).
(d) The mapping S → g S , where g S (T ) = tr[ST ], is an isometric linear bijection
from Ls (H) onto the real Banach space Ts (H)∗ .
Proof (a) This follows from the fact that f S (|ϕ ϕ|) = tr[S|ϕ ϕ|] = ϕ|Sϕ and
each T ∈ T (H), T ≥ 0, is of the form ti |ϕi ϕi |, ti ≥ 0 (see Theorem 3.10).
(b) The “if” part is obtained from (a) by representing a selfadjoint operator as
the difference of two positive operators. (In the case of L(H) see Proposition 2.4). If
D ∈ Ts (H), then D = D1 − D2 with both Di ∈ T (H)+
1 as can be seen by applying
Theorem 3.10. The equality ϕ|Sϕ = tr[S|ϕ ϕ|] for all ϕ ∈ H proves the “if” part.
The first norm equality was noted in Theorem 3.13, and in the latter case obviously
“≥” holds. On the other hand, for each ε > 0 there is some T ∈ T (H) satisfying
T 1 ≤ 1 and | f S (T )| ≥ f S − ε, and by multiplying T with a suitable complex
number c with |c| = 1 we may assume that f S (T ) ≥ 0. Then
fS
T + T∗
= 21 | f S (T ) + f S (T ∗ )| = f S (T ) ≥ f S − ε
(for f S (T ) = f S (T ∗ ), since f S (T ) + f S (T ∗ ) ∈ R and i f S (T ) − i f S (T ∗ ) = f S (i T −
i T ∗ ) ∈ R).
(c) Define f (T ) = g( 21 (T + T ∗ )) + ig( 2i1 (T − T ∗ )), T ∈ T (H). Then in view
of Theorem 3.13 f has the form f = f S for some S ∈ Ls (H). This S is uniquely
determined e.g. by the norm formula in (b).
(d) This follows from (b) and (c).
3.7 Linear Operators on Hilbert Tensor Products and the Partial Trace
55
3.7 Linear Operators on Hilbert Tensor Products
and the Partial Trace
In this section we give some results concerning the linear operators on the Hilbert
tensor product H ⊗ K of the Hilbert spaces H and K. Here and elsewhere, when
needed, to avoid trivialities we may assume that our Hilbert spaces contain nonzero
elements. All the familiar operator algebras such as the set L(H ⊗ K) of bounded
operators, the trace class T (H ⊗ K) and the set HS(H ⊗ K) of Hilbert–Schmidt
operators on H ⊗ K are defined in the usual way. We also use the notation P(H ⊗ K)
for the projection lattice of the Hilbert tensor product H ⊗ K. In what follows, we
show that one may construct operators on the tensor product space H ⊗ K out of
operators of the component spaces H and K.
Lemma 3.4 Let H and K be Hilbert spaces and S ∈ L(H) and T ∈ L(K). Then
n
n
Sϕ j ⊗ T ψ j ≤ S
j=1
ϕj ⊗ ψj
T
(3.12)
j=1
for all ϕ j ∈ H and ψ j ∈ K, j = 1, . . . , n, when n is an arbitrary natural number.
Proof First, suppose that T = IK . Let ϕ j ∈ H and ψ j ∈ K, j = 1, . . . , n. Consider
first the case when the set {ψ1 , . . . , ψn } is orthonormal. We now have
n
n
2
Sϕ j ⊗ ψ j
j=1
n
=
Sϕ j | Sϕk
ψ j | ψk =
j, k=1
j=1
n
n
≤ S
2
Sϕ j
ϕ j | ϕk
2
ψ j | ψk = S
2
ϕj ⊗ ψj
2
j, k=1
.
j=1
In the general case of a not necessarily orthonormal set {ψ1 , . . . , ψn } choose a finite
orthonormal set {η1 , . . . , ηm } ⊂ K such that ψ j = m
k=1 c jk ηk for all j = 1, . . . , n
with some complex numbers c jk , j = 1, . . . , n, k = 1, . . . , m. Then using the above
calculation we get
n
m
n
k=1
j=1
Sϕ j ⊗ ψ j =
j=1
S(c jk ϕ j ) ⊗ ηk
n
m
≤ S
n
c jk ϕ j ⊗ ηk = S
j=1 k=1
ϕj ⊗ ψj .
j=1
Assuming S = IH we obtain the same result for a general T ∈ L(K). Combining
these results gives us the inequality (3.12).
56
3 Classes of Compact Operators
Lemma 3.4 shows that whenever
n
n
j=1
m
j=1
⊗ ηk , then
n
Sξk ⊗ T ηk ≤ S
Sϕ j ⊗ T ψ j −
m
k=1 ξk
ϕj ⊗ ψj =
k=1
m
ξk ⊗ ηk = 0.
ϕj ⊗ ψj −
T
j=1
k=1
Thus the map nj=1 ϕ j ⊗ ψ j → nj=1 Sϕ j ⊗ T ψ j is well defined on a dense subspace of H ⊗ K and since it is clearly linear, we may extend it to a bounded linear
map on the whole of H ⊗ K. Let us formulate this result in a proposition.
Proposition 3.1 Let H and K be Hilbert spaces and S ∈ L(H) and T ∈ L(K).
There is a unique operator S ⊗ T ∈ L(H ⊗ K) such that
(S ⊗ T )(ϕ ⊗ ψ) = Sϕ ⊗ T ψ
for all ϕ ∈ H and ψ ∈ K.
The linear operator S ⊗ T of the above proposition is called the (Hilbert) tensor
product of the operators S and T . Next we list some basic properties of the tensor
product operators.
Proposition 3.2 Let H and K be Hilbert spaces and S, S1 , S2 ∈ L(H) and
T, T1 , T2 ∈ L(K). Then
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
α(S ⊗ T ) = (αS) ⊗ T = S ⊗ (αT ) for all α ∈ C;
(S1 + S2 ) ⊗ T = S1 ⊗ T + S2 ⊗ T ;
(S1 ⊗ T1 )(S2 ⊗ T2 ) = S1 S2 ⊗ T1 T2 ;
(S ⊗ T )∗ = S ∗ ⊗ T ∗ ;
if S and T are selfadjoint, then also S ⊗ T is selfadjoint;
if S and T are unitary, then also S ⊗ T is unitary;
if S ∈ P(H) and T ∈ P(K), then S ⊗ T ∈ P(H ⊗ K);
if S ∈ T (H) and T ∈ T (K), then S ⊗ T ∈ T (H ⊗ K) and
tr[S ⊗ T ] = tr[S]tr[T ];
(i) if ϕ1 , ϕ2 ∈ H and ψ1 , ψ2 ∈ K, then
|ϕ1 ⊗ ψ1 ϕ2 ⊗ ψ2 | = |ϕ1 ϕ2 | ⊗ |ψ1 ψ2 |.
Proof (a), (b) For all ϕ ∈ H and ψ ∈ K we have
α(S ⊗ T ) (ϕ ⊗ ψ) = α(Sϕ ⊗ Sψ) = α Sϕ ⊗ T ψ = (αS) ⊗ T (ϕ ⊗ ψ)
so that α(S ⊗ T ) = (αS) ⊗ T since finite sums of simple tensors form a dense
subspace of H ⊗ K. In the same way one sees that α(S ⊗ T ) = S ⊗ (αT ) and (S1 +
S2 ) ⊗ T = S1 ⊗ T + S2 ⊗ T .
(c) Again, for all ϕ ∈ H and ψ ∈ K we have (S1 ⊗ T1 )(S2 ⊗ T2 )(ϕ ⊗ ψ) =
(S1 ⊗ T1 )(S2 ϕ ⊗ T2 ψ) = S1 S2 ϕ ⊗ T1 T2 ψ so that the argument is concluded as
above.