According to quantum mechanics, a quantum system
S is described by a
-dimensional complex Hilbert space
(called the
state space of
S) with a right-linear inner product
and the states of the system are denoted by positive operators of trace 1 on
. The set of all states of
S is denoted by
. Thus,
where
is the
-algebra of all bounded linear operators on
. The elements of
are called the
mixed states of
S. A unit vector
in
is said to be a
pure state of
S and the set of all pure states of
S is denoted by
We also use
to denote the set
2.2. Separability Depending on Basis
Let
be an orthonormal basis (ONB) for
and
be the coordinate mapping:
for all
in
. We call the vector
the coordinate state of a state
of
.
We want to discuss the relationship between the separability of the coordinate state
and that of the original (abstract) state
. For convenience, we write
. Let
and
be the canonical
-bases for
and
, respectively. Then we get the canonical
-basis (with
)
for
. Thus, the definition (
2) of
U becomes
for all
in
. For example,
leading to
for all
, where
is the
matrix with
-entry
, called the density matrix of
under basis
and written as
due to the relation (
5).
Equation (
4) shows that the coordinate mapping
U transforms the basis states
as separable states
. Generally,
is not necessarily separable even if
is separable. For example, we let
and
and
for
and
, respectively, and define an ONB
for
by
Clearly,
is a separable state of
, but
which can not be written as a tensor product of two qubits and then not separable. If, however, we choose a “separable basis":
then the coordinate state of
reads
which is separable.
This observation leads to the following conclusion.
Theorem 1. Let where for some ONBs and for and , respectively. Then
(a) A pure state is a separable state of if and only if its coordinate state is a separable state of .
(b) A density operator is a separable state of if and only if its density matrix is a separable state of .
Proof. For every separable state
, it holds that
for all
. Writing
and
implies that
and so
is a separable state in
.
Conversely, let
be a separable state. Then Equation (
6) holds for some states
and
of
and
, respectively. Put
and
, then
and so
since
U is injective. Hence,
is separable. This shows that
is separable if and only if
is separable.
First, we let
be a separable state of
. Then
which is the tensor product of two density matrices. Thus,
is a separable density matrix. □
Conversely, we let
be a separable density matrix. Then
for some density matrices
and
where
and
. Define density operators
on
and
on
by
for all
, then we compute that
and so
which is separable. The proof is completed.
Similarly, one can check the following.
Theorem 2. Let , where for some ONBs and for and , respectively. Then
(a) A pure state is separable if and only if its coordinate state is separable.
(b) A density operator is separable if and only if its density matrix is separable.
The following theorem shows that the condition (resp. ) is necessary for the conditions and in Theorem 1 (resp. Theorem 2) to be satisfied.
Theorem 3. Let be an orthonormal basis for and and U be fined by Equation (2). Then the following statements are equivalent (TFSAE).
(a) The coordinate mapping U preserves separability of pure states in both directions, i.e., is separable if and only if is separable.
(b) The coordinate mapping U preserves separability of density operators in both directions, i.e., a density operator is separable if and only if its density matrix is separable.
(c) There exist unitary operators such that when , ; when , either , or , where is the swap operator: .
(d) There exist orthonormal bases and for and , respectively, such that when , for all ; When , either for all , or for all .
Proof.: Suppose that
holds. Let
be a convex combination of product states. It suffices to prove that
is separable for each
n. Fixed
n and write
then
which is separable using
. Hence,
is separable. It follows from Theorem 1
that the density matrix
is separable, i.e.,
is separable. Conversely, let
be separable. Then the density matrix
is separable and therefore the density operator
is separable (Theorem 1
). Thus, we can write
for some separable pure states
of
, where
with
. Since
and
is separable (using
) for all
n, we see that
is separable. Now,
follows. Clearly,
implies
. □
: Take ONBs
and
for
and
, respectively, and define unitary operators
and
by
and put
. Then we obtain a commutative diagram
Figure 1:
Let
be valid. Since both
U and
preserve separability of pure states in both directions, so does
V, i.e.,
is separable if and only if
is separable. It follows from [
47] that there are unitary operators
on
and
on
such that when
,
; when
, either
, or
. Thus, condition
follows by letting
(
). See
Figure 2 for the last case.
: Suppose that condition is satisfied. From the definition of U, we see that for all and where and are the canonical -bases for and , respectively. Put and for all .
When
, we have
for all
;
When
, if
, then we have
for all
; if
, then we have
for all
, here the fact that
and
were used. Now, condition
follows.
: Suppose that condition
is satisfied. Let
and
be any states of
and
, respectively. Then
When
for all
, we have
where
and
When
for all
, we have
and write
Thus,
where
and
This shows that
U maps any separable pure state as a separable pure state.
Conversely, we assume that
for some states
and
of
and
, respectively. Let us show that
is separable. To do this, we let
Then
and so
for all
Hence,
. Since either
for all
, or
and
for all
, we have either
or
This shows that is separable. Thus, U preserves separability of pure states in both directions and so condition is satisfied. The proof is completed.