In this page it's helpful to think carefully about the exact vector spaces involved, and so we start by treating the bra-ket notation rather more formally than usual. Let be a finite-dimensional Hilbert space and for define to be the linear map (where the inner product is conjugate-linear in the first place and linear in the second place) and to be the linear map . Thus, looked at this way, and .

The map is a conjugate-linear isometric isomorphism (see the Riesz Representation Theorem) and the map is a linear isometric isomorphism.

It's easily seen in terms of operator composition that is the operator in which maps to . Making the natural identification between and , this corresponds to , which justifies the notation . Likewise and maps to . This is the canonical form of a rank-1 operator in .

We now blur the distinction between and , and write
(the *dual* of ) for . Thus we regard as belonging
to and as belonging to .

The map is a bilinear map and extends by linearity to a map , which can be shown to be an isomorphism.

Likewise, for any and , the pair induce a linear map on by extending . This gives a map which can also be shown to be an isomorphism.

The map is bilinear on and so induces a linear map . Thus there is a unique linear map on which takes This map is our coordinate-free definition of the trace, and we can easily see its other familiar properties from this formulation. First note that and, by the same token, . Thus when and are elementary tensors and so, by linearity, this holds on all of .

Another useful property, when and , is that . This is easily seen when and because and the result follows in general by (bi)linearity.

Writing () for the standard basis of , note that and so which is the familiar formula for the trace as the sum of diagonal matrix entries.

The map is a bilinear map and so extends uniquely to a map such that Clearly for any , Thus by linearity, for any . Note also that . This follows easily for elementary tensors , and so in general by taking sums.

Another useful formula whuch generalizes the last one is . Again we see this easily for elementary tensors and obtain the general result by linearity:

To get the coordinatized formula for the partial trace, let and , and let for and be the standard basis for . Note that . Then for , and so In other words the entry in the position of the matrix of is .

A system in a mixed state is described by a density operator , is which is a positive, trace-1 operator. In more detail, if is a density operator and are the spectral projections of an observable , then the probability of outcome is given by The probabilities so obtained for all observables are a full picture of all the information which we can physically extract from the system in state .

Now suppose that is a mixed state on the system but that we can only make measurements in . Then we are restricted to measurement projections of the form where . The probabilities which we can extract from with such projections represent the sum total of all the information about which is available by means of measurements in alone. The probability of getting a measurement outcome with projection on is . However by properties of the partial trace seen above, this probability is Thus holds all the information about which is available to us from measurements in .

Furthermore, is a density operator on . To see it is positive, compute for an arbitrary unt vector , Similarly, , and so is a density operator, which contains all the physical information abouyt the system which is available to an observe measuring in .

In particular, even if is a pure state on then the part of available in is a mixed state, with density operator . As we shall see next, is pure (in ) if and only if is unentangled. Thus mixed states arise very naturally from pure states when considering partial measurements of entangled systems.

Definition 1. A state in is said to split if there are
and such that . Otherwise is
said to be *entangled*.

Recall the Schmidt Decomposition Theorem:

Theorem 2. (*Schmidt Decomposition*) Let be a unit vector in .
Then there are
orthonormal sequences in and
in , and positive numbers
satisfying
, such that
The sequence (in order) is unqiuely determined by and
.

Proof. See here.

As a consequence of the Schmidt Decomposition Theorem, if is a state (unit vector) on then it is entangled if and only if . Furthermore, Thus is pure in if and only if , which happens if and only if is unentangled.

From this calculation we can also observe that . Since the spectra of and are the same, this reveals the interesting fact that exactly the same information content is available from the compound in both of and separately.

Conversely, just as we have seen that mixed states arise naturally from pure
states when considering partial measurements of entangled systems, it also
turns out that *any* mixed state can be viewed as the restriction to a partial
trace of a pure state on a larger product space.

To see this, let be a density matrix on with spectral decomposition
where and
. Take to be any finite-dimensional Hilbert space of
dimension at least and let be an orthonormal sequence in .
Then the previous calculation shows that if is the pure state
in , then is the part
of measurable in . The state vector is called a *purification*
of .