I tried using the $\rho$ operator for this, and I got nothing. I approached it assuming I have two subsystems $A$ and $B$, both inside their own Hilbert space $H_A$ and $H_B$. And even defined $\rho = | \psi \rangle \langle \psi |$. But at the time of doing the extension, I am at a loss.
Prove that all extensions of a pure state are product states?
density-operatorhilbert-spacequantum mechanicsquantum-informationquantum-states
Related Solutions
Let me remind you that the Fock space of multiple fermions is defined to be the antisymmetric (fermionic) subspace of the full Fock space
$$ \Gamma_a=\bigoplus_{n=0}^{\infty}H^{\wedge n}, $$
where $\wedge$ stands for the antisymmetric tensor product
$$ v_1\wedge\ldots\wedge v_n=\frac{1}{n!}\sum_{p\in P_n}\sigma_p v_{p(1)}\wedge\ldots\wedge v_{p(n)}. $$
Here $\sigma_p$ is the sign of the permutation $p$ in the group of permutations $P_n$.
Thus the confusion here comes from the fact that $c_a^{\dagger}c_b^{\dagger}|0\rangle\neq|ab\rangle$ as you seem to state.
Recall that the creation and annihilation operators are defined within the occupation number representation, i.e. $c_a^{\dagger}c_b^{\dagger}|0\rangle=|11\rangle$, where the first number denotes the number of fermions in state $a$ while the second denotes the number of fermions in state $b$. On the other hand, states written in the occupation number representation are defined to be properly antisymmetrized (for fermions) many-body basis states, as forced upon us by the particle indistinguishability. Therefore they are defined within the fermionic Fock space. Any textbook will show so, take a look at the first chapter of Many-Body Quantum Theory in Condensed Matter Physics: An Introduction by Bruus and Flensberg for example. For two fermions described via a single particle basis $\{|a\rangle,|b\rangle\}$ one possible choice is:
$$ |11\rangle=\frac{1}{\sqrt{2}}\left(|ab\rangle-|ba\rangle\right). $$ Therefore
$$ c_a^{\dagger}c_b^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2-c_b^{\dagger}|0\rangle_1\otimes c_a^{\dagger}|0\rangle_2\right) $$
The familiar anticommutativity of these operators is now obvious from this from
$$ c_b^{\dagger}c_a^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(c_b^{\dagger}|0\rangle_1\otimes c_a^{\dagger}|0\rangle_2-c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2\right)=-c_a^{\dagger}c_b^{\dagger}|0\rangle $$
In fact one of the great advantages of creation and annihilation operators is that they include the antisymmetry (for fermions) of the wave function implicitly.
Dotting this with $\langle x_1x_2|$ we obtain:
$$ \langle x_1x_2|c_a^{\dagger}c_b^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(\phi_a(x_1)\phi_b(x_2)-\phi_b(x_1)\phi_a(x_2)\right). $$
Thus there is no inconsistency, both representations show that the particles are entangled.
On the other hand dotting $\langle x_1x_2|$ with $c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2$ would simply produce
$$ \psi(x_1,x_2)=\phi_a(x_1)\phi_b(x_2) $$
Therefore there is no inconsistency here also, however, as I said, the important thing to remember is that
$$ c_a^{\dagger}c_b^{\dagger}|0\rangle\neq c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2 $$
Let me first answer your questions:
Q1: No, the partial transposition acts on one part of the subsystem. To make this more clear, let me give a real definition of the criterion: Let $\mathcal{H}_1,\mathcal{H}_2$ be two Hilbert spaces, then the partial transpose on the second system is defined via
$$ ^{PT}: \mathcal{B}(\mathcal{H}_1\otimes \mathcal{H}_2)\to \mathcal{B}(\mathcal{H}_1\otimes \mathcal{H}_2);\quad A\otimes B\mapsto A\otimes B^T $$
Obviously, this only defines the partial trace on product states, but the rest is done by linearity. Now a qubit (!) state is entangled if and only if the partial trace on any subsystem is not positive. However, this is equivalent to saying that a state is entangled if and only if the partial trace on the second subsystem is nonnegative, because the partial trace on the first subsystem would map $A\otimes B\mapsto A^T\otimes B=(A\otimes B^T)^T$, i.e. if you compose the partial trace on the second subsystem with the transposition on the whole system, you obtain the partial trace on the first subsystem. Since the transposition leaves the eigenvalues invariant, if a state is not positive under the partial transposition of any one subsystem, it will remain nonpositive under the partial transposition of the other system.
Let me emphasize however (as you stated) that this criterion is not sufficient for proving separability on any larger system than a system comprised of qubits and qutrits!
Q2: Purity and entanglement are completely different concepts in the sense that purity is about a single system, whereas entanglement is about a bipartition of the system, so I wouldn't really expect a nice rank characterization. However, if you have a pure state $\rho=|\psi\rangle\langle \psi|$, then you can actually say something, because such a pure state is entangled if and only if its reduced density matrix is pure - and since this has a characterization with matrix ranks, you have such a connection. This connection is however false for mixed entangled/separable states and I don't know whether such a connection exists.
Let me give a short proof of the claim: Let $|\psi\rangle =\sum_{i=1}^k \lambda_i |\psi^1_i\rangle|\psi^2_i\rangle$ be an arbitrary mixed state in its Schmidt-decomposition (note that the state is separable iff $k=1$. Now the reduced density matrix of the first subsystem is easily seen to be:
$$ \rho_{red}= \sum_{i=1}^k \lambda_i^2 |\psi^1_i\rangle\langle \psi^1_i|$$
which has as rank the Schmidt-rank of $\rho$. Since a state is pure iff its density matrix is rank one, the reduced density matrix is pure iff the state was separable.
Actually, this characterization is exactly Exercise 2.78 in Nielsen&Chuang!
Q3: This was not actually stated, but you say that you have come across this condition that $\rho$ is pure iff it is rank one. Let me clarify the issue a bit: By definition, $\rho$ is pure iff it is of the form $|\psi\rangle\langle \psi|$ for some $\psi$. Obviously, this matrix is rank one, since it has one eigenvector with eigenvalue 1 (namely $|\psi\rangle$ - granted the state is normalized) and all other eigenvectors are 0 (all other states of an orthonormal basis comprising $|\psi\rangle$). On the other hand, if a state is rank 1, this means that it has only one eigenvector $|\psi\rangle$ with eigenvalue 1 and all other eigenvalues are zero. Since the density matrix is positive, via the spectral decomposition this implies that $\rho=|\psi\rangle\langle \psi|$.
Your Problem: Let me first point out a few misconceptions and misnotations and problems with the way you approach your problem. Later, I will give a solution.
You write that $\tau:\mathcal{H}_1\to\mathcal{H}_1$ is a linear map. This means that given the states $|x_i\rangle$, the map $\tau$ is actually represented by a matrix, which I am going to denote by $T$. Of course, $TT^*$ is then a positive operator and you could say that it is a "state", but I don't think one can ascribe the same phyiscal meaning. Then you write:
$$\sum_{i,j} |\tau x_i \rangle \langle \tau x_j | |y_i\rangle\langle y_j| = \sum_{i,j} |\rho x_i \rangle \langle x_j| |y_i\rangle \langle y_j|$$
This line is actually wrong. You can't just pull the $\tau$ to the other side like you want to. The problem is that the matrix on the right hand side is most likely not even Hermitian, while the matrix on the right hand side most definitely is! Therefore, everything that follows is wrong.
Still, there is one other problem that I'd like to address: You want to prove separability by showing that the partial transpose is positive. Let me repeat again: This is not possible. If the partial transpose is not positive, the state is entangled, but the other direction (If the state is entangled, the partial transpose is not positive, or else: If the partial transpose is positive, the state is separable) does not hold in dimensions other than $2\times 3$ and $2\times 2$. You sometimes do talk about "qubits", which would imply that $\mathcal{H}_1=\mathcal{H}_2=\mathbb{C}^2$, but you did not really specify this.
The final problem, and this is what I was getting at before: You did not specify the range of the indices of your sum in $|\Psi\rangle$. If it is the dimension of $\mathcal{H}_1$, the statement you want to prove is essentially correct, if it is not, however, the statement you want to prove is totally wrong.
My solution:
So we are given a state $|\Psi_0\rangle=\sum_{i=1}^k |x_i\rangle|y_i\rangle$ (note that the state is in Schmidt decomposition!). We want to say something about the separability of the state $|\Psi\rangle=\sum_{i=1}^k |\tau x_i\rangle|y_i\rangle$.
By definition $\tau:\mathcal{H}_1\to \mathcal{H}_1$, i.e. if we suppose that $\{x_i\}_{i=1}^n$ is a basis, we can define $\tau$ via $\tau(x_i)=\sum_j \lambda_{ji} x_j$. So far, nothing new.
We now ask: When is $\Psi$ separable? First note that whatever $\tau$ is, $|\Psi\rangle$ is still pure, because the map is defined on the level of pure states, e.g. we will not always have to go to density matrices.
Now let's have a look at the problem for different numbers of elements in the sum, i.e. for different $k$ in $|\Psi_0\rangle$. First cover the easy cases:
k=1: We have $|\Psi_0\rangle=|x_1\rangle|y_1\rangle$, which means that we start out with a separable state. In this case, $$|\Psi\rangle=|\tau x_1\rangle|y_1\rangle=\left(\sum_j \lambda_{1j}|x_j\rangle\right)\otimes |y_1\rangle $$ is still separable. This is completely independent of $\tau$.
k=n: This means we have essentially a maximally mixed state. To properly normalize it, we should write $|\Psi_0\rangle=\frac{1}{\sqrt{n}} \sum_{i=1}^n |x_i\rangle|y_i\rangle$. We now apply $\tau$ and we want the outcome to be separable, i.e. we want:
$$ |\Psi\rangle=\frac{1}{\sqrt{n}} \sum_{i=1}^n |\tau x_i\rangle|y_i\rangle=\frac{1}{\sqrt{n}} \sum_{i=1}^n \left(\sum_{j=1}^n \lambda_{ji}|x_j\rangle\right)|y_i\rangle \stackrel{!}{=} |\phi\rangle|\varphi\rangle $$ for some states $|\phi\rangle$ and $|\varphi\rangle$. This last bit is the criterion for separability. Now note that we have the sum and by multilinearity of the tensor product, the only way to get $|\phi\rangle$ and $|\varphi\rangle$ as above would be to have $|\varphi\rangle=\sum_{i=1}^n |y_i\rangle$, i.e. the first part must be independent of $i$. This means $\sum_{j=1}^n \lambda_{ji}|x_j\rangle$ must be equal to some state $|\phi\rangle$ independent of $i$, which means (since the $x_i$ form a basis) that $\tau$ must map $|x_i\rangle$ to $\lambda_i |\phi\rangle$ (there may be a factor depending on $i$ because $\lambda_i|\phi\rangle \otimes |\varphi\rangle = |\phi\rangle \otimes \lambda_i|\varphi\rangle$ always).
This means: $|\Psi\rangle$ is separable if and only if $\tau$ maps any state of the Hilbert space $\mathcal{H}_1$ to a state proportional to $|\phi\rangle$. The proportionality factor is dictated by linearity - in fact, most states must be mapped to zero. More to the point, the matrix representing $\tau$ must have rank one, because the image is one-dimensional. This means that the matrix $\tau\tau^*$ is also rank one if and only if $|\Psi\rangle$ is separable.
$1<k<n:$ This case lies in between the two extreme cases above. It is more cumbersome, because (if I am not mistaken) the result will be that $|\Psi\rangle$ is separable iff $\tau$ maps the vector space spanned by $\{|x_i\rangle\}_{i=1}^k$ to some arbitrary or fixed state $|\phi\rangle$, while the complement spanned by $\{|x_i\rangle\}_{k+1}^n$ (if we assume that all $x_i$ are orthogonal) is arbitrary.
Best Answer
I think the condition of determining the state of a composite register is the tensorial product of the states.
If you have two independent states,
$\sigma = \sum_{ij} \sigma_{ij} | v_i \rangle \langle v_j | \hspace{1mm}\in D(\mathcal{X})$
$\xi = \sum_{kl} \xi_{kl} | v_k \rangle \langle v_l | \hspace{1mm}\in D(\mathcal{Y})$
The mixed state is
$\rho = \sigma \otimes\xi = \left[ \sum_{ij} \sigma_{ij} | v_i \rangle \langle v_j | \right] \otimes \left[ \sum_{kl} \xi_{kl} | v_k \rangle \langle v_l | \right] = \sum_{ijkl} \sigma_{ij} \xi_{kl} | v_i \rangle \langle v_j | \otimes | v_k \rangle \langle v_l | $
And if you do the partial trace for example, $\text{Tr}_{\mathcal{Y}} (\rho) = \sigma$, it should help you to show that effectively, $\rho$ is an extension of $\sigma$. Oh, if you have an extension made from two pure states, that new extension is gonna be also a pure state. With this I think you can prove that the extension of a pure state can be written as tensorial products.