I'm a little confused about how we can distinguish between pure/mixed states and entangled/separable states and I would really appreciate some help!

I understand a density operator $\rho$ represents a pure state iff $\rho = | \psi \rangle \langle \psi |$ for some $\psi \in \mathcal{H}$. Now I understand that if $\psi \in \mathcal{H}_1 \bigotimes \mathcal{H}_2$ is a pure state then $\psi$ is separable iff $| \psi \rangle = |\psi_1\rangle|\psi_2\rangle$.

I know that we can tell if a density operator is pure by the condition $\rho$ is pure iff $Tr(\rho^2) = 1$ (as consequence of Cauchy-Schwarz inequality). And we can tell (for a two qubit system is entangled) by looking at whether the partial transpose of a subsystem has positive density operator or not.

I have a couple of questions about this:

  • When we use the partial transpose to check if a two qubit system is entangled, do we have to partial transpose on both subsystems and then the state is entangled iff both subsystems have a negative partial transpose?

  • I've also come across a condition that $\rho$ is pure iff it has rank 1. Is there a similar rank condition for telling whether a density operator represents an entangled/separable state - perhaps by tracing out one of the systems?

I ask these questions as I am trying to solve the following problem: I have a Hilbert space $\mathcal{H} = \mathcal{H}_1 \bigotimes \mathcal{H}_2$ (with bases $\{x_i\}$, $\{y_i\}$ respectively) and a linear operator $\tau: \mathcal{H}_1 \to \mathcal{H}_1$. I'm trying to show that $|\Psi\rangle = \sum |\tau x_i \rangle |y_i\rangle $ is separable iff the density operator $\rho = \tau \tau^*$ is pure. I haven't been able to make any real progress on this because I'm not sure I have the most convenient characterisations of pure/mixed and entangled/separable states.

My attempt at this question so far:

First of all suppose $\rho$ is pure hence $\rho = |\psi\rangle \langle \psi|$ for some $\psi \in \mathcal{H}_1$ so there are scalars $\lambda_i$ s.t. $\rho_ = \sum_{i,j} \lambda_i \lambda_j |x_i \rangle \langle x_j|$ Now we can look at the density matrix of $\Psi$ to determine if it is entangled. $$\rho_{\Psi}= | \Psi \rangle \langle \Psi| = \sum_{i,j} |\tau x_i \rangle \langle \tau x_j | |y_i\rangle\langle y_j| = \sum_{i,j} |\rho x_i \rangle \langle x_j| |y_i\rangle \langle y_j|$$

Now we can use the expression for $\rho$: $\rho |x_k \rangle = \sum_i \lambda_i \lambda_k |x_i>$ Hence $\rho_{\Psi}$ simplifies to:

$$\rho_{\Psi} = \sum_{i,j} \sum_k \lambda_k \lambda_i|x_k \rangle \langle x_j| |y_i\rangle \langle y_j|$$

Is this a separable state?

Alternatively we could find the partial transpose of this (on the first qubit) to see if it is entangled:

$$\rho_{\Psi}^T = \sum_{i,j} |x_j \rangle \langle \rho x_i| |y_i\rangle \langle y_j|$$

This is where I am stuck - how do we show that this has positive eigenvalues? I'm also unsure of how to get started on the other direction i.e. supposing $\Psi$ is separable and then showing $\rho$ is pure?

  • $\begingroup$ Thanks Martin, I was implicitly assuming the state $\psi$ was pure and just giving the definition of separability for a definite state - I'll make that a bit more clear in the post $\endgroup$
    – Wooster
    Apr 18, 2015 at 17:35
  • $\begingroup$ Saw this, and deleted the comment! Sorry for misunderstanding. $\endgroup$
    – Martin
    Apr 18, 2015 at 17:35
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/70436/2451 and links therein. $\endgroup$
    – Qmechanic
    Feb 13, 2019 at 22:19

1 Answer 1


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.

  • $\begingroup$ Martin, thank you for this very helpful and detailed answer! That's cleared up a lot of my confusion. I haven't actually covered the Schmidt decomposition in detail - do you know of a way to prove 2 without using it? I haven't been able to. I made an error in my question at the end which I've now corrected. Thanks $\endgroup$
    – Wooster
    Apr 18, 2015 at 18:23
  • $\begingroup$ @Wooster: It should be possible to find a way. I actually think that if you leave out the Schmidt decomposition, you could probably do the same calculation with a bit more effort in the exact same way (you'd probably have to choose your basis in a different way such that you know what it means for your state to be pure in the first place), but it's really not worth much to think about it. The Schmidt decomposition is really simple, it's basically just the singular value decomposition in disguise - nothing more! $\endgroup$
    – Martin
    Apr 19, 2015 at 0:20
  • $\begingroup$ Regarding your actual problem, I still don't really see why this should be true at all. Maybe you want $\tau$ to act on both systems? $\endgroup$
    – Martin
    Apr 19, 2015 at 0:20
  • $\begingroup$ I'll read a bit more on the Schmidt decomposition. I'm not sure $\tau$ really acts on anything, we just have a fixed state $|\Psi\rangle = \sum |\tau x_i \rangle |y_i\rangle $ and then a density operator $\rho$ defined by $\rho = \tau \tau^*$ and the question is whether $\psi$ is entangled or separable depending on purity of $\rho$ if that makes sense? $\endgroup$
    – Wooster
    Apr 19, 2015 at 9:52
  • $\begingroup$ Ah, now I saw that $\rho$ has to be pure instead of separable. Now the answer is still "no", if your sum has only one summand. It is also "no", if the sum has less elements than the dimension of the space. Otherwise, it's "yes". In principle, my explanation above covers this already, but if you want, I can clarify a bit more! $\endgroup$
    – Martin
    Apr 19, 2015 at 10:48

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