In general, an entangled state is one which cannot be decomposed as $\sum_{i}p_{i} \bigl(\rho_{i}^A\otimes\rho_{i}^B\bigr)$. But such an entangled state could still be mixed in principle.

How would one create a mixed entangled state? For instance, can one create such a state by tracing out one qubit or a subsystem of a pure state?

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    $\begingroup$ where are you getting this from? A (possibly mixed) entangled state is one that cannot be written as a convex decomposition of separable states, which has nothing to do with what you wrote. Also what do you mean by "construction" here? $\endgroup$ – glS Jul 22 at 17:22
  • $\begingroup$ From the definition of a mixed entangled state, one that cannot be written as a decomposition of mixed states. By construction I mean under what process would it emerge under. What I described above is exactly what I mean. If you trace out one qubit, you are left with $|\psi\rangle\langle\psi|$ as stated above. This state cannot be written as a decomposition of mixed states, so I am asking if this is a mixed entangled state. $|\psi\rangle\langle\psi|=\frac{1}{2}|00\rangle\langle00|+\frac{1}{2}|11\rangle\langle11|$ $\endgroup$ – GaussStrife Jul 22 at 18:52
  • $\begingroup$ Do you have a reference for the definition? $\endgroup$ – Rammus Jul 22 at 20:49
  • $\begingroup$ Please fix your definition or otherwise clearly cite a source that's causing this confusion. $\endgroup$ – keisuke.akira Jul 22 at 22:59
  • $\begingroup$ Here are two physics stack exchange posts on the matter physics.stackexchange.com/questions/468766/… physics.stackexchange.com/questions/171881/… $\endgroup$ – GaussStrife Jul 23 at 9:03

An example of a state which yeilds an entangled state when you trace one qubit out, is the 3-qubit "W" state: $$ \lvert W_3 \rangle = \tfrac{1}{\sqrt3} \Bigl( \lvert 100 \rangle + \lvert 010 \rangle + \lvert 001 \rangle \Bigr) $$ Taking the outer product with itself, we obtain $$ \lvert W_3 \rangle\!\langle W_3 \rvert = \tfrac{1}{3} \Bigl( \!\begin{aligned}[t]& \lvert 100 \rangle\!\langle 100 \rvert + \lvert 100 \rangle\!\langle 010 \rvert + \lvert 100 \rangle\!\langle 001 \rvert \phantom{\Big)} \\&+ \lvert 010 \rangle\!\langle 100 \rvert + \lvert 010 \rangle\!\langle 010 \rvert + \lvert 010 \rangle\!\langle 001 \rvert \\&+ \lvert 001 \rangle\!\langle 100 \rvert + \lvert 001 \rangle\!\langle 010 \rvert + \lvert 001 \rangle\!\langle 001 \rvert \Bigr) \end{aligned} $$ If we trace out the third qubit, we then obtain the state $$ \begin{align} \rho = \mathrm{tr}_3\Bigl(\lvert W_3 \rangle\!\langle W_3 \rvert\Bigr) &= \tfrac{1}{3} \Bigl( \lvert 10 \rangle\!\langle 10 \rvert + \lvert 10 \rangle\!\langle 01 \rvert + \lvert 01 \rangle\!\langle 10 \rvert + \lvert 01 \rangle\!\langle 01 \rvert + \lvert 00 \rangle\!\langle 00 \rvert \Bigr) \\&= \tfrac{1}{3} \lvert 00 \rangle\!\langle 00 \rvert + \tfrac{2}{3} \lvert \Psi^+ \rangle\!\langle \Psi^+ \rvert, \end{align} $$ where in particular $\lvert \Psi^+ \rangle = \tfrac{1}{\sqrt 2}\bigl( \lvert 01 \rangle + \lvert 10 \rangle \bigr)$ is a Bell state.

Note that $\rho$ is a rank-2 operator with different eigenvalues: any decomposition of $\rho$ as a convex combination of other density operators, can only involve terms whose eigenvectors are supported on $\mathrm{span}\,\{ \lvert 00 \rangle, \lvert \Psi^+ \rangle \}$. Any mixed tensor product density operator $\rho_A \otimes \rho_B$ has an eigenbasis consisting of two or more product states; but the only product state contained in $\mathrm{span}\,\{ \lvert 00 \rangle, \lvert \Psi^+ \rangle \}$ is the state $\lvert 00 \rangle$ itself. It follows that $\rho$ cannot be decomposed as a convex combination of products of possibly-mixed states, and is entangled.

More generally, if for $n > 1$ we define $\lvert W_n \rangle$ as the analogous $n$-term uniform superposition of standard basis states with a single 1, we may describe it as $$ \lvert W_n \rangle = \tfrac{\sqrt{n{-}1}}{\sqrt n} \lvert W_{n{-}1}\rangle\lvert0\rangle + \tfrac{1}{\sqrt n} \lvert00\cdots0\rangle\lvert1\rangle $$ so that $$ \mathrm{tr}_n\Bigl(\lvert W_n \rangle\!\langle W_n \rvert\Bigr) = \tfrac{n{-}1}{n} \lvert W_{n-1} \rangle\!\langle W_{n-1} \rvert + \tfrac{1}{n} \lvert 00\cdots0\rangle\!\langle 00\cdots 0\rvert $$ which for larger values of $n$ is closer and closer to being a pure entangled state, while still being mixed for any finite $n$.

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  • $\begingroup$ I take it that, as @DaftWullie stated, the PT of $\frac{n-1}{n}|W_{n-1}\rangle\langle W_{n-1}|+\frac{1}{n}|00...0\rangle\langle00...0|$ will yield negative eigenvalues? $\endgroup$ – GaussStrife Jul 29 at 15:15
  • $\begingroup$ The 'if and only if' condition that he mentions holds only two-qubit states, so we should consider whether $\rho = \tfrac{2}{3} \lvert \Psi^+ \rangle \!\langle \Psi^+ \rvert + \tfrac{1}{3} \lvert 00 \rangle\!\langle 00 \rvert$ has a positive partial transpose. From the expansion of $\rho$ that I give in the standard basis, we see that it's partial transpose (on either qubit) is $\tfrac{1}{3}( \lvert 10 \rangle\!\langle 10 \rvert + \lvert 00 \rangle\!\langle 11 \rvert + \lvert 11 \rangle\!\langle 00 \rvert + \lvert 00 \rangle\!\langle 00 \rvert )$, which has eigenvalues $\pm \tfrac{1}{3}$. $\endgroup$ – Niel de Beaudrap Jul 29 at 19:50

A mixed separable state is written in the form $$ \rho=\sum_ip_i\sigma^A_i\otimes\sigma^B_i $$ where the $\sigma_i$ are valid density matrices on a single site.

The example you give, say $\rho=\frac12|00\rangle\langle00|+\frac12|11\rangle\langle 11|$ is exactly of this form. Specifically, $$ p_0=p_1=\frac12,\qquad \sigma^A_0=\sigma^B_0=|0\rangle\langle 0|,\qquad \sigma^A_1=\sigma^B_1=|1\rangle\langle 1|. $$

In general, it can be tricky to definitively prove that such a decomposition does not exist. However, in the case of two-qubits, there's an if and only if condition: the state is entangled if and only if its partial transpose is not non-negative (i.e. contains a negative eigenvalue).

The classic example is the Werner state, $$ \rho=\frac{1-p}{4}I+p|\psi\rangle\langle\psi| $$ where $|\psi\rangle$ is a two-qubit Bell state. This can be written out as a matrix $$ \rho=\left(\begin{array}{cccc} \frac{1-p}{4} & 0 & 0 & 0 \\ 0 & \frac{1+p}{4} & -\frac{p}{2} & 0 \\ 0 & -\frac{p}{2} & \frac{1+p}{4} & 0 \\ 0 & 0 & 0 & \frac{1-p}{4} \end{array}\right) $$ If we take the partial transpose of this, we get $$ \rho^T=\left(\begin{array}{cccc} \frac{1-p}{4} & 0 & 0 & -\frac{p}{2} \\ 0 & \frac{1+p}{4} & 0 & 0 \\ 0 & 0 & \frac{1+p}{4} & 0 \\ -\frac{p}{2} & 0 & 0 & \frac{1-p}{4} \end{array}\right) $$ This has a negative eigenvalue if $p/2>(1-p)/4$.

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  • $\begingroup$ what do you mean "the tensor product of the two mixed states"? There are many mixed states, not just the maximally mixed state $I/2$. This actually includes, for example, all pure states. $\endgroup$ – DaftWullie Jul 23 at 10:17
  • $\begingroup$ I was under the impression that, like a pure state, the criteria for a mixed state being entangled is the inability to decompose it into the product of two mixed states, regardless of structure, ie the above example I gave. I did not realise that the sum of the product was an acceptable approach. $\endgroup$ – GaussStrife Jul 23 at 10:21
  • $\begingroup$ What about the definition of it mixed entangle state being a mixture of $\sum p_{i}|\psi_{i}\rangle\langle\psi_{i}|$ where $\psi_{i}$ is itself an entangled state? $\endgroup$ – GaussStrife Jul 23 at 10:23
  • $\begingroup$ That is certainly mixed, but there's no guarantee it's entangled. For example, an equal mixture of all 4 Bell states gives the maximally mixed state, which is certainly not entangled. $\endgroup$ – DaftWullie Jul 23 at 12:25
  • $\begingroup$ I assume $|\psi\rangle\langle\psi|$ in your example is $\frac12(|01\rangle-|10\rangle)(\langle01|-\langle10|)|$? $\endgroup$ – GaussStrife Jul 23 at 15:32

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