# How does one go about constructing a mixed entangled state?

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?

• 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?
– glS
Jul 22 '20 at 17:22
• 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. $|\psi\rangle\langle\psi|=\frac{1}{2}|00\rangle\langle00|+\frac{1}{2}|11\rangle\langle11|$ Jul 22 '20 at 18:52
• Do you have a reference for the definition? Jul 22 '20 at 20:49
• Please fix your definition or otherwise clearly cite a source that's causing this confusion. Jul 22 '20 at 22:59
• Here are two physics stack exchange posts on the matter physics.stackexchange.com/questions/468766/… physics.stackexchange.com/questions/171881/… Jul 23 '20 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 $$n$$.

• 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? Jul 29 '20 at 15:15
• 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 $\pm \tfrac{1}{3}$. Jul 29 '20 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 $$p/2>(1-p)/4$$.

• 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. Jul 23 '20 at 10:17
• 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. Jul 23 '20 at 10:21
• 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? Jul 23 '20 at 10:23
• 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. Jul 23 '20 at 12:25
• I assume $|\psi\rangle\langle\psi|$ in your example is $\frac12(|01\rangle-|10\rangle)(\langle01|-\langle10|)|$? Jul 23 '20 at 15:32