I Mixed states from entangled state

[ilikesquareobjects]

TL;DR Summary

Let p be a pure quantum state (entanglement entropy=0), and p1 and p2 two entangled sub-states, obtained by partially tracing over the other. We know the sum of the entropies of p1 and p2 is strictly positive (because they are entangled), hence at least one of the two is a mixed state. But can we prove both of them are mixed?

I'm an undergrad in physics, and have been asking myself the following question recently. Suppose you have a pure quantum state p (von neumann entropy=0), made of 2 sub-states p1 and p2 that are entangled. Because they are entangled, p \neq p1 x p2. Hence the entanglement entropy of p (=0) is strictly smaller than the sum of the entropies of p1 and p2. Then at least one of the 2 sub-states must be a mixed state, their entropies can't both be zero. Until now, everything makes sense.

However, I have the impression that p1 and p2 should both be mixed. It seems like it wouldn't physically make sense to have, as bi-products of a pure state p, a pure state p1 and a mixed state p2. It seems strange if they're entangled to one another. Unfortunately I don't know how to prove that both p1 and p2 are mixed. Perhaps there's a proof I can't seem to write, or maybe my intuition is wrong. Either way, I'd be grateful for some help! Thanks in advance :)

 

[stevendaryl]

This might be a terminology issue. Let's take a specific example, the canonical entangled state:

$|\psi\rangle = \frac{1}{\sqrt{2}} (|U, D\rangle - |D, U\rangle)$

where $U$ means spin-up and $D$ means spin-down (in the z-direction, for example).

More abstractly, $|\psi\rangle = \sum_{i j} C_{ij} |i, j\rangle$

I would say that neither the first particle nor the second particle is in a mixed state. I would say that neither particle has a state. Only the composite two-particle system has a state.

However, for the purposes of performing measurements on either particle independently of the other particle, we can form the corresponding mixed states:

Let $\rho = |\psi\rangle \langle \psi | = \sum_{iji'j'} C^*_{ij} C_{i'j'} |i,j\rangle \langle i',j'|$

Or as a matrix, $\rho_{iji'j'} = C^*_{ij} C_{i'j'}$

We form the mixed state $\rho(1)_{ii'}$ by setting $j=j'$ and summing:

Or as a matrix, $\rho(1)_{ii'} = \sum_j C^*_{ij} C_{i'j}$
Similarly, $\rho(2)_{jj'} = \sum_i C^*_{ij} C_{ij'}$

For our simple example, the results are:

$\rho(1)_{UU} = \rho(2)_{UU} =\rho(1)_{DD} =\rho(2)_{DD} =\frac{1}{2}$
$\rho(1)_{UD} = \rho(2)_{UD} =\rho(1)_{DU} =\rho(2)_{DU} = 0$

This is the same mixed state that you would get if you said that each particle has a 50% chance of being spin-up and 50% chance of being spin-down. Forming these mixed states throws away information about correlations, however. You can't look at the two mixed states to see that the particles are always anti-correlated.

I would not say that particle 1 is in the mixed state $\rho(1)$. I would say that $\rho(1)$ is the equivalent mixed state for particle 1. Particle 1 doesn't have a state, mixed or otherwise.

The significance of the mixed states $\rho(1)$ and $\rho(2)$ is that they are in some sense, the best possible way to approximate the entangled state $|\psi\rangle$ by two independent one-particle states. For a measurement that only involves particle 1, $\rho(1)$ gives you all the information you need to predict probabilities for outcomes. It's only when you look at correlations between the two particles that you will find that the mixed states leave out information.

 

[Demystifier]

However, I have the impression that p1 and p2 should both be mixed.

Yes, they are both mixed. Moreover, they have the same entropy.

 

[wle]

However, I have the impression that p1 and p2 should both be mixed.

They are not only both mixed but both have the same spectrum (eigenvalues).

This is a simple consequence of the Schmidt decomposition: given any bipartite pure state $\lvert \Psi \rangle$ living in some Hilbert space $\mathcal{H}_{\mathrm{A}} \otimes \mathcal{H}_{\mathrm{B}}$, it is always possible to find orthonormal bases $\{\lvert k \rangle_{\mathrm{A}}\}$ and $\{\lvert k \rangle_{\mathrm{B}}\}$ of $\mathcal{H}_{\mathrm{A}}$ and $\mathcal{H}_{\mathrm{B}}$ in which the state has the form $$\lvert \Psi \rangle = \sum_{k} c_{k} \lvert k \rangle_{\mathrm{A}} \lvert k \rangle_{\mathrm{B}}$$ and the coefficients $c_{k}$ (called the Schmidt coefficients) are nonnegative real numbers. From this you get the partial traces $$\begin{eqnarray*}
\rho_{\mathrm{A}} = \text{Tr}_{\mathrm{B}} \bigl[ \lvert \Psi \rangle \langle \Psi \rvert \bigr] &=& \sum_{k} {c_{k}}^{2} \lvert k \rangle \langle k \rvert_{\mathrm{A}} \,, \\
\rho_{\mathrm{B}} = \text{Tr}_{\mathrm{A}} \bigl[ \lvert \Psi \rangle \langle \Psi \rvert \bigr] &=& \sum_{k} {c_{k}}^{2} \lvert k \rangle \langle k \rvert_{\mathrm{B}} \,.
\end{eqnarray*}$$

 

[ilikesquareobjects]

This might be a terminology issue. Let's take a specific example, the canonical entangled state:

$|\psi\rangle = \frac{1}{\sqrt{2}} (|U, D\rangle - |D, U\rangle)$

where $U$ means spin-up and $D$ means spin-down (in the z-direction, for example).

More abstractly, $|\psi\rangle = \sum_{i j} C_{ij} |i, j\rangle$

I would say that neither the first particle nor the second particle is in a mixed state. I would say that neither particle has a state. Only the composite two-particle system has a state.

However, for the purposes of performing measurements on either particle independently of the other particle, we can form the corresponding mixed states:

Let $\rho = |\psi\rangle \langle \psi | = \sum_{iji'j'} C^*_{ij} C_{i'j'} |i,j\rangle \langle i',j'|$

Or as a matrix, $\rho_{iji'j'} = C^*_{ij} C_{i'j'}$

We form the mixed state $\rho(1)_{ii'}$ by setting $j=j'$ and summing:

Or as a matrix, $\rho(1)_{ii'} = \sum_j C^*_{ij} C_{i'j}$
Similarly, $\rho(2)_{jj'} = \sum_i C^*_{ij} C_{ij'}$

For our simple example, the results are:

$\rho(1)_{UU} = \rho(2)_{UU} =\rho(1)_{DD} =\rho(2)_{DD} =\frac{1}{2}$
$\rho(1)_{UD} = \rho(2)_{UD} =\rho(1)_{DU} =\rho(2)_{DU} = 0$

This is the same mixed state that you would get if you said that each particle has a 50% chance of being spin-up and 50% chance of being spin-down. Forming these mixed states throws away information about correlations, however. You can't look at the two mixed states to see that the particles are always anti-correlated.

I would not say that particle 1 is in the mixed state $\rho(1)$. I would say that $\rho(1)$ is the equivalent mixed state for particle 1. Particle 1 doesn't have a state, mixed or otherwise.

The significance of the mixed states $\rho(1)$ and $\rho(2)$ is that they are in some sense, the best possible way to approximate the entangled state $|\psi\rangle$ by two independent one-particle states. For a measurement that only involves particle 1, $\rho(1)$ gives you all the information you need to predict probabilities for outcomes. It's only when you look at correlations between the two particles that you will find that the mixed states leave out information.

Thank you very much for your clear explanation, I understand your point.
However, in a strictly mathematical sense, $\rho(1)$ and $\rho(2)$ are quantum states, in the sense that they are density matrices in $H_1$ and $H_2$. So the question "are they necessarily mixed" (is $\rho(1)^2=\rho(1)$ and $\rho(2)^2=\rho(2)$) should make sense, even though they are a very different type of state than the state $\rho$, because as you say they are only 'approximations' to the full entangled state.
But I think I have found the answer to my question, since $S(\rho(1))=S(\rho(2))$, then necessarily they must both be strictly positive and hence both $\rho(1)$ and $\rho(2)$ are mixed. Thanks again for your help!

 

[ilikesquareobjects]

They are not only both mixed but both have the same spectrum (eigenvalues).

This is a simple consequence of the Schmidt decomposition: given any bipartite pure state $\lvert \Psi \rangle$ living in some Hilbert space $\mathcal{H}_{\mathrm{A}} \otimes \mathcal{H}_{\mathrm{B}}$, it is always possible to find orthonormal bases $\{\lvert k \rangle_{\mathrm{A}}\}$ and $\{\lvert k \rangle_{\mathrm{B}}\}$ of $\mathcal{H}_{\mathrm{A}}$ and $\mathcal{H}_{\mathrm{B}}$ in which the state has the form $$\lvert \Psi \rangle = \sum_{k} c_{k} \lvert k \rangle_{\mathrm{A}} \lvert k \rangle_{\mathrm{B}}$$ and the coefficients $c_{k}$ (called the Schmidt coefficients) are nonnegative real numbers. From this you get the partial traces $$\begin{eqnarray*}
\rho_{\mathrm{A}} = \text{Tr}_{\mathrm{B}} \bigl[ \lvert \Psi \rangle \langle \Psi \rvert \bigr] &=& \sum_{k} {c_{k}}^{2} \lvert k \rangle \langle k \rvert_{\mathrm{A}} \,, \\
\rho_{\mathrm{B}} = \text{Tr}_{\mathrm{A}} \bigl[ \lvert \Psi \rangle \langle \Psi \rvert \bigr] &=& \sum_{k} {c_{k}}^{2} \lvert k \rangle \langle k \rvert_{\mathrm{B}} \,.
\end{eqnarray*}$$

Thanks a lot! I didn't know this proof! The result makes a lot of sense, but I must say I was trying to prove it using only the entanglement entropy

 

[wle]

I must say I was trying to prove it using only the entanglement entropy

I don't see how this makes sense. I mean, if you look up what "entanglement entropy" is on, e.g., the Wikipedia page, you'll find out that the entanglement entropy of a bipartite entangled pure state is nonzero and that the entropy is the same regardless of which reduced state $\rho_{\mathrm{A}}$ or $\rho_{\mathrm{B}}$ we calculate it on. But how do you think we know those things in the first place?